Risk Analysis Introduction and Overview · 2013-07-10 · 5 Terminology and Background •...

1

Risk Analysis Introduction and Overview

Thomas A. Mazzuchi Professor and Chairman

Department of Engineering Management and Systems Engineering

George Washington University

2

Terminology and Background

•  Risk - A measure of potential loss due to natural or

human activities - A combination of the probability or frequency of

the hazard and its consequence; e.g.,

•  Loss - Adverse consequences of such activities that

affect  Human life or health  Economics or property  The natural environment  Information , etc

3


•  Engineering Systems Losses Can Be -  Internal to the system; i.e,

 Damage to one of the system’s components - External to the system; i.e.,

 Damage to a component of the external environment in which the system must function; e.g.,  Humans  Organizations  Economic assets  Environmental assets

4


•  Risk Analysis -  Is the process of characterizing, managing, and

informing others about the existence, nature, magnitude, prevalence, contributing factors, and uncertainties that pertain to the potential losses - Other names for risk analysis

 Probabilistic Risk Analysis (PRA)  Quantitative Risk Analysis (QRA)  Probabilistic Safety Analysis (PSA)

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•  Importance of Risk Analysis - While formal methods for risk analysis have

been shaped by modern demands, the concept of risk analysis is not new; it is even ancient - People are living longer, healthier, more

prosperous lives and have more to loose - Today people expect greater protection than

before from industry and government, and they react with litigation when they feel let down

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•  Importance of Risk Analysis - Even as public concerns about risk exert

pressure on policy makers for regulations, engineering systems are increasing in complexity and autonomy  Simply making regulations without studying

their effects can be costly and suboptimal—even dangerous

- A proper risk analysis will adequately model the system, demonstrate the effect of mitigating measures, and communicate these to the public

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Elements of Risk Analysis

Risk Assessment

Risk Management

Risk Communication

National Research Council (1994)

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Elements of Risk Analysis

•  Risk Assessment - The process by which the probability or frequency of

loss by or to an engineering system is assessed, and the magnitude of the loss (consequences) estimated

•  Risk Management - The process by which the potential (probability or

frequency) for loss and/or the magnitude of loss is minimized and controlled

•  Risk Communication - The process by which information about the nature

and consequences of risk, as well as the risk assessment approach and the risk management options, are shared and discussed among decision makers and other stakeholders

9

Risk Assessment

•  Definition of Risk (Kaplan & Garrick, 1981) -  Risk addresses three basic questions:

  What can go wrong?   How likely is it to happen?   What are the losses (or consequences)?

-  A combination of hazard and likelihood -  A triple <Si,Pi,Ci>

  Si a specific scenario of a hazard   Pi probability of si (or frequency)   Ci consequence of si

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Risk Assessment

•  Modifications - Si may occur with a given

probability Pi or frequency fi -  Its occurrence may be static

or dynamic over time - Pi and Ci may be uncertain

and have probability distributions - These distributions may be a

function of time or Si or a combination of the two - These quantities may be

jointly distributed

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Quantitative Risk Assessment

•  Overview

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Important Risk Journals

•  Health, Risk and Society •  Journal of Risk and Insurance •  Journal of Risk and Uncertainty •  Journal of Risk: Health, Safety and Environment •  Journal of Risk Research •  Journal of Safety Research •  Journal of System Safety •  Risk Analysis, An International Journal •  Risk, Decision, and Policy •  Risk Management and Insurance Review •  Risk Management: An International Journal •  Safety Science •  The Journal of Risk •  Reliability Engineering and System Safety •  International Journal of Reliability, Quality, and Safety Eng

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Societies of Interest •  American Society of Mechanical Engineers •  Safety Engineering and Risk Analysis Division •  American Society of Safety Engineers •  American Statistical Association, Section on Risk Analysi •  IEEE Reliability Society •  International Association for Probabilistic Safety

Assessment and Management. •  Risk Assessment and Policy Association •  Risk Theory Society •  Society for Maintenance Reliability Professionals •  Society for Reliability Engineers •  Society for Risk Analysis •  System Safety Society •  The Safety and Reliability Society

• 

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Qualitative Risk Assessment: Risk Matrices

15

Introduction

•  Risk Matrix – a table that has several categories of probability,

likelihood or frequency on its rows (or columns) and several categories of severity, impact, or consequence on its columns (or rows)

–  It associates a recommended level of risk, urgency, priority, or management action associated with each column-row pair (i,e, cell)

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Introduction

Federal Highway Administration, 2006

Federal Aviation Administration, 2007

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Introduction

Qualitative Risk Assessment •  NASA Risk Management Reporting

Qualitative Risk Assessment •  NASA Risk Management Reporting

20 Problems with Risk Matrices and Matrix Design Cox (2008)

•  If Risk = probability * consequence

Consequence

Probability

Probability

Consequence

Risk

p*c=constant

21 Subjective Interpretations and Input Bias Smith et al (2009)

Consequence Li

kelih

ood

0 PROBABILITY 1

1 Objective Subjective

Utility Va

lue

Objective Subjective

Utility

Value

SRP

22 Extension of Cox for Opt. 5x5 Matrix Design Hong and Mazzuchi (2013)

23 Uncertainty Distribution for Portfolios of Risks Mazzuchi and Scolese (2014)

p

c

Quantitative Risk Analysis Scenario Analysis

25

Fault Trees

•  The Basics of Fault Trees - A fault tree develops a deterministic description of

the occurrence of the top event, in terms of the occurrence or not of intermediate events  Top events represent system-level failure

- Describes intermediate events further until, at a finer level of detail, basic events are obtained  Basic events represent component-level failure

- By itself, a fault tree is only a visual model of how a system failure can occur

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Fault Trees

1.  Identify undesirable TOP event

2.  Identify first contributors

3.  Link contributors to TOP event by logic gates

4.  Identify second level contributors

5.  Link second level contributors to TOP event by logic gate

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Fault Tree Construction

•  Symbols Event Symbols Transfer Symbols

Basic Event

Undeveloped Event Transfer In

External Event Transfer Out

Intermediate Event

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Fault Tree Construction

•  Symbols -  Gate Symbols

+

+ +

and gate: Output occurs if all input events occur or gate: Output occurs if any input event occurs exclusive or gate: Output occurs if exactly one input event occurs

priority and gate: Output occurs if all input events occur in a specific sequence inhibit gate: Output occurs if the single input occurs in the presence of an enabling condition not or gate: Output occurs if at least one input event does not occur not and gate: Output occurs if all input events do not occur

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Fault Tree Example 2 Example with Success Event

Isolation Valve VAL Permanent Ignition

Source I2

Pressure Relief Valve PRV

Possible Ignition Source

I1

Leak

Gas flowing through pipe, there is a leak after the isolation valve this valve should close but then the pressure relief vale must open to relieve local pressure

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Fault Tree Example 2 Example with Success Event

Explosion After Gas Leak Posterior to Isolation Valve

+

Explosion Prior to Isolation

Explosion Posterior to Isolation Valve

VAL Performs Correctly

PRV Fails

VAL Fails

I1 Present

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C

C

Sensing & Control System

AC Power Source

T1

V1

V2

V3

V4

V5

P1

P2

Fault Tree Example 3 Large Example

Pumping System Example

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No Water Delivered When Needed

+ No Water Delivered from V1

V3 Fails to Remain Open


No Water

from P2

No Water Delivered from P1 Branch




a +

P2 Fails to Function

AC Fails

a

+

AC Fails

S Fails

+


T1 Ruptures

S Fails to Send Signal

P1 Fails to

Function

AC Fails

+

b

b

No Water Delivered from P2 Branch

No Water

from P1

S Fails to Send Signal

Fault Tree Example 3

33

No Water Delivered When Needed




Pumping Branches Fail



+

P1 Fails to Function

AC Fails

S Fails


T1 Ruptures

P2 Fails to

Function

+


+

P2 Branch Fails

P1 Branch Fails

Alternative Construction

34

A

B

C

D

E

1

F

2

3

4

5

6

7

Fault Tree Example 4 Block Diagram Example

•  Circuit Block Diagram Example

35

No Current at Point F

+

Unit 7 Fails

No Current at D & E

No Current at Point E

+

No Current at Point C

No Current at Pnt A

+

Unit 2 Fails

Units 5 & 6 Fail

Unit 6 Fails

Unit 5 Fails

No Current at Point D

+

No Current at Point B

No Current at Pnt A

+

Unit 1 Fails

Units 3 & 4 Fail

Unit 4 Fails

Unit 3 Fails


36

Event Tree Method

•  The Event Tree Method is the primary technique used in PRA to generate risk scenarios

•  This method can be used when … - … Successful operation of a system depends on the

approximately chronological and discrete operation of its units - … Previous event tree model scenarios of

successive events have led to exposure to hazards, and ultimately to undesirable consequences

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Event Tree Method Example

Initiating Event A B C D E

Sequence Logic

System Results

Success ↑ Failure ↓

A B C E S

A B C E F

A B C D E S

A B C D E F

A B C D F

A B F Let A denote that subsystem A fails and A denote that it does not fail

Depends on sequence of events

Mutually Exclusive Events

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Event Tree Method

•  Building an event tree - Build from left to right - Start the sequence at the initiating event - Place protective barriers as the successive (binary)

events - Calculate branching probabilities (called split

fractions) from fault trees - Calculate the probability of the end mutually

exclusive events as the multiple of the path split fractions

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Event Tree Method Example 1

PUMP KLAXON

S

P K

A subgrade compartment containing important control equipment is protected from flooding using the above system. If the water rises it should close the float switch which operates a pump with separate power supply, A klaxon should also sound and alert operators to perform bailing.

B

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Water Rises

I

Float Switch

S Pump

P Bailing

B System Results Klaxon

K System Logic

ISP S

ISPKB S

ISPKB F ISPK F

IS F

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Attempted Illegal

Access by Hacker

I

Principal Firewall

F

Abnormal Signal Detected by

Operator O

System Results

Backup Firewall

Initiated by Operator

B

System Logic

IF S

IFOB S

IFOB F IFO F

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Event Tree Method

Split fractions are calculated using fault trees

Quantifying Scenario Analysis

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Quantifying Fault Trees and Event Trees

•  How Do You Quantify Fault Trees and Event Trees - A fault tree or an event tree by itself is only a visual

model of a system -  It can be a representation of Boolean logic, i.e. a

representation of the functioning (or not) of the system as a function of its components - Because the basic events are 0-1 (fail-no fail), we

can use Boolean algebra to reduce the system expression to the lowest terms -  In doing so we make the following assumptions

 All events are binary  The system is coherent

 I.e., failure of any component cannot improve the system

45

Boolean Reduction: Boolean Algebra

Notation Boolean Operator Set Theory t   X and Y X•Y X∩Y   X or Y X+Y = 1-(1-X)(1-Y) X∪Y   Not X X’ Xc

•  Important Laws   Commutative X•Y = Y•X X+Y = Y+X   Associative X•(Y•Z) = (X•Y)•Z X+(Y+Z)=(X+Y)+Z   Distributive X•(Y+Z) = X•Y+X•Z   Idempotent X•X = X X+X = X   Absorption X+X•Y = X   Complementation X+X’ = Ω (X’)’ = X   De Morgan’s (X•Y)’ = X’+Y’ (X+Y)’ = X’•Y’   Empty/Universal Set ∅’ = Ω

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T = E1•E2 = (A+E3) • (C+E4) = A•C + A•E4 + C•E3 + E3•E4 = A•C + A•(A•B) + C•(B+C) +

+ (B+C)•(A•B) = A•C + A•A•B + C•B + C•C +

+ B•A•B + C•A•B = A•C + A•B + B•C + C + A•B +

+ A•B•C = A•C + A•B + B•C +C + A•B•C = A•C + A•B + C + A•B•C = A•B + C + A•B•C = A•B + C

Reducing a Fault Tree Using Boolean Algebra

This is the reduced tree and reduced Boolean expression for the tree called Min Cut Set Representation

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Representing Systems in Terms of Their Components

Using the convention that

X•Y=X*Y and X+Y=1-(1-X)*(1-Y)

we may determine the state of the top event in terms of the component states.

From previous page T=A•B + C = 1-(1-A*B)*(1-C)

For example if A occurs and C occurs but B does not

T=1-(1-1*0)(1-1) = 1 (Then the top event occurs)

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•  Truth tables - Generate all possible component states and the

probabilities associated with each. - For m components, each can either function or not

(i.e. 2 states for each component) thus there are 2m possible states taking in to account all components. - Evaluate the system using the Boolean formula for

each state

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-  Generation of All Possible States

1st Col 2nd Col 3rd Col nth Col 0 1 0 1 0 1 0 1 : :

0 0 1 1 0 0 1 1 : :

0 0 0 0 1 1 1 1 : :

0 : :

0 1 : :

1 : :

20=1 20=1 21=2

21=2 22=4

22=4

2n-1

2n-1

…..

…..

…..

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•  Truth tables

Note that if all elements of {A,B} occur or all elements of {C} occur then the top event occurs

These are called Cut Sets

T=A•B + C = 1-(1-A*B)*(1-C)

=1-(1-0*0)(1-0) =1-(1-1*0)(1-0) =1-(1-0*1)(1-0) =1-(1-1*1)(1-0) =1-(1-0*0)(1-1) =1-(1-1*0)(1-1) =1-(1-0*1)(1-1) =1-(1-1*1)(1-1)

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•  Truth Tables in Excel T=A•B + C = 1-(1-A*B)*(1-C)

52

Some Important Definitions

•  Cut Set - A collection of basic events such that, if the events

occur together, the top event certainly occurs •  Min Cut Set

- A cut set such that, if any basic event is removed, the remaining set is no longer a cut set

•  Path Set - A collection of basic events that connect input and

output  A path set merely represents a path through the

graph •  Min Path Set

- A path set such that, if any basic event is removed, the remaining set is no longer a path set

53

Min Cut Set Representation for Fault Trees

•  What is it? - After Boolean reduction, the Boolean formula for any

fault tree will be in Min Cut Set Representation T = X11• X12• … • X1n1

+ X21• X22• … • X2n2+

….+ Xm1• Xm2• … • Xmnm

where {Xi1, Xi2, … , Xini} is the ith cut set and

Xij=1 if ith item failed and 0 otherwise, Letting Ci = Xi1• Xi2• … • Xini

where Ci is the ith cut set indicator

Ci =1 if all elements of the ith cut set fail Then T = C1+C2+….+ Cm

54

Min Cut Set Representation for Fault Trees

•  Converting Min Cut Set Representation to a Calculable Formula

T = C1+C2+….+ Cm Then we can write T = 1 – (1-C1)*(1-C2)*…*(1-Cm) And since Ci = Xi1• Xi2• … • Xini

We can write

T = 1 – (1-C1)*(1-C2)*…*(1-Cm) = 1 – (1- Πj=1,n1

X1j)*(1-Πj=1,n2X2j)…*(1-Πj=1,nm

Xmj)

55

Example

Consider the following Fault Tree

D+E B•C

(D+E)•B B•C+A

[(D+E)•B]•[B•C+A]

56

Example

T = [(D + E) • B] • [(B • C) + A] T = (B•D + B•E) • [(B•C) + A] T = (B•D•B•C) + (B•E•B•C) + (B•D•A) + (B•E•A) T = B•C•D + B•C•E + A•B•D + A•B•E

The minimal cut sets of the top event are thus C1 = {B, C, D} C2 = {B, C, E} C3 = {A, B, D} C4 = {A, B, E}

57

Example

Thus if A = 1 if component A fails and 0 otherwise and this is true for B,C,D,E as well we can write

T = 1-(1- B*C*D)*(1- B*C*E)*(1- A*B*D)*(1- A*B*E)

And if T=1 we have system failure and T=0 indicates system is functioning

58

Example

59

X1

Determining Boolean Representation for Series-Parallel Systems

X2

X3

X4

X5

X6

X7

X8

X1

X2*X3 X4

X5

X6

X7

X8

60


X1

X2*X3 X4

X5

X6

X7

X8

X1

X2*X3 X4

X5

X6*X7*X8

X1

1-(1-X2*X3)*(1-X4)

X5

X6*X7*X8

61


X1

1-(1-X2*X3)*(1-X4)

X5

X6*X7*X8

X1 [1-(1-X2*X3)*(1-X4)]X5 X6*X7*X8

1-(1-X1)*(1-[1-(1-X2*X3)*(1-X4)]X5)*(1-X6*X7*X8)

62

2

34

5

System Indicator = 1 – (1-X1)(1-(1-(1-X2X3)(1-X4))X5)(1-X6X7X8) =1-(1-X1)(1-X2X3X5-X4X5+X2X3X4X5)(1-X6X7X8) =1-(1-X1)(1-X2X3X5)(1-X4X5)(1-X6X7X8) since for binary variables (X5)2= X5 Which is called min cut representation (no Xi

n terms)

1

6

7

8


63

2

34

5

What is min cut set representation? 1-(1-X1)(1-X2X3X5)(1-X4X5)(1-X6X7X8) Note that for the sets of components {1}, {2,3,5}, {4,5}, {6,7,8} if all of the items in the sets fail, then the system fails – a cut set Also not that we can not reduce any set by even a single element and have it still be a cut set – a min cut set

1

6

7

8


64

2

34

5

What is a min path? Note that for the sets of components {1,5,6}, {1,5,7} {1,5,8}, {1,2,4,6}, {1,2,4,7}, {1,2,4,8}, {1,3,4,6}, {1,3,4,7}, {1,3,4,8}, if all of the items in the sets function, then the system functions (a path from beginning to end – a path set Also not that we can not reduce any set by even a single element and have it still be a path set – a min path set

1

6

7

8


65

Boolean Representation for General Systems

Z=1-(1-X1X2)(1-X1X3X5)(1-X4X5)(1-X2X3X4)

1

2

3

4

5

Non series-parallel structures

Use cut set representation

66


As structures get more complex this becomes difficult and we may have to resort to a Fault Tree

Determine the min path and min cut sets

in out

A

B

C

D

E

F

G

H

67 Boolean Representation for General Systems

No Flow to Out

H No Flow to H

No Flow From F No Flow From G

GNo Flow to G

No Flow From A No Flow From D

+

+

F No Flow to F

+

No Flow From C No Flow From E

+

A No Flow From “in”

+

D No Flow to D

+

B No Flow From “in”

+

C No Flow From “in”

+

E No Flow to E

+

B No Flow From “in” We will discount

this in our analysis

68 Boolean Representation for General Systems

[A•(B+D)+F] •[C•(B+E)+G]+H

H [A•(B+D)+F] •[C•(B+E)+G]

A•(B+D)+F C•(B+E)+G

GC•(B+E)

A B+D

+

+

F A•(B+D)

+

C B+E

A +

D B

B

C +

E B

B

69


Failure = [A●(B+D)+F]●[C●(B+E)+G]+H = [A●B + A●D + F] ● [C●B + C●E + G]+H = A●B●B●C+ A●B●C●E + A●B●G + A●D●B●C+

A●D●C●E + A●D●G + F●B●C+ F●C●E + F●G +H = A●B●C+ A●B●C●E + A●B●G + A●B●C●D+ A●C●D●E + A●D●G + B●C●F+ C●E●F + F●G +H = A●B●C + A●B●G + A●C●D●E + A●D●G + B●C●F

+ C●E●F + F●G + H

Cut Set: {A,B,C}, {A,B,G}, {A,C,D,E}, {A,D,G}, {B,C,F}, {C,E,F},{F,G},{H}

Using our indicator notation

T=1-(1-A*B*C)*(1-A*B*G)*(1-A*C*D*E)*(1-A*D*G) *(1-B*C*F)*(1-C*E*F)*(1-F*G)*(1-H)

70 Quantifying Event Trees (Using DeMorgan’s Laws)

Assume split fractions are calculated using fault trees A=b+c•d B=c+e C=b•d

A+

G1 b

c d

B+

e c

C

b d

I A B C ABC Scenario 1

ABC Scenario 2 AB Scenario 3 A Scenario 4

71

Quantifying Event Trees

Scenario 4 I • A = I • (b+c•d) Scenario 3 I • A • B = I • (b•c+b•d) • (c+e) = I • (b•c•e + b•c•d + b•d•e) Scenario 2 I • A • B • C = I • (b•c+b•d) • (c+e) • (b•d) = I • (b•c+b•d) • (c•e) • (b•d) ={ } Scenario 1 I • A • B • C = I • (b•c+b•d) • (c•e) • (b•d) = I • (b•c+b•d) • (c•e) • (b+d) = I • b•c•e

72

Calculating the Probability of the Top Event

•  Three Methods •  Converting Cut Set Formulation to Probability

Statements •  Using Truth Tables •  Using Binary Decision Diagrams

73

Calculating the Probability of the Top Event - Method 1

•  Additive Law for Events A1,…, An

P(A1∪…∪An) = ∑i=1,n P(Ai) – ∑i<j P(Ai∩Aj) + ∑i<j<k P(Ai∩ Aj∩Ak) +… + (–1)n+1* P(A1∩…∩An)

You know P(A1∪A2) = P(A1) + P(A2) - P(A1 ∩A2)

The above general formula is called the Inclusion-Exclusion Principle (as terms are added you overestimate then underestimate)

for example ∑i=1,n P(Ai) – ∑i<j P(Ai∩Aj) ≤ P(A1∪…∪An) ≤ ∑i=1,n P(Ai)

74

Calculating the Probability of the Top Event - Method 1

If a fault tree has minimal cut sets C1, C2, …, Cm, then T = C1 + C2 + … + Cm P(T=1) = P({C1 =1}∪ {C2 =1}∪ … ∪ {Cm=1}) and we can calculate P(T=1) = ∑P(Ci =1) - ∑P({Ci =1}∩ {Cj=1}) + … + + (–1)m+1 ∑P({C1 =1}∩{C2 =1}∩ … ∩ {Cm=1}) P(Ci=1)=P({Xi1=1}∩…∩{Xini

=1}) and we can calculate bounds

∑P1(Ci ) – ∑P1(Ci ∩ Cj) < P1(T) < ∑P1(Ci) where we use the notation henceforth P1(C)=P({C=1}) and P1(Ci ∩ Cj) = P({Ci =1}∩ {Cj=1})

75

Calculating the Probability of the Top Event – Method 1

•  Rare Event Approximation P1(T) ≈ ∑P1(Ci) (conservative) - Based on the notion that the simultaneous

occurrence of several rare events is negligible - Problematic when there is a large degree of

overlap in cut sets - An additional simplifying assumption is the

independence of components

P1(Ci) = P({Xi1=1}∩…∩{Xini=1})

= P1(Xi1) • P1(Xi2) • … • P1(Xini)

assuming independence

76 Calculating the Probability of the Top Event – Method 1

P1(A) = P1(B) = P1(C) = 0.1 and A,B,C mutually indep. P1(T) = P1(C∪A∩B) ≈ P1(C) + P1(A∩B)

≈ P1(C) + P1(A)*P1(B) = 0.110 (Bound) P1(T) = P1(C) + P1(A∩B) – P1(A∩B∩C) = P1(C) + P1(A)*P1(B) – P1(A)*P1(B)*P1(C) = 0.109 (Exact)

Example 1

C1= {A,B}

C2 = {C}

77

Calculating Probability of Top Event: Truth Tables – Method 2

  0 denotes that component does not fail

  1 denotes that component fails

  P(T) = 0.081 + 0.009 + 0.009 + 0.009 + 0.001 = 0.109

  Note: Independence of components is assumed

A B C P(A)P(B)P(C) System

0 0 0 (.9)(.9)(.9) = 0.729 0 1 0 0 (.1)(.9)(.9) = 0.081 0 0 1 0 (.9)(.1)(.9) = 0.081 0 0 0 1 (.9)(.9)(.1) = 0.081 1 1 1 0 (.1)(.1)(.9) = 0.009 1 1 0 1 (.9)(.1)(.1) = 0.009 1 0 1 1 (.9)(.1)(.1) = 0.009 1 1 1 1 (.1)(.1)(.1) = 0.001 1

78

Calculating Probability of Top Event: Binary Decision Diagrams – Method 3

A

B B

C C C C

0 1

0 1 0 1

0 1 0 1 0 1 0 10 1 0 1 0 1 1 1

 Tree represents all possible component states   Bottom of tree represents the truth table value for the tree path.  There are techniques to reduce the tree.

79

Calculating Probability of Top Event: Binary Decision Diagrams – Method 3

A

B B

C C C C

0 1

0 1 0 1

0 1 0 1 0 1 0 10 1 0 1 0 1 1 1

The tree is basically a physical representation of the truth table

80

Calculating Probability of Top Event: BDD’s – Method 3

A

B B

C C C C

.9 .1

.9 .1 .9 .1

.9 .1 0 1 0 1 0 1 1 1

  Calculate probability of top event by replacing the states with their probabilities, and folding back the tree  For example, 0.19 = 0.1 * 0.9 + 1 * 0.1

.9 .1 .9 .1 .9 .1 .1 .1 .1 1

.19 .10

.109 =.9*.10 + .1*.19

.19 =.9 *.1 + .1 * 1 0.10= .9*.1 + .1*.1

81

Putting it All Together Example

Consider the event tree and fault trees below:

Determine a Boolean equation representing each event tree scenario in terms of fault tree basic events (C1, C2, C3).

I B A

82

Putting it All Together Example

a) If the frequency of the initiating event I is 10-3 per year, and P1(C1) = 0.001, P1(C2) = 0.008, and

P1 (C3) = 0.005, calculate the risk (injuries per year).

83

Example

  Solution a)  The Boolean equations representing each of the

event tree scenarios in terms of the fault tree basic events (C1, C2, C3) are:

Scenario 1:

84

Example

Scenario 2:

Scenario 3:

85

Example Solution

86

Example: Solution

That is the rate of I

≈ 7.95x10-6+6.00x10-6

87

Example Solution

Advanced Probability Analysis

89

Probability of System Failure: Law of Total Probability

•  Notation We use the event Ci (S) to denote that component i (the system) fails and Ci’ (S’) that it does not.

We also use the indicator Xi=1 (Z=1) to indicate that component i (the system) fails and Xi=0 (Z=0) to indicate that component i (the system) does not fail

Thus P(Ci)=Pr(Xi=1) and P(S)=Pr(Z=1)

1

2 3

Z=1-(1-X1*X2)*(1-X3)

90


•  Use probability laws P(S) = P([C1∩C2 ]∪C3) = P(C1∩C2) + P(C3) – P(C1∩C2∩C3)

•  Or condition on component states

P(S|C1∩C2∩C3) * P(C1∩C2∩C3) + P(S|C1´∩C2∩C3) * P(C1´∩C2∩C3) + P(S|C1∩C2´∩C3) * P(C1∩C2´∩C3) + P(S|C1∩C2∩C3´) * P(C1∩C2∩C3´) +

P(S|C1´∩C2´∩C3) * P(C1´∩C2´∩C3) + … … + P(S|C1´∩C2´∩C3´) * P(C1´∩C2´∩C3´)

1

2 3

Z=1-(1-X1*X2)*(1-X3)

This side will be 0 or 1

This side will be the probability of a component state

91


1

2 3

Z=1-(1-X1*X2)*(1-X3)

Assuming Independence

Using SUMPRODUCT function

P(C1∩C2´∩C3)

92

Advanced Probability Laws: Conditional Probability

•  Conditional Probability - P(A|B) = P(A ∩ B) / P(B) , if P(B) > 0 - Conditional probability redefines the sample

space

AB

New Sample Space

Elements of A in the New Sample Space

93

•  Conditioning on component 2 failure: P(S|C2) = P(S∩C2) / P(C2) = P({[C1∩C2 ]∪C3}∩C2} / P(C2) = P([C1∩C2 ]∪[C3∩C2 ]) / P(C2) = {P(C1∩C2) + P(C3∩C2) - P(C1∩C2∩C3)} / P(C2)

•  If components are independent: = {P(C1)P(C2) + P(C3)P(C2) - P(C1)P(C2)P(C3)} / P(C2) = P(C1) + P(C3) - P(C1)P(C3)

Probability of System Failure: Conditional Probability

1

2 3

If component 2 fails what is the probability of System Failure –Measure Component Importance

94

Probability of System Failure: Conditional Probability

1

2 3

Note: Independence NOT Assumed

If component i fails what is the probability of System Failure

95

Probability of Component Failure: Conditional Probability

1

2 3

Note: Independence NOT Assumed

If system fails, what is the probability of component i failure – Maintenance Implications

96

Probability of Cut Set Causing Failure: Conditional Probability

Z=1-(1-X1X2)(1-X1X3X5)(1-X4X5)(1-X2X3X4)

1

2 3

4

5

Non series-parallel structures

Cut Sets: {1,2}, {1,3,5}, {4,5}, {2,3,4}

Cut Set Representation

97

Calculating Complex Structure Functions and Probability of Failure

=Pr(CS12∩Z}/Pr{Z} = Pr(CS12}/Pr{Z}

98

Importance Measures

•  Motivation − A key challenge in a PRA is to identify the elements

in the system that contribute most to the risk − Method to accomplish this is Importance Ranking − The many importance measures used for this

process can be categorized as either  Absolute

 Defines each risk element in terms of an absolute risk metric, such as the conditional frequency of a hazard exposure given the state of the element; or

 Relative  Compares risk contribution of each element to

that of another

99

Importance Measures

•  Formulation − Risk is usually composed of a collection of

scenarios that occur with a certain frequency or probability

− A series of cut sets can represent these scenarios − Wall, et al. (2001), represent total risk by a linear

function of any single risk element:

R = aP + b

100

Importance Measures

R = aP + b where

R: total System Risk a: total contribution from cut sets that involve a

particular element P: total risk contribution from a particular element b: total contribution from cut sets that do not

involve a particular element

− Wall, et al.’s, method is only useful for investigating one-at-a-time sensitivity to risk elements

101

Principles of Importance Measures

•  IB = a , RP=1 – RP=0 •  IFV = aP/(aP+b) , (Rbase – RP=0)/Rbase •  IC = aP/(aP+b) , (Rbase – RP=0)/Rbase •  II = aP ,Rbase – RP=0 •  IRRW = aP , Rbase – RP=0 (differential method) •  IRRW = (aP+b)/b , Rbase/RP=0 (fraction method) •  IRAW = a(1-P) , RP=1 – Rbase (differential method) •  IRAW = (a+b)/(aP+b) , RP=1/Rbase (fraction method) •  DIM1 , (R/Pi)/(Σj=1,nR/Pj) • DIM 2 , aiPi/Σj=1,naiPi

102

Safety Systems: k-out-of-n Systems

Consider a system where the system will function if k-out-of-n of its components function or will fail is n-k+1 or more components fail

Usually these are of identical components, each with probability of failure p, then the probability of system failure is

1

2

3

2-out-of-3 System Min Cut Sets {1,2}, {1,3}, {2,3} Prob of Failures

Why?

103

Modelling Dependent Failures

•  What is dependent failure? - Let Ci be the event that component i fails and let

P(Ci) denote its probability  If we have n components and their failures are

independent, then P(C1∩C2∩ … ∩Cn) = P(C1)P(C2) … P(Cn)  If their failures are not independent, then this is

not a simple multiplication, we use the Multiplicative Law

P(C1∩C2∩ … ∩Cn) = P(C1) • P(C2│C1) * P(C3│C1∩C2) *…* P(Cn│C1∩C2 ∩ … ∩Cn-1)  The probabilities of n joint dependent events on

the left side are usually greater than the corresponding independent probabilities

104


•  Example

105


•  What are Common Cause Failures - CCFs are considered to be the collection of all

sources of dependency, especially between components, that are not known or are difficult to model explicitly. - CCFs have been shown by many studies to

contribute significantly to the overall unreliability of complex systems; - CCFs have no unique or universal definitions. - A fairly general definition is given by Mosleh as: A

CCF is a subset of dependent events in which two or more component fault states exist at the same time, or in a short time interval, and are direct results of a shared cause.

106


•  Modelling CCFs: Two Components - As CCFs have no explicit definition, their

probabilities are modelled as possible joint combinations of failures of components - Consider a system with two redundant components

A, B; then  P(A fails) = P(AI) + P(CAB)  AI denotes A fails separately  BI denotes B fails separately  CAB denotes A & B fail together by common

cause

107


•  Modelling CCFs: Three Components - Consider a system with three redundant

components A, B and C  The total failure probability of A can be

expressed in terms of its independent failure AI and its dependent failures as follows:  CAB, CAC denote that (A,B) & (A,C) fail

together by common cause  CABC denotes that (A,B,C) fail together by

common cause - Component A fails if any of the events above occur

P(A fails) = P(AI) + P(CAB) + P(CAC) + P(CABC)

108


•  Modelling CCFs: Min Cut Representation - The equivalent Boolean representation of total

failure of component A is AT = AI+CAB+CAC+CABC -  If the success criterion for the system is “2 out of 3

components A, B and C succeed,” then failure of the system can be represented by the following cut sets: -  {AI,BI}, {AI,CI}, {BI,CI}, {CAB}, {CAC}, {CBC}, {CABC} - Thus the Boolean representation of system failure

will be S = (AI•BI) + (AI•CI) + (BI•CI) + CAB + CAC + CBC + CABC (why not include(AI•BI•CI?)

109


•  Modelling CCFs; Probability Representation

-  If independence is assumed, only the first four terms of the Boolean expression are used, i.e., P(CAB) = P(CAC) = P(CBC) = P(CABC) = 0;

 Otherwise, applying the Rare Event Approximation results

P(System Failure) = P(any 2 or 3 components fail)

QS ≈ P(AI)P(BI) + P(CAB) + P(AI)P(CI) + P(CAC) + P(BI)P(CI) + P(CBC) + P(CABC)

110


- Assume that components A, B, and C are similar, and define

Qi = Probability of i simultaneous component failures due to common cause

- and write QS = P(System Failure)

= P(any 2 or 3 components fail) = P(AI)P(BI) + P(CAB) + P(AI)P(CI) + P(CAC) + P(BI)P(CI) + P(CBC) + P(CABC)

= 3(Q1)2 + 3(Q2) + (Q3)

111


-  In general for a k out of n system to fail there must be n-k+1 or more failures

Example

112


- Generally, models for common cause failure derive expressions for Qk for a system of size m, 1 ≤ k ≤ m in terms of total probability of component failure (Qt)

113

Probability Models for Time Dependent Analysis

114

Previous Lecture: A Snap Shot in Time

► Use probability laws P(S) = P([C1∩C2 ]∪C3) = P(C1∩C2) + P(C3) – P(C1∩C2∩C3)

► Or condition on component states

P(S|C1∩C2∩C3) * P(C1∩C2∩C3) + P(S|C1´∩C2∩C3) * P(C1´∩C2∩C3) + P(S|C1∩C2´∩C3) * P(C1∩C2´∩C3) + P(S|C1∩C2∩C3´) * P(C1∩C2∩C3´) +

P(S|C1´∩C2´∩C3) * P(C1´∩C2´∩C3) + … … + P(S|C1´∩C2´∩C3´) * P(C1´∩C2´∩C3´)

1

2 3

Z=1-(1-X1*X2)*(1-X3)

This side will be 0 or 1

This side will be the probability of a component state

115 Random Variables: Time Dependent Behavior

► Random variables are important for describing system behavior as a function of time: TS is system life length, Ti life length of component i P(TS ≤ t) = P({T1 ≤ t} ∪ {T2 ≤ t}) (series system) P (TS ≤ t) = P({T1 ≤ t} ∩ {T2 ≤ t}) (parallel system)

Note: {Ti ≤ t} defines our previous notation, Ci , for a fixed value t but as t varies the probability is a function of time

When T takes values in [0, ∞), it is called a lifetime variable (used in reliability and risk analysis)

116 Important Functions for Random Variables

► Probability Distribution f(x) = Pr{X=x} for X discrete (called

pmf) f(x)dx ≈ Pr{x<X<x+dx} for X continuous

(called pdf) ► Cumulative Distribution Function:

F(x) = P(X ≤ x) = ∑ i≤x f(i) for X discrete = ∫ 0

x f(u)du for X continuous ► Reliability (Survival) Function

R(x) = P(X>x) =1– F(x) [F(x) or S(x) is often used in place of R(x)]



► Failure Rate Function (Continuous rv Only)  h(x) = Lim dx→0P(X ≤ x+dx|X>x}/dx h(x)dx ≈ P(x<X ≤ x+dx|X>x}

Denotes instantaneous probability of failure

► Cumulative Failure Rate Function (Continuous RV Only) H(x) = ∫ 0

x h(u)du (continuous only) Denotes cumulative wear or exposure

119

Classic Failure Rate Curve

Note: i. life lengths said to follow a bathtub failure rate with three phases: infant mortality, chance failure and wear out ii. if h(x) is nondecreasing, constant,

nonincreasing we say that X is IFR, CFR, or DFR for Increasing, Constant or Decreasing Failure Rate

Failure Rate

time

Infant Mortality

Chance Failure

Wear Out Failure

120

Classic Failure Rate Curve

Note: i. In practice we often only use one phase of the curve

ii. There are example phenomena from each phase (DFR –software, CFR-electronics,

IFR-mechanical devices)

Failure Rate

time

Infant Mortality

Chance Failure

Wear Out Failure

121 Parametric Families of Distributions

► When a distribution f(x) can be indexed by a set of parameters, say Θ, whose specification completely determines the distribution we say

that f(x|Θ) is a parametric family. ► Important Properties

 Failure Rate Behavior  Distribution of Minimums (for series systems) TS = Min{T1, …, Tn}  Distribution of Sums (for cold backup or switching

systems) TS = T1 +…+ Tn

122

Which Parametric Family to Use?

► Look at the data histogram

123 Use of Parametric Families: System Reliability as a Function

of Time ► Component Life Lengths

T1~Wei(2,10), T2~Wei(1,5) assume independence

► System Life TS P(TS ≤ t) = P({T1 ≤ t} ∪ {T2 ≤ t}) (series system) = P({T1 ≤ t}) + P({T2 ≤ t}) - P({T1 ≤ t})P({T2 ≤ t}) = (1 - e–(t/10)2) + (1 - e–(t/5))

- (1 - e–(t/10)2) (1 - e–(t/5)) P(TS ≤ t) = P({T1 ≤ t} ∩ {T2 ≤ t}) (parallel system) = (1 - e–(t/10)2) (1 - e–(t/5))

124

Making Risk Time Dependent

Some times you are lucky and the system lifelength distribution has a closed form

1

….

n

1 … n

1

n

….

Series Parallel Cold Standby (perfect switch)

TS=min{T1,…,Tn} TS=max{T1,…,Tn} TS=T1+…+Tn

If Ti~ Exp(λi) then Ts~ Exp(∑i=1,nλι)

No Distribution for Ti leads to a known form distribution for TS

If Ti~ gamma(νi,α) then Ts~ gamma(∑i=1,nνi ,α)

If Ti~ normal(µi,σi2)

then s~normal(∑i=1,nµi,∑i=1,nσi2)

125

Analyzing Serries Systems

1 … n

TS=min{T1,…,Tn}

System Failure = 1 - Pr{TS>t}

= 1- Pr{ min{T1,…,Tn}>t}

= 1 - Pr{T1>t, …., Tn>t}

=1 - ∏i=1,n Pr{Ti>t}

if components are independent

=1 - ∏i=1,n [1-Fi(t)]

126

Analyzing Parallel Systems

TS=max{T1,…,Tn}

System Failure = Pr{TS ≤ t}

= Pr{ max{T1,…,Tn} ≤ t}

= Pr{T1 ≤ t, …., Tn ≤ t}

= ∏i=1,n Pr{Ti ≤ t}

if components are independent

= ∏i=1,n Fi(t)

1

….

n

127


► Use probability laws (Cuts Set Rep) P(TS<t) = P([{T1<t} ∩{T2 <t}] ∪{T3<t}) = P({T1<t}∩{T2<t}) + P({T3<t}) – P({T1<t} ∩ {T2<t} ∩ {T3<t})

► Or condition on component states P({T1<t} ∩ {T2<t} ∩ {T3>t}) + P({T1>t} ∩ {T2>t} ∩ {T3<t}) +P({T1<t} ∩ {T2>t} ∩ {T3<t}) + P({T1>t} ∩ {T2<t} ∩ {T3<t}) + P({T1<t} ∩ {T2<t} ∩ {T3<t}) Assuming independent components such that the CDF of

component i is Fi(t) = Pr{Ti≤t} yields

1

2 3

Z=1-(1-X1*X2)*(1-X3)

128


► Use probability laws (Cuts Set Rep) P(TS<t) = F1(t)F2(t) + F3(t) - F1(t)F2(t)F3(t)

► Or condition on component states P(TS<t) = F1(t)F2(t)R3(t) + R1(t)R2(t)F3(t) + F1(t)R2(t)F3(t)

+ R1(t)F2(t)F3(t) + F1(t)F2(t)F3(t)

where Ri(t) = 1 – Fi(t) = Pr{Ti > t}

1

2 3

Z=1-(1-X1*X2)*(1-X3)

129

Statistical Inference for Time Dependent Model Parameters

130 Making Models Reflect Reality: Classical Estimation

► What is an estimator?  Given an unknown parameter θ and a random

sample X1, ..., Xn from (X|θ), what are some estimators Θ for θ?

 They are functions of the random sample  Θ(X) = (1/n) ∑i=1,n Xi), ,  Θ(X) = max(X1, ..., Xn )  Θ(X) = 3, ......

 An estimator is a random variable with a probability distribution and an estimate is a realization of that

random variable.   What is a good estimator?

 Look at its pdf

131

Classical Estimation

► What is a good estimator?

Unbiassedness E[Θ] = θ Minimum Variance VAR(Θ) as small as possible

(there is a Cremer-Rao Lower Bound)

Consistency Θn→ θ as n→∞

132


► Main Parametric Estimators: Given a random sample X1, ..., Xn from f(X|θ), with unknown parameter(s), θ  Method of Moments (ok properties but easy to use) Θ(X) is obtained as the solution to 1. E[X|θ] = (1/n) ∑i=1,n Xi, (θ has dimension one)

2. E[X|θ] = (1/n) ∑i=1,n Xi, VAR[X|θ] = S2 (θ has dimension two) more equations for higher dimensions

Example Exponential E[X] = 1/λ ⇒λ = 1/x Gamma E[X] = ν/α , VAR[X]= ν/α2

⇒ ν=x2/S2, α=x/S2

_ ^

_ _

^ ^

~

133


► Main Parametric Estimators: Method of Least Squares Θ(X) is obtained as the solution to Min ∑i=1,n {F(X(i)|θ) - i/n}2 , X(i) is the ith smallest Xi value and F is a particular parametric family

0

Selected F(x|θ) ~

134


 Method of Maximum Likelihood (Best Properties)

Θ(X) is obtained as that which maximizes the likelihood function, a function essentially describing the probability of observing what was observed

By selecting the values for the parameter that maximize the likelihood function, we select the parameter values which maximize the probability of observing what we oberved

There are several forms of likelihood functions

135 Formulating the Likelihood Function – Complete Samples

The Likelihood has many forms, based on the data

► Complete Samples: a random sample X1, ..., Xn L(θ|X) = Πi=1,n f(Xi|θ)

X

X

X

X

X

Exact failure times observed

136 Formulating the Likelihood Function - Censoring

► Right Censored Samples: A life test with n items that stops after time t*, if r failures are observed, let the observed failure times be denoted X(r) = X(1), ..., X(r) in addition we know X(i) > t* for i > r, L(θ|X(r),t*) ={Πi=1,r f(X(i)|θ)}R(t*|θ)n-r

~

X X

( (

X

0 t*

137 Formulating the Likelihood Function-Censoring

► Left Censored Samples: A life test with n items that begins at t = 0 but we do not get to observe the condition of the items until after time t*. Let r items be observed to be failed at t* and let the observed failure times be denoted X(n-r) = X(r+1), ..., X(n) in addition, we know X(i) ≤ t* for i ≤ r. L(θ|X) = {Πi=r+1,n f(X(i)|θ)}F(t*|θ)r

X X

)

X

0 t*

)

138 Formulating the Likelihood Function-Censoring

► Interval Censored Samples: A life test with n items begins at time t = 0 but observation of the state of the items (failed or surviving) is only at fixed time points 0 = t0 < t1< …. < tk < tk+1 = ∞. The test is stopped at tk. Let Xi, i = 1, ...,k denote the number of items observed failed in [tk-1,tk], Xk+1 is the number still surviving at tk L(θ|X) ∝ Πi=1,k+1 [F(ti|θ) - F(ti-1|θ)]Xi

( )

0 t1 t2 t3 t4

( )

( )

( )

139 Formulating the Likelihood Function

► Or any mixture

X

0 t1 t2 t3 t4

( )

)

( (

L(θ|Data)∝[F(t2|θ)]*[F(t2|θ)-F(t1|θ)]* R(t3|θ)*R(t4|θ)*f(t1|θ) Usually for numeric reasons we take the natural log and maximize

140 Formulating the Likelihood Function

► Example:Consider the following failure time data from an exponential distribution

t1=5, t2=12, t3=26, t4>10, t5>17, t6<4, t7∈[5,10], t8∈[5,10], t9∈[11,16], t10∈[20,30],

L = f(5)*f(12)*f(26)*R(10)*R(17)*F(4) *[F[10)-F(5)]2*[F(16)-F(11)]*[F(30)-F(20)]

or

141

Maximum Likelihood Estimation

- Weibull Likelihood Plots

142

Bayesian Statistical Inference for Time Dependent Model

Parameters

143

► Law of Total Probability  Given an event B and a collection of events A1, …, An which are mutually exclusive (Ai ∩ Aj =∅) and collectively exhaustive (∪Aj =Ω) then P(B) = ∑j=1,nP(B ∩ Aj) = ∑j=1,nP(B | Aj)P(Aj)

►  Bayes Law   Given an event B and a collection of events A1, …, An which are mutually exclusive (Ai ∩ Aj =∅) and collectively exhaustive (∪Aj =Ω) then P(Ai|B) = P(B | Ai)P(Ai)/ ∑j=1,nP(B | Aj)P(Aj)

Bayesian Statistical Inference

144

► Random Variables and The Law of Total Probability and Bayes Law  When a problem uses a random variable and

specifies its parameters conditioned on some physical act X~f(x|θ) where

 Unconditional questions about X – Law of TP

For example


145

 Questions about θ given observations on X –Bayes Law

For example


146

 Example: Products are produced by three separate machines. Machine 1,2, and 3 produce defective products with probability .1, .05 and .02 respectively and account for 10%, 40% and 50% of the total products produced.

Then X is number of defects, X|p~Bin(n,p) where n is a sample size and


147

a. If a box of 10 product are randomly selected and we do not know which machine manufactured the products in the box, what is the probability of no defects?

Pr{X =0} = Pr{X 0|p=.10}Pr{p=.10}+Pr{X=0|p=.05}PR{p=.05}

+ Pr{X =0|p=.02} Pr{p=.02} = (.90)10(.10)+(.95)10(.40)+(.98)10(.50) = .6829


148

b. If 1 defect is found what is the probability that machine 1 produced the box?


149


150


Note: the spreadsheet works for a single observation but may be used sequentially for multiple observations

Example Pr{p=.1|X1=1,X2>3}

Pr{p=.1} Pr{p=.1|X1=1}

Pr{p=.1|X1=1,X2>3}

151


► Bayes Theorem: Continuous Analogue  THM: Let X and Θ be continuous random vectors

with joint probability density f(x,θ) . Let f(x|θ) and f(θ|x) be the corresponding conditional densities and f(θ) = ∫ f(x,θ)dx be the marginal density of Θ. Then

f(θ|x) = f(x|θ)f(θ)/{∫ f(x|θ)f(θ)d θ}

 Proof: if f(θ) > 0 and f(x) > 0, f(θ|x) = f(x,θ)/f(x) ⇒ f(θ|x) = f(x,θ)/{∫f(x|θ)f(θ)dθ} = f(x|θ)f(θ)/{∫ f(x|θ)f(θ)dθ}

152


► Principals of Bayesian Inference  Description of uncertainty is via probability,  Uncertainty about unknown parameters α, β, γ, etc

for statistical models is expressed via probability distributions for the parameters

 Given a model f(x|Θ) with unknown Θ, a distribution, g(θ) is specified using expert judgment. This is called the prior distribution for Θ and describes our uncertainty about Θ

153


 If we wish to make probability statements about the random variable X, taking into account our uncertainty for Θ, we may do so using the

law of total probability f(x) = ∫ f(x|θ)g(θ)dθ This distribution is called the predictive distribution for x.

 Thus Pr{X∈A} = ∫Af(x)dx if X is continuous or using an appropriate summation if X is discrete

154


 If data becomes available we update our uncertainty distribution for Θ using Bayes Theorem.

 We use the probability model to describe the form of the data as a function of the parameter. This is called the likelihood function.

 There are many forms of the likelihood function depending on the form of the data, however for a complete random sample X1,…, Xn from f(x|θ), the likelihood is given as

L(θ|x1,…, xn ) = Πi=1,n f(xi|θ)

155


 Given the data, x = x1,…, xn the updated distribution which describes the uncertainty for Θ is given by Bayes Theorem as

g(θ|x) = L(θ|x)g(θ)/{∫ L(θ|x)g(θ)dθ}

This is called the posterior distribution for Θ and it describes our uncertainty for Θ in light of the data.

~

~

~ ~

156


 If we wish to make probability statements about the random variable X, taking into account our uncertainty for Θ AFTER the a random sample is observed, we may do so using the law of total probability

f(x|x) = ∫f(x|θ)g(θ|x)dθ

This is called the predictive distribution for X after observing x.

~ ~

~

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BEFORE DATA AFTER DATA

OBSERVABLE

PARAMETER

Prior Predictive f(x)

Posterior Predictive f(x|x)

Prior g(θ)

Posterior g(θ|x)

~

~

158


• Example (Effect of Prior on Posterior) Failure Data: (12,10,15,5,8)

159


• Example (Effect of Prior on Posterior) Failure Data: (.1,.5,.3,.1,.2)

160 Bayesian Statistics: Defining the Prior

► Methods  Conjugate Priors  Noninformative Priors  Maximum Entropy Priors  Empirical Bayes Priors

161 Bayesian Statistics: Prior Selection

PRIOR ASSESSMENT

Access to Experts

Any Prior

Access to Data

Access to Partial Information Moments

Access to Computer

Conjugate Prior

Yes

No

Empirical Bayes Prior

Maximum Information Prior

Noninformative Prior

Yes

Yes

Yes

No

No

No

162

Example Bayes Analysis

The number of non serious accidents at a plant is given by a Poisson process with rate λ per year. However, since λ is unknown, a prior distribution is constructed. The gamma distribution with υ=1 and α=5 is selected. Given the above, what point estimate would you use for λ?

163


The number of non serious accidents at a plant is given by a Poisson process with rate l per year. However, since λ is unknown, a prior distribution is constructed. The gamma distribution with υ=1 and α=5 is selected.

What is the probability that λ is less than .1? What is the probability of more than 2 accidents per year?

164


For a year period, we observe 3 accidents. Plot the prior and posterior distribution of λ. In a year we observe 3 accidents, what is the probability of more than 2 accidents in the following year?

165


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