Post on 21-Apr-2020
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
Terminology and Background
• 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
Terminology and Background
• 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)
5
Terminology and Background
• 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
6
Terminology and Background
• 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
7
Elements of Risk Analysis
Risk Assessment
Risk Management
Risk Communication
National Research Council (1994)
8
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
10
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
11
Quantitative Risk Assessment
• Overview
12
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
13
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
•
14
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)
16
Introduction
Federal Highway Administration, 2006
Federal Aviation Administration, 2007
17
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
26
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
27
Fault Tree Construction
• Symbols Event Symbols Transfer Symbols
Basic Event
Undeveloped Event Transfer In
External Event Transfer Out
Intermediate Event
28
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
29
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
30
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
31
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
32
No Water Delivered When Needed
+ No Water Delivered from V1
V3 Fails to Remain Open
V5 Fails to Remain Open
No Water
from P2
No Water Delivered from P1 Branch
+ No Water Delivered from V1
V4 Fails to Remain Open
V2 Fails to Remain Open
a +
P2 Fails to Function
AC Fails
a
+
AC Fails
S Fails
+
V1 Fails to Remain Open
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
+ No Water Delivered from V1
V3 Fails to Remain Open
V5 Fails to Remain Open
Pumping Branches Fail
V4 Fails to Remain Open
V2 Fails to Remain Open
+
P1 Fails to Function
AC Fails
S Fails
V1 Fails to Remain Open
T1 Ruptures
P2 Fails to
Function
+
Fault Tree Example 3
+
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
Fault Tree Example 4
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
37
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
38
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
39
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
40
Event Tree Method Example 1
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
41
Event Tree Method Example 2
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
42
Event Tree Method
Split fractions are calculated using fault trees
Quantifying Scenario Analysis
44
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
46
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
47
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)
48
Representing Systems in Terms of Their Components
• 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
49
Representing Systems in Terms of Their Components
- 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
…..
…..
…..
50
Representing Systems in Terms of Their Components
• 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)
51
Representing Systems in Terms of Their Components
• 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
Determining Boolean Representation for Series-Parallel Systems
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
Determining Boolean Representation for Series-Parallel Systems
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
Determining Boolean Representation for Series-Parallel Systems
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
Determining Boolean Representation for Series-Parallel Systems
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
Determining Boolean Representation for Series-Parallel Systems
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
Boolean Representation for General Systems
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
Boolean Representation for General Systems
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
Probability of System Failure: Law of Total Probability
• 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
Probability of System Failure: Law of Total Probability
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
Modelling Dependent Failures
• Example
105
Modelling Dependent Failures
• 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 Dependent Failures
• 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 Dependent Failures
• 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 Dependent Failures
• 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 Dependent Failures
• 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
Modelling Dependent Failures
- 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
Modelling Dependent Failures
- In general for a k out of n system to fail there must be n-k+1 or more failures
Example
112
Modelling Dependent Failures
- 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)]
117 Important Functions for Random Variables
118 Important Functions for Random Variables
► 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
Making Risk Time Dependent
► 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
Making Risk Time Dependent
► 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
Classical Estimation
► 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
Classical Estimation
► 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
Classical Estimation
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
Bayesian Statistical Inference
145
Questions about θ given observations on X –Bayes Law
For example
Bayesian Statistical Inference
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
Bayesian Statistical Inference
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
Bayesian Statistical Inference
148
b. If 1 defect is found what is the probability that machine 1 produced the box?
Bayesian Statistical Inference
149
Bayesian Statistical Inference
150
Bayesian Statistical Inference
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
Bayesian Statistical Inference
► 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
Bayesian Statistical Inference
► 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
Bayesian Statistical Inference
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
Bayesian Statistical Inference
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
Bayesian Statistical Inference
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.
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156
Bayesian Statistical Inference
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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157
Bayesian Statistical Inference
BEFORE DATA AFTER DATA
OBSERVABLE
PARAMETER
Prior Predictive f(x)
Posterior Predictive f(x|x)
Prior g(θ)
Posterior g(θ|x)
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158
Bayesian Statistical Inference
• Example (Effect of Prior on Posterior) Failure Data: (12,10,15,5,8)
159
Bayesian Statistical Inference
• 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
Example Bayes Analysis
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
Example Bayes Analysis
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
Example Bayes Analysis