Tutorials Complete

Title page Australian School of Business

Probability and StatisticsTutorial Exercises

Probability and Statistics

Exercises

Session 1, 2013

Overview Australian School of Business

Probability and StatisticsOverview

TutorialsStudents must attend the tutorial for which they are enrolled. Attendance will be recorded and count to-wards meeting the requirements to pass the course. If you wish to change your tutorial then you must lodgean application to change your tutorial time with the ASB student service centre.

In tutorials, we will implement interactive learning where collaborative group work is highly encouraged.To get the most out of the tutorials, students should read lecture notes and textbooks and references andcomplete assigned homework problems in advance of the tutorial.

Peer-Assisted-Support-Scheme (PASS)In addition, ASOC has an actuarial PASS program will be available to enhance student learning in thiscourse. It is highly recommended that you attend these sessions.ASOC website: www.asoc.unsw.edu.auPASS peer support class times:www.asoc.unsw.edu.au/index.php?option=com content&view=article&id=48&Itemid=42PASS material is available on: sites.google.com/a/asoc.unsw.edu.au/unsw-asoc/downloads

Organization of TutorialsThe more you read the more you know, but the more you practice the more you learn and understand. Sothe key to the understanding of this course is problem solving.

The purpose of tutorials is to enable you to raise questions about difficult topics or problems encounteredin their studies. Students must not expect another lecture, tutorials are a place where you have to learnby doing and subsequently also learn from mistakes, as that is the most efficient way to learn, especially inactuarial courses.

To benefit from the tutorial exercises provided, you have to make the attempt to the questions yourself.Copying the tutorial manual’s solution or the tutor’s solution is least effective for your learning process.Moreover, given that you already know the answer, the “copying learning strategy” prevent you from acquir-ing the knowledge of solving the question when trying to solve the exercise any subsequent time afterwards.

The object of the tutorials is to attempt the assigned exercises and topics covered in the course. Eachweek a document will be posted containing the exercises which are to be covered in tutorials. IN the scheduleon the next page the weekly questions are divided into three groups:

1. Exercises that should be done in the same week as the lecture. At the end of the week the solutionsto these questions will be uploaded on Blackboard. If there is enough interest there will be an onlinetutorial about those questions. Link and details will be provided via Blackboard.

2. Question done in the week after the lecture. The class tutorial will typically focus on these questions.After all tutorials, a new solution manual will be posted on Blackboard containing all the answers.

3. Some optional/additinal/exam preparation questions for the ones requiring more exercises than onlythe tutorial, pass class and book exercises when studying for the exam.

Tutorials are an integral component of the course; attendance and participation in your tutorial is crucialfor your successful completion of the course.

The Teaching Strategy (including the feedback loop) in this course is:The ”Get introduced” (explanation of course concepts in the lecture) and ”Try it out” (examples in the

lecture) are part of the lecture. If you are not able to make satisfactory attempt to the examples in thelecture, this is feedback that you should revise the lecture material in dept after the lecture. The ”Try again”(tutorial exercises) and ”Get feedback” (answers from tutor/ Wikispace / tutorial solutions) are part of thetutorial. The tutorial is designed to attempt the tutorial questions in groups (typically of 3-4 students). Thiscollaborative working is advised since it allows learning from peers and allows the more advanced studentwith a possibility to test whether s/he knows the material in dept and is thereby able to explain it to peers.

c© Katja Ignatieva School of Risk and Actuarial Studies, ASB, UNSW Page i of iii

http://www.asoc.unsw.edu.au/

http://www.asoc.unsw.edu.au/index.php?option=com_content&view=article&id=48&Itemid=42

https://sites.google.com/a/asoc.unsw.edu.au/unsw-asoc/downloads

ACTL2002 & ACTL5101 Probability and Statistics Overview

GetIntroduced

Try it out Get feedback Try again

Figure 1: Feedback loop

If the group is unable to solve a question, it can ask for help from the tutor. The tutor will not provide theanswer, but would help you in the direction of the solution. This is because you should practice and learnby doing, rather than seeing the solution. At the end of the week the tutorial solutions will be posted onBlackboard. It will only be posted at the end of the week to give you time to attempt the questions withouta solution manual. The solution manual should not be part of attempting a question, but to verify whetheryour attempt was correct.

Learning StrategyA required learning strategy for the tutorials (on which provision of the course materials is based) is:

1) Prior to make an attempt of the exercises, review your lecture notes.

2) Prior to the tutorial, make an attempt to the exercises you should make before the tutorial (see schedulebelow).

3) At the end of the week, verify your answers with the (partial) solution manual and/or the onlinetutorial on Blackboard. Revise the material that you did not know, did incorrectly.

4) During the tutorial, make an attempt to the exercises you should make in the tutorial (see schedulebelow).

5) After the tutorial, make an attempt to the exercises you should after in the tutorial (see schedulebelow).

6) If you have questions about the tutorial exercises, ask them at the Wikispace. If you think you havea good understanding of the material, you should try and answer the questions of your peers on theWikispace. This will give you feedback on your ability to explain the material and hence how well youknow the material.

7) Check your answer using the tutorial solution manual.

NOTE: tutorials are for making tutorial exercises, not for reviewing material or working onassignment or anything else!

Schedule of Tutorial Exercises

c© Katja Ignatieva School of Risk and Actuarial Studies, ASB, UNSW Page ii of iii

ACTL2002 & ACTL5101 Probability and Statistics Overview

Exercises Before tuto-rial

During Tutorial After Tutorial Additional

Week 1 1, 2 10, 11, 12, 13 14, 15, 16, 17 3-9Week 2 1, 2 3, 4, 5 6, 7, 8Week 3 1, 2, 3 4, 6, 7, 5 8, 9, 10, 11 12Week 4 1 3abc, 2, 4, 3d,e 5, 6Week 5 1, 2 3, 4, 5, 6 7, 8, 9 10, 11, 12, 13, 14,

15, 16Week 6 1, 2 4, 3, 6, 5 7, 8, 9, 10 11, 12, 13, 14Week 7 1 2, 3, 4, 6 5, 7Week 8 Working on

assignment1, 2, 8, 7 Working on assign-

ment3, 5, 6

Week 9 Working onassignment

1, 2a, 5 Working on assign-ment

3a-d, 4, 7a-d


2-6 Working on assign-ment


5-14 Working on assign-ment

1, 7, 8

Week 12 15 of week 11tutorial

1-3 Revision for exam 1-4

Sample Exam Revision forexam

Revision for exam

Table 1: Schedule of tutorial exercises.

c© Katja Ignatieva School of Risk and Actuarial Studies, ASB, UNSW Page iii of iii

Week 1 Australian School of Business

Probability and StatisticsTutorial Exercises, Week 1, 2013

1. An urn contains one black ball and one gold ball while a second urn contains one white and one goldball. One ball is selected at random from each urn.

(a) Describe the sample space for this experiment.

(b) Describe a σ-algebra for this experiment.

(c) Describe the event that both balls will be of the same colour. What is the probability of thisevent?

2. A box contains 100 Christmas balls: 49 are red, 34 are gold, and 17 are silver. Three balls are to bedrawn without replacement. Determine the probability that:

(a) all 3 balls are red;

(b) the balls are drawn in the order: red, gold, and silver;

(c) the third ball is a silver, given that the first 2 are red and gold (not necessarily in that order);and

(d) the first 2 are red, given that the third ball is a silver;

3. Let A and B be two independent events. Prove that the following pairs are also independent:

(a) A and BC

(b) AC and B

(c) AC and BC

4. A pair of events A and B cannot be simultaneously mutually exclusive and independent. Assume thattheir probabilities are strictly positive, i.e., Pr (A) > 0 and Pr (B) > 0. Prove the following:

(a) If A and B are mutually exclusive, then they cannot be independent.

(b) If A and B are independent, then they cannot be mutually exclusive.

5. This exercise shows that independence does not imply pairwise independence. Consider a randomexperiment which consists of tossing two dice. Define the following events:

E1 = doubles appearE2 = the sum is between (and includes) 7 and 10E3 = the sum is 2 or 7 or 8

(a) Show that E1, E2 and E3 are independent.

(b) Show that E1 and E2 are not pairwise independent.

(c) Show that E2 and E3 are not pairwise independent.

(d) What about E1 and E3—are they pairwise independent?

6. In an undergraduate statistics class, three students A, B, and C submitted exactly (word-for-word)the same solution to a homework problem. It is the policy of the lecturer to give zero marks for thosewho copy homework problems. Believing that there must be one of the three who actually did thework, the lecturer will pardon one of the three and chooses at random the student to pardon.However, the lecturer will only inform the students at the end of the semester who among them hasbeen pardoned.The next day, A tries to get the lecturer to tell him who had been pardoned. The lecturer refuses. Athen asks which of B or C will not be pardoned. The lecturer thinks for a while, then tells A that Bis not to be pardoned.

c© Katja Ignatieva School of Risk and Actuarial Studies, ASB, UNSW Page 1 of 4

ACTL2002 & ACTL5101 Probability and Statistics Tutorial Exercises, Week 1

• Lecturer’s reasoning: Each student has a 1/3 chance of being pardoned. Clearly, either B or Cmust not be pardoned, so I have given A no information about whether A will be pardoned.

• A’s reasoning: Given that B will not be pardoned, then either A or C will be pardoned. Mychance of being pardoned has risen to 1/2.

(a) Evaluate the lecturer’s reasoning, i.e., explain whether his reasoning is justified.

(b) Evaluate student A’s reasoning, i.e., explain whether his reasoning is justified.

7. Two airlines serving some of the same cities in Australia have merged. Management has decided toeliminate some of the repetitious daily flights. On the Perth-Sydney route, one airline originally hadfive daily flights (each at different a time) and the other had six daily flights (each at different a time).Determine the number of ways:

(a) four flights can be eliminated.

(b) the first airline can eliminate two of its scheduled five flights.

(c) the second airline can eliminate two of its scheduled six flights.

(d) two flights can be eliminated from each airline.

8. Three boxes are numbered 1, 2 and 3. For k = 1, 2 and 3, box k contains k blue marbles and 5 − kred marbles. In a two-step experiment, a box is selected and 2 marbles are drawn from it withoutreplacement. If the probability of selecting box k is proportional to k, what is the probability that thetwo marbles drawn have different colors?

9. The probability function of a certain discrete random variable on the non-negative integers satisfiesthe following:

• Pr(0) = Pr(1)

• Pr(k + 1) = Pr(k)/k for k = 1, 2, 3, . . ..

Find Pr(0).

10. Consider X , a continuous random variable with density function

fX(x) = ce−x, x > 1, and zero otherwise.

Find

(a) all c such that fx is a random variable, and

(b) Pr(X < 3 |X > 2).

11. The distribution function for a discrete random variable X is given by:

FX(x) =

0 if x < −11/3 if − 1 ≤ x < 2/31 if x ≥ 2/3.

(a) Specify the probability mass function pX(x).

(b) Sketch the graphs of pX(x) and FX(x).

12. Let X be a random variable with density:

fX (x) =1√2πσ

exp

[−1

2

(x− µσ

)2]

, for −∞ < x <∞.

Here, X is called a normally distributed random variable.

(a) Find an expression for the moment generating function, MX (t) of X .

(b) Now define S (t) = log [MX (t)]. Show that, in general,

d

dtS (t)

∣∣∣∣t=0

= E [X ] andd2

dt2S (t)

∣∣∣∣t=0

= V ar (X) .



(c) Use the above result to prove that, with the normal density, we have

E (X) = µ and V ar (X) = σ2.

(d) How do we call the function S (t)?

13. Let X be a random variable with parameters α, β, θ, and ν ∈ ℜ, and have the following momentgenerating function: MX(t) = αt+ βt2 + θt3 + νt4.

(a) How many distribution functions corresponds to this m.g.f. for given values of the parameters?

(b) Determine the first five non-central moments of X .

(c) Determine the first five central moments of X .

(d) Determine the mean, variance, skewness, and kurtosis of X .

(e) Let X represents the claim sizes, i.e., a higher value is “bad” for the insurer. Insurer A and Bi

ask a quote for reinsuring a tail risk (for example: the reinsurer makes a payment to the insurerif the loss is larger than $1 million). Based on the mean, variance, skewness and kurtosis, whichof the two would receive a higher quote for reinsuring the risk, and why, if:

i) A’s parameters are: α = 1, β = 2, θ = 1, and ν = 1 and B1 parameters are α = 1,β = 1,θ = 0.5384, and ν = 0.2606;

ii) A’s parameters are: α = 1, β = 2, θ = 1, and ν = 1 and B2 parameters are α = 1, β = 2,θ = 2, and ν = 2;

iii) A’s parameters are: α = 1, β = 2, θ = 1, and ν = 1 and B2 parameters are α = 1, β = 2,θ = 1, and ν = 1.625.

14. The probability density function for a continuous random variable X is given by:

fX(x) =

2/x3 for x ≥ 10 otherwise.

(a) Determine a formula for the cumulative distribution function FX(x).

(b) Determine the probability that X ≥ 4.

(c) Sketch the graphs of fX(x) and FX(x).

15. Let X be a random variable with probability density function:

fX (x) =

12λe

−λx, if x ≥ 0;

12λe

λx, if x < 0.

(a) Verify that fX (·) is a pdf.

(b) Find expression for the cdf FX (x).

(c) Find the moment generating function and the probability generating function of X .

(d) Suppose λ = 1. Evaluate Pr (|X | < 3/4).

16. Actuaries often model the age-at-death as a non-negative random variable X and define the ‘force ofmortality’ as follows:

µ (x) = limh→0

FX (x+ h)− FX (x)

h (1− FX (x)),

where FX (·) denotes the cdf of X .

(a) Using this definition, prove that:

FX (x) = 1− exp

(−∫ x

0

µ (z)dz

).

(b) Show that for a non-negative random variable:

E [X ] =

∫ ∞

0

[1− FX (z)] dz.

Use this result to show that:

E [X ] = E

[1

µ (X)

].



17. A random variable X has a probability density function of the form:

fX (x) = ax(1− bx2

), for 0 ≤ x ≤ 1, and zero otherwise,

where a and b are positive constants.

(a) Determine the value of a in terms of b and show that b ≤ 1.

(b) For the case b = 1, determine the mean and variance of X .

-End of week 1 Tutorial Exercises-




1. For each of the following situations, specify the type of distribution that best models the randomvariable X and give the parameters of the distribution chosen (where possible):

(a) This year, there are 100 students enrolled in an introductory actuarial studies course. For themid-session test for this course, the papers are marked by a team of tutors; however, a sampleof these papers is examined by the course professor for marking consistency. Experience suggeststhat 1% of all papers will be improperly marked. The professor selects 10 papers at random fromthe 100 papers and examines them for marking inconsistencies. X is the number of papers in thesample that are improperly marked.

(b) A standard drug has been known to be effective in 90% of the cases in which it is used. Tore-evaluate the effectiveness of this same drug, a clinical trial will be performed where 20 hasvolunteered. X is the number of cases where the drug has been found effective.

(c) An immunologist is studying blood disorders exhibited by people with rare blood types. It isestimated that 10% of the population has the type of blood being investigated. Volunteers whoseblood type is unknown are tested until 100 people with the desired blood type are found. X is

the number of people tested who do not have the desired rare blood type.

(d) Customers arrive at a fastfood restaurant independently and at random. During lunch hour,where more customers are often expected to arrive, customers arrive at the fastfood restaurantat the rate of two per minute on the average. X is the number of people who arrive between 12:15

p.m. and 12:30 p.m.

(e) A set of 25 multiple-choice questions was asked in an examination. It has been determined,according to experience, that the proportion of the questions which are guessed and answeredcorrectly is 35%. X is the number of questions guessed and answered correctly by a particularstudent who wrote for the examination.

2. For each of the following moment generating functions of discrete random variables X , identify thedistribution and specify the associated parameters.

(a) MX (t) =et

2− et

(b) MX (t) =

(et + 1

2

)3

(c) MX (t) = exp

(1

2et − 1

2

)

(d) MX (t) =

(et

2− et)4

(e) MX (t) =

(3et + 1

4

)5

3. Poisson approximation to the binomial. This exercise is to show that binomial probabilities canbe approximated using the Poisson probabilities, which are generally easier to calculate. Let X ∼Binomial(n, p) and Y ∼ Poisson(λ) where λ = np. The approximation states that

Pr (X = x) ≈ Pr (Y = x) ,

for large n and small np. This can be proven using convergence of mgf’s. Denote the respective mgf’sby MX (t) and MY (t).



(a) Prove that limn→∞MX (t) =MY (t).

Hint: use limn→∞

(1 + xn )

n = exp(x) = limh→0

(1 + hx)1h .

(b) Another method to prove this approximation is as follows: First, establish that the Poissondistribution satisfies the relation

Pr (Y = x)

Pr (Y = x− 1)=λ

xfor x = 1, 2, . . . .

Second, a similar relation can be approximated for the binomial distribution:

limp→0

Pr (X = x)

Pr (X = x− 1)=np

x.

Hint: first show that Pr(X=x)Pr(X=x−1) = n−x+1

x · p1−p , then take lim

p→0and use that lim

p→0(np) = λ. Then

show that:

Pr (Y = 0) ≈ Pr (X = 0) ,

for large n.

(c) A typesetter, on the average makes one error in every 400 words typeset. A typical page contains300 words. Use the Poisson approximation to the binomial to compute the probability that therewill be more than 3 errors in 10 pages.

4. An insurance company receives 200 claims per day on the average. Claims arrive independently andat random at the company office. Of the claims, 95% are for amounts less than $100 and are processedimmediately; the remaining 5% are examined more closely to verify their accuracy and eligibility.

(a) Determine the probability of getting no claims over $100 in a given day.

(b) Determine the probability of getting at most two claims over $100 in a given day.

(c) How many claims for amounts less than $100 should this company expect to receive in a week (5business days)?

5. Let X have a Gamma(α, β) distribution.

(a) Prove that the mgf of X can be expressed as:

(β

β − t

)α

for t < β.

(b) Establish also that for any positive constant r

E [Xr] = β−r Γ (r + α)

Γ (α).

6. Suppose that you have $1 000 to invest for a year. You are currently evaluating two investments:Investment A and Investment B, with annual rates of return, respectively denoted by RA and RB.Assume:

RA ∼ Normal (0.05, 0.1) and RB ∼ Normal (0.10, 0.5) .

(a) Under Investment A, compute the probability that your investment will be below $1 000 in a year.

(b) Under Investment A, compute the probability that your investment will exceed $1 200 in a year.

(c) Under Investment B, compute the probability that your investment will below $1 000 in a year.

(d) Under Investment B, compute the probability that your investment will exceed $1 200 in a year.

7. A city engineer has studied the frequency of accidents at two busy intersections. He has determinedthat the time T in months between accidents at each intersection has an exponential distribution. Theparameters for these two distributions are 2 and 2.5. Assume that the occurrence of accidents at theseintersections is independent.

(a) Determine the probability that there are no accidents at either intersection in the next month.



(b) Determine the probability that there will be no accidents for at least one of the intersections inthe next month.

8. The Pareto distribution is very commonly used to model certain insurance loss amounts. We say Xhas a Pareto distribution if its density can be expressed as:

fX (x) =α

β

(β

x

)α+1

for x > β,

and zero otherwise.

(a) Find expressions for the mean and variance of X .

(b) Find expression for the quantile of X . The quantile function is f(u) = F−1X (u) hence, one should

solve u = FX

(F−1X (u)

).

(c) What is then its median (i.e., u = 0.5)?

(d) An insurance policy has a deductible1 of $5. The random variable for the loss amount (beforedeductible) on claims filed has a Pareto distribution with α = 3.5 and β = 4. Find:

1. the mean loss amount;

2. the expected value of the amount of a single claim; and

3. the variance of the amount of a single claim.


1A deductible is that the policy only makes no payment if the loss amount is smaller than the deductible; and the claimamount equals the loss amount minus the deductible if the loss amount is larger than the deductible.




1. The claim amounts (in dollars, to the nearest $10) for a sample of 24 recent claims for storm damageto private homes in a particular town are as follows:2

2 710 670 2 380 4 670 1 220 6 780 1 590 3 110960 8 230 3 320 3 380 2 490 1 940 3 710 4 630

4 270 4 210 1 880 3 880 1 490 5 400 2 430 850

(a) Construct a stem-and-leaf display of these claim amounts.

(b) Find the mean and median of the claim amounts. What can you say about the skewness of thedistribution?

(c) Find the interquartile range of the claim amounts.

(d) Evaluate F24 (1 000) where F24 (·) denotes the ecdf.

2. Data were collected on 100 consecutive days for the number of claims, X , arising from a group ofinsurance policies. This resulted in the following frequency distribution:

observed claims from policy (x): 0 1 2 3 4 ≥ 5frequency: 14 25 26 18 12 5

Calculate the following sample statistics for these data:

(a) mode

(b) median

(c) interquartile range

(d) Suppose the average value for 5 claims or more is 7.5. Calculate the sample mean.

3. For a set of 32 observations, you are given:

32∑

k=1

xk = 13 337.6 and

32∑

k=1

x2k = 5 667 388.7.

The largest of the observations is 605. Suppose you are interested in measuring the impact of thelargest observation on the mean and standard deviation.

(a) Calculate the sample mean and the sample standard deviation.

(b) Calculate the sample mean and the sample standard deviation, with the largest observationdeleted.

(c) What is the percentage change in the mean?

(d) What is the percentage change in the standard deviation?

4. Let X and Y be two discrete random variables whose joint probability function is given by:

Pr(X = x, Y = y) X = x0 1 2 3

Y = y 1 0.05 0.20 0.15 0.052 0.20 0.15 0.12 0.08

2Modified Institute of Actuaries exam question.



Calculate:

(a) E [X ]

(b) E [Y ]

(c) E [X |Y = 1]

(d) V ar (Y |X = 3)

(e) E [XY ] and Cov(X,Y).

5. Let X and Y be two discrete random variables whose joint probability mass function is given by:

Pr(X = x, Y = y) X = x1 2 3 4

1 0.10 0.05 0.02 0.02Y = y 2 0.05 0.20 0.05 0.02

3 0.02 0.05 0.20 0.044 0.02 0.02 0.04 0.10

(a) Find the marginal probability mass functions of X and Y .

(b) Find the conditional probability mass of X given Y = 2 and of Y given X = 2.

(c) Find E[XY ] and Cov(X,Y ).

6. Let X and Y have the joint density:

fX,Y (x, y) =6

7(x+ y)2 , for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1,

and zero otherwise.

(a) By integrating over the appropriate regions, find:

1. Pr (X < Y )

2. Pr (X + Y ≤ 1)

3. Pr(X ≤ 1

2

)

(b) Find the marginal densities of X and Y .

(c) Find the two conditional densities.

7. Two independent measurements, X and Y , are taken of a quantity µ. We are given the means areequal, E [X ] = E [Y ] = µ, but the variances σ2

X and σ2Y are not equal. The two measurements are then

combined by means of a weighted average to give:

Z = αX + (1− α) Y,

where α is a constant between 0 and 1, i.e., 0 ≤ α ≤ 1.

(a) Show that E [Z] = µ.

(b) Find α in terms of σX and σY to minimise V ar (Z).

(c) Under what circumstances is it better to use the average (X + Y ) /2 than either X or Y alone todetermine µ? Hint: a smaller variance would give a better estimate of the population mean.

(d) Now, suppose X and Y are instead not independent and have covariance:

Cov (X,Y ) = σXY .

Find α in terms of σX , σY and σXY to minimise V ar (Z).

8. Let xn and s2n denote the sample mean and variance for the sample x1, x2, . . . , xn. Let xn+1 and s2n+1

denote these quantities when an additional observation xn+1 is added to the sample.

(a) Show how xn+1 can be computed from xn and xn+1.

(b) Show that:

s2n+1 =

(n− 1

n

)s2n +

1

n+ 1(xn+1 − xn)2

so that s2n+1 can be computed from xn, xn+1, and s2n.



9. Suppose X and Y are two continuous random variables. Prove that:

E [Y ] =

∫ ∞

−∞E [Y |X = x ] fX (x) dx.

10. You are given:

• X1 ∼ Uniform[0, 1]

• Conditional on X1, X2 ∼ Uniform[0, X1]

(a) Find the joint distribution function of X1 and X2.

(b) Find the marginal distribution function of X2.

11. Suppose that the joint distribution function of X1 and X2 is given by

FX1,X2 (x1, x2) =

0, if x1 < 0 or x2 < 0;x1x2

[1 + 1

2 (1− x1) (1− x2)], if 0 ≤ x1 ≤ 1 and 0 ≤ x2 ≤ 1;

Fx1(x1), if x2 > 1;Fx2(x2), if x1 > 1,

(a) Find the joint density.

(b) Find the marginal distribution functions of X1 and X2. Can you recognise these distributions?

(c) Find the correlation coefficient of X1 and X2.

12. We have the joint probability density function:

fX1,X2 (x1, x2) =

k(1− x2), if 0 ≤ x1 ≤ x2 ≤ 1;0, else.

(a) Determine the value k for which this function is a density.

(b) Determine the region for the integral for determining Pr(X1 ≤ 3/4, X2 ≥ 1/2).

(c) Calculate Pr(X1 ≤ 3/4, X2 ≥ 1/2).





1. Compound Distribution. In a portfolio of insurance policies, the amount of claim is a random variable

Xk which has an exponential distribution with mean1

θ, for k = 1, 2, . . . The number of claims N in a

single period is also a random variable but with a Poisson(λ) . The total claims then in the portfolioduring the period is given by:

S = X1 +X2 + . . .+XN .

(a) Find the mean of S, E[S].

(b) Find the variance of S, V ar(S).

(c) Find the moment generating function of S, MS (t).

2. Let X1, X2 and X3 be i.i.d. with common density:

fX (x) = e−x, x ≥ 0.

(a) Find the joint density of X(1), and X(3).Hint: First find the joint density of X(1), X(2), and X(3), i.e., fX(1),X(2),X(3)

(y1, y2, y3) = . . .Second you find the distribution of only X(1) and X(3) by integrating over the other randomvariable (similar to finding the marginal distribution). Be careful by the limits for X(2), what arethe lowest and highest numbers it can take?

(b) Compute E[X(1)

]and E

[X(3)

].

(c) Compute V ar(X(1)

)and V ar

(X(3)

).

(d) Compute E[X(1)X(3)

]and the correlation coefficient ρ

(X(1), X(3)

).

3. Let X ∼ Gamma(α, 1) and Y ∼ Gamma(β, 1) be independent random variables. Define U = X + Yand V = X/(X + Y ).

(a) Use the moment generating function technique to find the distribution of U .

(b) Use the Jacobian transformation technique to find the joint distribution of U and V .

(c) Show that U and V are independent.Hint: You do not need to do any additional calculations to show this.

(d) Find the marginals of U and V using their joint distribution derived in Question 3b. Demonstratethe the marginal of U is consistent with that derived from Question 3a.

(e) Use Question 3c. and Question 3d. to find the mean and variance of V .

4. Let X1, X2 and X3 be three independent and identically distributed as Exp(1) random variables. Find:

(a) E[X(3)

∣∣X(1) = x]

(b) E[X(1)

∣∣X(3) = x]

(c) fX(1),X(3)(x, y)

(d) fR (r), where R = X(3) −X(1) is the range.

5. Let X1 and X2 be i.i.d. (independent and identically distributed) N (0, 1) random variables.

(a) Show that X1 +X2 has a normal distribution and specify its parameters.

(b) Show that X1 −X2 has the same distribution as X1 +X2.

(c) Suppose X1 and X2 are no longer independent but each still has N (0, 1) distribution. WillX1 +X2 and X1 −X2 be still independent?



(d) Let X ∼Gamma(α, β) distributed.

1. Find the p.d.f. of an Inverse Gamma Distribution, i.e., find the p.d.f. of Y = 1X .

2. Find the c.d.f. of the inverse gamma distribution as function of the c.d.f. of the gammadistribution.

6. I Let Z1 and Z2 be two independentN (0, 1) random variables and let V1 ∼ χ2 (r1) and V2 ∼ χ2 (r2)be two independent chi-squared random variables. Which of the following random variables hasa t-distribution with degrees of freedom (r1 + r2)?

(A)Z1 + Z2√

(V1 + V2) /(r1 + r2)

(B)Z1 + Z2√

(V1 /r1 ) + (V2 /r2 )

(C)Z1 + Z2√

2 (V1 + V2) /(r1 + r2)

(D)Z1 − Z2√

(V1 + V2) /(r1 + r2)

(E)Z1√V1 /r1

+Z2√V2 /r2

II Let Z1 and Z2 be two independent standard normal random variables. Which of the followingcombinations of the two has also a standard normal random variable?

(A) (Z1 + Z2) /2

(B) Z1 + Z2

(C) Z1/Z2

(D) Z1 − Z2

(E) (Z1 − Z2) /√2

III Let Z1 ∼ N (0, 1) and Z2 ∼ N (0, 1) be two random variables with correlation coefficient

ρ (Z1, Z2) = ρ,

where −1 ≤ ρ ≤ 1. Let V be a χ2 (r) random variable independent of Z1 and Z2.Which of the following has a t-distribution with r degrees of freedom?

i.√rZ1V

−1/2

ii.√rZ2V

−1/2

iii.

√r

2(Z1 + Z2)V

−1/2

iv.

√r

2 (ρ+ 1)(Z1 + Z2)V

−1/2

(A) All but i

(B) All but ii

(C) All but iii

(D) All but iv

(E) All

IV Let X1, X2, . . . , Xn be i.i.d. (independent and identically distributed) Exp(λ) random variables

(m.g.f.: MXi(t) =(1− t

λ

)−1). Which of the following describes the distribution of the sample

mean:

X =1

n

n∑

k=1

Xk?

(A) X ∼Exp(λ)(B) X ∼Exp(nλ)(C) X ∼Exp(λ/n)(D) X ∼Gamma(n, λ · n)(E) X ∼Gamma

(n, λn

)

Note: m.g.f. of Gamma: MXi(t) =(1− t

λ

)−n).



V Let X1, . . . , Xn be n independent and identically distributed Poisson random variables with meanλ. Describe the distribution of the sum of these random variables:

S =

n∑

k=1

Xk.

(A) S ∼ Poisson(1)

(B) S ∼ Poisson(λ)

(C) S ∼ Poisson(λ/n)

(D) S ∼ Poisson(nλ)

(E) Cannot be determined from the given information

VI Suppose X1, X2, . . . , X20 are twenty independent random variables and are identically distributedas Exp(2). Determine Pr

(X(20) > 1

).

(A) Pr(X(20) > 1

)= 0.94

(B) Pr(X(20) > 1

)= 0.95

(C) Pr(X(20) > 1

)= 0.96

(D) Pr(X(20) > 1

)= 0.97

(E) Pr(X(20) > 1

)= 0.98

VII Let X1, X2, . . . , Xn be n i.i.d. (independent and identically distributed) random variables eachwith density:

fX (x) = 2x, for 0 < x < 1,

and zero otherwise.Determine E

[X(n)

].

(A) E[X(n)

]= n/(n+ 2)

(B) E[X(n)

]= n/(n+ 1)

(C) E[X(n)

]= 1

(D) E[X(n)

]= 2n/(2n+ 1)

(E) E[X(n)

]= 2n/(n+ 1)

VIII In a 100-meter Olympic race, the running times are considered to be uniformly distributed between8.5 and 10.5 seconds. Suppose there are 8 competitors in the finals. The current world record is9.9 seconds.Determine the probability that the loser of the race will not break the world record.

(A) 0.54

(B) 0.64

(C) 0.74

(D) 0.84

(E) 0.94





1. An insurance company has a portfolio of 100 insurance contracts. The company’s losses on thesecontracts are independent and identically distributed. Each loss X has an exponential distributionwith mean 5 000 and each policyholder pays a premium of 5 050. Notice that each policyholder paysan amount larger than its expected loss. Determine the probability that the aggregate loss of theinsurance company will exceed the total premiums collected. Use the normal approximation.

2. Assume that X1, X2, . . . , Xn is a random sample from a population with density:

fX (x|θ) =

2(θ − x)θ2

, for 0 < x < θ;

0, otherwise.

Find an estimator for θ using the method of moments.

3. Let X1, X2, . . . be a sequence of independent random variables with common mean E[Xk] = µ butdifferent variance V ar(Xk) = σ2

k. Suppose:

1

n2

n∑

k=1

σ2k → 0, as n→∞.

Prove Xp→ µ in probability.

4. A drunkard executes a “random walk” in the following manner: each minute, he takes a step north orsouth, with probability 1

2 each, and his successive step directions are independent. Each step he takesis of length 50 cm. Use the central limit theorem to approximate the probability distribution of hislocation after one hour. Where is he most likely to be?

5. Consider N independent random variables each having a binomial distribution with parameters n = 3and θ so that:

Pr (Xi = k) =

(3

k

)θk (1− θ)n−k

,

for i = 1, 2, . . . , N and k = 0, 1, 2, 3, and zero otherwise. Assume that of these N random variablesn0 take the value 0, n1 take the value 1, n2 take the value 2, and n3 take the value 3 with N =n0 + n1 + n2 + n3.

(a) Use maximum likelihood to develop a formula to estimate θ.

(b) Assume that when you go to the races that you always bet on 3 races. You have taken a randomsample of your last 20 visits to the races and find that you had no winning bets on 11 visits, onewinning bet on 7 visits, and two winning bets on 2 visits. Estimate the probability of winning onany single bet.

6. Assume that we have n independent observations y⊤ = [y1, y2, . . . , yn], each with the Pareto p.d.f.given by:

fYi|α(yi|α;A) =αAα

yα+1i

,

where 0 < α <∞ and 0 < A < yi <∞, and zero otherwise. You are now told the value of A, leavingα as the only unknown parameter.



(a) Explain why the likelihood function L(α; y,A) can be written as:

αnAnα

Gn(α+1),

where G = (y1y2 · . . . · yn)1/n is the geometric mean of the observations.

(b) Explain why we can express the relationship between the posterior distribution, prior distributionand likelihood function as follows:

π(α|y;A) ∝ fY |α(y|α;A)π(α).

(c) We assume our prior pdf for α is such that log(α) is uniformly distributed, implying:

π(α) ∝ 1

α, 0 < α <∞.

Show that the posterior pdf for α is:

π(α|y;A) ∝ αn−1e−anα,

where a = log(G/A).

(d) Explain why the posterior pdf is given by:

π(α|y;A) = (an)n

Γ(n)αn−1e−anα, 0 < α <∞.

(e) Calculate the Bayes estimator of α, αB .

7. Using moment generating functions:

(a) show that as n → ∞, p → 0 and np → λ, the binomial distribution with parameters n and ptends to the Poisson distribution.

(b) show that as α → ∞, the gamma distribution with parameters α and β, properly standardised,tends to the standard Normal distribution.

8. A random variable X with p.d.f.

fX (x) =1

π (1 + x2), for −∞ < x <∞

is said to have a Cauchy distribution. It is well-known that for Cauchy distribution, its mean does notexist. Furthermore, suppose X1, X2, . . . , Xn are n independent Cauchy random variables, then it canbe shown that the sample mean:

Xn =1

n

n∑

k=1

Xk

also has a Cauchy distribution.3 Deduce then that from these results, the Cauchy violates the law oflarge numbers. Explain why.

9. Given that there are n realizations of xi,where i = 1, 2, . . . , n. We know that xi|p ∼Ber(p) andp ∼ U(0, 1).

(a) Find the Bayesian estimator for p.

(b) Find the Bayesian estimator for p(1− p).(c) Why might we be interested in the Bayesian estimator for p(1−p)? Hint: consider the case when

n is large.

10. Let X1, X2, . . . be independent random variables with common density:

fX (x) = αx−(α+1), for x > 1,

where α > 0. Define a new sequence of random variables:

Yn =1

n1/αX(n),

where X(n) is the highest observation of n i.i.d. r.v. X1, . . . , Xn.Show that Yn converges in distribution as n→∞ and find the limiting distribution.

3Proofs of these results are not expected for this course.



11. (Problem from Rice) Suppose that X1, X2, . . . , X20 are independent random variables with densityfunctions:

fX (x) = 2x, for 0 ≤ x ≤ 1,

and zero otherwise. Let S = X1 + . . .+X20. Use the central limit theorem to approximate

Pr (S ≤ 10) .

12. Suppose that X follows a geometric distribution, with probability mass function:

Pr(X = k) = p · (1− p)k−1 , if k = 1, 2, . . ., and zero otherwise,

and assume a sample of size n.

(a) Find the method of moments estimator of p.

(b) Find the maximum likelihood estimator of p.

13. The Pareto distribution is often used in economics as a model for a density function with a slowlydecaying tail. Its density is given by:

fX(x|θ) = θ · xθ0 · x−θ−1, x ≥ x0, θ > 1,

and zero otherwise. Assume that x0 > 0 is given and that x1, . . . , xn is a sample from this distribution.

(a) Find the method of moments estimate of θ.

(b) Find the maximum likelihood estimator of θ.

14. Using the p.d.f. of a chi-squared distribution with one degree of freedom:

fY (y) =exp(−y/2)√

2πy, if y > 0,

and zero otherwise, prove that the moment generating function of Y is given by:

MY (t) = (1− 2t)−1/2.

15. Prove that:

tn−1d→ N(0, 1) as n→∞,

where you might use that:

limn→∞

Γ(n+12

)

Γ(n2

) =

√n

2.

16. Prove that the p.d.f. of a Snecdor’s F distribution, given by the transformation:

F =U/n1

V/n2,

where U ∼ χ2(n1) and V ∼ χ2(n2), is given by:

fF (f) = nn1/21 · nn2/2

2 · Γ((n1 + n2)/2)

Γ(n1/2) · Γ(n2/2)· fn1/2−1

(n2 + fn1)(n1+n2)/2

.





1. Let X1, X2, . . . , Xn be a random sample from an exponential distribution with:

fX (x|θ) = θe−θx, x > 0,

and zero otherwise, where θ > 0. Find the value of a so that the interval from 0 to a/X provides a

95% confidence interval for the parameter θ.

2. Consider a random sampling from a normal distribution with mean µ and variance σ2. Derive a100 (1− α)% confidence interval of σ2 when µ is known.

3. This exercise aims to show that if we sample from a continuous distribution, a pivotal quantity alwaysexists. Let X1, X2, . . . , Xn be a random sample from a continuous distribution fX (x|θ). Denote thecorresponding cumulative distribution function by:

FX (x|θ) =∫ x

−∞fX (z|θ) dz.

(a) Show that FX (X |θ) ∼ U (0, 1).Hint: Show that Pr(FX (X |θ) ≤ x) = x using the quantile function (inverse of the c.d.f.),then explain why x (representing a probability, taking values between 0 and 1) would have thisdistribution.

(b) Show that W = log (1/FX (X |θ)) has an exponential distribution with mean 1. To do so, firstfind the c.d.f c.d.f. W .

(c) From (b), deduce thatn∑

k=1

log (1/FX (Xk|θ)) has a Gamma distribution. Specify its parameters.

(d) Use (c) to prove that there will always be a pivotal quantity when sampling from a continuousdistribution.

4. (modified based on a past Institute of Actuaries exam.) Let X1, X2, . . . , Xn denote a random sampleof a Gamma(3, λ) and X is the sample mean.

(a) Describe the distribution of the sample mean X.

(b) Use (a) to construct a lower 95% confidence interval for λ, of the form (0, U) .

(c) Use (a) to construct an upper 95% confidence interval for λ, of the form (L,∞).

(d) Use (a) to construct a 95% confidence interval for λ, of the form (L,U) where L and U are notnecessarily equal to those found in (b) and (c).

(e) Evaluate the intervals in (b), (c) and (d) in the case for which the total of a random sample of 20

observations yielded a value of∑20

k=1 xk = 98.2.

5. A local health club advertises that its members lose at least 10 pounds on the average during a 30-dayweight loss programme. After receiving a number of complaints from people who were enticed to jointhe club, the Better Business Bureau sends out a representative to the club to check out the claim.



The representative sampled the following nine (9) people who are enrolled in the program:

Person Before-Weight After-Weight Diffrence1 157 150 72 174 167 73 198 187 114 205 198 75 147 146 16 165 153 127 212 199 138 169 171 -29 158 156 2∑9

i=1 xi 1,585 1,527 58∑9i=1 x

2i 283,457 262,465 590

The representative of Better Business Bureau reported its findings in terms of a confidence interval.Construct the appropriate 95% confidence interval for the average weight loss for participants in theprogramme.

6. (Past Institute of Actuaries Exam Question) Independent random samples of size n1 and n2 are takenfrom the normal populations N

(µ1, σ

21

)and N

(µ2, σ

22

). Let the sample means be X1 and X2 and

the sample variances be S21 and S2

2 . You may assume that Xl and S2l , l = 1, 2 are independent and

distributed as follows:

Xk ∼ N

(µk,

σ2k

nk

)and

(nk − 1)S2k

σ2k

∼ χ2 (nk − 1) for k = 1, 2.

(a) It is required to construct a confidence interval for (µ1 − µ2), the difference between the populationmeans.

i. Suppose that σ21 and σ2

2 are known. State the distribution of(X1 −X2

)and write down a

suitable pivotal quantity together with its sampling distribution. Hence, write down a 95%confidence interval for (µ1 − µ2).

ii. Suppose that σ21 and σ2

2 are unknown but are known to be equal. State the definition of atk variable in terms of independent N(0, 1) and χ2

k variables and use it to develop a suitablepivotal quantity. Hence, write down a 95% confidence interval for (µ1 − µ2).

(b) It is required to construct a confidence interval forσ21

σ22

, the ratio of the population variances.

State the definition of an Fk,l variable in terms of independent χ2k and χ2

l variables and use it to

develop a suitable pivotal quantity. Hence, obtain a 90% confidence interval forσ21

σ22

.

(c) A regional newspaper included a consumer rights article comparing the cost of shopping in “cornershops” and “supermarkets”. The researchers investigated the price of a standard “selection” ofhousehold goods in a sample of 10 corner shops selected at random from the region, and in asample of 10 supermarkets selected at random from the region. The data yielded the followingvalues:

Sample Mean Sample S.D.Corner Shops 22.55 1.22Supermarkets 19.72 0.96

i. Use the result in part (a)(ii) to calculate a 95% confidence interval for (µ1 − µ2), the differencebetween the population means (1 = corner shops, 2 = supermarkets).

ii. Use your result in part (b) to calculate a 90% confidence interval forσ21

σ22

, the ratio of the

population variances. Use this result to comment briefly on the assumption of equal variancesrequired for the confidence interval in part (c)(i).

7. (IoA, Subject CT3, April 2005, No.6) In a survey conducted by a mail order company a random sampleof 200 customers yielded 172 who indicated that they were highly satisfied with the delivery time oftheir orders.Calculate an approximate 95% confidence interval for the proportion of the company’s customers whoare highly satisfied with delivery times.



8. (IoA, Subject CT3, April 2005, No.8) The distribution of claim size under a certain class of policy ismodelled as a normal random variable, and previous years records indicate that the standard deviationis £120.

(a) Calculate the width of a 95% confidence interval for the mean claim size if a sample of size 100 isavailable.

(b) Determine the minimum sample size required to ensure that a 95% confidence interval for themean claim size is of width at most £10.

(c) Comment briefly on the comparison of the confidence intervals in (a) and (b) with respect towidths and sample sizes used.

9. (IoA, Subject CT3, April 2005, No.12 (partial))

(a) A random variable Y has a Poisson distribution with parameter but there is a restriction thatzero counts cannot occur. The distribution of Y in this case is referred to as the zero-truncatedPoisson distribution.

1. Show that the probability function of Y is given by:

pY (y) =θye−θ

y!(1− e−θ), for y = 1, 2, 3, . . . ,

and zero otherwise.

2. Show that E[Y ] =θ

1− e−θ.

(b) Answer the following.

1. Let y1, . . . , yn denote a random sample from the zero-truncated Poisson distribution. Showthat the maximum likelihood estimate of θ may be determined by the solution to the followingequation:

y − θ − θe−θ

1− e−θ= 0,

and deduce that the maximum likelihood estimate is the same as the method of momentsestimate.

2. Obtain an expression for the Cramer-Rao lower bound for the variance of an unbiased esti-mator of θ.

10. (IoA, Subject 101, April 2004, No.12) For the estimation of a bernoulli probability p = Pr(success), aseries of n independent trials are performed and X represents the number of successes observed.

(a) Write down the likelihood function L(p;x) and show that the maximum likelihood estimator(MLE) of p is p = X/n.

(b) Answer the following.

1. Determine the Cramer-Rao lower bound for the estimation of p.

2. Show that the variance of the MLE is equal to the Cramer-Rao lower bound.

3. Write down an approximate sampling distribution for p valid for large n.

(c) In order to develop an approximate 95% confidence interval for p for large n, the following pivotalquantity is to be used:

p − p√p(1− p)

n

≈ N(0, 1).

Assuming that this pivotal quantity is monotonic in p, show that rearrangement of the inequality:

−1.96 <p − p

√p(1− p)

n

< 1.96

leads to a quadratic inequality in p, and hence determine an approximate 95% confidence intervalfor p.



(d) A simpler and more widely used approximate confidence interval is obtained by using the followingpivotal quantity

p − p√p(1− p)

n

≈ N(0, 1).

Determine the resulting approximate 95% confidence interval using this.

(e) In two separate applications the following data were observed:

1. 4 successes out of 10 trials

2. 80 successes out of 200 trials

In each case calculate the two approximate confidence intervals from parts (c) and (d) and com-ment briefly on your answers.

11. A random sample of 16 values, x1, x2, . . . , x16, was drawn from a normal population and gave thefollowing summary statistics:

16∑

i=1

xi = 51.2

16∑

i=1

x2i = 243.19

Calculate a 95% confidence interval for the population mean.

12. Consider a random sample of size n from a normal distribution N(µ, σ2) and let S2 denote the samplevariance.

(a) State the sampling distribution for(n− 1)S2

σ2and specify an approximate sampling distribution

for this expression when n is large.

(b) For n = 101 calculate an approximate value for the probability that S2 exceeds σ2 by more thana factor of 10%, i.e. Pr(S2 > 1.1σ2).

13. A group of 500 insurance policies gave rise to a total of 83 claims during the last year. Assuming aPoisson model for the occurrence of claims, calculate an approximate 95% confidence interval for λ,the claim rate per policy per year.

14. Let Xi, i = 1, . . . , n denote a random sample of size n from a population with a uniform distributionon the interval (0, θ). Let X(n) = maxX1, . . . , Xn and define U = (1/θ)X(n).

(a) Show that U has distribution function:

FU (u) =

0, if u < 0;un, if 0 ≤ u ≤ 1;1, if u > 1.

(b) Because the distribution of U does not depend on θ, U is a pivotal quantity. Find the 95% lowerconfidence bound for θ.





1. Explain carefully the distinction between each of the following pairs of terms:

(a) null and alternative hypotheses;

(b) one-tailed and two-tailed hypotheses;

(c) simple and composite hypotheses;

(d) Type I and Type II errors;

2. Let X1, X2, . . . , X10 be a random sample of size 10 from a Poisson distribution with mean λ. Considerthe critical region C defined by:

C =

(x1, x2, . . . , x10) :

10∑

k=1

xk ≥ 3

.

(a) Show that C is a best critical region for testing H0 : λ = 0.1 against Ha : λ = 0.5.

(b) Determine the level of significance for this test.

3. Let X1, X2, . . . , Xn be a random sample from the density function:

fX (x|θ) = 1√2π

exp

(−1

2(x− θ)2

).

At a level of significance α, find the best critical region (or most powerful test) for testing the simplenull H0 : θ = 0 against the simple alternative Ha : θ = 1.

4. Let X1, X2, . . . , Xn be a random sample from a Poisson(λ) distribution. In testing the simple nullH0 : λ = λ0 against the simple alternative Ha : λ = λ1, where λ1 > λ0 :

(a) Find the best critical region (or most powerful test).

(b) Determine the distribution of the test statistic under the null hypothesis.

5. Past Institute exam

(a) A manufacturing company produces screws of a particular size which are put into boxes of 150. Ona particular day a random sample of such boxes is taken from each of the morning and afternoonproduction runs. The number of defective screws found in each sampled box are given in thefollowing table:

Morning 28 17 18 16 20 12 11 10 18 17 20 25Afternoon 19 15 22 21 9 14 17 13 22 9

Table 2: Number of defectives per box

1. Test for a difference between the mean number of defectives produced in the morning andafternoon (you may assume that the underlaying population variances are equal).

2. Plot the data in an appropriate and simple way and comment briefly on the validity of thetest of part ii).



(b) On another day screws are put into boxes of 100. The table below gives the number of defectivesin twenty boxes sampled from this day’s production run.

5 15 18 12 8 7 9 14 11 106 18 14 9 18 12 11 5 18 12

Table 3: Number of defectives per box of 100 screws

1. Carry out a test to establish whether there is a difference between the proportions of defectivesproduced on the two days.

2. Carry out a test to establish whether the proportion of defectives in boxes of 100 screws ismore than 9%.

6. Past Institute examWhen comparing the mean premiums for policies issued by two companies, a two-sample t test ispreformed assuming equal population variances. The sample sizes and sample variances are given by:n1 = 25, s21 = 139.7n2 = 30, s22 = 76.6

Preform an approximate F test at the 5% level to investigate the validity of the equal variance as-sumption.

7. Past Institute examThe following data refers to an outbreak of botulism, a form of food poisoning that may be fatal. Eachsubject is a person who contracted botulism in the outbreak. The variables recorded are the subject’sage in years, the time in hours between eating the infected food and the first signs of illness (incubationperiod) and whether the subject survived (denoted by survival category Y) or died (denoted by survivalcategory N).

Subject 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18Age (x) 29 39 44 37 42 17 38 43 51 30 32 59 33 31 32 32 36 50Incubation 13 46 43 34 20 20 18 72 19 36 48 44 21 32 86 48 28 16period (y)Survival N Y Y N N Y N Y N N N Y N N Y N Y N

Died:∑x = 405

∑y = 305

∑x2 = 15517

∑y2 = 10035

Survived:∑x = 270

∑y = 339

∑x2 = 11396

∑y2 = 19665

(a) A scatterplot of incubation period against age is given below, in which different symbols are usedfor subjects who died and for subjects who survived.



15 20 25 30 35 40 45 50 55 6010

20

30

40

50

60

70

80

90

Age

Incu

b P

er

A Plot of Incubation Period against age

DiedSurvived

Comment briefly on any relationships between age and incubation period for those subjects whodied and for those who survived.

(b) Construct suitable dotplots to investigate any relationship between:

1. age and survival, and

2. incubation period and survival

and make a brief informal comparison of the died and survived groups based on these dotplots.

(c) Construct a 95% and 99% confidence intervals for the mean difference between the incubationperiod for subjects who survived and subjects who dies (i.e., take the mean incubation period forsubjects who survived minus the mean incubation period for subjects who died).Comment briefly on these confidence intervals.

(d) 1. Construct a test to investigate whether the variances of the incubation period for subjectswho died and subjects who survived are equal.

2. Comment on the validity of the assumptions that are required for the confidence intervalsgiven in part c) to be approximate.

8. Past Institute examIt is desired to investigate the level of premium charged by two companies for contents policies forhouses in a certain area. Random samples of 10 houses insured by company A are compared with 10similar houses insured by company B. The premiums charged in each case are as follows:

Company A 117 154 166 189 190 202 233 263 289 331Company B 142 160 166 188 221 241 276 279 284 302

For these data:∑A = 2, 134,

∑A2 = 494, 126,

∑B = 2, 259,

∑B2 = 541, 463.

(a) Illustrate the data given above on a suitable diagram and hence comment briefly on the validityof the assumptions required for a two-sample t test for the premiums of these two companies.

(b) Assuming that the premiums are normally distributed, carry out a formal test to check that it isappropriate to apply a two-sample t test to these data.

(c) Test whether the level of premiums charged by company B was higher than that charged bycompany A. State your p-value and conclusions clearly.



(d) Calculate a 95% confidence interval for the difference between the proportions of premiums ofeach company that are in excess of £200. Comment briefly on your result.

(e) The average premium charged by company A in the previous year was £170. Formally testwhether company A appears to have increased its premium since the previous year.





1. Explain carefully the distinction between the significance level and power.

2. Let X have a Bernoulli distribution where θ = Pr (X = 1). Take a random sample of size n = 10 fromthis Bernoulli distribution and consider the test:

H0 : θ ≤ 1/2 versus Ha : θ > 1/2.

Using the critical region C =

(X1, X2, . . . , X10) :

10∑k=1

xk ≥ 6

:

(a) Find the power function and sketch it.

(b) Find the size of this test.

3. Recall from week 7 tutorial material question 2:Let X1, X2, . . . , X10 be a random sample of size 10 from a Poisson distribution with mean λ. Considerthe critical region C defined by:

C =

(x1, x2, . . . , x10) :

10∑

k=1

xk ≥ 3

.

Determine the power of the test under Ha.

4. Recall from week 7 tutorial material question 3:Let X1, X2, . . . , Xn be a random sample from the density function:

fX (x|θ) = 1√2π

exp

(−1

2(x− θ)2

).

Determine the power of this test.

5. Prove:

k∑

j=1

nj∑

i=1

(xij − x)2 =

k∑

j=1

nj∑

i=1

(xij − xj)2 +k∑

j=1

nj(xj − x)2

where

xj =

nj∑

i=1

xijnj

x =

k∑

j=1

nj∑

i=1

xijN

=

k∑

j=1

njxjN

Hint: 1) Rewrite the left side by adding and subtracting within the squares xj ;2) Rewrite is using Binomial expansion (see F&T page 2).

6. Given is that:

κ =m4 − 4m1m3 + 6m2m

21 − 3m3

1

(m2 −m21)

2= h(m1,m2,m3,m4)

using the m.g.f. one can easily show that:

E[[m1 m2 m3 m4]⊤] = [0 1 0 3]⊤

Thus:√n · [m1 m2 − 1 m3 m4 − 3]→ N(0,Σ).



(a) Find Σ = E[n · [m1 m2 − 1 m3 m4 − 3] · [m1 m2 − 1 m3 m4 − 3]⊤

].

(b) Find ∇h(m1,m2,m3,m4).

(c) Show ∇h(0, 1, 0, 3) = [0 − 6 0 1]⊤.

(d) Let Xi ∼Bin(n = 10, p = 0.2) be i.i.d. for i = 1, . . . , n. In the table below are some sum-mary statistics of four samples. For which samples can you reject that the sample is normallydistributed? Use the chi-squared approximation.

sample 1 sample 2 sample 3 sample 4n 10 25 50 100x 1.6 2.04 1.8 1.92∑n

i=1 x2i 24.4 34.96 56 167.36∑n

i=1 x3i 31.92 14.8 6 116.22∑n

i=1 x4i 161.39 140.3 157.28 822.76∑n

i=1(xi − x)2 15.25 17.14 31.11 87.17∑ni=1(xi − x)3 19.95 7.25 3.33 60.53∑ni=1(xi − x)4 100.87 68.77 87.38 428.52

(e) For another sample with n = 100 observations we get the value of the test statistic equals 5.4.Find the p-value of this test.

Note that if Z ∼N(0,1) then: E [Z] = 0, E[Z2]= 1, E

[Z3]= 0, E

[Z4]= 3, E

[Z5]= 0, E

[Z6]= 15,

E[Z7]= 0, E

[Z8]= 105.

7. The following observations represent weight loss (in pounds) of men of similar physique, metabolicactivity, and so on, after a certain amount of time on three types of diet programs: A, B, and C.

Diet ProgramA B C3 2 77 4 104 6 85 6 96 5 4- 3 8- 4 -

Test for the differences in the mean weight loss between the three diet programs. State any assumptionsyou make. Provide the point estimates estimates of the mean losses, and the ANOVA table used topartition the various sources of variation.

8. Past Institute examThe following test concerning the mean claim amount (µ) for a certain class of policy:

H0 : µ = £200 v.s. H1 : µ 6= £200,

is to be preformed. A random sample of 50 claims is examined and yields a mean amount of £207 anda standard deviation of £42.Calculate the approximate p-value for the test.





1. A group of 1, 725 school children were cross-classified according to their intelligence and their mannerof clothing. A result of this classification is given below:

dull intelligent very capablevery well clothed 81 322 233well clothed 141 457 153poorly clothed 127 163 48

Test for independence using a 1% level of significance.

2. (a) Past Institute examA 2× 2 contingency table was set up to investigate whether or not two classifications criteria areindependent and resulted in the following data:

I IIA 22 28 50B 28 22 50

50 50 100

Calculate the observed χ2 test statistic and state an approximate conclusion concerning the in-dependence of the two criteria.

(b) (Added to the past Institute exam) Preform the Pearson’s chi-square test using R.

3. Continued from previous question. Using the Fisher’s exact test:

(a) Write down the corresponding hypothesis and test statistic.

(b) Calculate the probability mass function of a Hypergeometric distribution with N = 100, M = 50,n = 50 and x = 22.

(c) Use R to calculate show that the cumulative density function of a Hypergeometric distributionwith N = 100, M = 50, n = 50 and x = 22 equals 0.15867.

(d) Preform the hypothesis testing.

(e) Check your answer using R;

4. Compare the results in question 2. and 3. and explain the differences/similiraties.

5. Past Institute examA particular area in a town suffers a high burglary rate. A sample of 100 streets is taken, and in eachof the sample streets, a sample of six similar houses is taken. The table below shows the number ofsampled houses, which have had burglaries during the last six months.

No. of houses burgled x 0 1 2 3 4 5 6No. of streets f 39 38 18 4 0 1 0

(a) 1. State any assumptions needed to justify the use of a binomial model for the number of sampledhouses per street which have been burgled during the last six months.

2. Derive the maximum likelihood estimator of p, the probability that a house of the typesampled has been burgled during the last six months.

3. Fit the binomial model using your estimate of p, and, without doing a formal test, commenton the fit.



(b) An insurance company works on the basis that the probability of a house being burgled over asix month period is 0.18. Carry out a test to investigate whether the binomial model with thisvalue of p provides a good fit for the data.

6. Check your answer of question 5b) using R.

7. Does education really make a difference in how much money you will earn?4 Researchers randomlyselected 100 people from each of three income categories—‘marginally rich’, ‘comfortably rich’, and“super rich”—and recorded their education levels. The data are summarised in the table that follows.

Highest Marginally ComfortablyEducation Level Rich Rich Super Rich Total

No college 32 20 23 75Some college 13 16 1 30

Undergraduate degree 43 51 60 154Postgraduate study 12 13 16 41

Total 100 100 100 300

(a) Describe the independent multinomial populations whose proportions are compared in the χ2

analysis.

(b) Provide a table with the observed proportions.

(c) Do the data indicate that the proportions in the various education levels differ for the three incomecategories? Test at the α = 0.01 level.

(d) Construct a 95% confidence interval for the difference in proportions with at least an undergrad-uate degree for individuals who are marginally and super rich. Interpret the interval.

(e) Use R to check the answer in c) and d).


4Extended version of [W+] 14.25




1. Consider the exponential regression model with one independent variable:

Yi = α′β′xieǫi for each i = 1, 2, . . . , n,

where the ǫi’s are independent and identically distributed normal random variables with E[ξi] = 0 andV ar(ǫi) = σ2.

(a) Rewrite the exponential regression model as a linear regression model with parameters α and βand describe the relationship between α and α′ and the relationship between β and β′.Derive the following from the linear regression model:

(b) β =∑n

i=1 ci log(yi) where ci = (xi − x)/Sxx and Sxx =∑n

i=1(xi − x)2.(c) E[β|X = x] = β

(d) V ar(β|X = x] = σ2/Sxx

(e) (β|X = x) ∼ N(β1, σ2/Sxx)

(f) E[α|X = x] = α

(g) V ar(α|X = x) = σ2

(1

n+

x2

Sxx

)

(h) (α|X = x) ∼ N

(α, σ2

(1

n+

x2

Sxx

))

(i) Cov(α, β|X = x) = −σ2xSxx

(j) What are the distributions of Y , α′ and β′ conditional on (X = x) using the LSE estimates inthe linear regression model and the relationship found in question (a)?

2. Forensic scientists use various methods for determining the likely time of death from post-mortemexamination of human bodies. A recently suggested objective method uses the concentration of acompound (3-methoxytyramine or 3-MT) in a particular part of the brain.In a study of the relationship between post-mortem interval and the concentration of 3-MT, samplesof the approximate part of the brain were taken from coroners cases for which the time of death hadbeen determined form eye-witness accounts. The intervals (x; in hours) and concentrations (y; inparts per million) for 18 individuals who were found to have died from organic heart disease are givenin the following table. For the last two individuals (numbered 17 and 18 in the table) there was noeye-witness testimony directly available, and the time of death was established on the basis of otherevidence including knowledge if the individuals’ activities.



Observation Interval Concentrationnumber (x) (y)1 5.5 3.262 6.0 2.673 6.5 2.824 7.0 2.805 8.0 3.296 12.0 2.287 12.0 2.348 14.0 2.189 15.0 1.9710 15.5 2.5611 17.5 2.0912 17.5 2.6913 20.0 2.5614 21.0 3.1715 25.5 2.1816 26.0 1.9417 48.0 1.5718 60.0 0.61

∑x = 337

∑x2 = 9854.5

∑y = 42.98

∑y2 = 109.7936

∑xy = 672.8

In this investigation you are required to explore the relationship between concentration (regarded theresponds/dependent variable) and interval (regard as the explanatory/independent variable).

(a) Construct a scatterplot of the data. Comment on any interesting features of the data and discussbriefly whether linear regression is appropriate to model the relationship between concentrationof 3-MT and the interval from death.

(b) Calculate the correlation coefficient for the data, and use it to test the null hypothesis that thepopulation correlation coefficient is equal to zero.

(c) Calculate the equation of the least-squares fitted regression line and use it to estimate the con-centration of 3-MT:

1. after 1 day and

2. after 2 days.

Comment briefly on the reliability of these estimates.

(d) Calculate a 99% confidence interval for the slope of the regression line. Using this confidenceinterval, test the hypothesis that the slope of the regression line is equal to zero. Comment onyour answer in relation to the answer given in part b) above.

3. Past Institute exam.Consider a linear regression model in which responses Yi are uncorrelated and have expectations βxiand common variance σ2 (i = 1, . . . , n), i.e. Yi is modelled as a linear regression through the origin:

E[Yi|xi] = βxi and V (Yi|xi) = σ2 (i = 1, . . . , n).

(a) 1. Show that the least squares estimator of β is β1 =∑n

i=1 xiYi/∑n

i=1 x2i .

2. Derive the expectation and variance of β1 under the model.

(b) An alternative to test the least squares estimator in this case is:

β2 =

n∑

i=1

Yi/

n∑

i=1

xi = Y /x.

1. Derive the expectation and variance of β2 under the model.

2. Show that the variance of the estimator β2 is at least as large as that if the least squaresestimator β1.

(c) Now consider an estimator β3 of β which is a linear function of the responses, i.e. an estimator

which has the form β3 =∑n

i=1 aiYi, where a1, . . . , an are constants.

1. Show that β3 is unbiased for β if∑n

i=1 aixi = 1, and that the variance of β3 is∑n

i=1 a2iσ

2.



2. Show that the estimators β1 and β2 above may be expressed in the form β3 =∑n

i=1 aiYi and

hence verify that β1 and β2 satisfy the condition for unbiasedness in c)i).

3. It can be shown that, subject to condition∑n

i=1 aixi = 1, the variance of β3 is minimised bysetting ai = xi/

∑ni=1 x

2i . Comment on this result.

4. A university wishes to analyse the performance of its students on a particular degree course. It recordsthe scores obtained by a sample of 12 students at the entry to the course, and the scores obtained intheir final examinations by the same students. The results are as follows:

Student A B C D E F G H I J K LEntrance exam score x (%) 86 53 71 60 62 79 66 84 90 55 58 72Final paper score y (%) 75 60 74 68 70 75 78 90 85 60 62 70

∑x = 836

∑y = 867

∑x2 = 60, 016

∑y2 = 63, 603

∑(x− x)(y − y) = 1, 122

(a) Calculate the fitted linear regression equation of y on x.

(b) Assuming the full normal model, calculate an estimate of the error variance σ2 and obtain a 90%confidence interval for σ2.

(c) By considering the slope parameter, formally test whether the data is positively correlated.

(d) Find a 95% confidence interval for the mean finals paper score corresponding to an individualentrance score of 53.

(e) Test whether this data come form a population with a correlation coefficient equal to 0.75.

(f) Calculate the proportion of variance explained by the model. Hence, comment on the fit of themodel.

5. Complete the following ANOVA table for a simple linear regression with 60 observations:

Source D.F. Sum of Squares Mean Squares F-Ratio

Regression

Error 8.2

Total 639.5

6. Suppose you are interested in relating the accounting variable EPS (earnings per share) to the marketvariable STKPRICE (stock price). Then, a regression equation was fitted using STKPRICE as theresponse variable with EPS as the regressor variable. Following is the computer output from yourfitted regression. You are also given that: x = 2.338, y = 40.21, Sx = 2.004, and Sy = 21.56.

Regression Analysis

The regression equation is

STKPRICE = 25.044 + 7.445 EPS

Predictor Coef SE Coef T p

Constant 25.044 3.326 7.53 0.000

EPS 7.445 1.144 6.51 0.000

Analysis of Variance

SOURCE DF SS MS F p

Regression 1 10475 10475 42.35 0.000

Error 46 11377 247

Total 47 21851

(a) Calculate the correlation coefficient of EPS and STKPRICE.

(b) Estimate the STKPRICE given an EPS of $2. Provide a 95% confidence interval of your estimate.

(c) Provide a 95% confidence interval for the slope coefficient β.

(d) Compute s and R2.

(e) Describe how you would check if the errors have constant variance.

(f) Perform a test of the significance of EPS in predicting STKPRICE at a level of significance of 5%.



(g) Test the hypothesis H0 : β = 24 against Ha : β > 24 at a level of significance of 5%.

7. (Modified from an Institute of Actuaries exam problem)An insurance company issues house buildings policies for houses of similar size in four different post-code regions A, B, C, and D. An insurance agent takes independent random samples of 10 housebuildings policies for houses of similar size in each of the four regions. The annual premiums (indollars) were as follows:

Region A : 229 241 270 256 241 247 261 243 272 219(∑x = 2, 479,

∑x2 = 617, 163

)

Region B : 261 269 284 268 249 255 237 270 269 257(∑x = 2, 619,

∑x2 = 687, 467

)

Region C : 253 247 244 245 221 229 245 256 232 269(∑x = 2, 441,

∑x2 = 597, 607

)

Region D : 279 268 290 245 281 262 287 257 262 246(∑x = 2, 677,

∑x2 = 718, 973

)

Perform a one-way analysis of variance at the 5% level to compare the premiums for all four regions.State briefly the assumptions required to perform this analysis of variance.

8. You are given the following one-way ANOVA model:

Yij = µ+ αi + εij , for i = 1, . . . , I and j = 1, . . . , J

where the error terms εij are i.i.d. normal random variables with mean 0 and common variance σ2.Using fundamental principles of maximum likelihood, derive the maximum likelihood estimates for allparameters in the model.

9. For the one-way ANOVA model derive the following maximum likelihood estimators:

(a) µ = Y =

I∑

i=1

ni∑

j=1

Yij

I∑

i=1

ni

=

I∑

i=1

ni∑

j=1

Yij

N

(b) αi = Y i. − Y =

ni∑

j=1

Yij

ni

− Y

10. Suppose that Y represents a single observation from the probability density given by:

fY (y|θ) =

θyθ−1, 0 < y < 10, elsewhere.

Find the most powerful test with significance level α = 0.05 to test H0 : θ = 2 against Ha : θ = 1.

11. Past Institute Exam (April 2005)As part of an investigation into health service funding a working party was concerned with the issue ofwhether mortality could be used to predict sickness rates. Data on standardised mortality rates andstandarised sickness rates collected for a sample of 10 regions and are shown in the table below:

Region Mortality rate m (per 100,000) Sickness rate s (per 100,000)1 125.2 206.82 119.3 213.83 125.3 197.24 111.7 200.65 117.3 189.16 100.7 183.67 108.8 181.28 102.0 168.29 104.7 165.210 121.1 228.5



Data summaries:∑m = 1136.1,

∑m2 = 129, 853.03,

∑s = 1934.2,

∑s2 = 377, 700.62, and

∑ms = 221, 022.58.

(a) Calculate the correlation coefficient between the mortality rates and the sickness rates and de-termine the probability-value for testing whether the underlaying correlation coefficient is zeroagainst the alternative that it is positive.

(b) Noting the issue under investigation, draw an appropriate scatterplot for these data and commenton the relationship between the two rates.

(c) Determine the fitted linear regression of sickness rate on mortality rate and test whether theunderlaying slope coefficient can be considered to be as large as 2.0.

(d) For a region with mortality rate 115.0, estimate the expected sickness rate and calculate 95%confidence limits for this expected rate.

12. Past Institute Exam (September 2005)The data given in the following table are the number of deaths from AIDS in Australia for 12 consecutivequarters starting from the second quarter of 1983.

Quarter (i) 1 2 3 4 5 6 7 8 9 10 11 12Number of deaths (ni) 1 2 3 1 4 9 18 23 31 20 25 37

(a) 1. Draw a scatterplot of the data.

2. Comment on the nature of the relationship between the number of deaths and the quater inthis early phase of the epidemic.

(b) A statistician has suggested that a model of the form:

E[ni] = γi2

might be appropriate for these data, where γ is a parameter to be estimated from the data above.She has proposed two methods for estimating γ, and these are given in part i. and ii. below.

1. Show that the least squares estimate of γ, obtained by minimising q =∑12

i=1(ni − γi2)2 isgiven by:

γ =

∑12i=1 i

2ni∑12i=1 i

4.

2. Show that an alternative (weighted) least squares estimate of γ, obtained by minimising

q∗ =∑12

i=1(ni−γi2)2

i2 is given by:

γ =

∑12i=1 ni∑12i=1 i

2.

3. Noting that∑12

i=1 i4 = 60, 710 and

∑12i=1 i

2 = 650, calculate γ and γ for the data above.

(c) To assess whether the single parameter model which was used in part b) is appropriate for thedata, a two parameter model is considered. The model is of the form:

E[Ni] = γiθ

for i = 1, . . . , 12.

1. To estimate the parameters γ and θ, a simple linear regression model

E[Yi] = α+ βxi

is used, where xi = log(i) and Yi = log(Ni) for i = 1, . . . , 12. Relate the parameters γ and θto the regression parameters α and β.

2. The least squares estimates of α and β are -0.6112 and 1.6008 with standard errors 0.4586and 0.2525 respectively (you are not asked to verify these results).Using the value for the estimate β, conduct a formal statistical test to assess whether theform of the model suggested in (b) is adequate.



13. Past institute ExamConsider the following data, which comprise of four groups sizes (y), each comprising four observations,In scenario I, information is also given on the sum assured under the policy concerned - the sum assuredis the same for all four policies in a group. In scenario II, we regard the policies in the different groupsas having been issued by four different companies - the policies in a group are all issued the samecompany.All monetary amounts are in units of £10, 000. Summaries of the claim sizes in each group are givenin a second table.

Group 1 2 3 4Claim sizes y 0.11 0.46 0.52 1.43 1.48 2.05 1.52 2.36

0.71 1.45 1.84 2.47 2.38 3.31 2.95 4.08I: sum assured x 1 2 3 4II: Company A B C D

Summaries of claim sizes:

Group 1 2 3 4∑y 2.73 6.26 9.22 10.91∑y2 2.8303 11.8018 23.0134 33.2289

(a) In scenario I, suppose we adopt the linear regression model

Yi = α+ βxi + ǫi

where Yi is the ith claim size and xi is the corresponding sum assured, i = 1, . . . , 16.

1. Calculate the total sum of squares and its partition into the regression (model) sum of squaresand the residual (error) sum of squares.

2. Fit the model and calculate the fitted values for the first claim size of group 1 (namely 0.11)and the last claim size of group 4 (namely 4.08).

3. Consider a test of the hypothesis H0 : β = 0 against a two-sided alterative. By preforming ap-propriate calculations, assess the strength of the evidence against this “no linear relationship”hypothesis.

(b) In scenario II, suppose we adopt the analysis of variance model

Yij = µ+ τi + eij

where Yij is the jth claim size for company i and τi is the ith company effect, i = 1, 2, 3, 4 andj = A,B,C,D.

1. Calculate the partition of the total sum of squared into the “between companies” (model)sum of squares and the “within companies” (residual/error) sum of squares.

2. Fit the model.

3. Calculate the fitted values for the first claim size of group 1 and the last claim size of group4.

4. Consider a test of hypothesis H0 : τi = 0, i = A,B,C,D against a general alternative.By preforming appropriate calculations, assess the strength of the evidence against this “nocompany effects” hypothesis.





1. Consider the regression model

Yk = βxk + εk, for each k = 1, 2, . . . , n,

that is, the regression with one regressor variable variable but without the intercept term. This modelis called regression through the origin because the true regression line passes through the point (0,0).Derive the least squares estimate of β.Now, consider the quadratic regression model passing through the origin;

Yk = βx2k + εk , for each k = 1, 2, . . . , n.

Use the previous result to determine the least squares estimate of β.

2. Use the following steps to establish a relationship between the coefficient of determination and thecorrelation coefficient:

(a) Show that:

yk − y = β · (xk − x) .

(b) Use this result to show that:

SSM =

n∑

k=1

(yk − y)2 = β2s2x (n− 1) .

where s2x is the sample variance of X .

(c) Use the previous result to establish:

R2 = β2 s2x

S2y

= r2.

where s2x, s2y is the sample variance of X and Y , respectively.

3. In the regression model Yk = α+ βxk + εk, use algebra to establish the following results:

(a) R2 = 1− n− 2

n− 1

s2

s2y, where s2y is the sample variance of Y .

(b) s = sy

√(1− r2) n− 1

n− 2, where sy is the sample standard deviation of Y .

(c) t(β)=

β

se(β) =

√n− 2

√r2

1− r2

4. (a) Write down the design matrix for the simple linear regression model.

(b) Write out the matrix X⊤X for the simple linear regression model.

(c) Write out the matrix X⊤Y for the simple linear regression model.

(d) Write out the matrix (X⊤X)−1 for the simple linear regression model.

(e) Calculate β = (X⊤X)−1X⊤Y using your results above.



5. The following model was fitted to a sample of supermarkets in order to explain their profit levels:

y = β0 + β1x1 + β2x2 + β3x3 + ε

where

y = profits, in thousands of dollars

x1 = food sales, in tens of thousands of dollars

x2 = nonfood sales, in tens of thousands of dollars, and

x3 = store size, in thousands of square feet.

The estimated regression coefficients are given below:

β1 = 0.027 and β2 = −0.097 and β3 = 0.525.

Which of the following is TRUE?

(A) A dollar increase in food sales increases profits by 2.7 cents.

(B) A 2.7 cent increase in food sales increases profits by a dollar.

(C) A 9.7 cent increase in nonfood sales decreases profits by a dollar.

(D) A dollar decrease in nonfood sales increases profits by 9.7 cents.

(E) An increase in store size by one square foot increases profits by 52.5 cents.

6. In a regression model of three explanatory variables, twenty-five observations were used to calculatethe least squares estimates. The total sum of squares and regression sum of squares were found to be666.98 and 610.48, respectively. Calculate the adjusted coefficient of determination.

(A) 89.0%

(B) 89.4%

(C) 89.9%

(D) 90.3%

(E) 90.5%

7. In a multiple regression model given by:

y = β0 + β1x1 + . . .+ βp−1xp−1 + ε,

which of the following gives a correct expression for the coefficient of determination?

I. SSMSST

II. SST−SSESST

III. SSMSSE

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only

8. The ANOVA table output from a multiple regression model is given below:

ANOVA TableSource D.F. SS MS F-Ratio Prob(> F)Regression 5 13326.1 2665.2 13.13 0.000

Error 42 8525.3 203.0

Total 47 21851.4

Compute the adjusted coefficient of determination.



(A) 52%

(B) 56%

(C) 61%

(D) 63%

(E) 68%

9. You have information on 62 purchases of Ford automobiles. In particular, you have the amount paidfor the car (y) in hundreds of dollars, the annual income of the individuals (x1) in hundreds of dollars,the sex of the purchaser (x2, 1 = male and 0 = female), and whether or not the purchaser graduatedfrom college (x3, 1 = yes and 0 = no) . After examining the data and other information available, youdecide to use the regression model:

y = β0 + β1x1 + β2x2 + β3x3 + ε.

You are given that:

(X⊤X

)−1=

0.109564 −0.000115 −0.035300 −0.026804−0.000115 0.000001 −0.000115 −0.000091−0.035300 −0.000115 0.102446 0.023971−0.026804 −0.000091 0.023971 0.083184

and the mean square error for the model is s2 = 30106. Calculate se(β2

).

(A) 0.17

(B) 17.78

(C) 50.04

(D) 55.54

(E) 57.43

10. Suppose in addition to the information in question 9., you are given:

X⊤Y =

9 5584 880 937

7 3966 552

.

Calculate the expected difference in the amount spent to purchase a car between a person who gradu-ated from college and another one who did not.

(A) 233.5

(B) 1 604.3

(C) 2 195.3

(D) 4 920.6

(E) 6 472.1

11. A regression model of y on four independent variables x1, x2, x3 and x4 has been fitted to a dataconsisting of 212 observations and the computer output from estimating this model is given below:

Regression Analysis

The regression equation is

y = 3894 - 50.3 x1 + 0.0826 x2 + 0.893 x3 + 0.137 x4

Predictor Coef SE Coef T

Constant 3893.8 409.0 9.52

x1 -50.32 9.062 -5.55

x2 0.08258 0.02133 3.87

x3 0.89269 0.04744 18.82

x4 0.13677 0.05303 2.58

Which of the following statement is NOT true?



(A) All the explanatory variables has a positive influence on y.

(B) The variable x1 is a significant variable.

(C) The variable x2 is a significant variable.

(D) The variable x3 is a significant variable.

(E) The variable x4 is a significant variable.

12. In a multiple regression model, which of the following gives a correct expression for the unbiasedestimate of σ2?

(A) 1n−p+1

(Y −Xβ

)⊤ (Y −Xβ

)

(B) 1n−p+1

(Y − Y

)⊤ (Y − Y

)

(C) 1n−1Y

⊤Y

(D) 1n−1

(Y − Y

)⊤ (Y − Y

)

(E) 1n−p

(Y −Xβ

)⊤ (Y −Xβ

)

Note: p is the rank of X .

13. The estimated regression model of fitting life expectancy from birth (LIFE EXP) on the country’sgross national product (in thousands) per population (GNP) and the percentage of population livingin urban areas (URBAN%) is given by:

LIFE EXP = 48.24 + 0.79 GNP + 0.154 URBAN%.

For a particular country, its URBAN% is 60 and its GNP is 3.0. Calculate the estimated life expectancyat birth for this country.

(A) 49

(B) 50

(C) 57

(D) 60

(E) 65

14. What is the use of the scatter plot of the fitted values and the residuals?

(A) to examine the normal distribution assumption of the errors

(B) to examine the goodness of fit of the regression model

(C) to examine the constant variation assumption of the errors

(D) to test whether the errors have zero mean

(E) to examine the independence of the errors

15. For the case of the multiple regression model, show:

(a) E[β|X = x] = β

(b) V ar(β|X = x) = σ2(X⊤X)−1

16. (a) Suggest why H = X(X⊤X)−1X⊤ is called the ‘hat’ matrix.

(b) Show that HH⊤ = H2 = H .

(c) Explain why:

yi = hiiyi +∑

j 6=i

hijyj,

where hij is the (i, j)th element of H .



(d) Show that the (i, j)th element of H is given by:

1

n+

(xi − x)(xj − x)Sxx

for the special case of the simple linear regression model.

(e) Using H , write down an expression for the vector of residuals e = Y − Y . (We’re back in themultiple linear regression setting.)

(f) Using H , calculate E[e|X = x].

(g) Using H , calculate V ar(e|X = x).

(h) Explain why the ith standardised residual in a multiple regressionmodel is given by ei/(s√1− hii),

where

s =

√√√√ 1

n− p

n∑

j=1

e2j .

17. Use R or Excel5 to carry out the following regression calculations.

The accompanying dataset gives, for 18 countries, data on total expenditures on education as a per-centage of gross national product (y), per capita income (x1), median educational attainment (in years)of the population over 25 years of age (x2), and the ratio of the population aged 0–14 to the totalpopulation (x3). Estimate the multiple regression model:

Y = β0 + β1x1 + β2x2 + β3x3 + ε.

Carry out all appropriate tests to examine the significance of each predictor/regressor/independentvariable and to test for the validity of assumptions typical in a regression model.

Name of Dataset in Excel format: EDUC.xls


5To apply linear regression in Excel you have to add a toolbox. Proceed in the following way. Go to menu in the upper leftof Excel, choose “Excel options” (lower right, besides “Exit Excel”), choose “Add-Ins” in the menu left, choose “Go...” on thebottom line, check the box “Analysis Toolpak” and click on “OK”. Now you can go to “Data”, “Data Analysis” (just created),“Regression”. Alternatively: use the functions: MINVERSE, TRANSPOSE and MMULT with “Ctrl+Shift+Enter”




1. Consider the Excel file “Exercises week 12.xlsx”, tab “Exercise 1”.

(a) Run a linear regression and analyse the residuals?

(b) How would you model this?


(a) Use linear regression to explain the percent of successful free throws.

(b) Explain whether the variables have a direct effect or a confounding effect

(c) Check for collinearity, if collinearity is observed, how to deal with it.


(a) Explain salary using the data.

(b) Use dummy and categorical variables, which one would you prefer?



Sample ExamAustralian School of Business

Probability and StatisticsSample Exam, Session 1, 2013

Question 1 - [15 marks]Suppose the joint probability density function of X1 and X2 is given by

fX1,X2(x1, x2) =

x21 +

x1x23

0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 2,

0 otherwise.

(a) [6 marks] Find the conditional density of X2 given X1 = x1.(b) [5 marks] Determine E[X2|X1 = x1].(c) [4 marks] Verify that E [E[X2|X1] = E[X2].

Question 2 - [18 marks]Let X be a random variable with an Erlang distribution, with parameters k = 1, 2, 3, . . . and λ > 0. This

distribution has the following probability density function:

fX(x) =λkxk−1e−λx

(k − 1)!if x > 0,

and zero otherwise.

(a) [4 marks] Show that the moment generation function is equal to MX(t) = (1− t/λ)−k.

What restrictions are there on the moment generating function in order for it to exist?(b) [3 marks] Describe the relationship of the Erlang distribution with the exponential distri-

bution. Using this relationship and the first two central moments of the exponential distribution provethat:

E [X ] =k

λand V ar (X) =

k

λ2.

Make clear what assumptions are needed to go from one step to another.(c) [5 marks] We have a sample of 100 observations. From this sample we have the following

information:∑100

i=1 xi = 150 and∑100

i=1 x2i = 300. Determine the Method of Moment estimator of λ and

prove that the Method of Moment estimator of parameter k is exactly equal to 3.(d) [6 marks] Next, we have given that the Maximum Likelihood estimator of k is 3 and∑100

i=1 xi = 150. Determine the Maximum Likelihood estimator of λ.

Question 3 - [20 marks]A UNSW actuarial student will either become an fiaa or will not become an fiaa. This depends on her orhis ‘actuarity’, a peculiar mental ability which is traditionally measured by a number lying (continuously)between zero and one. Without testing, we do not know the actuarity of any particular person; we do knowthat students with higher actuarity coefficients are more likely to become actuaries. As a result of painstakingresearch by UNSW staff, it has been established that a reasonable prior distribution for actuarity is Betawith parameters (α, β).

Researchers have also established that, given a student’s actuarity Θ, her or his ability to become anactuary is Bernoulli distributed,

(I|Θ) ∼ Bernoulli(Θ).

(a) [5 marks] For this model, what is the marginal distribution of I?(b) [10 marks] What is the posterior density, π(θ|i)?(c) [5 marks] Calculate the Bayesian estimator of Θ.

Question 4 - [18 marks] The table below gives 105 observations of the times (in minutes) between callsto the emergency services telephone number. These times have been grouped into nine minute intervals (theleft-hand column, ‘Class’).


ACTL2002 & ACTL5101 Probability and Statistics Sample Exam, Session 1, 2013

Class Frequency Probability ExpectedA1 = [0, 9] 41 0.362 38.052 0A2 = (9, 18] 22 0.231 24.265 5A3 = (18, 27] 11 0.147 15.466 5A4 = (27, 36] 10 0.094 9.870 0A5 = (36, 45] 9 0.060 6.289 5A6 = (45, 54] 5 0.038 4.011 0A7 = (54, 63] 2 0.024 2.562 0A8 = (63, 72] 3 0.016 1.627 5A9 = (72,∞) 2 0.027 2.866 5

(a) [10 marks] A colleague suggests to you that the distribution of times is exponential with a meanof 20. Test whether this is the case.

(b) [5 marks] Above, you have assumed a mean of 20. If, instead, you had to estimate the mean, inwhat way would your testing procedure differ?

(c) [3 marks] Suggest some graphical ways of investigating whether these data are exponential withmean 20.

Question 5 - [25 marks]Aggregate national data for a particular industry are provided below. You have been asked to fit a Cobb-Douglas production function to these data.

Country B C D E F G HOutput, Q 80 150 135 165 95 130 110Labour, L 60 100 100 120 70 90 80Capital, K 50 100 80 100 60 80 70

(a) [1 mark] A Cobb-Douglas production function takes the form Q = ALαKβ. How will you fitthis model to the data?

(b) [5 marks] For the model you have chosen to fit, you are given (in standard notation) that

(X⊤X)−1 =

62.780 946 07 −55.746 515 58 24.179 864 75−55.746 515 58 204.981 847 2 −181.993 220 724.179 864 75 −181.993 220 7 175.080 437 9

X⊤X =

7 13.559 763 03 13.128 399 2713.559 763 03 26.330 018 58 25.496 918 4513.128 399 27 25.496 918 45 24.696 210 72

X⊤Y =

14.560058628.2725888127.38094584

Determine β .(c) [5 marks] Given

∑y2i = 30.360 112 93

∑y2i = 30.359 956 6 y = 2.080 008 372

calculate SST and SSM.(d) [3 marks] If the SSE were 0.000 156 327, give estimates of the variance of α and β and their

covariance.(e) [5 marks] Carefully detailing your steps, test for the significance of α and β at the 5% level.(f) [2 marks] Calculate the adjusted R2 of your model.(g) [4 marks] Test whether your model is homogeneous of degree one, that is, whether α + β = 1,

at a 5% level of significance.

-End of Sample Exam-


Australian School of Business

Probability and StatisticsAdditional material


Additional material

Includes:

- UNSW TV: Back to Basics: Basic Mathematical Tools for Actuarial Students

- Greek alphabet

- List of some Excel functions

- Mathematics (Integrals, differentiation & mathematic rules)

- Linear Algebra

- Formulae and Tables

Session 1, 2013

ACTL2002 & ACTL5101 Probability and Statistics UNSW TV&Greek alphabet

UNSW TV: Back to Basics: Basic Mathematical Tools for Actuarial Students

Back to Basics: List of all videos

http://tv.unsw.edu.au/collection/BBA28340-2B1D-11DF-AA9D123139020041/mediaType/unswVideo

Back to Basics: Introduction

http://tv.unsw.edu.au/video/back-to-basics-introduction

Back to Basics: Topic 1: Sum of geometric series

http://tv.unsw.edu.au/video/back-to-basics-topic-1

Back to Basics: Topic 2: Exponential and logarithm function


Back to Basics: Topic 3: Quadratic formula


Back to Basics: Topic 4: Taylor series


Back to Basics: Topic 5: Derivative


Back to Basics: Topic 6: The product rule, the quotient rule and the chain rule (derivatives)


Back to Basics: Topic 7: Integration by parts


Back to Basics: Topic 8: Differential of an integral


Back to Basics: Topic 9: Newton-Ralphson approximation


Greek alphabetName Lower

caseUppercase

Name Lowercase

Uppercase

alpha α A nu ν Nbeta β B xi ξ Ξgamma γ Γ omikron o Odelta δ ∆ pi π, Πepsilon ǫ, ε E rho ρ, Pzeta ζ Z sigma σ, ς Σeta η H tau τ Ttheta θ, ϑ Θ upsilon υ Υiota ι I phi φ, ϕ Φkappa κ K chi χ Xlambda λ Λ psi ψ Ψmu µ M omega ω Ω


http://tv.unsw.edu.au/collection/BBA28340-2B1D-11DF-AA9D123139020041/mediaType/unswVideo

http://tv.unsw.edu.au/video/back-to-basics-introduction










ACTL2002 & ACTL5101 Probability and Statistics List of some Excel functions

List of general Excel functions:Function Explanation Function Explanationexp(1) returns e correl(a1:a4,b1:b4) returns the correlation

coefficient between ar-ray A1-A4 and B1-B4

ln natural log covar(a1:a4,b1:b4) returns the covariancebetween array A1-A4and B1-B4

pi() returns pi var(a1:a4) returns the variance ofarray A1-A4

exp(gammaln(alpha)) gamma function evalu-ated in alpha

stdev(a1:a4) returns the standarddeviation of array A1-A4

fact(a1) returns the factorial ofcell a1

skew(a1:a4) returns the skewness ofarray A1-A4

combin(n,r) returns n choose r kurt(a1:a4) returns the kurtosis ofarray A1-A4

sqrt(a1) returns the square rootof cell a1

sum(a1:b4) returns the sum of thecells A1-B4

abs(a1) returns the absolutevalue of cell A1

sumsq(a1:b4) returns the sum of thesquared values of cellsA1-B4

round(a1,r) rounds using r digits

sumproduct(a1:a4,b1:b4) returns the sum ofproduct a1b1-a4b4

floor(a1,r) rounds down using rdigits

average(a1:a4) returns the average ofarray A1-A4

ceiling(a1,r) rounds up using r dig-its

trimmean(a1:a4,alpha) returns the alpha-thtrimmed mean of thearray A1-A4

min(a1:a4) returns the minimumof array A1-A4

median(a1:a4) returns the median ofthe array A1-A4

max(a1:a4) returns the maximumof array A1-A4

count(a1:a4) counts the number ofcells with a number inthe array A1-A4

countif(a1:a4,alpha) counts the number ofcells equal to alpha inthe array A1-A4

percentile(a1:a4,alpha) returns the alpha(≤1)-th percentile of thearray A1-A4

quartile(a1:a4,alpha) returns the alpha(=0, 1, 2, 3, 4)-th quartileof the array A1-A4

Function Explanationif(test,value true, value false) produces value true if condition is met and value false otherwiseand(test1, test2, test3) produces only TRUE if all conditions are met, otherwise FALSErand() returns a random number between 0 and 1randbetween(a,b) returns a random number between a and bvlookup(a1,a1:c4,r) looks for value of cell a1 in the array A1-A4 and returns the r-th

columnhlookup(a1,a1:c4,r) looks for value of cell a1 in the array A1-C1 and returns the r-th

row


ACTL2002 & ACTL5101 Probability and Statistics List of some Excel functions

List of statistical Excel functions:Be aware of the differences in the definition of distributions and check which definition it is, and adjust yourparameters to it.

Continuous distributions Discrete distributionsFunction Explanation Function Explanationnormdist(x,mu,sd,1) returns normal cdf poisson(x,lambda,1) returns poisson cdfnormdist(x,mu,sd,0) returns normal pdf poisson(x,lambda,0) returns poisson pmfnorminv(p,mu,sd) returns the p-th quan-

tile of a normal r.v.Poisson inverse make a table with the

poisson c.d.f. and youif function to find theinverse?

lognormdist(x,mu,sd) returns Lognormal cdf binomdist(x,n,p,1) returns Binomial cdfloginv(p,mu,sd) returns the p-th quan-

tile of a Lognormal r.v.binomdist(x,n,p,0) returns Binomial pmf

critbinom(alpha,n,p) returns the alpha-thquantile of a Binomialr.v.

expondist(x,lambda,1) returns exponential cdf negbinomdist(x,n,p) returns Negative Bino-mial pmf

expondist(x,lambda,0) returns exponentialpdf

Geometric distribution use relationship be-tween NB and GEO

use relation exponentialand gamma distributionto find the p-th quantile ofan exponential r.v

NBin inverse make a table with theNBin c.d.f. and you iffunction to find the in-verse?

gammadist(x,alpha,beta,1) returns gamma cdf hypgeodist(x,n,M,N) returns Hypergeomet-ric cdf

gammadist(x,alpha,beta,0) returns gamma pdf Hypgeometric inverse make a table with theHypgeometric c.d.f.and you if function tofind the inverse?

gammainv(p,alpha,beta) returns the p-th quan-tile of a gamma r.v.

weibull(x,alpha,beta,1) returns weibulll cdfweibull(x,alpha,beta,0) returns weibull pdf

betadist(x,alpha,beta) returns beta cdfbetainv(p,alpha,beta) returns the p-th quan-

tile of a beta r.v.

chidist(x,df) returns chi-squared cdfchiinv(p,df) returns the p-th quan-

tile of a chi-squaredr.v.

tdist(x,df) returns student-t cdftinv(1-p/2,df) returns the 1-p-th

quantile of a student-tr.v.

fdist(x, df1,df2) returns F cdffinv(p, df1,df2) returns the p-th quan-

tile of a F r.v.


ACTL2002 & ACTL5101 Required Pre-Knowledge Mathematics



Required Pre-knowledge Mathematics

List of integrals

∫ b

ac · xddx =

[c

d+1 · xd+1]ba, for a, b, c, d ∈ ℜ, d 6= −1;

∫ b

ac

d·x+edx =[cd · log (d · x+ e)

]ba, for a, b, c, d, e ∈ ℜ, d 6= 0;∫ b

a log (c · x) dx = [x · log (c · x)− x]ba , for a, b, c ∈ ℜ, c 6= 0;∫ b

aexp (c · x) dx =

[exp(c·x)

c

]ba, for a, b, c ∈ ℜ, c 6= 0;

∫ b

ax · exp (c · x) dx =

[exp(c·x)

c2 · (c · x− 1)]ba, for a, b, c ∈ ℜ, c 6= 0;

∫ b

a x2 · exp (c · x) dx =

[exp (c · x) ·

(x2

c − 2xc2 + 2

c3

)]ba, for a, b, c ∈ ℜ, c 6= 0;

∫ b

a sin (x) dx = [− cos (x)]ba for a, b ∈ ℜ;∫ b

acos (x) dx = [sin (x)]ba for a, b ∈ ℜ;∫∞

0 xα−1 · exp (−x) dx = Γ(α) for α > 0, week 2 Prob & Stat;∫ 1

0xα−1 · (1− x)β−1 dx = Γ(α)·Γ(β)

Γ(α+β) for α, β > 0, week 2 of Prob & Stat;∫∞0

exp(−x2/2

)dx =

√π/2 week 2 of Prob & Stat (Gaussian).

Integral rules

∫ b

a udvdxdx = [u · v]ba −

∫ b

a vdudxdx, Integration by parts (see page 3 F&T);∫ b

a f(x)dx = −∫ a

b f(x)dx, Reversing limits;∫ b

af (g (x)) · g′ (x) dx =

∫ g(b)

g(a)f (x) dx, Integration by substitution;

∂∫ baf(x)dx

∂b = f(b);∂∫

baf(x)dx

∂a = −f(a), Differentiation of integral.

Differentiation rules

∂f(x)·g(x)∂x = ∂f(x)

∂x · g (x) + f (x) · ∂g(x)∂x , Product rule;∂∂x

(f(x)g(x)

)=

∂f(x)∂x ·g(x)−f(x)· ∂g(x)

∂x

(g(x))2, Quotient rule, g (x) 6= 0;

∂f(g(x))∂x = f ′ (g (x)) · g′ (x) , Chain rule.


ACTL2002 & ACTL5101 Required Pre-Knowledge Mathematics

Mathematical rules

x! = x · (x− 1)!, 0! = 1, Factorial function, x ∈ N+ = 0, 1, 2, . . .;(

nk

)= n!

(n−k)!·k! , Binomial coefficient, n, k ∈ N+;

exp (x) =∑∞

i=0xi

i! , Exponential function (see page 2 F&T);

log (1 + x) =∑∞

i=1(−1)i+1 · xi

i , Natural log function, −1 < x ≤ 1 (see page 2 F&T);(a+ b)

n=

∑ni=0

(ni

)· ai · bn−i, Binomial expansion, n ∈ N

+ (see page 2 F&T);

f (x+ h) =∑∞

i=0hi

i! · f (i)(x), Taylor series expansion (see page 3 F&T);∑nk=0 x = n(n+ 1)/2 for n ∈ N ;∑n

k=0 x2 = n(n+ 1)(2n+ 1)/6 for n ∈ N ;∑∞

k=0 xk = 1

1−x Geometric series, for −1 < x < 1;∑n−1k=0 x

k = 1−xn

1−x∑∞k=n x

k = xn

1−x∑∞k=1

1k2k

= log(2)exp (a+ b) = exp (a) · exp (b)exp (a− b) = exp (a) / exp (b)

exp (a)n= exp (n · a)

log (a · b) = log (a) + log (b)log (a/b) = log (a)− log (b)log (an) = n · log (a) .

ax2 + bx+ c = 0⇒ x = −b±√b2−4ac2a abc formula.




Required Pre-knowledge Linear Algebra

Terminology

Matrix

We denote a matrix:

A =

A11 A12 . . . A1m

A21 A22 . . . A2m

......

. . ....

An1 An2 . . . Anm

,

where Ai,j is the (i, j)-element of the matrix A, i.e., the ith row, jth column element of the matrix A. Thematrix is said to have size (n×m), i.e., n rows and m columns.

Vector

We denote a vector:

x =

x1x2...xn

,

where xi is the ith element of the vector x. The length of the vector is n, i.e., the number of elements of the

vector. The size is n× 1, i.e., a vector is a matrix with only one column

Transpose

The transpose of a matrix A with size (n×m) is given by:

A⊤ =

A11 A21 . . . An1

A12 A22 . . . An2

......

. . ....

A1m A2m . . . Anm

, where A =

A11 A12 . . . A1m

A21 A22 . . . A2m

......

. . ....

An1 An2 . . . Anm

.

Thus the A⊤ (which is also denoted by A′) is a matrix of size (m× n) where the (i,j) element of the matrixA⊤ is the (j,i) element of the matrix A.

Matrix multiplication

Vector and vector

Multiplication of two vectors x and y with same length n:

x⊤

[1×n]

y[n×1]

=

n∑

i=1

xiyi

[1×1]

,

where [n×m] denotes n row and m columns.Note: this vector multiplication is only possible if vectors x and y have the same length (number of

elements). The result of the product of the “row vector” x⊤ and vector y is a constant/number ∈ R or a1× 1 matrix.

ACTL2002 & ACTL5101 Required Pre-Knowledge Linear Algebra

Matrix and matrix

Multiplication of two matrices A with size (n× k) and B with size (k ×m):

A[n×k]

B[k×m]

= C[n×m]

,

where

Cij =

k∑

l=1

AilBlj ,

hence, the (i, j) element of the matrix C is given by multiplying the ith row of A with the jth column of B.Note: number of columns of matrix A should be equal to the number of rows of matrix B. The matrix

C has the number of rows equal to the number of rows of matrix A and the number of columns of matrix Cis equal to the number of columns of matrix B.

Matrix and vector

Multiplication of a matrix A with size (n×m) with a vector x with length m (i.e., size (m× 1):

A[n×m]

x[m×1]

= y[n×1]

where

yi =

m∑

l=1

Ailxl,

hence, the ith element of the vector y is obtained by multiplying the ith row of the matrix A with the vectorx.

Note that the number of columns of the matrix A should be equal to the length of the vector x. Thelength of vector y will be equal to the number of rows of vector A.

Vector and vector II

In Section 3 we multiplied two vectors (a “row vector” with a “column vector”) which resulted in a constant.Alternatively, one can multiply a “column vector” with a “row vector” to obtain a matrix. Note that thisdoes not occur often.

Alternative multiplication of two vectors x with length n and y with length m:

x[n×1]

y⊤

[1×m]

= A[n×m]

,

where

Aij = xiyi,

hence, the (i, j) element of A is the product of the ith element of vector x and the jth element of vector y.

Matrix and solving system of equations

For solving the following system of equations of n equations with m parameters:

a11x1 + a12x2 + . . . + a1mxm = y1a21x1 + a22x2 + . . . + a2mxm = y2

......

. . ....

...an1x1 + an2x2 + . . . + anmxm = yn

we can use linear algebra. Denoting in matrices we have:

A[n×m]

x[m×1]

= y[n×1]

, (1)



where,

A =

A11 A12 . . . A1m

A21 A22 . . . A2m

......

. . ....

An1 An2 . . . Anm

, x =

x1x2...xm

, and y =

y1y2...yn

To solve equation (1), using n = m (i.e., number of equations equals the number of parameters) we proceedas follows:

1. Create the matrix:

[A | y] =

A11 A12 . . . A1n y1A21 A22 . . . A2n y2...

.... . .

......

An1 An2 . . . Ann yn

2. Using elementary row operations (see below) create a matrix:

[In | z] =

1 0 . . . 0 z10 1 . . . 0 z2...

.... . .

......

0 0 . . . 1 zn

,

where In is an (n× n) identity matrix.

3. Then, we have that x = z.

Elementary row operations

Let A be a (n ×m) matrix. Let Ri be the ith row of the matrix A. Then, the elementary row operationsare:

1. Row switching:A row within the matrix A can be switched with another row in the matrix:

A =

R1

...Ri

...Rj

...Rn

∼

R1

...Rj

...Ri

...Rn

A =

A11 A12 . . . A1m

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

Aj1 Aj2 . . . Ajm

......

. . ....

An1 An2 . . . Anm

∼

A11 A12 . . . A1m

......

. . ....

Aj1 Aj2 . . . Ajm

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

An1 An2 . . . Anm

.

2. Row multiplication (c 6= 0):Each element of a row can be multiplied with a non-zero constant c ∈ R\0:

A =

R1

...Ri

...Rn

∼

R1

...c ·Ri

...Rn

A =

A11 A12 . . . A1m

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

An1 An2 . . . Anm

∼

A11 A12 . . . A1m

......

. . ....

cAi1 cAi2 . . . cAim

......

. . ....

An1 An2 . . . Anm

.

3. Row addition:



A row can be replaced by the row itself plus another row:

A =

R1

...Ri

...Rj

...Rn

∼

R1

...Ri

...Rj +Ri

...Rn

=

A11 A12 . . . A1m

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

Aj1 Aj2 . . . Ajm

......

. . ....

An1 An2 . . . Anm

∼

A11 A12 . . . A1m

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

Ai1 +Aj1 Ai2 +Aj2 . . . Aim +Ajm

......

. . ....

An1 An2 . . . Anm

.

Note, combining all three we have:

A =

R1

...Ri

...Rj

...Rn

∼

Ri + cRj

...R1

...Rj

...Rn

=

A11 A12 . . . A1m

......

. . ....

Ai1 Ai2 . . . Aim

......

. . ....

Aj1 Aj2 . . . Ajm

......

. . ....

An1 An2 . . . Anm

∼

Ai1 +Aj1 Ai2 +Aj2 . . . Aim +Ajm

......

. . ....

A11 A12 . . . A1m

......

. . ....

Aj1 Aj2 . . . Ajm

......

. . ....

An1 An2 . . . Anm

.

Example Gaussian Elimination

In a matrix:

- a leading row is one which is not all zeros;

- in a leading row, the leading entry is first (leftmost) non-zero entry;

- a leading column contains the leading entry for some row.

Example:

1 2 30 2 40 0 1

In this case, elements a11, a22, and a33 are leading entries, thus all rows are leading rows and all columnsare leading columns.

Gaussian elimination: Given an augmented matrix C = [A|y] with n rows/equations. Solve:

3x1 + 2x2 =1

2x1 − 1x2 =3



The augmented matrix is:[

3 2 12 −1 3

]

Then solve it by:

1. Multiply first row by 1/3 ⇒ R1 ← R1/3;

2. Subtract twice the (new) first row from the second row ⇒ R2 ← R2 − 2R1

Hence we now have:[

3 2 12 −1 3

]∼[

3/3 2/3 1/32− 2 −1− 4/3 3− 2/3

]=

[1 2/3 1/30 −7/3 7/3

]

The next steps are:

1. Multiply second row by -3/7 ⇒ R2 ← R2/− 7/3;

2. Subtract 2/3 the (new) second row from the first row ⇒ R1 ← R1 − 2/3R2

Hence we now have:[

1 2/3 1/30 −7/3 7/3

]∼[

1 2/3 1/30 1 −1

]∼[

1 2/3− 2/3 1/3− (−2/3)0 1 −1

]=

[1 0 10 1 −1

]

Thus we have that x1 = 1 and x2 = −1.

Inverse Matrices

An inverse of a matrix A only exists for square matrices (i.e., the size of matrix A is n×n). For non-singularsquare matrices (i.e., matrices with an inverse) we have the following:

A−1A = In and we have: AA−1 = In,

where In is an identity matrix of size n× n.

For 2× 2 matrices

Denote the matrix 2× 2 matrix A:

A =

[a bc d

].

Then, the inverse of the matrix A is obtained by:

A−1 =1

det(A)

[d −b−c a

],

where det(A) = a · d − b · c. Note that the matrix A is non-singular if and only if det(A) 6= 0. Hence, ifdet(A) = 0 then the matrix A is singular and its inverse does not exist, which can be observed from the factthat you cannot divide by zero.

For 3× 3 matrices

Denote the matrix 3× 3 matrix A:

A =

a b cd e fg h i

.

Then, the inverse of the matrix A is obtained by:

A−1 =1

det(A)

(ei− fh) (ch− bi) (bf − ce)(fg − di) (ai− cg) (cd− af)(dh− eg) (gb− ah) (ae− eg)

,

where det(A) = a · (ei− fh)− b · (id− fg)+ c · (dh− eg). Note that the matrix A is non-singular if and onlyif det(A) 6= 0. Hence, if det(A) = 0 then the matrix A is singular and its inverse does not exist, which canbe observed from the fact that you cannot divide by zero.



For n× n matrices

To find the inverse of an n× n matrix A consists of the following steps:

1. Create the following n× 2n matrix:

[A In] =

A11 A12 . . . A1n 1 0 . . . 0A21 A22 . . . A2n 0 1 . . . 0...

.... . .

......

.... . .

...An1 An2 . . . Ann 0 0 . . . 1

2. Using Gaussian elimination (also referred to as Gauss-Jordan) (i.e., elementary row operations, seesection 3) create the matrix [In B].

3. Now, the matrix B is the inverse of the matrix A, i.e., B = A−1.

Blockwise inversion

Let X be a symmetric (n+m)× (n+m) matrix, A be a symmetric n× n, D be a symmetric m×m, B bea n×m matrix, and C be a m× n matrix, then we have:

X−1 =

[A BC D

]−1

=

[A−1 +A−1B(D − CA−1B)−1CA−1 −A−1B(D − CA−1B)−1

(D − CA−1B)−1CA−1 (D − CA−1B)−1

].

Eigenvalues and Eigenvectors

Let A be a symmetric n× n matrix, λ -a scaler- the eigenvalue, c -a vector- the eigenvector if:

Ac = λc.

In general, there are n solutions λ1, . . . , λn called the eigenvalues (characteristic roots) of A, correspondingwith n eigenvectors c1, . . . , cn called the eigenvectors (characteristic vectors)

A matrix is positive definite if all its eigenvalues are positive. A matrix is positive semi-definite if all itseigenvalues are non-negative. The determinant of a symmetric matrix equals the product of its n eigenvalues.The rank of a symmetric matrix corresponds to the number of non-zero eigenvalues. If an eigenvalue is zero,the matrix A is not of full rank and thus singular.

Differentiation

Let x be an n-dimensional column vector, c be an n-dimensional column vector, then c⊤x is a scalar.Consider:

∂c⊤x

∂x= c.

This is a column vector of n derivatives the typical element being ci. More generally, we have for a vectorialfunction Ax (where A is a matrix) that:

∂Ax

∂x= A⊤.

The element in column i, row j in this matrix is the derivative of the jth element in the function Ax withrespect to xi. Further:

∂x⊤Ax

∂x= 2Ax

for a symmetric matrix A. If A is not symmetric, we have

∂x⊤Ax

∂x= (A+A⊤)x.

All these results follow from collecting the results from an element-by-element differentiation.



Least Squares Manipulations

Let xi = [xi1, xi2, . . . , xiK ]⊤

with xi1 ≡ 1 and β = [β1, β2, . . . , βK ]⊤, then:

x⊤i β = β1 + β2xi2 + . . .+ βKxiK .

The matrix:

N∑

i=1

xix⊤i =

N∑

i=1

xi1xi2...

xiK

[xi1, xi2, . . . , xiK ]

=

∑Ni=1 x

2i1

∑Ni=1 xi2xi1 . . .

∑Ni=1 xiKxi1

...∑N

i=1 x2i2

......

. . ....∑N

i=1 xi1xiK∑N

i=1 xi2xiK . . .∑N

i=1 x2iK

is a K ×K symmetric matrix containing sums of squares and cross-products. The vector:

N∑

i=1

xiyi =

∑Ni=1 xi1yi∑Ni=1 xi2yi

...∑Ni=1 xiKyi

has length K, so that the system:

(N∑

i=1

xix⊤i

)β =

N∑

i=1

xiyi

is a system of K equations with K unknowns. With matrix notation, the N ×K matrix X is defined as:

X =

x11 x12 . . . x1K...

.... . .

...xN1 xN1 . . . xNK

and y = [y1, y2, . . . , yN ]⊤. From this it can easily be verified that:

X⊤X =N∑

i=1

xix⊤i

X⊤y =

N∑

i=1

xiyi


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1 Mathematical Methods1.1 Series

Exponential function

exp(x) = ex = 1 + x+ x2

2! +x3

3! + . . .Natural log function

log(1 + x) = ln(1 + x) = x− x2

2 + x3

3 − . . . (−1 < x ≤ 1)Binomial expansion

(a+ b)n = an +(n1

)an−1b+

(n2

)an−2b2 + . . .+ bn

where n is a positive integer

(1 + x)p = 1 + px+ p(p−1)2! x2 + p(p−1)(p−2)

3! x3 + . . .(−1 < x ≤ 1)

1.2 CalculusTaylor series (one variable)

f(x+ h) = f(x) + hf ′(x) + h2

2! f′′(x) + . . .

Taylor series (two variables)f(x+ h, y + k) = f(x, y) + hf ′

x(x, y) + kf ′y(x, y)

+ 12!

(h2f ′′

xx(x, y) + 2hkf ′′xy(x, y) + k2f ′′

yy(x, y))+ . . .

Integration by parts∫ b

au dvdxdx = [uv]

ba −

∫ b

av dudxdx

Double integrals (changing the order of integration)∫ b

a

(∫ x

af(x, y)dy

)dx =

∫ b

a

(∫ b

yf(x, y)dx

)dy or

∫ b

a dx∫ x

a dyf(x, y) =∫ b

a dy∫ b

y dxf(x, y)

The domain of integration here is the set of values (x, y) for whicha ≤ y ≤ x ≤ b.Differentiating an integral

ddy

∫ b(y)

a(y)f(x, y)dx = b′(y)f [b(y), y]− a′(y)f [a(y), y]

+∫ b(y)

a(y)∂∂yf(x, y)dx

1.3 Solving EquationsNewton-Ralphson methodIf x is a sufficient good approximation to a root of the equationf(x) = 0 then (provided convergence occurs) a better approximationis

x⋆ = x− f(x)f ′(x)

Integrating factorsThe integrating factor for solving the differential equationdydx + P (x)y = Q(x) is:

exp(∫P (x)dx

)

Second order difference equationsThe general solution of the difference equationaxn+2 + bxn+1 + cxn = 0 is:if b2 − 4ac > 0: xn = Aλn1 +Bλn2

(distinct real roots, λ1 6= λ2)if b2 − 4ac = 0: xn = (A+Bn)λn

(equal real roots, λ1 = λ2 = λ)if b2 − 4ac < 0: xn = rn(A cosnθ +B sinnθ)

(complex roots, λ1 = λ2 = reiθ)where λ1 and λ2 are the roots of the quadratic equationaλ2 + bλ+ c = 0

1.4 Gamma functionDefinition

Γ(x) =∫∞0 tx−1e−tdt, x > 0

PropertiesΓ(x) = (x− 1)Γ(x− 1)Γ(n) = (n− 1)!, n = 1, 2, 3, . . .Γ(1/2) =

√π

1.5 Bayes’ formulaLet A1, A2, . . . , An be a collection of mutually exclusive and exhaustiveevents withP (Ai) 6= 0, i = 1, 2, . . . , n.For any event B such that P (X) 6= 0:

P (Ai|B) = P (B|Ai)P (Ai)n∑

j=1

P (B|Aj)P (Aj), i = 1, 2, . . . , n.

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2 Statistical distributionsNotationPF =Probability function, p(x)PDF =Probability density function, f(x)DF =Distribution function, F (x)PGF =Probability generating function, G(s)MGF =Moment generating function, M(t)Note. Where formulae have been omitted below, this indicates that(a) there is no simple formula or (b) the function does not have afinite value or (c) the function equals zero.

2.1 Discrete distributionsBinomial distributionParameters: n, p (n=positive integer, 0 < p < 1 with q = 1− p)PF: p(x) =

(nx

)pxqn−x, x = 0, 1, 2, . . . , n

DF: The distribution function is tabulate in the statisticaltables section.

PGF: G(s) = (q + ps)n

MGF: M(t) = (q + pet)n

Moments: E[X ] = np, var(X) = npqCoefficientof skewness: q−p√

npq

Bernoulli distributionThe Bernoulli distribution is the same as the binomial distributionwith parameter n = 1.Poisson distributionParameter: µ (µ > 0)

PF: p(x) = e−µ·µx

x! , x = 0, 1, 2, . . .DF: The distribution function is tabulate in the statistical

tables section.PGF: G(s) = exp (µ · (s− 1))MGF: M(t) = exp (µ · (et − 1))Moments: E[X ] = µ, V ar(X) = µCoefficientof skewness: 1√

µ

Negative binomial distribution - Type 1Parameters: k, p (k=positive integer, 0 < p < 1 with q = 1− p)PF: p(x) =

(x−1k−1

)pkqx−k, x = k, k + 1, k + 2, . . .

PGF: G(s) =(

ps1−qs

)k

MGF: M(t) =(

pet

1−qet

)k

Moments: E[X ] = kp , V ar(X) = kq

p2

Coefficient

of skewness: 2−p√kq

Negative binomial distribution - Type 2Parameters: k, p (k > 0, 0 < p < 1 with q = 1− p)PF: p(x) = Γ(k+x)

Γ(x+1)Γ(k)pkqx−k, x = 0, 1, 2, . . .

PGF: G(s) =(

p1−qs

)k

MGF: M(t) =(

p1−qet

)k

Moments: E[X ] = kqp , V ar(X) = kq

p2

Coefficient

of skewness: 2−p√kq

Geometric distributionThe geometric distribution is the same as the negative binomialdistribution with parameter k = 1.

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Uniform distribution (discrete)Parameters: a, b, h (a < b, h > 0, b− a is a multiple of h)PF: p(x) = h

b−a+h , x = a, a+ h, a+ 2h, . . . , b− h, bPGF: G(s) = h

b−a+h

(sb+h−sa

sh−1

)

MGF: M(t) = hb−a+h

(e(b+h)t−eat

eht−1

)

Moments: E[X ] = 12 (a+ b), var(X) = 1

12 (b − a)(b− a+ 2h)2.2 Continuous distributions

Standard Normal distribution - N(0,1)Parameters: none

PF: f(x) = 1√2πe−

12x

2

, −∞ < x <∞DF: The distribution function is tabulate in the statistical

tables section.

MGF: M(t) = e12 t

2

Moments: E[X ] = 0, var(X) = 1

E[Xr] = 12r/2

Γ(1+r)

Γ(1+ r2 ), r = 2, 3, 6, . . .

Normal (Gaussian) distribution - N(µ, σ2)Parameters: µ, σ2 (σ > 0)

PF: f(x) = 1σ√2π

exp(− 1

2

(x−µσ

)2), −∞ < x <∞

MGF: M(t) = eµt+12σ

2t2

Moments: E[X ] = µ, var(X) = σ2

Exponential distributionParameter: λ (λ > 0)PDF: f(x) = λe−λx, x ≥ 0DF: F (x) = 1− e−λx

MGF: M(t) =(1− t

λ

)−1, t < λ

Moments: E[X ] = 1λ , V ar(X) = 1

λ2

E[Xr] = Γ(1+r)λr r = 1, 2, 3, . . .

Coefficientof skewness: 2

Gamma distributionParameter: α, λ (α > 0, λ > 0)

PDF: f(x) = λα

Γ(α) · xα−1 · e−λx, x ≥ 0

DF: When 2α is an integer, probabilities for the gammadistribution can be found using the relationship:2λX ∼ χ2

2α

MGF: M(t) =(1− t

λ

)−α, t < λ

Moments: E[X ] = αλ , V ar(X) = α

λ2

E[Xr] = Γ(α+r)Γ(α)λr r = 1, 2, 3, . . .

Coefficientof skewness: 2√

α

Chi-squared distribution - χ2ν

The chi-squared distribution with ν degrees of freedom is the same asthe gamma distribution with parameters α = ν

2 and λ = 12 .

The distribution function for the chi-squared distribution is tabulated inthe statistical tables section

Uniform distribution (continuous) - U(a,b)Parameters: a, b (a < b)PDF: f(x) = 1

b−a , a < x < b

DF: F (x) = x−ab−a

MGF: M(t) = 1(b−a)

1t

(ebt − eat

)

Moments: E[X ] = 12 (a+ b), V ar(X) = 1

12 (b − a)2E[Xr] = 1

(b−a)1

r+1

(br+1 − ar+1

), r = 1, 2, 3, . . .

Beta distributionParameters: α, β (α > 0, β > 0)

PDF: f(x) = Γ(α+β)Γ(α)Γ(β)x

α−1(1− x)β−1, 0 < x < 1

Moments: E[X ] = αα+β , V ar(X) = αβ

(α+β)2(α+β+1)

E[Xr] = Γ(α+β)Γ(α+r)Γ(α)Γ(α+β+r) , r = 1, 2, 3, . . .

Coefficient

of skewness: 2(β−α)(α+β+2)

√α+β+1

αβ

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Lognormal distributionParameters: µ, σ2 (σ > 0)

PDF: f(x) = 1σ√2π

1x exp

(− 1

2

(log x−µ

σ

)2)

Moments: E[X ] = eµ+12σ

2

, var(X) = e2µ+σ2(eσ

2 − 1)

E[Xr] = erµ+12 r

2σ2

, r = 1, 2, 3, . . .Coefficient

of skewness:(eσ

2

+ 2)√

eσ2 − 1

Pareto distribution (two parameter version)Parameters: α, λ, (α > 0, λ > 0)

PDF: f(x) = αλα

(λ+x)α+1 , x > 0

DF: F (x) = 1−(

λλ+x

)α

Moments: E[X ] = λα−1 (α > 1), var(X) = αλ2

(α−1)2(α−2) α > 2)

E[Xr] = Γ(α−r)Γ(1+r)Γ(α) λr, r = 1, 2, 3, . . . , r < α

Coefficient

of skewness: 2(α+1)(α−3)

√α−2α (α > 3)

Pareto distribution (three parameter version)Parameters: α, λ, k, (α > 0, λ > 0, k > 0)

PDF: f(x) = Γ(α+k)λαxk−1

Γ(α)Γ(k)(λ+x)α+k , x > 0

Moments: E[X ] = kλα−1 (α > 1), var(X) = k(k+α−1)λ2

(α−1)2(α−2) α > 2)

E[Xr] = Γ(α−r)Γ(k+r)Γ(α)Γ(k) λr , r = 1, 2, 3, . . . , r < α

Weibull distributionParameters: c, γ (c > 0, γ > 0)PDF: f(x) = cγxγ−1e−cxγ

, −∞ < x <∞DF: F (x) = 1− e−cxγ

Moments: E[Xr] = γ(1 + r

γ

)1

cr/γ, r = 1, 2, 3, . . .

Burr distributionParameters: α, λ, γ (α > 0, λ > 0,γ > 0)

PDF: f(x) = αγλαxγ−1

(λ+xγ)α+1 , x ≥ 0

DF: F (x) = 1−(

αλ+xγ

)α

Moments: E[Xr] = Γ(α− r

γ

)Γ(α = r

γ

)λr/γ

Γ(α) r = 1, 2, 3, . . . , r < αγ

2.3 Compound distributionsConditional expectation and variance

E[Y ] = E[E[Y |X ]]V ar(Y ) = V ar(E[Y |X ]) + E[V ar(Y |X)]

Moments of a compound distributionIf X1, X2, . . . are i.i.d random variables with mgf MX(t) and N isan independent nonnegative integer-valued random variable, thenS = X1 + . . .+XN (with S = 0 when n = 0) has the followingproperties:Mean: E[S] = E[N ] · E[X ]

Variance: V ar(S) = E[N ] · V ar(X) + V ar(N) · (E[X ])2

MGF: MS(t) =MN (log(MX(t)))Compound Poisson distributionMean: λm1

Variance: λm2

Third central moment: λm3

where λ = E[N ] and mr = E[Xr]

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3 Statistical Methods3.1 Sample mean and varianceThe random sample (x1, x2, . . . , xn) has the following samplemoments:

Sample mean: x = 1n

n∑i=1

xi

Sample variance: s2 = 1n−1

(n∑

i=1

x2i − nx2)

3.2 Parametric inference (normal model)One sample

For a single sample of size n under the normal model X ∼ N(µ, σ2):X−µS/

√n∼ tn−1 and (n−1)S2

σ2 ∼ χ2n−1

Two samplesFor two independent samples of sizes m and n under the normalmodels X ∼ N(µX , σ

2X) and Y ∼ N(µY , σ

2Y ):

S2X/σ2

X

S2Y /σ2

Y∼ Fm−1,n−1

Under the additional assumption that σ2X = σ2

Y :(X−Y )−(µx−µY )

Sp

√1m+ 1

n

∼ tm+n−2

where S2p = 1

m+n−2

((m− 1)S2

X + (n− 1)S2Y

)is the pooled sample

variance.3.3 Maximum likelihood estimators

Asymptotic distribution

If θ is the maximum likelihood estimator of parameter θ based on

a sample X, then θ is asymptotically normally distributed with meanθ and variance equal to the Cramer-Rao lower bound

CRLB(θ) = −1/E

[∂2

∂θ2 logL(θ,X)]

Likelihood ratio test

−2(ℓp − ℓp+q) = −2 log(

maxH0

L

maxH0∪H1

L

)∼ χ2

q approximately (under H0)

where ℓp = maxH0

logL is the maximum log-likelihood for the

model under H0 (in which there arep free parameters)

and ℓp+q = maxH0∪H1

logL is the maximum log-likelihood for the

model under H0 ∪H1 (in which thereare p+ q free parameters)

3.4 Linear regression model with normal errorsModelYi ∼ N(α+ βxi, σ

2), i = 1, 2, . . . , nIntermediate calculations

sxx =n∑

i=1

(xi − x)2 =n∑

i=1

x2i − nx2

syy =n∑

i=1

(yi − y)2 =n∑

i=1

y2i − ny2

sxy =n∑

i=1

(xi − x)(yi − y) =n∑

i=1

xiyi − nxyParameter estimates

α = y − βx, β =sxy

sxx

σ2 = 1n−2

n∑i=1

(yi = yi)2 = 1

n−2

(syy − s2xy

sxx

)

Distribution of ββ−β√σ2/sxx

∼ tn−2

Variance of predicted mean response

var(α + βx0) =(

1n + (x0−x)2

sxx

)σ2

An additional σ2 must be added to obtain the variance of thepredicted individual response.Testing the correlation coefficientr =

sxy√xxxsyy

If ρ = 0, then r√n−2√1−r2

∼ tn−2.

Fisher Z transformation

zr ∼ N(zρ,

1n−3

)approximately

where zr = tanh−1 r = 12 log

(1+r1−r

)and zρ = tanh−1 ρ = 1

2 log(

1+ρ1−ρ

).

Sum of squares relationshipn∑

i=1

(yi − y)2 =n∑

i=1

(yi − y)2 +n∑

i=1

(yi − y)2

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3.5 Analysis of VarianceSingle factor normal modelYij ∼ N(µ+ τi, σ

2), i = 1, 2, . . . , k, j = 1, 2, . . . , ni

where n =∑k

i=1 ni, with∑k

i=1 niτi = 0Single factor normal model

Total: SST =k∑

i=1

ni∑j=1

(yij − y••)2 =k∑

i=1

ni∑j=1

y2ij −y2••

n

Between treaments: SSB =k∑

i=1

ni(yi• − y••)2 =k∑

i=1

y2i•

ni− y2

••

n

Residual: SSR = SST − SSB

Variance estimate

σ2 = SSR

n−k

Statistical testUnder the appropriate null hypothesis:

SSB

k−1

/SSR

n−k ∼ Fk−1,n−k

3.7 Bayesian methodsRelationship between posterior and prior distribution

Posterior ∝ Prior × Likelihood

The posterior distribution f(θ|x) for the parameter θ is related tothe prior distributionf(θ) via the likelihood function f(x|θ):f(θ|x) ∝ f(θ)× f(x|θ)

Normal / normal modelIf x is a random sample of size n from a N(µ, σ2) distribution,where σ2 is known, and the prior distribution for the parameter µ isN(µ0, σ

20), then the posterior distribution for µ is:

µ|x ∼ N(µ⋆, σ2⋆)

where µ⋆ =(

nxσ2 + µ0

σ20

)/(nσ2 + 1

σ20

)and σ2

⋆ = 1/(

nσ2 + 1

σ20

)

Statistical tablesPercentage Points for the standard normal distribution:

Formulae and Tables page 162Cumulative density function for the standard normal distribution:

Formulae and Tables page 160-161Percentage Points for the t distribution:

Formulae and Tables page 163Probabilities for the χ2 distribution:

Formulae and Tables page 164-166Percentage Points for the χ2 distribution:

Formulae and Tables page 168-169Percentage Points for the F distribution:

Formulae and Tables page 171-174

Percentage Points for standard normal distribution

P x P x P x P x P x P x50% 0.0000 5.0% 1.6449 3.0% 1.8801 2.0% 2.0537 1.0% 2.3263 0.10% 3.090245% 0.1257 5.0% 1.6646 2.9% 1.8957 1.9% 2.0749 0.9% 2.3656 0.09% 3.121440% 0.2533 4.8% 1.6849 2.8% 1.9110 1.8% 2.0969 0.8% 2.4089 0.08% 3.155935% 0.3853 4.6% 1.7060 2.7% 1.9268 1.7% 2.1201 0.7% 2.4573 0.07% 3.194730% 0.5244 4.4% 1.7279 2.6% 1.9431 1.6% 2.1444 0.6% 2.5121 0.06% 3.2389

25% 0.6745 4.0% 1.7507 2.5% 1.9600 1.5% 2.1701 0.5% 2.5758 0.05% 3.290520% 0.8416 3.8% 1.7744 2.4% 1.9774 1.4% 2.1973 0.4% 2.6521 0.01% 3.719015% 1.0364 3.6% 1.7991 2.3% 1.9954 1.3% 2.2262 0.3% 2.7478 0.005% 3.890610% 1.2816 3.4% 1.8250 2.2% 2.0141 1.2% 2.2571 0.2% 2.8782 0.001% 4.26495% 1.6449 3.2% 1.8522 2.1% 2.0335 1.1% 2.2904 0.1% 3.0902 0.0005% 4.4172

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Cumulative density function for the standard normal distribution

x Φ(x) x Φ(x) x Φ(x) x Φ(x) x Φ(x)0.00 0.50000 0.40 0.65542 0.80 0.78814 1.20 0.88493 1.60 0.945200.01 0.50399 0.41 0.65910 0.81 0.79103 1.21 0.88686 1.61 0.946300.02 0.50798 0.42 0.66276 0.82 0.79389 1.22 0.88877 1.62 0.947380.03 0.51197 0.43 0.66640 0.83 0.79673 1.23 0.89065 1.63 0.948450.04 0.51595 0.44 0.67003 0.84 0.79955 1.24 0.89251 1.64 0.94950

0.05 0.51994 0.45 0.67364 0.85 0.80234 1.25 0.89435 1.65 0.950530.06 0.52392 0.46 0.67724 0.86 0.80511 1.26 0.89617 1.66 0.951540.07 0.52790 0.47 0.68082 0.87 0.80785 1.27 0.89796 1.67 0.952540.08 0.53188 0.48 0.68439 0.88 0.81057 1.28 0.89973 1.68 0.953520.09 0.53586 0.49 0.68793 0.89 0.81327 1.29 0.90147 1.69 0.95449

0.10 0.53983 0.50 0.69146 0.90 0.81594 1.30 0.90320 1.70 0.955430.11 0.54380 0.51 0.69497 0.91 0.81859 1.31 0.90490 1.71 0.956370.12 0.54776 0.52 0.69847 0.92 0.82121 1.32 0.90658 1.72 0.957280.13 0.55172 0.53 0.70194 0.93 0.82381 1.33 0.90824 1.73 0.958180.14 0.55567 0.54 0.70540 0.94 0.82639 1.34 0.90988 1.74 0.95907

0.15 0.55962 0.55 0.70884 0.95 0.82894 1.35 0.91149 1.75 0.959940.16 0.56356 0.56 0.71226 0.96 0.83147 1.36 0.91309 1.76 0.960800.17 0.56749 0.57 0.71566 0.97 0.83398 1.37 0.91466 1.77 0.961640.18 0.57142 0.58 0.71904 0.98 0.83646 1.38 0.91621 1.78 0.962460.19 0.57535 0.59 0.72240 0.99 0.83891 1.39 0.91774 1.79 0.96327

0.20 0.57926 0.60 0.72575 1.00 0.84134 1.40 0.91924 1.80 0.964070.21 0.58317 0.61 0.72907 1.01 0.84375 1.41 0.92073 1.81 0.964850.22 0.58706 0.62 0.73237 1.02 0.84614 1.42 0.92220 1.82 0.965620.23 0.59095 0.63 0.73565 1.03 0.84849 1.43 0.92364 1.83 0.966380.24 0.59483 0.64 0.73891 1.04 0.85083 1.44 0.92507 1.84 0.96712

0.25 0.59871 0.65 0.74215 1.05 0.85314 1.45 0.92647 1.85 0.967840.26 0.60257 0.66 0.74537 1.06 0.85543 1.46 0.92785 1.86 0.968560.27 0.60642 0.67 0.74857 1.07 0.85769 1.47 0.92922 1.87 0.969260.28 0.61026 0.68 0.75175 1.08 0.85993 1.48 0.93056 1.88 0.969950.29 0.61409 0.69 0.75490 1.09 0.86214 1.49 0.93189 1.89 0.97062

0.30 0.61791 0.70 0.75804 1.10 0.86433 1.50 0.93319 1.90 0.971280.31 0.62172 0.71 0.76115 1.11 0.86650 1.51 0.93448 1.91 0.971930.32 0.62552 0.72 0.76424 1.12 0.86864 1.52 0.93574 1.92 0.972570.33 0.62930 0.73 0.76730 1.13 0.87076 1.53 0.93699 1.93 0.973200.34 0.63307 0.74 0.77035 1.14 0.87286 1.54 0.93822 1.94 0.97381

0.35 0.63683 0.75 0.77337 1.15 0.87493 1.55 0.93943 1.95 0.974410.36 0.64058 0.76 0.77637 1.16 0.87698 1.56 0.94062 1.96 0.975000.37 0.64431 0.77 0.77935 1.17 0.87900 1.57 0.94179 1.97 0.975580.38 0.64803 0.78 0.78230 1.18 0.88100 1.58 0.94295 1.98 0.976150.39 0.65173 0.79 0.78524 1.19 0.88298 1.59 0.94408 1.99 0.97670

0.40 0.65542 0.80 0.78814 1.20 0.88493 1.60 0.94520 2.00 0.97725

Cumulative density function for the standard normal distribution

x Φ(x) x Φ(x) x Φ(x) x Φ(x) x Φ(x) x Φ(x)2.00 0.97725 2.40 0.99180 2.80 0.99744 3.20 0.99931 3.60 0.99984 4.00 0.999972.01 0.97778 2.41 0.99202 2.81 0.99752 3.21 0.99934 3.61 0.99985 4.01 0.999972.02 0.97831 2.42 0.99224 2.82 0.99760 3.22 0.99936 3.62 0.99985 4.02 0.999972.03 0.97882 2.43 0.99245 2.83 0.99767 3.23 0.99938 3.63 0.99986 4.03 0.999972.04 0.97932 2.44 0.99266 2.84 0.99774 3.24 0.99940 3.64 0.99986 4.04 0.99997

2.05 0.97982 2.45 0.99286 2.85 0.99781 3.25 0.99942 3.65 0.99987 4.05 0.999972.06 0.98030 2.46 0.99305 2.86 0.99788 3.26 0.99944 3.66 0.99987 4.06 0.999982.07 0.98077 2.47 0.99324 2.87 0.99795 3.27 0.99946 3.67 0.99988 4.07 0.999982.08 0.98124 2.48 0.99343 2.88 0.99801 3.28 0.99948 3.68 0.99988 4.08 0.999982.09 0.98169 2.49 0.99361 2.89 0.99807 3.29 0.99950 3.69 0.99989 4.09 0.99998

2.10 0.98214 2.50 0.99379 2.90 0.99813 3.30 0.99952 3.70 0.99989 4.10 0.999982.11 0.98257 2.51 0.99396 2.91 0.99819 3.31 0.99953 3.71 0.99990 4.11 0.999982.12 0.98300 2.52 0.99413 2.92 0.99825 3.32 0.99955 3.72 0.99990 4.12 0.999982.13 0.98341 2.53 0.99430 2.93 0.99831 3.33 0.99957 3.73 0.99990 4.13 0.999982.14 0.98382 2.54 0.99446 2.94 0.99836 3.34 0.99958 3.74 0.99991 4.14 0.99998

2.15 0.98422 2.55 0.99461 2.95 0.99841 3.35 0.99960 3.75 0.99991 4.15 0.999982.16 0.98461 2.56 0.99477 2.96 0.99846 3.36 0.99961 3.76 0.99992 4.16 0.999982.17 0.98500 2.57 0.99492 2.97 0.99851 3.37 0.99962 3.77 0.99992 4.17 0.999982.18 0.98537 2.58 0.99506 2.98 0.99856 3.38 0.99964 3.78 0.99992 4.18 0.999992.19 0.98574 2.59 0.99520 2.99 0.99861 3.39 0.99965 3.79 0.99992 4.19 0.99999

2.20 0.98610 2.60 0.99534 3.00 0.99865 3.40 0.99966 3.80 0.99993 4.20 0.999992.21 0.98645 2.61 0.99547 3.01 0.99869 3.41 0.99968 3.81 0.99993 4.21 0.999992.22 0.98679 2.62 0.99560 3.02 0.99874 3.42 0.99969 3.82 0.99993 4.22 0.999992.23 0.98713 2.63 0.99573 3.03 0.99878 3.43 0.99970 3.83 0.99994 4.23 0.999992.24 0.98745 2.64 0.99585 3.04 0.99882 3.44 0.99971 3.84 0.99994 4.24 0.99999

2.25 0.98778 2.65 0.99598 3.05 0.99886 3.45 0.99972 3.85 0.99994 4.25 0.999992.26 0.98809 2.66 0.99609 3.06 0.99889 3.46 0.99973 3.86 0.99994 4.26 0.999992.27 0.98840 2.67 0.99621 3.07 0.99893 3.47 0.99974 3.87 0.99995 4.27 0.999992.28 0.98870 2.68 0.99632 3.08 0.99896 3.48 0.99975 3.88 0.99995 4.28 0.999992.29 0.98899 2.69 0.99643 3.09 0.99900 3.49 0.99976 3.89 0.99995 4.29 0.99999

2.30 0.98928 2.70 0.99653 3.10 0.99903 3.50 0.99977 3.90 0.99995 4.30 0.999992.31 0.98956 2.71 0.99664 3.11 0.99906 3.51 0.99978 3.91 0.99995 4.31 0.999992.32 0.98983 2.72 0.99674 3.12 0.99910 3.52 0.99978 3.92 0.99996 4.32 0.999992.33 0.99010 2.73 0.99683 3.13 0.99913 3.53 0.99979 3.93 0.99996 4.33 0.999992.34 0.99036 2.74 0.99693 3.14 0.99916 3.54 0.99980 3.94 0.99996 4.34 0.99999

2.35 0.99061 2.75 0.99702 3.15 0.99918 3.55 0.99981 3.95 0.99996 4.35 0.999992.36 0.99086 2.76 0.99711 3.16 0.99921 3.56 0.99981 3.96 0.99996 4.36 0.999992.37 0.99111 2.77 0.99720 3.17 0.99924 3.57 0.99982 3.97 0.99996 4.37 0.999992.38 0.99134 2.78 0.99728 3.18 0.99926 3.58 0.99983 3.98 0.99997 4.38 0.999992.39 0.99158 2.79 0.99736 3.19 0.99929 3.59 0.99983 3.99 0.99997 4.39 0.99999

2.40 0.99180 2.80 0.99744 3.20 0.99931 3.60 0.99984 4.00 0.99997 4.40 0.99999

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Percentage Points for the t distribution

P= 40% 30% 25% 20% 15% 10% 5% 2.5% 1% 0.5% 0.1% 0.05%ν1 0.3249 0.7265 1.0000 1.376 1.963 3.078 6.314 12.71 31.82 63.66 318.3 636.62 0.2887 0.6172 0.8165 1.061 1.386 1.866 2.920 4.303 6.965 9.925 22.33 31.603 0.2767 0.5844 0.7649 0.9785 1.250 1.638 2.353 3.182 4.541 5.841 10.21 12.924 0.2707 0.5686 0.7407 0.9410 1.190 1.533 2.132 2.776 3.747 4.604 7.173 8.610

5 0.2672 0.5594 0.7267 0.9195 1.156 1.476 2.015 2.571 3.365 4.032 5.894 6.8696 0.2648 0.5534 0.7176 0.9057 1.134 1.440 1.943 2.447 3.143 3.707 5.208 5.9597 0.2632 0.5491 0.7111 0.8960 1.119 1.415 1.895 2.365 2.998 3.499 4.785 5.4088 0.2619 0.5459 0.7064 0.8889 1.108 1.397 1.860 2.306 2.896 3.355 4.501 5.0419 0.2610 0.5435 0.7072 0.8834 1.100 1.383 1.833 2.262 2.821 3.250 4.297 4.781

10 0.2602 0.5415 0.6998 0.8791 1.093 1.372 1.812 2.228 2.764 3.169 4.144 4.58711 0.2596 0.5399 0.6974 0.8755 1.088 1.363 1.796 2.201 2.718 3.106 4.025 4.43712 0.2590 0.5386 0.6955 0.8726 1.083 1.356 1.782 2.176 2.681 3.055 3.930 4.31813 0.2586 0.5375 0.6938 0.8702 1.079 1.350 1.771 2.160 2.650 3.012 3.852 4.22114 0.2582 0.5366 0.6924 0.8681 1.076 1.345 1.761 2.145 2.624 2.977 3.787 4.140

15 0.2579 0.5357 0.6912 0.8662 1.074 1.341 1.753 2.131 2.602 2.947 3.733 4.07316 0.2576 0.5350 0.6901 0.8647 1.071 1.337 1.746 2.120 2.583 2.921 3.686 4.01517 0.2573 0.5344 0.6892 0.8633 1.069 1.333 1.740 2.110 2.567 2.898 3.646 3.96518 0.2571 0.5338 0.6884 0.8620 1.067 1.330 1.734 2.101 2.552 2.878 3.610 3.92219 0.2569 0.5333 0.6876 0.8610 1.066 1.328 1.729 2.093 2.539 2.861 3.579 3.883

20 0.2567 0.5329 0.6870 0.8600 1.064 1.325 1.725 2.086 2.528 2.845 3.522 3.85021 0.2566 0.5325 0.6864 0.8591 1.063 1.323 1.721 2.080 2.518 2.831 3.527 3.81922 0.2564 0.5321 0.6858 0.8583 1.061 1.321 1.717 2.074 2.508 2.819 3.505 3.79223 0.2563 0.5317 0.6853 0.8575 1.060 1.319 1.714 2.069 2.500 2.807 3.485 3.76824 0.2562 0.5314 0.6848 0.8569 1.059 1.318 1.711 2.064 2.492 2.797 3.467 3.745

25 0.2561 0.5312 0.6844 0.8562 1.058 1.316 1.708 2.060 2.485 2.787 3.450 3.72526 0.2560 0.5309 0.6840 0.8557 1.058 1.315 1.706 2.056 2.479 2.779 3.435 3.70727 0.2559 0.5306 0.6837 0.8551 1.057 1.314 1.703 2.052 2.473 2.771 3.421 3.68928 0.2558 0.5304 0.6834 0.8546 1.056 1.313 1.701 2.048 2.467 2.763 3.408 3.67429 0.2557 0.5302 0.6830 0.8542 1.055 1.311 1.699 2.045 2.462 2.756 3.396 3.660

30 0.2556 0.5300 0.6828 0.8538 1.055 1.310 1.697 2.042 2.457 2.750 3.385 3.64632 0.2555 0.5297 0.6822 0.8530 1.054 1.309 1.694 2.037 2.449 2.738 3.365 3.62234 0.2553 0.5294 0.6818 0.8523 1.052 1.307 1.691 2.032 2.441 2.728 3.348 3.60136 0.2552 0.5291 0.6814 0.8517 1.052 1.306 1.688 2.028 2.434 2.719 3.333 3.58238 0.2551 0.5288 0.6810 0.8512 1.051 1.304 1.686 2.024 2.429 2.712 3.319 3.566

40 0.2550 0.5286 0.6807 0.8507 1.050 1.303 1.684 2.021 2.423 2.704 3.307 3.55150 0.2547 0.5278 0.6794 0.8489 1.047 1.299 1.676 2.009 2.403 2.678 3.261 3.49660 0.2545 0.5272 0.6786 0.8477 1.045 1.296 1.671 2.000 2.390 2.660 3.232 3.460120 0.2539 0.5258 0.6765 0.8446 1.041 1.289 1.658 1.980 2.358 2.617 3.160 3.373∞ 0.2533 0.5244 0.6745 0.8416 1.036 1.282 1.645 1.960 2.326 2.576 3.090 3.291

Probabilities for the χ2 distribution

ν = 1 1 2 2 3 3x x x x x x

0.0 0.0000 4.0 0.9545 0.0 0.0000 4.0 0.8647 0.0 0.0000 4.0 0.73850.1 0.2482 4.1 0.9571 0.1 0.0488 4.1 0.8713 0.1 0.0082 4.1 0.74910.2 0.3453 4.2 0.9596 0.2 0.0952 4.2 0.8775 0.2 0.0224 4.2 0.75930.3 0.4161 4.3 0.9619 0.3 0.1393 4.3 0.8835 0.3 0.0400 4.3 0.76920.4 0.4729 4.4 0.9641 0.4 0.1813 4.4 0.8892 0.4 0.0598 4.4 0.7786

0.5 0.5205 4.5 0.9661 0.5 0.2212 4.5 0.8946 0.5 0.0811 4.5 0.78770.6 0.5614 4.6 0.9680 0.6 0.2592 4.6 0.8997 0.6 0.1036 4.6 0.79650.7 0.5972 4.7 0.9698 0.7 0.2953 4.7 0.9046 0.7 0.1268 4.7 0.80490.8 0.6289 4.8 0.9715 0.8 0.3297 4.8 0.9093 0.8 0.1505 4.8 0.81300.9 0.6572 4.9 0.9731 0.9 0.3624 4.9 0.9137 0.9 0.1746 4.9 0.8207

1.0 0.6827 5.0 0.9747 1.0 0.3935 5.0 0.9179 1.0 0.1987 5.0 0.82821.1 0.7057 5.1 0.9761 1.1 0.4231 5.1 0.9219 1.1 0.2229 5.1 0.83541.2 0.7267 5.2 0.9774 1.2 0.4512 5.2 0.9257 1.2 0.2470 5.2 0.84231.3 0.7458 5.3 0.9787 1.3 0.4780 5.3 0.9293 1.3 0.2709 5.3 0.84891.4 0.7633 5.4 0.9799 1.4 0.5034 5.4 0.9328 1.4 0.2945 5.4 0.8553

1.5 0.7793 5.5 0.9810 1.5 0.5276 5.5 0.9361 1.5 0.3177 5.5 0.86141.6 0.7941 5.6 0.9820 1.6 0.5507 5.6 0.9392 1.6 0.3406 5.6 0.86721.7 0.8077 5.7 0.9830 1.7 0.5726 5.7 0.9422 1.7 0.3631 5.7 0.87281.8 0.8203 5.8 0.9840 1.8 0.5934 5.8 0.9450 1.8 0.3851 5.8 0.87821.9 0.8319 5.9 0.9849 1.9 0.6133 5.9 0.9477 1.9 0.4066 5.9 0.8834

2.0 0.8427 6.0 0.9857 2.0 0.6321 6.0 0.9502 2.0 0.4276 6.0 0.88842.1 0.8527 6.1 0.9865 2.1 0.6501 6.1 0.9526 2.1 0.4481 6.1 0.89322.2 0.8620 6.2 0.9872 2.2 0.6671 6.2 0.9550 2.2 0.4681 6.2 0.89772.3 0.8706 6.3 0.9879 2.3 0.6834 6.3 0.9571 2.3 0.4875 6.3 0.90212.4 0.8787 6.4 0.9886 2.4 0.6988 6.4 0.9592 2.4 0.5064 6.4 0.9063

2.5 0.8862 6.5 0.9892 2.5 0.7135 6.5 0.9612 2.5 0.5247 6.5 0.91032.6 0.8931 6.6 0.9898 2.6 0.7275 6.6 0.9631 2.6 0.5425 6.6 0.91422.7 0.8997 6.7 0.9904 2.7 0.7408 6.7 0.9649 2.7 0.5598 6.7 0.91792.8 0.9057 6.8 0.9909 2.8 0.7534 6.8 0.9666 2.8 0.5765 6.8 0.92142.9 0.9114 6.9 0.9914 2.9 0.7654 6.9 0.9683 2.9 0.5927 6.9 0.9248

3.0 0.9167 7.0 0.9918 3.0 0.7769 7.0 0.9698 3.0 0.6084 7.0 0.92813.1 0.9217 7.1 0.9923 3.1 0.7878 7.1 0.9713 3.1 0.6235 7.1 0.93123.2 0.9264 7.2 0.9927 3.2 0.7981 7.2 0.9727 3.2 0.6382 7.2 0.93423.3 0.9307 7.3 0.9931 3.3 0.8080 7.3 0.9740 3.3 0.6524 7.3 0.93713.4 0.9348 7.4 0.9935 3.4 0.8173 7.4 0.9753 3.4 0.6660 7.4 0.9398

3.5 0.9386 7.5 0.9938 3.5 0.8262 7.5 0.9765 3.5 0.6792 7.5 0.94243.6 0.9422 7.6 0.9942 3.6 0.8347 7.6 0.9776 3.6 0.6920 7.6 0.94503.7 0.9456 7.7 0.9945 3.7 0.8428 7.7 0.9787 3.7 0.7043 7.7 0.94743.8 0.9487 7.8 0.9948 3.8 0.8504 7.8 0.9798 3.8 0.7161 7.8 0.94973.9 0.9517 7.9 0.9951 3.9 0.8577 7.9 0.9807 3.9 0.7275 7.9 0.9519

4.0 0.9545 8.0 0.9953 4.0 0.8647 8.0 0.9817 4.0 0.7385 8.0 0.9540

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Tables

ν = 4 5 6 7 8 9 10 11 12 13 14x

0.5 0.0265 0.0079 0.0022 0.0006 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.00001.0 0.0902 0.0374 0.0144 0.0052 0.0018 0.0006 0.0002 0.0001 0.0000 0.0000 0.00001.5 0.1734 0.0869 0.0405 0.0177 0.0073 0.0029 0.0011 0.0004 0.0001 0.0000 0.00002.0 0.2642 0.1509 0.0803 0.0402 0.0190 0.0085 0.0037 0.0015 0.0006 0.0002 0.0001

2.5 0.3554 0.2235 0.1315 0.0729 0.0383 0.0191 0.0091 0.0042 0.0018 0.0008 0.00033.0 0.4422 0.3000 0.1912 0.1150 0.0656 0.0357 0.0186 0.0093 0.0045 0.0021 0.00093.5 0.5221 0.3766 0.2560 0.1648 0.1008 0.0589 0.0329 0.0177 0.0091 0.0046 0.00224.0 0.5940 0.4506 0.3233 0.2202 0.1429 0.0886 0.0527 0.0301 0.0166 0.0088 0.00454.5 0.6575 0.5201 0.3907 0.2793 0.1906 0.1245 0.0780 0.0471 0.0274 0.0154 0.0084

5.0 0.7127 0.5841 0.4562 0.3400 0.2424 0.1657 0.1088 0.0688 0.0420 0.0248 0.01425.5 0.7603 0.6421 0.5185 0.4008 0.2970 0.2113 0.1446 0.0954 0.0608 0.0375 0.02246.0 0.8009 0.6938 0.5768 0.4603 0.3528 0.2601 0.1847 0.1266 0.0839 0.0538 0.03356.5 0.8352 0.7394 0.6304 0.5173 0.4086 0.3110 0.2283 0.1620 0.1112 0.0739 0.04777.0 0.8641 0.7794 0.6792 0.5711 0.4634 0.3629 0.2746 0.2009 0.1424 0.0978 0.0653

7.5 0.8883 0.8140 0.7229 0.6213 0.5162 0.4148 0.3225 0.2427 0.1771 0.1254 0.08638.0 0.9084 0.8438 0.7619 0.6674 0.5665 0.4659 0.3712 0.2867 0.2149 0.1564 0.11078.5 0.9251 0.8693 0.7963 0.7094 0.6138 0.5154 0.4199 0.3321 0.2551 0.1904 0.13839.0 0.9389 0.8909 0.8264 0.7473 0.6577 0.5627 0.4679 0.3781 0.2971 0.2271 0.16899.5 0.9503 0.9093 0.8527 0.7813 0.6981 0.6075 0.5146 0.4242 0.3403 0.2658 0.2022

10.0 0.9596 0.9248 0.8753 0.8114 0.7350 0.6495 0.5595 0.4696 0.3840 0.3061 0.237810.5 0.9672 0.9378 0.8949 0.8380 0.7683 0.6885 0.6022 0.5140 0.4278 0.3474 0.275211.0 0.9734 0.9486 0.9116 0.8614 0.7983 0.7243 0.6425 0.5567 0.4711 0.3892 0.314011.5 0.9785 0.9577 0.9259 0.8818 0.8251 0.7570 0.6801 0.5976 0.5134 0.4310 0.353612.0 0.9826 0.9652 0.9380 0.8994 0.8488 0.7867 0.7149 0.6364 0.5543 0.4724 0.3937

12.5 0.9860 0.9715 0.9483 0.9147 0.8697 0.8134 0.7470 0.6727 0.5936 0.5129 0.433813.0 0.9887 0.9766 0.9570 0.9279 0.8882 0.8374 0.7763 0.7067 0.6310 0.5522 0.473513.5 0.9909 0.9809 0.9643 0.9392 0.9042 0.8587 0.8030 0.7381 0.6662 0.5900 0.512414.0 0.9927 0.9844 0.9704 0.9488 0.9182 0.8777 0.8270 0.7670 0.6993 0.6262 0.550314.5 0.9941 0.9873 0.9755 0.9570 0.9304 0.8944 0.8486 0.7935 0.7301 0.6604 0.5868

15.0 0.9953 0.9896 0.9797 0.9640 0.9409 0.9091 0.8679 0.8175 0.7586 0.6926 0.621815.5 0.9962 0.9916 0.9833 0.9699 0.9499 0.9219 0.8851 0.8393 0.7848 0.7228 0.655116.0 0.9970 0.9932 0.9862 0.9749 0.9576 0.9331 0.9004 0.8589 0.8088 0.7509 0.686616.5 0.9976 0.9944 0.9887 0.9791 0.9642 0.9429 0.9138 0.8764 0.8306 0.7768 0.716217.0 0.9981 0.9955 0.9907 0.9826 0.9699 0.9513 0.9256 0.8921 0.8504 0.8007 0.7438

17.5 0.9985 0.9964 0.9924 0.9856 0.9747 0.9586 0.9360 0.9061 0.8683 0.8226 0.769518.0 0.9988 0.9971 0.9938 0.9880 0.9788 0.9648 0.9450 0.9184 0.8843 0.8425 0.793218.5 0.9990 0.9976 0.9949 0.9901 0.9822 0.9702 0.9529 0.9293 0.8987 0.8606 0.815119.0 0.9992 0.9981 0.9958 0.9918 0.9851 0.9748 0.9597 0.9389 0.9115 0.8769 0.835119.5 0.9994 0.9984 0.9966 0.9932 0.9876 0.9787 0.9656 0.9473 0.9228 0.8916 0.8533

20 0.9995 0.9988 0.9972 0.9944 0.9897 0.9821 0.9707 0.9547 0.9329 0.9048 0.869921 0.9997 0.9992 0.9982 0.9962 0.9929 0.9873 0.9789 0.9666 0.9496 0.9271 0.898422 0.9998 0.9995 0.9988 0.9975 0.9951 0.9911 0.9849 0.9756 0.9625 0.9446 0.921423 0.9999 0.9997 0.9992 0.9983 0.9966 0.9938 0.9893 0.9823 0.9723 0.9583 0.939724 0.9999 0.9998 0.9995 0.9989 0.9977 0.9957 0.9924 0.9873 0.9797 0.9689 0.9542

25 0.9999 0.9999 0.9997 0.9992 0.9984 0.9970 0.9947 0.9909 0.9852 0.9769 0.965426 1.0000 0.9999 0.9998 0.9995 0.9989 0.9980 0.9963 0.9935 0.9893 0.9830 0.974127 1.0000 0.9999 0.9999 0.9997 0.9993 0.9986 0.9974 0.9954 0.9923 0.9876 0.980728 1.0000 1.0000 0.9999 0.9998 0.9995 0.9990 0.9982 0.9968 0.9945 0.9910 0.985829 1.0000 1.0000 0.9999 0.9999 0.9997 0.9994 0.9988 0.9977 0.9961 0.9935 0.9895

30 1.0000 1.0000 1.0000 0.9999 0.9998 0.9996 0.9991 0.9984 0.9972 0.9953 0.9924

Probabilities for the χ2 distribution

ν = 15 16 17 18 19 20 21 22 23 24 25x

3 0.0004 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00004 0.0023 0.0011 0.0005 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

5 0.0079 0.0042 0.0022 0.0011 0.0006 0.0003 0.0001 0.0001 0.0000 0.0000 0.00006 0.0203 0.0119 0.0068 0.0038 0.0021 0.0011 0.0006 0.0003 0.0001 0.0001 0.00007 0.0424 0.0267 0.0165 0.0099 0.0058 0.0033 0.0019 0.0010 0.0005 0.0003 0.00018 0.0762 0.0511 0.0335 0.0214 0.0133 0.0081 0.0049 0.0028 0.0016 0.0009 0.00059 0.1225 0.0866 0.0597 0.0403 0.0265 0.0171 0.0108 0.0067 0.0040 0.0024 0.0014

10 0.1803 0.1334 0.0964 0.0681 0.0471 0.0318 0.0211 0.0137 0.0087 0.0055 0.003311 0.2474 0.1905 0.1434 0.1056 0.0762 0.0538 0.0372 0.0253 0.0168 0.0110 0.007112 0.3210 0.2560 0.1999 0.1528 0.1144 0.0839 0.0604 0.0426 0.0295 0.0201 0.013413 0.3977 0.3272 0.2638 0.2084 0.1614 0.1226 0.0914 0.0668 0.0480 0.0339 0.023514 0.4745 0.4013 0.3329 0.2709 0.2163 0.1695 0.1304 0.0985 0.0731 0.0533 0.0383

15 0.5486 0.4754 0.4045 0.3380 0.2774 0.2236 0.1770 0.1378 0.1054 0.0792 0.058616 0.6179 0.5470 0.4762 0.4075 0.3427 0.2834 0.2303 0.1841 0.1447 0.1119 0.085217 0.6811 0.6144 0.5456 0.4769 0.4101 0.3470 0.2889 0.2366 0.1907 0.1513 0.118218 0.7373 0.6761 0.6112 0.5443 0.4776 0.4126 0.3510 0.2940 0.2425 0.1970 0.157619 0.7863 0.7313 0.6715 0.6082 0.5432 0.4782 0.4149 0.3547 0.2988 0.2480 0.2029

20 0.8281 0.7798 0.7258 0.6672 0.6054 0.5421 0.4787 0.4170 0.3581 0.3032 0.253221 0.8632 0.8215 0.7737 0.7206 0.6632 0.6029 0.5411 0.4793 0.4189 0.3613 0.307422 0.8922 0.8568 0.8153 0.7680 0.7157 0.6595 0.6005 0.5401 0.4797 0.4207 0.364323 0.9159 0.8863 0.8507 0.8094 0.7627 0.7112 0.6560 0.5983 0.5392 0.4802 0.422424 0.9349 0.9105 0.8806 0.8450 0.8038 0.7576 0.7069 0.6528 0.5962 0.5384 0.4806

25 0.9501 0.9302 0.9053 0.8751 0.8395 0.7986 0.7528 0.7029 0.6497 0.5942 0.537626 0.9620 0.9460 0.9255 0.9002 0.8698 0.8342 0.7936 0.7483 0.6991 0.6468 0.592427 0.9713 0.9585 0.9419 0.9210 0.8953 0.8647 0.8291 0.7888 0.7440 0.6955 0.644128 0.9784 0.9684 0.9551 0.9379 0.9166 0.8906 0.8598 0.8243 0.7842 0.7400 0.692129 0.9839 0.9761 0.9655 0.9516 0.9340 0.9122 0.8860 0.8551 0.8197 0.7799 0.7361

30 0.9881 0.9820 0.9737 0.9626 0.9482 0.9301 0.9080 0.8815 0.8506 0.8152 0.775731 0.9912 0.9865 0.9800 0.9712 0.9596 0.9448 0.9263 0.9039 0.8772 0.8462 0.811032 0.9936 0.9900 0.9850 0.9780 0.9687 0.9567 0.9414 0.9226 0.8999 0.8730 0.842033 0.9953 0.9926 0.9887 0.9833 0.9760 0.9663 0.9538 0.9381 0.9189 0.8959 0.868934 0.9966 0.9946 0.9916 0.9874 0.9816 0.9739 0.9638 0.9509 0.9348 0.9153 0.8921

35 0.9975 0.9960 0.9938 0.9905 0.9860 0.9799 0.9718 0.9613 0.9480 0.9316 0.911836 0.9982 0.9971 0.9954 0.9929 0.9894 0.9846 0.9781 0.9696 0.9587 0.9451 0.928437 0.9987 0.9979 0.9966 0.9948 0.9921 0.9883 0.9832 0.9763 0.9675 0.9562 0.942338 0.9991 0.9985 0.9975 0.9961 0.9941 0.9911 0.9871 0.9817 0.9745 0.9653 0.953739 0.9994 0.9989 0.9982 0.9972 0.9956 0.9933 0.9902 0.9859 0.9802 0.9727 0.9632

40 0.9995 0.9992 0.9987 0.9979 0.9967 0.9950 0.9926 0.9892 0.9846 0.9786 0.970841 0.9997 0.9994 0.9991 0.9985 0.9976 0.9963 0.9944 0.9918 0.9882 0.9833 0.977042 0.9998 0.9996 0.9993 0.9989 0.9982 0.9972 0.9958 0.9937 0.9909 0.9871 0.982043 0.9998 0.9997 0.9995 0.9992 0.9987 0.9980 0.9969 0.9953 0.9931 0.9901 0.986044 0.9999 0.9998 0.9997 0.9994 0.9991 0.9985 0.9977 0.9965 0.9947 0.9924 0.9892

45 0.9999 0.9999 0.9998 0.9996 0.9993 0.9989 0.9983 0.9973 0.9960 0.9942 0.991646 0.9999 0.9999 0.9998 0.9997 0.9995 0.9992 0.9987 0.9980 0.9970 0.9956 0.993647 1.0000 0.9999 0.9999 0.9998 0.9996 0.9994 0.9991 0.9985 0.9978 0.9967 0.995148 1.0000 1.0000 0.9999 0.9998 0.9997 0.9996 0.9993 0.9989 0.9983 0.9975 0.996349 1.0000 1.0000 0.9999 0.9999 0.9998 0.9997 0.9995 0.9992 0.9988 0.9981 0.9972

50 1.0000 1.0000 1.0000 0.9999 0.9999 0.9998 0.9996 0.9994 0.9991 0.9986 0.9979

c©Katja

Ignatieva

SchoolofRisk

andActu

aria

lStudies,

ASB,UNSW

Page1108of1113

ACTL2002&

ACTL5101

Form

ulaeand

Tables

Percentage points for the χ2 distribution

P= 99.95% 99.9% 99.5% 99% 97.5% 95% 90% 80% 70% 60%

ν

1 3.927E-07 1.571E-06 3.927E-05 1.571E-04 9.821E-04 0.003932 0.01579 0.06418 0.1485 0.27502 0.001000 0.00200 0.01003 0.02010 0.05064 0.1026 0.2107 0.4463 0.7133 1.0223 0.01528 0.02430 0.07172 0.1148 0.2158 0.3518 0.5844 1.005 1.424 1.8694 0.06392 0.09080 0.2070 0.2971 0.4844 0.7107 1.064 1.649 2.195 2.753

5 0.1581 0.2102 0.4117 0.5543 0.8312 1.145 1.610 2.343 3.000 3.6556 0.2994 0.3811 0.6757 0.8721 1.237 1.635 2.204 3.070 3.828 4.5707 0.4849 0.5985 0.9893 1.239 1.690 2.167 2.833 3.822 4.671 5.4938 0.7104 0.8571 1.344 1.646 2.180 2.733 3.490 4.594 5.527 6.4239 0.9717 1.152 1.735 2.088 2.700 3.325 4.168 5.380 6.393 7.357

10 1.265 1.479 2.156 2.558 3.247 3.940 4.865 6.179 7.267 8.29511 1.587 1.834 2.603 3.053 3.816 4.575 5.578 6.989 8.148 9.23712 1.934 2.214 3.074 3.571 4.404 5.226 6.304 7.807 9.034 10.1813 2.305 2.617 3.565 4.107 5.009 5.892 7.042 8.634 9.926 11.1314 2.697 3.041 4.075 4.660 5.629 6.571 7.790 9.467 10.82 12.08

15 3.108 3.483 4.601 5.229 6.262 7.261 8.547 10.31 11.72 13.0316 3.536 3.942 5.142 5.812 6.908 7.962 9.312 11.15 12.62 13.9817 3.980 4.416 5.697 6.408 7.564 8.672 10.09 12.00 13.53 14.9418 4.439 4.905 6.265 7.015 8.231 9.390 10.86 12.86 14.44 15.8919 4.912 5.407 6.844 7.633 8.907 10.12 11.65 13.72 15.35 16.85

20 5.398 5.921 7.434 8.260 9.591 10.85 12.44 14.58 16.27 17.8121 5.896 6.447 8.034 8.897 10.28 11.59 13.24 15.44 17.18 18.7722 6.404 6.983 8.643 9.542 10.98 12.34 14.04 16.31 18.10 19.7323 6.924 7.529 9.260 10.20 11.69 13.09 14.85 17.19 19.02 20.6924 7.453 8.085 9.886 10.86 12.40 13.85 15.66 18.06 19.94 21.65

25 7.991 8.649 10.52 11.52 13.12 14.61 16.47 18.94 20.87 22.6226 8.538 9.222 11.16 12.20 13.84 15.38 17.29 19.82 21.79 23.5827 9.093 9.803 11.81 12.88 14.57 16.15 18.11 20.70 22.72 24.5428 9.656 10.39 12.46 13.56 15.31 16.93 18.94 21.59 23.65 25.5129 10.23 10.99 13.12 14.26 16.05 17.71 19.77 22.48 24.58 26.48

30 10.80 11.59 13.79 14.95 16.79 18.49 20.60 23.36 25.51 27.4432 11.98 12.81 15.13 16.36 18.29 20.07 22.27 25.15 27.37 29.3834 13.18 14.06 16.50 17.79 19.81 21.66 23.95 26.94 29.24 31.3136 14.40 15.32 17.89 19.23 21.34 23.27 25.64 28.73 31.12 33.2538 15.64 16.61 19.29 20.69 22.88 24.88 27.34 30.54 32.99 35.19

40 16.91 17.92 20.71 22.16 24.43 26.51 29.05 32.34 34.87 37.1350 23.46 24.67 27.99 29.71 32.36 34.76 37.69 41.45 44.31 46.8660 30.34 31.74 35.53 37.48 40.48 43.19 46.46 50.64 53.81 56.6270 37.47 39.04 43.28 45.44 48.76 51.74 55.33 59.90 63.35 66.4080 44.79 46.52 51.17 53.54 57.15 60.39 64.28 69.21 72.92 76.19

90 52.28 54.1552 59.20 61.75 65.65 69.13 73.29 78.56 82.51 85.99100 59.90 61.9179 67.33 70.06 74.22 77.93 82.36 87.95 92.13 95.81

Percentage points for the χ2 distribution

P= 50% 40% 30% 20% 10% 5% 2.5% 1% 0.5% 0.1% 0.05%

ν

1 0.4549 0.7083 1.074 1.642 2.706 3.841 5.024 6.635 7.879 10.83 12.122 1.386 1.833 2.408 3.219 4.605 5.991 7.378 9.210 10.60 13.82 15.203 2.366 2.946 3.665 4.642 6.251 7.815 9.348 11.34 12.84 16.27 17.734 3.357 4.045 4.878 5.989 7.779 9.488 11.14 13.28 14.86 18.47 20.00

5 4.351 5.132 6.064 7.289 9.236 11.07 12.83 15.09 16.75 20.52 22.116 5.348 6.211 7.231 8.558 10.64 12.59 14.45 16.81 18.55 22.46 24.107 6.346 7.283 8.383 9.803 12.02 14.07 16.01 18.48 20.28 24.32 26.028 7.344 8.351 9.524 11.03 13.36 15.51 17.53 20.09 21.95 26.12 27.879 8.343 9.414 10.66 12.24 14.68 16.92 19.02 21.67 23.59 27.88 29.67

10 9.342 10.47 11.78 13.44 15.99 18.31 20.48 23.21 25.19 29.59 31.4211 10.34 11.53 12.90 14.63 17.28 19.68 21.92 24.72 26.76 31.26 33.1412 11.34 12.58 14.01 15.81 18.55 21.03 23.34 26.22 28.30 32.91 34.8213 12.34 13.64 15.12 16.98 19.81 22.36 24.74 27.69 29.82 34.53 36.4814 13.34 14.69 16.22 18.15 21.06 23.68 26.12 29.14 31.32 36.12 38.11

15 14.34 15.73 17.32 19.31 22.31 25.00 27.49 30.58 32.80 37.70 39.7216 15.34 16.78 18.42 20.47 23.54 26.30 28.85 32.00 34.27 39.25 41.3117 16.34 17.82 19.51 21.61 24.77 27.59 30.19 33.41 35.72 40.79 42.8818 17.34 18.87 20.60 22.76 25.99 28.87 31.53 34.81 37.16 42.31 44.4319 18.34 19.91 21.69 23.90 27.20 30.14 32.85 36.19 38.58 43.82 45.97

20 19.34 20.95 22.77 25.04 28.41 31.41 34.17 37.57 40.00 45.31 47.5021 20.34 21.99 23.86 26.17 29.62 32.67 35.48 38.93 41.40 46.80 49.0122 21.34 23.03 24.94 27.30 30.81 33.92 36.78 40.29 42.80 48.27 50.5123 22.34 24.07 26.02 28.43 32.01 35.17 38.08 41.64 44.18 49.73 52.0024 23.34 25.11 27.10 29.55 33.20 36.42 39.36 42.98 45.56 51.18 53.48

25 24.34 26.14 28.17 30.68 34.38 37.65 40.65 44.31 46.93 52.62 54.9526 25.34 27.18 29.25 31.79 35.56 38.89 41.92 45.64 48.29 54.05 56.4127 26.34 28.21 30.32 32.91 36.74 40.11 43.19 46.96 49.64 55.48 57.8628 27.34 29.25 31.39 34.03 37.92 41.34 44.46 48.28 50.99 56.89 59.3029 28.34 30.28 32.46 35.14 39.09 42.56 45.72 49.59 52.34 58.30 60.73

30 29.34 31.32 33.53 36.25 40.26 43.77 46.98 50.89 53.67 59.70 62.1632 31.34 33.38 35.66 38.47 42.58 46.19 49.48 53.49 56.33 62.49 65.0034 33.34 35.44 37.80 40.68 44.90 48.60 51.97 56.06 58.96 65.25 67.8036 35.34 37.50 39.92 42.88 47.21 51.00 54.44 58.62 61.58 67.99 70.5938 37.34 39.56 42.05 45.08 49.51 53.38 56.90 61.16 64.18 70.70 73.35

40 39.34 41.62 44.16 47.27 51.81 55.76 59.34 63.69 66.77 73.40 76.0950 49.33 51.89 54.72 58.16 63.17 67.50 71.42 76.15 79.49 86.66 89.5660 59.33 62.13 65.23 68.97 74.40 79.08 83.30 88.38 91.95 99.61 102.770 69.33 72.36 75.69 79.71 85.53 90.53 95.02 100.4 104.2 112.3 115.680 79.33 82.57 86.12 90.41 96.58 101.9 106.6 112.3 116.3 124.8 128.3

90 89.33 92.76 96.52 101.1 107.6 113.1 118.1 124.1 128.3 137.2 140.8100 99.33 102.9 106.9 111.7 118.5 124.3 129.6 135.8 140.2 149.4 153.2

c©Katja

Ignatieva

SchoolofRisk

andActu

aria

lStudies,

ASB,UNSW

Page1109of1113

ACTL2002&

ACTL5101

Form

ulaeand

Tables

10% Points for the F distribution

ν1 = 1 2 3 4 5 6 7 8 9 10 12 24 ∞

ν2

1 39.86 49.50 53.59 55.83 57.24 58.20 58.91 59.44 59.86 60.19 60.71 62.00 63.332 8.526 9.000 9.162 9.243 9.293 9.326 9.349 9.367 9.381 9.392 9.408 9.450 9.4913 5.538 5.462 5.391 5.343 5.309 5.285 5.266 5.252 5.240 5.230 5.216 5.176 5.1344 4.545 4.325 4.191 4.107 4.051 4.010 3.979 3.955 3.936 3.920 3.896 3.831 3.761

5 4.060 3.780 3.619 3.520 3.453 3.405 3.368 3.339 3.316 3.297 3.268 3.191 3.1056 3.776 3.463 3.289 3.181 3.108 3.055 3.014 2.983 2.958 2.937 2.905 2.818 2.7227 3.589 3.257 3.074 2.961 2.883 2.827 2.785 2.752 2.725 2.703 2.668 2.575 2.4718 3.458 3.113 2.924 2.806 2.726 2.668 2.624 2.589 2.561 2.538 2.502 2.404 2.2939 3.360 3.006 2.813 2.693 2.611 2.551 2.505 2.469 2.440 2.416 2.379 2.277 2.159

10 3.285 2.924 2.728 2.605 2.522 2.461 2.414 2.377 2.347 2.323 2.284 2.178 2.05511 3.225 2.860 2.660 2.536 2.451 2.389 2.342 2.304 2.274 2.248 2.209 2.100 1.97212 3.177 2.807 2.606 2.480 2.394 2.331 2.283 2.245 2.214 2.188 2.147 2.036 1.90413 3.136 2.763 2.560 2.434 2.347 2.283 2.234 2.195 2.164 2.138 2.097 1.983 1.84614 3.102 2.726 2.522 2.395 2.307 2.243 2.193 2.154 2.122 2.095 2.054 1.938 1.797

15 3.073 2.695 2.490 2.361 2.273 2.208 2.158 2.119 2.086 2.059 2.017 1.899 1.75516 3.048 2.668 2.462 2.333 2.244 2.178 2.128 2.088 2.055 2.028 1.985 1.866 1.71817 3.026 2.645 2.437 2.308 2.218 2.152 2.102 2.061 2.028 2.001 1.958 1.836 1.68618 3.007 2.624 2.416 2.286 2.196 2.130 2.079 2.038 2.005 1.977 1.933 1.810 1.65719 2.990 2.606 2.397 2.266 2.176 2.109 2.058 2.017 1.984 1.956 1.912 1.787 1.631

20 2.975 2.589 2.380 2.249 2.158 2.091 2.040 1.999 1.965 1.937 1.892 1.767 1.60721 2.961 2.575 2.365 2.233 2.142 2.075 2.023 1.982 1.948 1.920 1.875 1.748 1.58622 2.949 2.561 2.351 2.219 2.128 2.060 2.008 1.967 1.933 1.904 1.859 1.731 1.56723 2.937 2.549 2.339 2.207 2.115 2.047 1.995 1.953 1.919 1.890 1.845 1.716 1.54924 2.927 2.538 2.327 2.195 2.103 2.035 1.983 1.941 1.906 1.877 1.832 1.702 1.533

25 2.918 2.528 2.317 2.184 2.092 2.024 1.971 1.929 1.895 1.866 1.820 1.689 1.51826 2.909 2.519 2.307 2.174 2.082 2.014 1.961 1.919 1.884 1.855 1.809 1.677 1.50427 2.901 2.511 2.299 2.165 2.073 2.005 1.952 1.909 1.874 1.845 1.799 1.666 1.49128 2.894 2.503 2.291 2.157 2.064 1.996 1.943 1.900 1.865 1.836 1.790 1.656 1.47829 2.887 2.495 2.283 2.149 2.057 1.988 1.935 1.892 1.857 1.827 1.781 1.647 1.467

30 2.881 2.489 2.276 2.142 2.049 1.980 1.927 1.884 1.849 1.819 1.773 1.638 1.45632 2.869 2.477 2.263 2.129 2.036 1.967 1.913 1.870 1.835 1.805 1.758 1.622 1.43734 2.859 2.466 2.252 2.118 2.024 1.955 1.901 1.858 1.822 1.793 1.745 1.608 1.41936 2.850 2.456 2.243 2.108 2.014 1.945 1.891 1.847 1.811 1.781 1.734 1.595 1.40438 2.842 2.448 2.234 2.099 2.005 1.935 1.881 1.838 1.802 1.772 1.724 1.584 1.390

40 2.835 2.440 2.226 2.091 1.997 1.927 1.873 1.829 1.793 1.763 1.715 1.574 1.37760 2.791 2.393 2.177 2.041 1.946 1.875 1.819 1.775 1.738 1.707 1.657 1.511 1.291120 2.748 2.347 2.130 1.992 1.896 1.824 1.767 1.722 1.684 1.652 1.601 1.447 1.193∞ 2.706 2.303 2.084 1.945 1.847 1.774 1.717 1.670 1.632 1.599 1.546 1.383 1.000


ν1 = 1 2 3 4 5 6 7 8 9 10 12 24 ∞

ν2

1 161.4 199.5 215.7 224.6 230.2 234.0 236.8 238.9 240.5 241.9 243.9 249.1 254.32 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38 19.40 19.41 19.45 19.503 10.13 9.552 9.277 9.117 9.013 8.941 8.887 8.845 8.812 8.786 8.745 8.639 8.5264 7.709 6.944 6.591 6.388 6.256 6.163 6.094 6.041 5.999 5.964 5.912 5.774 5.628

5 6.608 5.786 5.409 5.192 5.050 4.950 4.876 4.818 4.772 4.735 4.678 4.527 4.3656 5.987 5.143 4.757 4.534 4.387 4.284 4.207 4.147 4.099 4.060 4.000 3.841 3.6697 5.591 4.737 4.347 4.120 3.972 3.866 3.787 3.726 3.677 3.637 3.575 3.410 3.2308 5.318 4.459 4.066 3.838 3.687 3.581 3.500 3.438 3.388 3.347 3.284 3.115 2.9289 5.117 4.256 3.863 3.633 3.482 3.374 3.293 3.230 3.179 3.137 3.073 2.900 2.707

10 4.965 4.103 3.708 3.478 3.326 3.217 3.135 3.072 3.020 2.978 2.913 2.737 2.53811 4.844 3.982 3.587 3.357 3.204 3.095 3.012 2.948 2.896 2.854 2.788 2.609 2.40412 4.747 3.885 3.490 3.259 3.106 2.996 2.913 2.849 2.796 2.753 2.687 2.505 2.29613 4.667 3.806 3.411 3.179 3.025 2.915 2.832 2.767 2.714 2.671 2.604 2.420 2.20614 4.600 3.739 3.344 3.112 2.958 2.848 2.764 2.699 2.646 2.602 2.534 2.349 2.131

15 4.543 3.682 3.287 3.056 2.901 2.790 2.707 2.641 2.588 2.544 2.475 2.288 2.06616 4.494 3.634 3.239 3.007 2.852 2.741 2.657 2.591 2.538 2.494 2.425 2.235 2.01017 4.451 3.592 3.197 2.965 2.810 2.699 2.614 2.548 2.494 2.450 2.381 2.190 1.96018 4.414 3.555 3.160 2.928 2.773 2.661 2.577 2.510 2.456 2.412 2.342 2.150 1.91719 4.381 3.522 3.127 2.895 2.740 2.628 2.544 2.477 2.423 2.378 2.308 2.114 1.878

20 4.351 3.493 3.098 2.866 2.711 2.599 2.514 2.447 2.393 2.348 2.278 2.082 1.84321 4.325 3.467 3.072 2.840 2.685 2.573 2.488 2.420 2.366 2.321 2.250 2.054 1.81222 4.301 3.443 3.049 2.817 2.661 2.549 2.464 2.397 2.342 2.297 2.226 2.028 1.78323 4.279 3.422 3.028 2.796 2.640 2.528 2.442 2.375 2.320 2.275 2.204 2.005 1.75724 4.260 3.403 3.009 2.776 2.621 2.508 2.423 2.355 2.300 2.255 2.183 1.984 1.733

25 4.242 3.385 2.991 2.759 2.603 2.490 2.405 2.337 2.282 2.236 2.165 1.964 1.71126 4.225 3.369 2.975 2.743 2.587 2.474 2.388 2.321 2.265 2.220 2.148 1.946 1.69127 4.210 3.354 2.960 2.728 2.572 2.459 2.373 2.305 2.250 2.204 2.132 1.930 1.67228 4.196 3.340 2.947 2.714 2.558 2.445 2.359 2.291 2.236 2.190 2.118 1.915 1.65429 4.183 3.328 2.934 2.701 2.545 2.432 2.346 2.278 2.223 2.177 2.104 1.901 1.638

30 4.171 3.316 2.922 2.690 2.534 2.421 2.334 2.266 2.211 2.165 2.092 1.887 1.62232 4.149 3.295 2.901 2.668 2.512 2.399 2.313 2.244 2.189 2.142 2.070 1.864 1.59434 4.130 3.276 2.883 2.650 2.494 2.380 2.294 2.225 2.170 2.123 2.050 1.843 1.56936 4.113 3.259 2.866 2.634 2.477 2.364 2.277 2.209 2.153 2.106 2.033 1.824 1.54738 4.098 3.245 2.852 2.619 2.463 2.349 2.262 2.194 2.138 2.091 2.017 1.808 1.527

40 4.085 3.232 2.839 2.606 2.449 2.336 2.249 2.180 2.124 2.077 2.003 1.793 1.50960 4.001 3.150 2.758 2.525 2.368 2.254 2.167 2.097 2.040 1.993 1.917 1.700 1.389120 3.920 3.072 2.680 2.447 2.290 2.175 2.087 2.016 1.959 1.910 1.834 1.608 1.254∞ 3.841 2.996 2.605 2.372 2.214 2.099 2.010 1.938 1.880 1.831 1.752 1.517 1.000

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2.5% Points for the F distribution

ν1 = 1 2 3 4 5 6 7 8 9 10 12 24 ∞

ν2

1 647.8 799.5 864.2 899.6 921.8 937.1 948.2 956.7 963.3 968.6 976.7 997.2 10182 38.51 39.00 39.17 39.25 39.30 39.33 39.36 39.37 39.39 39.40 39.41 39.46 39.503 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.47 14.42 14.34 14.12 13.904 12.22 10.65 9.979 9.605 9.364 9.197 9.074 8.980 8.905 8.844 8.751 8.511 8.257

5 10.01 8.434 7.764 7.388 7.146 6.978 6.853 6.757 6.681 6.619 6.525 6.278 6.0156 8.813 7.260 6.599 6.227 5.988 5.820 5.695 5.600 5.523 5.461 5.366 5.117 4.8497 8.073 6.542 5.890 5.523 5.285 5.119 4.995 4.899 4.823 4.761 4.666 4.415 4.1428 7.571 6.059 5.416 5.053 4.817 4.652 4.529 4.433 4.357 4.295 4.200 3.947 3.6709 7.209 5.715 5.078 4.718 4.484 4.320 4.197 4.102 4.026 3.964 3.868 3.614 3.333

10 6.937 5.456 4.826 4.468 4.236 4.072 3.950 3.855 3.779 3.717 3.621 3.365 3.08011 6.724 5.256 4.630 4.275 4.044 3.881 3.759 3.664 3.588 3.526 3.430 3.173 2.88312 6.554 5.096 4.474 4.121 3.891 3.728 3.607 3.512 3.436 3.374 3.277 3.019 2.72513 6.414 4.965 4.347 3.996 3.767 3.604 3.483 3.388 3.312 3.250 3.153 2.893 2.59514 6.298 4.857 4.242 3.892 3.663 3.501 3.380 3.285 3.209 3.147 3.050 2.789 2.487

15 6.200 4.765 4.153 3.804 3.576 3.415 3.293 3.199 3.123 3.060 2.963 2.701 2.39516 6.115 4.687 4.077 3.729 3.502 3.341 3.219 3.125 3.049 2.986 2.889 2.625 2.31617 6.042 4.619 4.011 3.665 3.438 3.277 3.156 3.061 2.985 2.922 2.825 2.560 2.24718 5.978 4.560 3.954 3.608 3.382 3.221 3.100 3.005 2.929 2.866 2.769 2.503 2.18719 5.922 4.508 3.903 3.559 3.333 3.172 3.051 2.956 2.880 2.817 2.720 2.452 2.133

20 5.871 4.461 3.859 3.515 3.289 3.128 3.007 2.913 2.837 2.774 2.676 2.408 2.08521 5.827 4.420 3.819 3.475 3.250 3.090 2.969 2.874 2.798 2.735 2.637 2.368 2.04222 5.786 4.383 3.783 3.440 3.215 3.055 2.934 2.839 2.763 2.700 2.602 2.331 2.00323 5.750 4.349 3.750 3.408 3.183 3.023 2.902 2.808 2.731 2.668 2.570 2.299 1.96824 5.717 4.319 3.721 3.379 3.155 2.995 2.874 2.779 2.703 2.640 2.541 2.269 1.935

25 5.686 4.291 3.694 3.353 3.129 2.969 2.848 2.753 2.677 2.613 2.515 2.242 1.90626 5.659 4.265 3.670 3.329 3.105 2.945 2.824 2.729 2.653 2.590 2.491 2.217 1.87827 5.633 4.242 3.647 3.307 3.083 2.923 2.802 2.707 2.631 2.568 2.469 2.195 1.85328 5.610 4.221 3.626 3.286 3.063 2.903 2.782 2.687 2.611 2.547 2.448 2.174 1.82929 5.588 4.201 3.607 3.267 3.044 2.884 2.763 2.669 2.592 2.529 2.430 2.154 1.807

30 5.568 4.182 3.589 3.250 3.026 2.867 2.746 2.651 2.575 2.511 2.412 2.136 1.78732 5.531 4.149 3.557 3.218 2.995 2.836 2.715 2.620 2.543 2.480 2.381 2.103 1.75034 5.499 4.120 3.529 3.191 2.968 2.808 2.688 2.593 2.516 2.453 2.353 2.075 1.71736 5.471 4.094 3.505 3.167 2.944 2.785 2.664 2.569 2.492 2.429 2.329 2.049 1.68738 5.446 4.071 3.483 3.145 2.923 2.763 2.643 2.548 2.471 2.407 2.307 2.027 1.661

40 5.424 4.051 3.463 3.126 2.904 2.744 2.624 2.529 2.452 2.388 2.288 2.007 1.63760 5.286 3.925 3.343 3.008 2.786 2.627 2.507 2.412 2.334 2.270 2.169 1.882 1.482120 5.152 3.805 3.227 2.894 2.674 2.515 2.395 2.299 2.222 2.157 2.055 1.760 1.310∞ 5.024 3.689 3.116 2.786 2.566 2.408 2.288 2.192 2.114 2.048 1.945 1.640 1.000


ν1 = 1 2 3 4 5 6 7 8 9 10 12 24 ∞

ν2

1 4052 4999 5403 5625 5764 5859 5928 5981 6022 6056 6106 6235 63662 98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 99.40 99.42 99.46 99.503 34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 27.23 27.05 26.60 26.134 21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 14.55 14.37 13.93 13.46

5 16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 10.05 9.888 9.466 9.0206 13.75 10.92 9.780 9.148 8.746 8.466 8.260 8.102 7.976 7.874 7.718 7.313 6.8807 12.25 9.547 8.451 7.847 7.460 7.191 6.993 6.840 6.719 6.620 6.469 6.074 5.6508 11.26 8.649 7.591 7.006 6.632 6.371 6.178 6.029 5.911 5.814 5.667 5.279 4.8599 10.56 8.022 6.992 6.422 6.057 5.802 5.613 5.467 5.351 5.257 5.111 4.729 4.311

10 10.04 7.559 6.552 5.994 5.636 5.386 5.200 5.057 4.942 4.849 4.706 4.327 3.90911 9.646 7.206 6.217 5.668 5.316 5.069 4.886 4.744 4.632 4.539 4.397 4.021 3.60212 9.330 6.927 5.953 5.412 5.064 4.821 4.640 4.499 4.388 4.296 4.155 3.780 3.36113 9.074 6.701 5.739 5.205 4.862 4.620 4.441 4.302 4.191 4.100 3.960 3.587 3.16514 8.862 6.515 5.564 5.035 4.695 4.456 4.278 4.140 4.030 3.939 3.800 3.427 3.004

15 8.683 6.359 5.417 4.893 4.556 4.318 4.142 4.004 3.895 3.805 3.666 3.294 2.86816 8.531 6.226 5.292 4.773 4.437 4.202 4.026 3.890 3.780 3.691 3.553 3.181 2.75317 8.400 6.112 5.185 4.669 4.336 4.102 3.927 3.791 3.682 3.593 3.455 3.084 2.65318 8.285 6.013 5.092 4.579 4.248 4.015 3.841 3.705 3.597 3.508 3.371 2.999 2.56619 8.185 5.926 5.010 4.500 4.171 3.939 3.765 3.631 3.523 3.434 3.297 2.925 2.489

20 8.096 5.849 4.938 4.431 4.103 3.871 3.699 3.564 3.457 3.368 3.231 2.859 2.42121 8.017 5.780 4.874 4.369 4.042 3.812 3.640 3.506 3.398 3.310 3.173 2.801 2.36022 7.945 5.719 4.817 4.313 3.988 3.758 3.587 3.453 3.346 3.258 3.121 2.749 2.30523 7.881 5.664 4.765 4.264 3.939 3.710 3.539 3.406 3.299 3.211 3.074 2.702 2.25624 7.823 5.614 4.718 4.218 3.895 3.667 3.496 3.363 3.256 3.168 3.032 2.659 2.211

25 7.770 5.568 4.675 4.177 3.855 3.627 3.457 3.324 3.217 3.129 2.993 2.620 2.16926 7.721 5.526 4.637 4.140 3.818 3.591 3.421 3.288 3.182 3.094 2.958 2.585 2.13127 7.677 5.488 4.601 4.106 3.785 3.558 3.388 3.256 3.149 3.062 2.926 2.552 2.09728 7.636 5.453 4.568 4.074 3.754 3.528 3.358 3.226 3.120 3.032 2.896 2.522 2.06429 7.598 5.420 4.538 4.045 3.725 3.499 3.330 3.198 3.092 3.005 2.868 2.495 2.034

30 7.562 5.390 4.510 4.018 3.699 3.473 3.304 3.173 3.067 2.979 2.843 2.469 2.00632 7.499 5.336 4.459 3.969 3.652 3.427 3.258 3.127 3.021 2.934 2.798 2.423 1.95634 7.444 5.289 4.416 3.927 3.611 3.386 3.218 3.087 2.981 2.894 2.758 2.383 1.91136 7.396 5.248 4.377 3.890 3.574 3.351 3.183 3.052 2.946 2.859 2.723 2.347 1.87238 7.353 5.211 4.343 3.858 3.542 3.319 3.152 3.021 2.915 2.828 2.692 2.316 1.837

40 7.314 5.179 4.313 3.828 3.514 3.291 3.124 2.993 2.888 2.801 2.665 2.288 1.80560 7.077 4.977 4.126 3.649 3.339 3.119 2.953 2.823 2.718 2.632 2.496 2.115 1.601120 6.851 4.787 3.949 3.480 3.174 2.956 2.792 2.663 2.559 2.472 2.336 1.950 1.381∞ 6.635 4.605 3.782 3.319 3.017 2.802 2.639 2.511 2.407 2.321 2.185 1.791 1.000

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7 Stochastic processes7.1 Markov “jump” processes

Kolmogorov differential equationsForward equation: ∂

∂tpij(s, t) =∑k∈S

pik(s, t) · σkj(t)

Backward equation: ∂∂tpij(s, t) = −∑

k∈S

σik(s) · pkj(s, t)

where σij(t) is the transition rate from state i to state j (j 6= i) attime t, and σii = −

∑j 6=i

σij .

Expected time to reach a subsequent state kmi =

1λi

+∑

j 6=i,j 6=k

σij

λimj , where λi =

∑j 6=i

σij

7.2 Brownian motion and related processesMartingales for standard Brownian motionIf Bt, t ≥ 0 is a standard Brownian motion, then the followingprocesses are martingales:Bt, B

2t − t and exp(λBt − 1

2λ2t)

Distribution of the maximum value

P

[max0≤s≤t

(Bs + µs) > y

]= Φ

(−y+µt√

t

)+ e2µyΦ

(−y−µt√

t

), y > 0

Hitting timesIf τy = min

s≥0s : Bs + µs = y where µ > 0 and y < 0, then

E[e−λτy ] = ey(µ+√

µ2+2λ, λ > 0Ornstein-Uhlenbeck processdXt = −γXtdt+ σdBt, γ > 0

7.3 Monte Carlo methodsBox-Muller formulaeIf U1 and U2 are independent random variables form the U(0, 1)distribution thenZ1 =

√−2 logU1 cos(2πU2) and Z2 =√−2 logU1 sin(2πU2)

are independent standard normal variables.Polar methodIf V1 and V2 are independent random variables from the U(−1, 1)distribution and S = V 2

1 + V 22 then, conditional on 0 < S ≤ 1,

Z1 = V1

√−2 logS

S and Z2 = V2

√−2 logS

S

are independent standard normal variables.Psuedorandom values from the U(0, 1) distribution and the N(0, 1)distribution are included in the statistical tables section.

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8 Time series8.1 Time series - Time domain

Sample autocovariance and autocorrelation function

Autocovariance: γk = 1n

n∑t=k+1

(xt − µ) · (xt−k − µ), where µ = 1n

n∑t=1

xt

Autocorrelation: ρk = γk

γ0

Autocorrelation function for ARMA(1,1)For the process Xt = αXt−1 + et + βet−1:

ρk = (1+αβ)·(α+β)1+β2+2αβ · αk−1, k = 1, 2, 3, . . .

Partial autocorrelation function

φ1 = ρ1 φ2 =ρ2−ρ2

1

1−ρ21

φk =det(P⋆

k )det(Pk)

, k = 2, 3, . . .,

where Pk =

1 ρ1 ρ2 . . . ρk−1

ρ1 1 ρ1 . . . ρk−2

ρ2 ρ1 1 . . . ρk−3

......

.... . .

...ρk−1 ρk−2 ρk−3 . . . 1

and P ⋆k equals Pk, but with the last column replaced with

[ρ1, ρ2, ρ3, . . . , ρk]⊤.

Partial autocorrelation function for MA(1)For the process Xt = µ+ et + βet−1:

φk = (−1)k+1 · (1−β2)βk

1−β2·(k+1) , k = 1, 2, 3, . . .

8.2 Time series - Frequency domainSpectral density function

f(ω) = 12π

∞∑k=−∞

e−ikωγk, −π < ω < π

Inversion formula

γk =π∫−π

eikωf(ω)dω

Spectral density function for ARMA(p, q)The spectral density function of the process φ(B)(Xt − µ) = θ(B)et,where var(et) = σ2, is

f(ω) = σ2

2πθ(eiω)θ(e−iω)φ(eiω)φ(e−iω)

Linear filters

For the linear filter Yt =∞∑

k=−∞akXt−k:

fY (ω) = |A(ω)|2 fX(ω)

where A(ω) =∞∑

k=−∞e−ikωak is the transfer function for the filter.

8.3 Time series - Box-Jenkins methodologyLjung and Box “Portmanteau” test of the residuals for anARMA(p, q) model

n(n+ 2)m∑

k=1

r2kn−k ∼ χ2

m−(p+q)

where rk (k = 1, 2, . . . ,m0 is the estimated value of the k thautocorrelation coefficient of the residuals and n is the number ofdata values used inn the ARMA(p, q) seriesTurning point testIn a sequence of n independent random variables the number ofturning points T is such that:

E[T ] = 23 (n− 2) and var(T ) = 16n−29

90

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