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RPRA 2. Elements of Probability Theory 1

Engineering Risk Benefit Analysis

1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82,ESD.72, ESD.721

RPRA 2. Elements of Probability Theory

George E. Apostolakis

Massachusetts Institute of Technology

Spring 2007

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Probability: Axiomatic Formulation

The probability of an event A is a number that satisfies the

following axioms (Kolmogorov):

0 P(A) 1P(certain event) = 1

For two mutually exclusive events A and B:

P(A or B) = P(A) + P(B)

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Relative-frequency interpretation

Imagine a large number n of repetitions of theexperiment of which A is a possible outcome.

If A occurs k times, then its relative frequency is:

It is postulated that:

n

k

)A(Pn

klim

n

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Degree-of-belief (Bayesian) interpretation

No need for identical trials.

The concept of likelihood is primitive, i.e., it is meaningful

to compare the likelihood of two events.

P(A) < P(B) simply means that the assessor judges B to be

more likely than A.

Subjective probabilities must be coherent, i.e., must satisfy

the mathematical theory of probability and must be

consistent with the assessors knowledge and beliefs.

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Basic rules of probability: Negation

S

EVenn Diagram___

E SEE =

==+ 1)S(P)E(P)E(P

)E(P1)E(P =

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Basic rules of probability: Union

( ) ( )( )

+

+

= + I

U

KN

1 i

1N

1N

1i

N

1ijji

N

1ii

N

1i

AP1

AAPAPAP

( )=

N

1i

i

N

1

i APAP U

Rare-Event Approximation:

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Union (contd)

For two events: P(AB) = P(A) + P(B) P(AB)

For mutually exclusive events:

P(AB) = P(A) + P(B)

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Example: Fair Die

Sample Space: {1, 2, 3, 4, 5, 6} (discrete)

Fair: The outcomes are equally likely (1/6).

P(even) = P(2 4 6) = (mutually exclusive)

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Union of minimal cut sets

=+= + + N1i i1N1N 1i N 1ij jiN1i iT MMMMX 1)(...

Rare-event approximation: ( ) ( )N1

iT MPXP

From RPRA 1, slide 15

+ =+= + Ni iNNi Nij jiNi iT MP...)MM(P)M(P)X(P )( 1111 11 1

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Upper and lower bounds

( ) ( )N1

iT MPXP

= + 1N 1i N 1ij jiN1i iT )MM(P)M(P)X(P

=

+

= +==

++=N

1i

i1N

1N

1i

N

1ij

ji

N

1i

iT )M(P...)MM(P)M(P)X(P )1(

The first term, i.e., sum,gives an upper bound.

The first two terms, give

a lower bound.

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Conditional probability

( ) ( ) ( ) ( ) ( )APA/BPBPBAPABP ==

( ) ( )( )BPABP

BAP

For independent events:

( ) ( )BPA/BPBPAPABP

==

Learning that A is true has no impact on our probability of B.

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Example: 2-out-of-4 System

1

2

3

4

M1 = X1 X2 X3 M2 = X2 X3 X4

M3 = X3 X4 X1 M4 = X1 X2 X4

XT = 1 (1 M1) (1 M2) (1 M3) (1 M4)

XT = (X1 X2 X3 + X2 X3 X4 + X3 X4 X1 + X1 X2 X4) - 3X1 X2 X3X4

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2-out-of-4 System (contd)

P(XT = 1) = P(X1 X2 X3 + X2 X3 X4 + X3 X4 X1 + X1 X2 X4)

3P(X1 X2 X3X4)

Assume that the components are independent and nominally

identical with failure probability q. Then,

P(XT = 1) = 4q3 3q4

Rare-event approximation: P(XT = 1) 4q3

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Updating probabilities (1)

The events, Hi, i = 1...N, are mutually exclusive and exhaustive, i.e.,

HiHj= , for ij, Hi = S, the sample space. Their probabilities are P(Hi).

Given an event E, we can always write

)H(P)H/E(P)E(P i

N

1i=

H1

H2

H3

E

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Updating probabilities (2)

Evidence E becomes available.

What are the new (updated) probabilities P(Hi/E)?

Start with the definition of conditional probabilities,

slide 11.

Using the expression on slide 14 for P(E), we get

== )E(P)E/H(P)H(P)H/E(P)EH(P iiii

)E(P

)H(P)H/E(P

)E/H(P

ii

i=

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Bayes Theorem

PosteriorProbability

Prior

Probability

Likelihood of the

Evidence

( ) ( ) ( )( ) ( )= N1

ii

iii

HPHEP

HPHEPEHP

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Example: Lets Make A Deal

Suppose that you are on a TV game show and the host has offered you

what's behind any one of three doors. You are told that behind one ofthe doors is a Ferrari, but behind each of the other two doors is a Yugo.You select door A.

At this time, the host opens up door B and reveals a Yugo. He offers youa deal. You can keep door A or you can trade it for door C.

What do you do?

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Lets Make A Deal: Solution (1)

Setting up the problem in mathematical terms:A = {The Ferrari is behind Door A}

B = {The Ferrari is behind Door B}

C = {The Ferrari is behind Door C}

The events A, B, C are mutually exclusive and exhaustive.

P(A) = P(B) = P(C) = 1/3

E = {The host opens door B and a Yugo is behind it}

What is P(A/E)? Bayes Theorem

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( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )CPCEPBPBEPAPAEP APAEP

EAP

+=

But

P(E/B) = 0 (A Yugo is behind door B).

P(E/C) = 1 (The host must open door B, if the

Ferrari is behind door C; he

cannot open door A under

any circumstances).

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Let P(A/E) = x and P(E/A) = p

Bayes' theorem gives:

p1

px +

Therefore

For P(E/A) = p = 1/2 (the host opens door B randomly, if

the Ferrari is behind door A)

P(A/E) = x = 1/3 = P(A) (the evidence has had no

impact)

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Since P(A/E) + P(C/E) = 1 P(C/E) = 1 - P(A/E) = 2/3

The player should switch to door C

For P(E/A) = p = 1 (the host always opens door B, if theFerrari is behind door A)

P(A/E) = 1/2 P(C/E) = 1/2, switching to door C does notoffer any advantage.

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Random Variables

Sample Space: The set of all possible outcomes of an experiment.

Random Variable: A function that maps sample points onto the realline.

Example: For a die S = {1,2,3,4,5,6} For the coin: S = {H, T} {0, 1}

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Events

-

3 5421 6

3.6

0

We say that {X x} is an event, where x is any numberonthe real l ine.

For example (die experiment):

{X 3.6} = {1, 2, 3} {1 or 2 or 3}{X 96} = S (the certain event){X -62} = (the impossible event)

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Sample Spaces

The SS for the die is an example of adiscrete sample space and X is adiscrete

random variable (DRV).

A SS isdiscrete if it has a finite or countably infinite number of sample points.

A SS iscontinuous if it has an infinite (and uncountable) number of samplepoints. The corresponding RV is acontinuous random variable (CRV).

Example:

{T t} = {failure occurs before t}

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Cumulative Distribution Function (CDF)

Thecumulative distribution function (CDF) is

F(x) Pr[X x] This is true for both DRV and CRV.

Properties:

1. F(x) is a non-decreasing function of x.

2. F(-) = 03. F() = 1

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CDF for the Die Experiment

F(x)

1/6

x65

4321

1

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Probability Mass Function (pmf)

For DRV: probability mass

function

) ii pxXP =

( ) xxallforpxF ii ,1p)S(P

i

i = normalizationExample: For the die, pi = 1/6 and

3

1

6

1

6

1p)21(P)3.2(F

2

1

i =+===

1p6

1i =

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Probability Density Function (pdf)

f(x)dx = P{x < X < x+dx}

( ) ( )dx

xdFxf =

= 1ds)s(f)(F)S(P=x

ds)s(f)x(F

normalization

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Example of a pdf (1)

Determine k so that

( ) 1x0forkxxf 2 ,( ) ,0xf = otherwiseis a pdf.

Answer:

The normalization condition gives:

=10

2 3k1dxkx

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Example of a pdf (2)

3xxF =)(=dxx3)75.0(F)875.0(F 875.0

75.0

2

= 0.67 - 0.42 = 0.25 =

= P{0.75 < X < 0.875}

1

1

f(x)

F(x)

3

0.67

0.42

0.75 0.875

1

M t

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Moments

Expected (or mean, or average) value

[ ] ( )

j

jj DRVpx

CRVdxxfxmXE

Variance (standard deviation )

( ) ] ( ) ( )( )

=

j

j2

j

2

22

DRVpmx

CRVdxxfmxmXE

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Percentiles

Median: The value xm

for which

F(xm) = 0.50

For CRV we define the 100 percentile asthat value of x for which

( ) x

dxxf

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Example

=1

0

3 750dxx3m .

( ) =10

222 03750dxx750x3 .. 1940.

( ) 790x50xxF m3mm .. =370x050x 050

3050 .. .. =

980x950x 9503

950 .. .. =

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