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Introduction Randomness Probability Probability rules Probability distributions Discrete Continuous

STAT 151 Introduction to Statistical Theory

August 21, 2014

STAT 151 Class 1 Slide 1
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Instructor and TAs (weekly consultation location and hours, byappointments only)

DenisHY LEUNGSOE5047

email:[email protected]: 68280396Tues 4-6pm, Wed/Thurs 9-11am

TAs SOE Grp Study Rm 4-7 (Rm 4023)

Mon 12-3pm DANG VU Dieu [email protected]

Wed 12-3pm HO Yan [email protected]

Thu 4-7pm Muhammad Hafiz Bin [email protected]

Thu 12-3pm LAU Jin [email protected]

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Assessments

Class Participation (10%)

Projects (40%)

1 project with presentations 16%

2 projects without presentations 24% (2 12%)

Each projects grade includes 6% individual assessment(quizzes)

Exam (50%) Closed book but one 2-sided A-4 cheat sheet is allowed


I d i R d P b bili P b bili l P b bili di ib i Di C i
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Structure of each class

New materials (

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Data

Survival time (in years) in 70 cancer patients

0.67 0.01 0.481.060.85 0.45 0.19 0.78 1.23 0.180.372.170.15 1.19 0.581.220.72 0.671.420.020.12 0.50 0.15 0.81 0.64 0.22 0.05 0.55 0.46 0.830.46 0.56 0.82 0.072.280.34 0.64 0.09 0.77 0.260.45 0.41 0.23 0.16 0.39 0.29 0.62 1.09 0.14 0.49

0.66 0.89 0.99 0.98 0.95 0.14 0.03 0.012.620.990.081.391.31 0.50 0.74 1.19 0.15 0.14 1.18 1.53Orange - Subtype I Black - Subtype II

Outcome (H vs. T) in tossing a coin 10 times

H H T H T T H T T T

Occurrence of financial crises

Mexican

82

S&L

84

Black Mon.

87

Comm. RE

91

Asian

97

LTCM

98

Dotcom

00

Subprime

07

Euro

12?

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Randomness

Observation fromsurvival timedata - Whydo some patientslivelonger thanothers?

Subtype I - 1.06 2.17 1.22 1.42 2.28 2.62 1.39 1.18 1.53average 1.65

Subtype II - 0.67 0.01 0.48 0.85 0.45 0.19 0.78 1.23 0.18 0.37 0.151.19 0.58 0.72 0.67 0.02 0.12 0.50 0.15 0.81 0.64

0.22 0.05 0.55 0.46 0.83 0.46 0.56 0.82 0.07 0.340.64 0.09 0.77 0.26 0.45 0.41 0.23 0.16 0.39 0.290.62 1.09 0.14 0.49 0.66 0.89 0.99 0.98 0.95 0.140.03 0.01 0.99 0.08 1.31 0.50 0.74 1.19 0.15 0.14average 0.50

I has a better chance of survival because it has a higher average than II

There are still variations within each subtype - inevitable in the real world

We attribute (accommodate) these (unexplained) variations as random

Chance and randomness can be described using probability theory



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Definition of probability - Coin tossing data

Toss 1 2 3 4 5 6 7 8 9

Outcome H T T H H T H T H

Proportion

ofh

eads

0.0

0.2

0.4

0.6

0.8

1.0

1 2 3 4 9 50 100 500 1000

probability = 0.5

Number of tosses

The long run proportion (frequency) of heads is the probability of heads.



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Insight from coin tossing experiment

Probability is the long run frequency of an outcome

Probability cannot predict individual outcomes

However, it can be used to predict long run trends

Probability always lies between 0 and 1, with a value closer to1 meaning a higher frequency of occurrence

Probability is numeric in value so we can use it to:

- compare the relative chance between different outcomes(events)

- carry out calculations


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y y y

Probability Axioms - Urn model (1)- drawing marbles from an urn(with replacement)

12

4

3 5Probability

35

25

Draws

1 2 3 4 5 6 7 8 9 ......


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y y y

Urn model (2)

Five possible Outcomes: 1 2 43 5

Interested in Event A:

A={ 1 4 5 }; hence an event is a collection ofoutcomes

P(A) =3

5

= 0.6(60%) = Number of marbles in A

Total number of marbles


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y y y

Complementary events

Marbles in urn: 1 2 43 5

Interested in A: (Not A)

A

= 2 3 A, sometimes written as AC, is called the complementaryevent ofA

Chance of = 1chance of

P(A) = 1 P(A) = 1 3

5=

2

5


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Joint probability - independent events - Drawing a blue in draw 1 anda green in draw 2 (with replacement)

Aand B areindependentmeans the occurrence of one event doesnot change the chance of the other

12

4

3 5

Draws

1 2

A= { in 1st draw}

B= { in 2nd draw}

P(A) =3

5

; P(B) =2

5P(Aand B) =

35

25

= 6

25= P(A)P(B)



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Joint probability and disjoint events

Two events A and B aredisjoint or sometimes called mutuallyexclusive if they cannot occur simultaneously: P(A and B)=0

Example

P( and in a singledraw) = 0


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Conditional probability

Conditional probability isauseful quantification of howtheassessment ofchance changed due to new information: IfA happened, what is the chance

ofB?

The conditional probability of B given A is written as P(B|A)

Example Drawing marbles WITHOUT replacement

?A= { in 1st draw}

B= { in 2nd draw}

P(B) =2

5

P(B|A) =24

2

5

2

4because has been drawn

P(A and B) =2

4

3

5= P(B|A)P(A) =

6

20= P(B)P(A)

;


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Conditional probability and independence

The following relationships hold ifA and Bare independent

P(A|B) = P(A)

P(B|A) = P(B)

P(A and B) = P(A)P(B)

Independence is NOT the same as mutually exclusive (disjoint),which is P(Aand B) = 0.


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The multiplication rule

P(AB) = P(A|B)P(B) = P(B|A)P(A) Multiplication Rule

Example

Marbles in urn: 1 2 43 5

P( and Odd) = P( |Odd)P(Odd) =

1

3

3

5

=

1

5

= P(Odd| )P( ) =

1

2

2

5

=

1

5

Rearranging the multiplication rule:

P(A|B) = P(AB)

P(B)


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Partition rule

P( ) = P( and odd) + P( and even)

= P({ 1 5 }) + P({ 4 })

= 2

5+

1

5

= 3

5


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Union of events (1)

Unionof events can sometimes be bestvisualized using aVenn diagram(JohnVenn, 1834-1923)Example What is the probability of drawing

a or an odd number ?

12

4

3 5

UrnGreen

Odd

12

4

35

;


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Union of events (2)

P( or odd) = P( ) + P(Odd) P( and odd)

= 2

5+

3

5

1

5

= 45

In general, ifAand B are:

disjoint, then P(A or B) = P(A) + P(B)

not disjoint, then P(Aor B) = P(A) + P(B) P(AB)


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Probability tree

Probability tree is useful for studying combination of events. Branches of a tree areconditionalprobabilities.

ExampleDrawing two marbles from urn without replacement:

Draw 1

Draw 2

P( ) = 35 24

24

= P( | )

P( ) = 35 24

24 = P( | )

35

Draw 2

P( ) = 25 34

34

= P( | )

P( ) = 25 14

14 = P( | )

25


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Bayes Theorem (Thomas Bayes, 1701-1761)

Trees are useful for visualizing P(B|A) when Bfollows from A in a natural (time)order. Manyproblemsrequire P(A|B), Bayes Theorem provides an answer.

ExampleTesting for an infectious disease.

Disease

Test

P(D T) = 99100 910

T

910

= P(T|D)

P(D T) = 99100 110T

1

10 =P

(T|D

)

D

99100

Test

P(D T) = 1100 110

T1

10

= P(T|D)

P(D T) = 1100 910T

910 = P(T|D)

D

1100

What is P(D|T) or P(D|T)?


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Data revisited- Discrete vs. Continuous Data

Discrete - countable number of possible values

Examples

(1) Coin Tossing: H H T H T T H T T TTwo possible values: H or T

(2) Financial crisis: 1982, 1984, 1987,... (No. of crises in adecade)Many possible values: 0, 1, 2, 3, 4,...

Continuous- values fall in an interval (a, b), acould be andbcould be

ExampleSurvival time in cancer patients: 0.67 0.01 0.48 1.06 0.85 0.45 ...0 < survival time < a positive number


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First look - summary statistics

Nave methods

Numerical summary: Min, max, mean, variance, etc.Graphical summary: Histogram, pie chart, bar graph, etc.Tabular summary: Tables of frequencies

Attributes

Quick and dirtyIdentify potential problems, errors

Flaws

Difficult to generalize especially when the dataset iscomplicatedNot easily portable cannot describe a chart or graph unlessyou can see it!


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What do we do about data?

Financial crises data

Mexican

82

S&L

84

Black Mon.

87

Comm. RE

91

Asian

97

LTCM

98

Dotcom

00

Subprime

07

Euro

12?

We may be interested in X, the number of crises in the next decade. Possible questions:

What is the probability ofX=0, or 1, or 2, or 3 or...?Probability distribution - tells us the long run frequencies (chances) of the possibleoutcomes

On average, how many crises in a decade?Expectation (Expected value) - tells us the weighted average of the possibleoutcomes, where the weights are the probabilities

How likely will Xdiffer from the expected?Standard deviation (Variance) - tells us the weighted average of deviations of thepossible outcomes from the expected value


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Discrete distributions - Tossing a coin

(H) or (T) ?

Long run frequencies

H T

12

12

The long run frequencies are 1/2 each forH and T there is equal chance for H or T

;


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Discrete distributions - Drawing a marble

1 24

3 5

1 2 3 4 5 ?

1 2 3 4 515

15

15

15

15

35

25

;

The long run frequencies tell us there is equal chance for 1, 2, 3, 4and 5 but there is a higher chance for blue than green



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Discrete probability distributions

X H T

P(X) 1212

X 1 2 3 4 5

P(X) 1515

15

15

15

35

25

X a1 a2 a3 ...

P(X) P(a1) P(a2) P(a3) ...

Probability distribution function

;

A probability distribution summarizes the long run frequencies (chances) of thepossible outcomes ofX



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Probability distribution for a discrete random variable

X is the unknown outcome.X is called a discreterandomvariable ifits value can onlycomefrom a countable

number of possible values: a1, a2, ...,ak.Examples

Coin toss: X= H or T (2 possible values)Marbles: X= 1, 2, 3, 4, or 5 (5 possible values)Financial crisis: X= 0, 1, 2, 3,... (Infinite but countable number of possiblevalues)

P(X = ai) gives the probability X = aiand is called a probability distributionfunction.A valid probability distribution function must satisfy the following rules:

P(X = ai) must be between 0 and 1

We are certain that one of the values will appear, therefore:

P(X = a1 or X = a2 or ...X = ak) = P(X = a1) + P(X = a2) + ... + P(X = ak) X=a1,X=a2,...are disjoint events

= 1



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Example: Probability distribution of the number from drawing a marble

X 1 2 3 4 5P(X) 0.2 0.2 0.2 0.2 0.2

P(X= 1) = 0.2

P(X 3) = P(X= 3) + P(X= 4) + P(X= 5) = 0.6

P(X = 1.5) = 0

P(X

>5) = 0

1 = P(X= 1)+P(X= 2)+P(X= 3)+P(X= 4)+P(X = 5)

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Probability distribution for a continuous random variable

A continuous randomvariable, X, is a random variable withoutcomethat fallswithin a range or interval, (a, b)

Examples

X= survival time of a patient: (a, b) = [0, 150] yearsX= return from an investment of 10000; (a, b) = [10000,) dollarsX= per capita income; (a, b) = [0, 1000000000000] dollars

The probability distribution ofX

is defined by a (probability) densityfunction (PDF), f(x). The interpretation of f(x) is similar to probability butthere are subtle differences:

f(x) 0 for any value ofx in (a, b)f(x) = P(X = x) = 0 for any value of xP(r X s) = P(r < X < s)

since P(X=r)=P(X=s)=0

probability ofX between r and s

P(a X b) = 1 since Xmust fall within (a, b)

Thecumulative distribution function (CDF for short), written as F(x), isdefined as P(X x)



Example survival data
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Example - survival data

Density of survival time

X (Timeinyears)

Den

sity

0 1 2 3 4 5

f(x)= 1.5e1.5x

f(x) =ex, >0 is called an

exponential density

Different values ofcan be usedto model different populations

All exponential densities have the

same shape but different gradientsso it is easy to describe to others

Allows us to make probabilitystatements about survival times wehave and have notobserved

The area under f(x) is 1 whichmeans the probability of dyingbetween time 0 and is 1



Some calculations
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Some calculations

X (Timeinyears)

De

nsity

0 1 2 3 4 5

f(x)= 1.5e1.5x

P(X >2) = 0.049 is theredarea under the curve and can beobtained by analytical or numerical (computer) analysis



Some calculations

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Some calculations

X (Timeinyears)

De

nsity

0 1 2 3 4 5

f(x)= 1.5e1.5x

P(X 1) = P(0< X 1) = 0.776 is theredarea under the curveand can be obtained by analytical or numerical analysis.



Some calculations
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Some calculations

P(X 2|X >1) = P(1< X and X 2)

P(X >1)

conditional probability

= P(1< X 2)P(X >1)

= 1 P(X 1) P(X >2)

1 P(X 1)

= 1 0.776 0.049

1 0.776= 0.781

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