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ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
ACTL2002/ACTL5101 Probability and Statistics
c© Katja Ignatieva
School of Risk and Actuarial StudiesAustralian School of BusinessUniversity of New South Wales
Week 1 Video Lecture NotesProbability: Week 1 Week 2 Week 3 Week 4
Estimation: Week 5 Week 6 Review
Hypothesis testing: Week 7 Week 8 Week 9
Linear regression: Week 10 Week 11 Week 12
Video lectures: Week 2 VL Week 3 VL Week 4 VL Week 5 VL
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Links to UNSW TV
Click on the topic to go to the online recording:
Week 1: Probability space, Calculating with probabilities,Counting.
Week 2: Bernoulli distribution, Binomial distribution,Geometric distribution, Negative Binomial distribution.
Week 3: Numerical methods to summarize data, Graphicalprocedures to summarize data.
Week 4: Sampling with and without replacement, Propertiesof sample mean and variance.
Week 5: Chi-squared distribution, Student-t distribution,Snecdor’s F distribution, Distribution of sample mean andvariance.
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Sample Space & σ-algebra
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Sample Space & σ-algebra
Sample Space & σ-algebra
In a random experiment, the outcomes cannot be predictedwith certainty in advance.
The set of all possible outcomes is called the sample space,denoted by Ω. An element ω of Ω is a sample point.
A family F of subsets of the sample space Ω is said to be aσ-algebra (σ-field) if the following conditions hold:
1. E ∈ F implies E c ∈ F ;2. E1,E2, . . . are pairwise disjoint sets in F , that is, Ei ∩ Ej = ∅
(null/empty set) for all i 6= j , then∞⋃k=1
Ek ∈ F .
An element E of F is called an event. It is nothing but asubset of Ω.Note: A sample point is a simple event.
2/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Sample Space & σ-algebra
If there are N elements of the sample space, a σ-algebra F ofevents may consist of 2N elements.
Example:
Consider the experiment of selecting a card at random from adeck of four cards with four different suits, and noting its suit:hearts (H), diamonds (D), spades (S), or clubs (C ).
The sample space is: Ω = H,D, S ,C.
All the sets of possible events are a σ-algebra of F .
3/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Sample Space & σ-algebra
The following collection of 24 = 16 sets is a σ-algebra F :
(H) (H,D) (H,D,S) (H,D, S ,C )(= Ω) ∅(D) (H, S) (H,D,C )(S) (H,C ) (H, S ,C )(C ) (D,S) (D,S ,C )
(D,C )(S ,C )
Each element in the σ-algebra is called an event.
Note that a standard deck has 52 elements in the samplespace, so one σ-algebra has 252 elements.
4/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Sample Space & σ-algebra
Sample Spaces
5/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Probability Measure
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Probability space
Probability Measure
Probability Measure
A probability measure on ω is a function Pr from the subsetsof ω to < that satisfies the following axioms:
1. For all events Ei , 0 ≤ Pr (Ei ) ≤ 1;2. Pr (Ω) = 1;3. For any events E1,E2, . . . where Ei ∩ Ej = ∅ for every i 6= j ,
then:
Pr
( ∞⋃k=1
Ek
)= Pr (E1) + Pr (E2) + . . . =
∞∑k=1
Pr (Ek) .
A random experiment is therefore described as a probabilitytriple (or probability space): Ω,F ,P.
6/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Definitions
Union of two events: C = A ∪ Bis an event that both A and/or B occurs, i.e., ω ∈ C iff ω ∈ Aor ω ∈ B.
Intersection of two events: C = A ∩ Bis an event that both A and B occurs, i.e., ω ∈ C iff ω ∈ Aand ω ∈ B.
Complement of an event: B = Ac
is the event that A does not occur, i.e., ω ∈ Ac iff ω /∈ A.
Events are disjoint: if they have no outcomes in communi.e., A and C are disjoint iff A ∩ C = ∅.
7/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Useful laws
Commutative laws:A ∪ B = B ∪ Aand A ∩ B = B ∩ A.
Associative laws:(A ∪ B) ∪ C = A ∪ (B ∪ C )and (A ∩ B) ∩ C = A ∩ (B ∩ C ).
Distributive laws:(A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C )and (A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C ).
DeMorgan’s laws:(A ∩ B)c = Ac ∪ Bc and (A ∪ B)c = Ac ∩ Bc .
8/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Properties of the Probability Measure
Complement: Pr(E c) = 1− Pr(E ).
Null / Empty Set: Pr(∅) = 0.
Subsets: If E1 and E2 are two events such that E1 ⊆ E2, thenPr (E1) ≤ Pr (E2).
Additive: If E1 and E2 are any two events, then:
Pr (E1 ∪ E2) = Pr (E1) + Pr (E2)− Pr (E1 ∩ E2) .
9/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Properties of the Probability Measure
10/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
In the case of three events, we have:
Pr (E1 ∪ E2 ∪ E3) = Pr ((E1 ∪ E2) ∪ E3)∗= Pr (E1 ∪ E2) + Pr (E3)
−Pr ((E1 ∪ E2) ∩ E3)∗∗= Pr (E1 ∪ E2) + Pr (E3)
−Pr ((E1 ∩ E3) ∪ (E2 ∩ E3))∗∗∗= Pr (E1) + Pr (E2) + Pr (E3)
−Pr (E1 ∩ E2)− Pr (E1 ∩ E3)
−Pr (E2 ∩ E3) + Pr (E1 ∩ E2 ∩ E3) .
* using additive rule: Pr (A ∪ B) = Pr (A) + Pr (B)− Pr (A ∩ B) withA = (E1 ∪ E2) and B = E3.** using distributive law: (A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C ) with A = E1,B = E2, C = E3.*** using additive rule: Pr (A ∪ B) = Pr (A) + Pr (B)− Pr (A ∩ B) withA = E1 and B = E2 & A = (E1 ∩ E3) and B = (E2 ∩ E3).11/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Inequality rules
Using the definition of probability measure one can prove thefollowing inequalities:
Boole’s Inequality: If E1,E2, . . . ,En are any n events, then
Pr
(n⋃
k=1
Ek
)≤
n∑k=1
Pr (Ek) ,
which follows the additive law of probability.
Bonferroni’s Inequality: If E1,E2, . . . ,En are any n events,then:
Pr (E1 ∩ . . . ∩ En) ≥ 1−n∑
k=1
Pr (E ck ) .
12/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Properties of the Probability Measure
Inequality rules
13/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Conditional Probability
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Conditional Probability
Conditional Probability
The conditional probability of A, given B, as:
Pr (A |B ) =Pr (A ∩ B)
Pr (B)
provided Pr (B) > 0, otherwise Pr (A |B ) = 0.
The multiplication rule immediately follows:
Pr (A ∩ B) = Pr (A |B ) · Pr (B) .
The following properties are also immediate:1. Pr (A |B ) ≥ 0;
2. Pr (A |A ) = 1;
3. If A1,A2, . . . are mutually disjoint events, then
Pr
( ∞⋃k=1
Ak |B)
=∞∑k=1
Pr (Ak |B ).
14/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Conditional Probability
Conditional Probability
15/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Conditional Probability
Law of total probabilityLaw of total probability: If E1,E2, . . . are mutually disjoint(Ei ∩ Ej = ∅ for i 6= j) and comprise the entire sample space(⋃∞
k=1 Ek = Ω), then for any event A ∈ F , we have:
Pr (A) =∞∑k=1
Pr (A |Ek ) · Pr (Ek) .
16/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Independence
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Independence
Independence
Events A and B are said to be independent if:
Pr (A ∩ B) = Pr (A) · Pr (B) .
Equivalently, we have events A and B independent if:
Pr (A |B ) = Pr (A) and Pr (B |A) = Pr (B) .
Note that we say the collection of events E1,E2, . . . ,En areindependent if:
Pr (E1 ∩ E2 ∩ . . . ∩ En) = Pr (E1) · Pr (E2) · . . . · Pr (En) .
For a collection of several events E1,E2, . . . ,En, we say thatthe events are mutually independent if for any sub-collectionEi1 ,Ei2 , . . . ,Eim , we have:
Pr (Ei1 ∩ Ei2 ∩ . . . ∩ Eim) = Pr (Ei1) · Pr (Ei2) · . . . · Pr (Eim) .17/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Independence
18/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Independence
Mutually Exclusive Events: A and B are said to be mutuallyexclusive if they are mutually disjoint, i.e., A ∩ B = ∅ so thatPr (A ∩ B) = 0 and
Pr (A ∪ B) = Pr (A) + Pr (B) .
We can generalize this to several events as follows: eventsE1,E2, . . . ,En are said to be mutually exclusive events if notwo have an element in common (mutually disjoint) and that
Pr (E1 ∪ E2 ∪ . . . ∪ En) = Pr (E1) + Pr (E2) + . . .+ Pr (En) .
19/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Independence
Example
Consider the experiment of selecting n cards at random from adeck of 52 cards with 13 hearts (H), 13 diamonds (D), 13 spades(S), and 13 clubs (C ) cards.
False For n = 2, the second card is independent of the first card.
False For n = 14, the 14th card is mutually independent of the first13 cards.
True Previous two questions, but now when the card is put back inthe stock before the next one is selected.
True For n = 3, A=“at least 2 H” and B=“at least 2 C” aremutually exclusive events.
False For n = 4, A=“at least 2 H” and B=“at least 2 C” aremutually exclusive events.
20/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Bayes Theorem
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Bayes Theorem
Bayes TheoremBayes Theorem: Suppose E1,E2, . . . represent a completepartitioning of the sample space Ω, then for any non-emptyevent A ∈ F , we have:
Pr (Ek |A) =Pr (A |Ek ) · Pr (Ek)∑∞j=1 Pr (A |Ej ) · Pr (Ej)
,
for any k = 1, 2, . . ..
21/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Bayes Theorem
ExampleAn insurance company classifies its policyholders according tothree risk classes: L (low risk), M (medium risk) and H (highrisk). The proportion of H policyholders is 20% and theproportion of L policyholders is 50%. For each of the riskclasses, the probability of a claim is 0.01 for L, 0.02 for M,and 0.04 for H.
a. Question: If a claim occurs, what is the probability that it isfrom a L (low risk) policyholder?
a. Solution: Let C = “be the event that there is a claim”, thenwe have:Pr (L|C ) = Pr(L∩C)
Pr(C) = 26%, where
Pr (L ∩ C ) = Pr (C |L) · Pr (L) = 0.01 · 0.5 = 0.005Pr (C ) = Pr (C |L)·Pr (L)+Pr (C |M)·Pr (M)+Pr (C |H)·Pr (H)= 0.01 · 0.5 + 0.02 · 0.3 + 0.04 · 0.2 = 0.019 (using LTP).22/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Calculating with probabilities
Bayes Theorem
Example (cont.)
b. Question: If a claim occurs, what is the probability that it isfrom a M (medium risk) policyholder?
c. Question: If a claim occurs, what is the probability that it isfrom a H (high risk) policyholder?
b. Solution: Similar to a.,Pr (M|C ) = Pr(M∩C)
Pr(C) = Pr(C |M)·Pr(M)Pr(C) = 0.3·0.02
0.019 = 32%.
c. Solution: Similar to a. and b.,Pr (H|C ) = Pr(H∩C)
Pr(C) = Pr(C |H)·Pr(H)Pr(C) = 0.2·0.04
0.019 = 42%.
23/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Counting Principles
Multiplication Rule: Suppose S1, S2, . . . ,Sm are m sets withrespective number of elements n1, n2, . . . , nm. The number ofways of choosing one element from each set is given by:
n1 · n2 · . . . · nm.
Permutation: The number of ways of arranging n distinctobjects is given by:
n! ≡n · (n − 1) · (n − 2) · . . . · 2 · 1n! =n · (n − 1)!
0! ≡1.
24/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Combination: The number of ways of choosing r objects fromn distinct objects, where r ≤ n, is given by:(
n
r
)≡ n!
r ! · (n − r)!.
Multinomial: The number of ways that n objects can begrouped into r classes with nk in the kth class, where
k = 1, 2, . . . , r andr∑
k=1
nk = n, is given by:
(n
n1, n2, . . . , nr
)≡ n!
n1! · n2! · . . . · nr !.
25/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Example
Question: An airline has 6 flights daily from Sydney toHonolulu, and 8 flights daily from Honolulu to Los Angeles.The airline offers no direct flight from Sydney to Los Angeles.If the flights are to be made on separate days, how manydifferent flight arrangements can the airline offer from Sydneyto Los Angeles?
Solution: Use the multiplication rule. Let:
S1 = “Flight from Sydney to Honolulu”, with n1 = 6;S2 = “Flight from Honolulu to Los Angeles”, with n2 = 8,
then we have n1 · n2 = 6 · 8 = 48 different flight arrangements.
26/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Question: A committee is to consist of 4 academics and 2practitioners, to be selected from a larger group of 8academics and 5 practitioners. How many ways can you forma committee if:
a. there are no additional restrictions.b. two of the chosen academics must be the two female members
of the group of 8.
a. Solution: We use combinations & multiplication.For the academics (“S1”) we have n = 8 distinct ones and weneed to choose r = 4 academics, possible number of ways:(nr
)=(84
)= 70.
For the practitioners (“S2”) we have n = 5 distinct ones andwe need to choose r = 2 practitioners, possible number ofways:
(nr
)=(52
)= 10.
Total ways of forming a committee: n1 · n2 = 70 · 10 = 700.27/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
b. Solution: We use combinations & multiplication again.
- For the female academics (“S1”) we have n = 2 distinct onesand we need to choose r = 2 female academics, possiblenumber of ways:
(nr
)=(22
)= 1.
- For the male academics (“S2”) we have n = 6 distinct onesand we need to choose r = 2 male academics, possiblenumber of ways:
(nr
)=(62
)= 15.
- For the practitioners (“S3”) we have n = 5 distinct ones andwe need to choose r = 2 practitioners, possible number ofways:
(nr
)=(52
)= 10.
Total ways of forming a committee:n1 · n2 · n3 = 1 · 15 · 10 = 150.
Note: probability of having two female academic members is:150/700 = 3/14 ≈ 0.21428/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Question: How many different letter arrangements can beobtained from the letters of the word mississippi, using allthe letters?
Solution: Let- class one be “letter m”, with n1 = 1;
- class two be “letter i”, with n2 = 4;
- class three be “letter s”, with n3 = 4;
- class four be “letter p”, with n4 = 2.
Note: we have n = n1 + n2 + n3 + n4 = 11.
We use multinomial, hence there are(n
n1, n2, n3, n4
)=
n!
n1! · n2! · n3! · n4!= 34650
different letter arrangements.29/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Counting Principles
Question: An actuarial consulting company has four projectsto do. Two of the projects require 3 actuaries, one requires 2actuaries, and the other requires 4 actuaries. The companycurrently has 15 actuaries, of which 5 are females. How manydifferent ways are there to assign the actuaries to the projects?
Solution: We have 15 distinct actuaries. Use combinations.For the first project we need to choose r = 3 actuaries fromn = 15 distinct actuaries, hence
(nr
)=(153
)= 455 ways. For
the second project we need to choose r = 3 actuaries from then = 12 left distinct actuaries, hence
(nr
)=(123
)= 220 ways.
For the third project we need to choose r = 2 actuaries fromthe n = 9 left distinct actuaries, hence
(nr
)=(92
)= 36 ways.
For the last project we need to choose r = 4 actuaries fromthe n = 7 left distinct actuaries, hence
(nr
)=(74
)= 35 ways.
The total number of ways to assign the actuaries to theprojects is 455 · 220 · 36 · 35 = 126, 126, 000.
30/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Computing Probabilities
Introduction in Probability
Probability spaceSample Space & σ-algebraProbability Measure
Calculating with probabilitiesProperties of the Probability MeasureConditional ProbabilityIndependenceBayes Theorem
CountingCounting PrinciplesComputing Probabilities
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Computing Probabilities
Computing ProbabilitiesIf the elements of the sample space Ω all have equalprobability and assuming: (a) there are N elements in Ω and(b) the event A can occur in any of n mutually exclusive ways,then:
Pr (A) =n
N.
Question: A drawer contains 10 pairs of socks. If 6 socks aretaken at random and without replacement, compute theprobability that there is at least one matching pair amongthese 6 socks.
Solution: Let M = “at least one matching pair among 6socks”. We have Pr (M) = 1− Pr (Mc), where Mc = “nomatching pair among 6 socks”.Pr (Mc) = 1 · 1819 ·
1618 ·
1417 ·
1216 ·
1015 = 0.3467. Hence,
Pr (M) = 1− Pr (Mc) = 1− 0.3467 = 0.6533.31/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Computing Probabilities
Question: A committee of four individuals is to be formedfrom a group of 5 males and 6 females. What is theprobability that the committee formed has both sexes equallyrepresented?
Solution: Let E = “sexes are equally represented in thecommittee”.
Number of combinations: E =MMFF, MFMF, MFFM, FMMF, FMFM, FFMM,i.e., n = 4 and r = 2:
(nr
)=(42
)= 6.
Each combination has equal probability, i.e.,Pr (MMFF ) = . . . = Pr (FFMM) = 5
11 ·410 ·
69 ·
58 = 0.0758.
Combing we have:Pr (E ) =
(42
)· 511 ·
410 ·
69 ·
58 = 6 · 0.0758 = 5
11 .
32/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1 Video Lecture Notes
Counting
Computing Probabilities
Odds and probabilities
Sometimes, we get confused between odds and probabilities.The odds that an event will occur is the ratio of theprobability that the event will occur to the probability it willnot occur, provided neither probability is zero. That is, if E isthe event, the odds that E will occur is:
Pr (E )
Pr (E c)=
Pr (E )
1− Pr (E ).
Often, odds are quoted in terms of positive integers. Forexample, the odds are a to b that an event will occur. Thenthey mean that the probability it will occur is:
Pr (E ) =a
a + b.
33/33
ACTL2002/ACTL5101 Probability and Statistics: Week 1
ACTL2002/ACTL5101 Probability and Statistics
c© Katja Ignatieva
School of Risk and Actuarial StudiesAustralian School of BusinessUniversity of New South Wales
Week 1Probability: Week 2 Week 3 Week 4
Estimation: Week 5 Week 6 Review
Hypothesis testing: Week 7 Week 8 Week 9
Linear regression: Week 10 Week 11 Week 12
Video lectures: Week 1 VL Week 2 VL Week 3 VL Week 4 VL Week 5 VL
ACTL2002/ACTL5101 Probability and Statistics: Week 1
101/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction
Course introduction
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction
Course introduction
Course overview
Week 1 General introduction in probability.Week 2-4 Distribution functions:
- univariate & multivariate special distributions;- joint distributions;- functions of distributions.
Week 5-6 Parameter estimation.Week 7-9 Hypothesis testing.Week 10-12 Linear regression.
102/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction
Course introduction
This week
Describing distribution using measures of location anddispersion:
- mean;- variance;- skewness;- kurtosis.
Calculating these measures:
- central moments;- non-central moments;- generating functions.
103/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction in probability
Exercise
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction in probability
Exercise
Sample spaces
An insurer offers health insurance. An individual can have fileeither one, two, three, or no claim in a given quarter. Forsimplicity, it is not possible to issue more than three claims.
Questions:a. State all possible events.
b. Give the sample space.
c. Give the σ−algebra.
Solutions:a. E ∈ 0, 1, 2, 3, 0, 1, 0, 2, 0, 3, 1, 2, 1, 3,2, 3, 0, 1, 2, 0, 1, 3, 0, 2, 3, 1, 2, 3, 0, 1, 2, 3, ∅.
b. Ω = 0, 1, 2, 3.
c. 0, 1, 2, 3, 0, 1, 0, 2, 0, 3, 1, 2, 1, 3, 2, 4,0, 1, 2, 0, 1, 3, 0, 2, 3, 1, 2, 3, 0, 1, 2, 3, ∅.
104/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction in probability
Exercise
Calculating with probabilities
We have the following probabilities: Pr(C = 0) = 0.8,Pr(C = 1) = 0.1, Pr(C = 2) = 8/90, and Pr(C = 3) = 1/90.
Questions:a. What is the probability of 2 or 3 claims given that there are
odd number of claims. Same for even number of claims.
b. Using a., Pr (even) = 8/9, Pr (odd) = 1/9, and the law oftotal probability, find Pr(C ≥ 2).
c. Are “odd number of claims” and “two or three claims”independent?
d. Using a. and Bayes theorem find the probability of odd numberof claims given that the number of claims is two or three.
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ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction in probability
Exercise
Solutions:a. Pr (C = 2, 3|C = 1, 3) = Pr(C=2,3∩C=1,3)
Pr(C=1,3) =Pr(C=3)
(Pr(C=1)+Pr(C=3)) = 1/10, and
Pr (C = 2, 3|C = 0, 2) = Pr(C=2)(Pr(C=0)+Pr(C=2)) = 1/10.
b. Pr (2, 3) = Pr (2, 3| odd) · Pr (odd) + Pr (2, 3| even) ·Pr (even) = 1
9 ·110 + 8
9 ·110 = 1
10 .
c. Pr (C = 2, 3|C = 1, 3) = 1/10 = Pr (2, 3), thusindependent.
d. Pr (C = 1, 3|C = 2, 3) =Pr(C=2,3|C=1,3)·Pr(C=1,3)
Pr(C=2,3|C=1,3)·Pr(C=1,3)+Pr(C=2,3|C=0,2)·Pr(C=0,2) =110 ·
19
110 ·
19+
110 ·
89
= 19 .
106/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Introduction in probability
Exercise
Counting principlesThe history of an individual’s claiming record is had he had 3quarters with 2 claims, 2 quarters with 1 claim and 15quarters without a claim.
Questions:a. What is the probability that the insured had first 15 quarters
without a claim and then 5 quarters with at least one claim?b. What is the probability that the insured had first 15 quarters
without a claim and then 2 quarters with one claim and then 3quarters with two claims?
c. Comment on your results.
Solutions:a. Use Combination, n = 20, r = 5, number of ways choosing
objects:(205
)= 15, 504. Thus, probability is 1/15,504.
b. Use Multinomial, n = 20, r1 = 15, r2 = 2, r3 = 3, number ofways choosing objects: 20!
15!·2!·3! = 155, 040. Thus, probability is1/155,040.107/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Random variables and distributions
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Random variables and distributions
Random variables and distributions
A random variable is generally a quantity X whose valuedepends on the outcome of a random experiment. It is amapping from the sample space Ω to the set of real numbers<. That is,
X : Ω→ <.
The cumulative distribution function, abbreviated c.d.f., of Xis defined by:
FX (x) = Pr (X ≤ x) , for all x .
The survival function of X is defined by:
SX (x) = 1− FX (x) = Pr(X > x).
108/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Random variables and distributions
Distribution function
Properties of a distribution function:1. FX (·) is a non-decreasing function, i.e., FX (x1) ≤ FX (x2)
whenever x1 ≤ x2;
2. FX (·) is right-continuous, that is for all x ,limε→0+
FX (x + ε) = FX (x);
3. FX (−∞) = 0;
4. FX (+∞) = 1.
Types of Random Variables:- continuous;
- discrete;
- mixed.109/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Random variables and distributions
−10 −5 0 5 100
0.2
0.4
0.6
0.8
1
1.2
x
F(x
)
Which one is a distribution function?
ABCDEF
Which oneis/aredistributionfunctions?Solution:B and F.Not non-decreasing:A;Not right-continuous:C;NotF (−∞) = 0:D;
Not
F (∞) = 1:
E.110/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Random variables and distributions
The probability mass function (p.m.f.) is defined by:
pX (xk) = Pr (X = xk) = FX (xk)− FX (xk−1) .
The probability density function (p.d.f.) is defined by:
fX (x) =∂
∂xFX (x) .
Note that:
p.m.f. satisfies∞∑k=0
pX (xk) = 1;
p.m.f. requires the right-continuous property;p.d.f. satisfies
∫∞−∞ fX (x) dx = 1.
111/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
Measures of location: probability distributions
Let X be a r.v. with pdf pX (if discrete) or fX (if continuous).
The expected value of X is:
µX = E[X ] =
∑all x
x · pX (x) ,
if X is discrete;∫ ∞−∞
x · fX (x) dx ,
if X is continuous.
Exercise: Calculate E [X ] when X is a r.v. with pdf:
fX (x) =
1, if 0 ≤ x ≤ 1;0, else.
Solution:E [X ] =
∫∞−∞ x · fX (x)dx =
∫ 10 x · fX (x)dx =
[12x
2]10
= 12 .
112/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
Mathematical expectation of functions of a r.v.
Let X be a r.v. with pdf pX (if discrete) or fX (if continuous)and let h(X ) be a real-valued function.
E[h(X )] =
∑all x
h (x) · pX (x) ,
if X is discrete;∫ ∞−∞
h (x) · fX (x) dx ,
if X is continuous.
Exercise: Let X be a r.v. with pdf: pX (x) = 110 for
x = 0, 1, 2, . . . , 9 and zero otherwise, let h(X ) = X 2.Calculate E [h(X )].Solution:E [h(X )] =
∑all x h(x) · pX (x) =
∑9x=0 x
2 110 = 28.5.
113/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
Properties of the expected value operator:Let X and Y be random variables, and m, b ∈ <, we have:
E[mX + b] = mE[X ] + b; E[X + Y ] = E[X ] + E[Y ];E[X · Y ] = E[X ] · E[Y ], only if X ,Y independent.
Proof for discrete r.v.: (* using X and Y independent)
E[mX + b] =∑
x (mx + b) pX (x) =m∑
x (xpX (x)) + b =mE[X ] + b
E[X + Y ] =∑x ,y
(x + y) pX ,Y (x , y) =∑x
∑y
xpX ,Y (x , y) + ypX ,Y (x , y)
=∑x ,y
xpX ,Y (x , y) +∑x ,y
ypX ,Y (x , y) = E[X ] + E[Y ]
E[X · Y ] =∑x ,y
(x · y) · pX ,Y (x , y)∗=∑x
∑y
(x · pX (x)) · (y · pY (y))
=∑x
(x · pX (x)) ·∑y
(y · pY (y)) = E[X ] · E[Y ].
114/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
ExampleAn insurance company offers motor vehicle insurance. Theprobability that an insured files a claim is 20%. Assume thatthe insured files not more than one claim.
a. Question: What is the probability mass function?
b. Question: What is the expected number of claims.
c. Question: Assumes that there are 150 insured. What is theexpected number of claims.
The claims will be paid at the end of the year. The requiredcapital will depends on investment return. A 10% increase inasset value occurs w.p. 20% and a decrease of 5% occursw.p. 15%. The claim value is $1,000 for each claim.
d. Question: What is the expected value the insurer needs tocover the claims?115/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of location: probability distributions
Solutiona. pX (x) =
0.2, if x = 1;0.8, if x = 0,
and zero otherwise.
b. E [X ] =∑
all x pX (x) · x = 0.2 · 1 + 0.8 · 0 = 0.2.
c. E [150 · X ] = 150 · E [X ] = 150 · 0.2 = 30.
d. Let Y be the random variable of the asset value. We have:
pY (y) =
0.2, if y = 1/1.1;0.65, if y = 1;0.15, if y = 1/0.95,
and zero otherwise.
E [1000 · Y ] =1000 · E [Y ]
=1000 ·(
0.2 · 1
1.1+ 0.65 · 1 + 0.15 · 1
0.95
)=1000 · 0.99689 = 996.89.
E [(150 · X ) · (1000 · Y )]∗=E [150 · X ] · E [1000 · Y ]
=30 · 996.89 = 29, 907.
* using independence between X and Y .
116/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Measures of dispersion
X − µX gives deviation from the expected value.
Question: Can we use expected deviation as measure ofdispersion: E [X − µX ]?
Solution: No, we have that:E [X − µX ] = E [X ]− E [µX ] = µX − µX = 0.
One measure of dispersion: Mean absolute deviation:
MAD(X ) = E [|X − µX |] .
Note: the mean absolute deviation minimized when µX is themedian of the distribution.
117/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Variance of X
MAD is not easily to calculate, and has not the niceproperties (not related to moments of distribution).
Another measure of dispersion: Let X be a random variable,the variance is given by:
σ2 = Var (X ) =E[(X − µX )2
]=E
[X 2]− µ2X .
The function E[(X − α)2
]is minimized when α = µX .
The standard deviation of X is given by:
σ =√
Var (X ).118/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Properties of the variance:Let X and Y be random variables, and m, b, c ∈ <, we have:
Var (c) = 0; Var (mX + b) = m2 · Var (X ) ;Var (X + Y ) = Var (X ) + Var (Y ) , only if X ,Y independent.
Proof:
Var (c) =E[(c − µc)2
]= E
[(c − c)2
]= 0
Var (mX + b) =E[(mX + b − µmX+b)2
]= E
[(mX + b −mµX − b)2
]=m2 · E
[(X − µX )2
]= m2 · Var (X )
Var (X + Y ) =E[((X − µX ) + (Y − µY ))2
]=E
[(X − µX )2 + (Y − µY )2 + 2(X − µX ) · (Y − µY )
]∗=Var (X ) + Var (Y ) ,
* using independence between X and Y .119/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Exercises
The claims of the motor vehicle insurance are itself stochastic.The distribution of the claim value of an insured is:
fX (x) =
0, if x < 0;λ5 · exp (−λ · x) , if x > 0;
pX (0) =0.8.
a. Question: Find the expected value of the claims size.
b. Question: Find the variance of the claims size.
c. Question: The price of an insurance contract is the expectedvalue plus half the standard deviation. Find the price of theMVI contract.
d. Question: Same as c., but now for the 150 contract together.120/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Solution
a. E [X ] = 0 · 0.8 +∫∞0
λ5 exp (−λ · x) · xdx
= 0 + λ5 ·[exp(−λ·x)(−λ)2 · (−λ · x − 1)
]∞0
= λ5 ·(0−
(1λ2· (−1)
))= 1
5·λ .
b. E[X 2]
=02 · 0.8 +
∫ ∞0
λ
5exp (−λ · x) · x2dx
=0 +λ
5·[
exp(−λ · x) ·(
x2
−λ− 2x
(−λ)2+
2
(−λ)3
)]∞0
=λ
5·(
0−(
2
−λ3
))=
2
5 · λ2.
Var(X ) =E[X 2]− (E [X ])2 =
2
5λ2− 1
25λ2=
9
25λ2.
c. Price = E [X ] + 0.5 ·√Var(X ) = 1
5·λ + 0.5 · 35·λ = 1
2·λ .
d. Price = E[∑150
i=1 Xi
]+ 0.5 ·
√Var(
∑150i=1 Xi ) =
150·E [X ]+0.5·√
150 · Var(X ) = 1505·λ +0.5· 3·
√150
5·λ =√913.5λ ≈ 30.22
λ .121/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Skewness
Let X be a random variable. The skewness of X is given by:
γX = E
[(X − µXσX
)3].
Coefficient of skewness:
α3 =E[X 3]
(E [X 2])3/2.
Properties of skewness (σX+Y is the s.d. of X + Y ):
γX = 0, if the distribution of X is symmetric;γm·X+b = sign(m) · γX ;
γX+Y =γX ·σ3
X+γY ·σ3Y
σ3X+Y
, if X ,Y are independent.
122/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Which statements are true?Note: E[X ] = 0 and σ = 1 for i) and ii)
and E[X ] = −4 and σ = 3 for iii).
−1 0 20
1/61/31/2
x
Pr(
X=
x)
distribution i)
−2 0 10
1/61/31/2
x
Pr(
X=
x)
distribution ii)
−10 −4 −10
1/61/31/2
x
Pr(
X=
x)
distribution iii)
a. Question: Distribution i) has apositive skewness.
a. Solution: True: skewness is 1.
b. Question: Distribution ii) has apositive skewness.
b. Solution: False: skewness is -1.
c. Question: Distribution iii) has asmaller skewness than distribution ii).
c. Solution: False: both have the SAMEskewness.
123/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
0 2 40
0.2
0.4
0.6
0.8
1
x
prob
abilit
y den
sity f
uncti
on
Snedecor‘s F p.d.f.
v1= 20, v
2= 40
0 2 40
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
cumu
lative
dens
ity fu
nctio
n
Snedecor‘s F c.d.f.
v1= 20, v
2= 40
Question: Positive/negative skew (skewed to the right/left)?
Solution: Positive skew (skewed to the right).124/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
0 0.5 10
0.5
1
1.5
2
x
prob
abilit
y den
sity f
uncti
on
Beta p.d.f.
a= 4, b= 2
0 0.5 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
cumu
lative
dens
ity fu
nctio
n
Beta c.d.f.
a= 4, b= 2
Question: Positive/negative skew (skewed to the right/left)?Solution: Negative skew (skewed to the left).125/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Kurtosis
Let X be a random variable. The (excess) kurtosis of X isgiven by:
κX = E
[(X − µXσX
)4]− 3.
Measures the peakedness (positive) or flatness (negative) of arandom variable.
Kurtosis coefficient:
α4 =E[X 4]
(E [X 2])2.
126/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Which statements are true?Note: E[X ] = 0, σ = 1, and γ = 0 for i) and ii)
and E[X ] = −4, σ = 3 and γ = 0 for iii).
−3 0 31/18
8/9
x
Pr(
X=
x)
distribution i)
−2 0 21/8
3/4
x
Pr(
X=
x)
distribution ii)
−10 −4 21/8
3/4
x
Pr(
X=
x)
distribution iii)
a. Question: Distribution ii) has asmaller excess kurtosis thandistribution i).
a. Solution: True: excess kurtosis for i)is 6 and for ii) is 1.
c. Question: Distribution iii) has asmaller excess kurtosis thandistribution ii).
c. Solution: False: both have the SAMEexcess kurtosis.
127/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
−2 0 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x
proba
bility
dens
ity fu
nctio
n
p.d.f.
−2 0 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
cumu
lative
dens
ity fu
nctio
n
c.d.f.
Uniform(−√ 3,√ 3)Normal(0,1)Laplace(0,1/√ 2)
Question: Positive/none/negative excess kurtosis?Solution: Positive: green; none: red; negative: blue excess kurtosis.
Note: E[X ] = 0, Var(X ) = 1 and γX = 0 for all distributions.128/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Exercises
An insurance company is pricing its policies using thestandard deviation pricing principle.
The regulator requires that insurers have enough reserves inorder to reduce the probability of ruin to 0.5%.
Holding capital is a cost.
a. Would the insurer company prefer claims with a distributionI. with mean $100 and standard deviation of $5.
II. with mean $50 and standard deviation of $10.
b. Would the insurer company prefer claims with a distributionI. with mean $100 and standard deviation of $5 and a skewness
of $5.II. with mean $100 and standard deviation of $5 and a skewness
of -$5.129/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Measures of dispersion
Exercises
c. Would the insurer company prefer claims with a distributionI. with mean $100 and standard deviation of $5 and a skewness
of $0 and a kurtosis of $5.II. with mean $100 and standard deviation of $5 and a skewness
of $0 and a kurtosis of $2.
d. Would the insurer company prefer claims with a distributionI. with mean $100 and standard deviation of $5, a positive
skewness and a negative kurtosis.II. with mean $100 and standard deviation of $5, a negative
skewness and a positive kurtosis.
Solution:a. II.b. II.c. II.d. Cannot say from the question.
130/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
r th central/non-central moments
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
r th central/non-central moments
r th central moments
Let X be a r.v., the rth central moment is given by:
E [(X − µX )r ] =
∑all x
(x − µX )r · pX (x) ,
if X is discrete;∫ ∞−∞
(x − µX )r · fX (x) dx ,
if X is continuous.
Relation central & non-central moments:
E [(X − µX )r ]∗=
r∑k=0
(r
k
)· E[X k]· (−µX )r−k .
* using binomial expansion.131/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
r th central/non-central moments
(Non-central) MomentsLet X be a r.v., the rth (non-central) moment is given by:
E [X r ] =
∑all x
x r · pX (x) ,
if X is discrete;∫ ∞−∞
x r · fX (x) dx ,
if X is continuous.
Consider the Motor Vehicle insurer from slide 115.
Recall the mean of the claim size from b. on slide 116.
a. Question: Find the second central moment for an insured.
a. Solution: Var(X ) =E[X 2]− (E [X ])2 = (0.2 · 12 + .08 · 02)− 0.22 = 0.16 or
E[(X − µX )2
]= (1− 0.2)2 · 0.2 + (0− 0.2)2 · 0.8 = 0.16.
132/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
r th central/non-central moments
Exercisesb. Question: Find the skewness using the third central moment.b. Solution: Start with the third central moment:
E[(X − µX )3
]=0.2 · (1− 0.2)3 + 0.8 · (0− 0.2)3=0.1024− 0.0064=0.096.
γX =E
[(X − µX )3
σ3
]=
E[(X − µX )3
]σ3
=0.096
0.163/2= 1.5.
c. Question: Find the second and third non-central moments.c. Solution: E
[X 2]
= 12 · 0.2 + 02 · 0.8 = 0.2 andE[X 3]
= 13 · 0.2 + 03 · 0.8 = 0.2.d. Question: Find the skewness using only non-central moments.d. Solution:
E[(X − µX )3
]= E
[X 3 − 3µXX
2 + 3µ2XX − µ3X]
=
E[X 3]− 3µXE
[X 2]
+ 3µ2XE [X ]− 3µ3X =0.2− 0.12 + 0.024− 0.008=0.096.133/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
r th central/non-central moments
Knowledge of the mean, variance, skewness and kurtosis canprovide useful knowledge of the distribution.
Theorem: The complete set of all moments is required tocharacterize an arbitrary distribution, i.e., every distinctdistribution function has an unique set of moments.
Proof: Using Taylor series:
1 + E [X ] · t + E[X 2]· t
2
2!+ E
[X 3]· t
3
3!+ . . .
= E[
1 + X · t + X 2 · t2
2!+ X 3 · t
3
3!+ . . .
]∗= E
[eXt],
* using exp(x) = 1 + x + x2
2! + x3
3! + . . ., with x = X · t.134/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Moment generating function (mgf) of a r.v.
The moment generating function of a r.v. X is defined as:
MX (t) =E[eX ·t
]=1 + E [X ] · t + E
[X 2]· t
2
2!+ E
[X 3]· t
3
3!+ . . . .
Properties of m.g.f.:
M(r)X (0) = E [X r ] , for r = 0, 1, 2, 3, . . . ;
Mm·X+b(t) = MX (m · t) · eb·t , for constants m, b;MX+Y (t) = MX (t) ·MY (t), only if X ,Y are independent.
135/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Use of moment generating function
Relation m.g.f. and non-central moments: we can write them.g.f. as an infinite series of the moments as follows:
MX (t) = E[eX ·t
]=∞∑k=0
µk ·tk
k!.
Generating non-central moments using the m.g.f.: we cangenerate the moments from the m.g.f. using the relationship:
µr = E [X r ] = M(r)X (t)
∣∣∣t=0
.
Function of random variables: week 4.
136/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Proof: To prove the above result, consider the continuouscase (similar proof in the discrete case):
M(r)X (t) =
∂r
∂trE[eX ·t
]=
∂r
∂tr
∫ ∞−∞
ex ·t · fX (x) dx
=
∫ ∞−∞
(∂r
∂trex ·t)· fX (x) dx
=
∫ ∞−∞
(x r · ex ·t
)· fX (x) dx
=E[X r · eX ·t
].
Set t = 0 and you get the desired result.
Remark: If the m.g.f. exists for t in an open intervalcontaining zero, then it uniquely determines the probabilitydistribution.
137/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Example: Consider the MVI from slide 115. The m.g.f. is:
MX (t) = E[eXt]
=∑all x
pX (x) · ext = 0.2 · e1·t + 0.8 · e0·t = 0.8 + 0.2et .
The first k non-central moments are:
E [X ] =∂MX (t)
∂t
∣∣∣∣t=0
= 0.2et∣∣t=0
= 0.2
E[X k]
=∂kMX (t)
∂tk
∣∣∣∣t=0
= 0.2et∣∣t=0
= 0.2.
The kurtosis is given by:
Var(X ) =E[X 2]− (E [X ])2 = 0.2− 0.22 = 0.16.
E[(X − µX )4
]=E
[X 4]− 4µXE
[X 3]
+ 6µ2XE[X 2]− 4µ3XE [X ] + µ4X
=0.2− 0.16 + 0.048− 0.0064 + 0.00032 = 0.08192.
κX =E[(X − µX )4
]/Var(X )2 − 3 = 0.08192/0.162 − 3 = 0.2.
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ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
ExercisesConsider the Motor Vehicle insurer from slide 120.
a. Question: Determine the m.g.f. of the claim size.
b. Question: Use the m.g.f. to determine the skewness.
a. Solution: We have:
MX (t) = E[et·X
]=0.8 · e0·t +
∫ ∞0
λ
5· e−λ·x · et·xdx
=0.8 +
∫ ∞0
λ
5· e(t−λ)·xdx
=0.8 +
[λ
5 · (t − λ)· e(t−λ)·x
]∞0
∗=0.8− λ
5 · (t − λ).
* note m.g.f. exists if t − λ < 0.139/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
b. Solution: Using MX (t) = 0.8− λ5·(t−λ) .
Find the non-central moments using the derivatives:
E [X ] =∂MX (t)
∂t
∣∣∣∣t=0
=−λ5· −1 · (t − λ)−2
∣∣∣∣t=0
=1
5λ
E[X 2]
=∂2MX (t)
∂t2
∣∣∣∣t=0
=λ
5· −2 · (t − λ)−3
∣∣∣∣t=0
=2
5λ2
E[X 3]
=∂3MX (t)
∂t3
∣∣∣∣t=0
=−2λ
5· −3 · (t − λ)−4
∣∣∣∣t=0
=6
5λ3.
Var(X ) =E[X 2]− (E [X ])2 =
2
5λ2− 1
(5λ)2=
9
25λ2
γX =E[(X − µX )3
]Var(X )3/2
=E[X 3]− 3µXE
[X 2]
+ 3µ2XE [X ]− µ3XVar(X )3/2
=6
5λ3− 3 · 1·2
25λ3+ 2 · 1·1
125λ3
(3/(5λ))3=
6 · 52
33− 3 · 2 · 5
33+ 2 · 1
33= 122/27.
140/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Probability generating function (p.g.f.) of a r.v.
Let Y be an integer-valued random variable withPr(Y = i) = pi for i = 0, 1, 2, . . ., the p.g.f. is defined as:
PY (t) = E[tY]
=∞∑i=1
pY (i) · t i .
Properties of p.g.f.:
- The relationship between p.g.f. and m.g.f. is as follows:
PY (t) = MY (log(t)) .
- Probabilities: Pr(Y = r) = P(r)Y (t)
∣∣∣t=0
/r !
- Take the k th derivative:P
(k)Y (1) = E [Y · (Y − 1) · (Y − 2) · . . . · (Y − k + 1)] ≡ µ[k].
141/144
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Mathematical methods
Generating functions
Cumulant generating function (c.g.f.) of a r.v.The cumulant generating function CX (t) for a randomvariable is given by:
CX (t) = log(MX (t)) =∞∑i=1
1
i !· hi · κi or MX (t) = eCX (t),
where κi = C(i)X (0) is the i th cumulant.
Properties of c.g.f.:
κi = i th central moment, for r = 2, 3;Cm·X+b(t) = CX (m · t) + b · t, for constants m, b;CX+Y (t) = CX (t) + CY (t), X ,Y independent.
Note that C(r)X (0) 6= E [(X − µX )r ] , for r = 4, 5, 6, . . ..
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ACTL2002/ACTL5101 Probability and Statistics: Week 1
Summary
Summary
Moments and measures of dispersion
IntroductionCourse introduction
Introduction in probabilityExercise
Mathematical methodsRandom variables and distributionsMeasures of location: probability distributionsMeasures of dispersion
r th central/non-central momentsGenerating functions
SummarySummary
ACTL2002/ACTL5101 Probability and Statistics: Week 1
Summary
Summary
Mathematical Expectation
The mathematical expectation of h (X ) is:
E [h (X )] =
∑all x
h (x) · pX (x) , if X is discrete;
∫ ∞−∞
h (x) · fX (x) dx , if X is continuous.
Mean: µ = E [X ].
Moments: µr = E [X r ] refers to the r th (non-central)moment.
Central moments: E [(X − µx)r ] refers to the r th centralmoment.
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ACTL2002/ACTL5101 Probability and Statistics: Week 1
Summary
Summary
Dispersion
Variance: σ2 = Var(X ) = E[(X − µ)2
]= E
[X 2]− µ2.
Skewness: γX = E
[(X − µXσX
)3]
refers to the skewness. It
measures the lack of symmetry in the p.d.f..
Kurtosis: κX = E
[(X − µXσX
)4]− 3 refers to the kurtosis.
It measures the peakedness or flatness of the p.d.f..
Moment generating function: MX (t) = E[eX ·t
], it
uniquely defines a density function. It is useful to calculate
moments: µr = E [X r ] = M(r)X (t)
∣∣∣t=0
.
144/144