Mohan Madgulkar-Compensation Management/2006 Compensation Management Mohan Madgulkar.
Uai12 Mohan Pearl
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Graphical Models for Causal
Inference
Karthika Mohan and Judea Pearl
University of California, Los Angeles
August 14, 2012
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Introduction
Why do we need graphs?
Figure: Motivating Example
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Introduction
Figure: Motivating Example
Variables in the study:
Season
Sprinkler Rain
Wetness of pavement(Wet)
Slipperiness of
pavement(Slippery)
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Introduction
Figure: Motivating Example
# Variables Table size5 326 64
7 1288 2569 512
10 1, 02420 1, 048, 576
30 1, 073, 741, 824
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Introduction
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG Representation
Conditional ProbabilityDistributions
P(X1) : 2 P(X3|X1) : 4
P(X2|X1) : 4
P(X4|X2, X3) : 8
P(X5|X4) : 4
Total # of Table Entries = 22
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Graphs: Notations
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Adjacent Nodes
Root and Leaf Nodes Skeleton
Path
Kinship Terminology
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Graphs: Notations
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Chain X1X3X4X5X1X3X4X5
Fork X3X1X2
Collider X3X4X2
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Background Factors & Bi-directed Edges
UX UY
X Y
UX UY
(a) (b) (c)X Y X Y
Figure: (a) Causal Model with background factors (b) & (c) CausalModel with correlated background factors
Note: Figure (a) expresses the assumption: Ux Uy and Figure(b)& (c) express the assumption Ux Uy
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Decomposing joint distribution-P(V)
How would you decompose joint distribution P(V) into smallerdistributions?
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Decomposing joint distribution-P(V)
How would you decompose joint distribution P(V) into smallerdistributions?
By applying Chain rule
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Decomposing joint distribution-P(V)
How would you decompose joint distribution P(V) into smallerdistributions?
By applying Chain rule
Let X1, X2,...,Xn be any arbitrary ordering of nodes in a DAG.P(x1, x2,...,xn) =
jP(xj |x1,...,xj1)
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Decomposing joint distribution-P(V)
How would you decompose joint distribution P(V) into smallerdistributions?
By applying Chain rule
Let X1, X2,...,Xn be any arbitrary ordering of nodes in a DAG.P(x1, x2,...,xn) =
jP(xj |x1,...,xj1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
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Decomposing joint distribution-P(V)
How would you decompose joint distribution P(V) into smallerdistributions?
By applying Chain rule
Let X1, X2,...,Xn be any arbitrary ordering of nodes in a DAG.P(x1, x2,...,xn) =
jP(xj |x1,...,xj1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
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Markovian Parents
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Markovian Parents
X1:
X2:{X1}X3:{X1}X4:{X2, X3}X5:{X4}
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Markovian Parents
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Markovian Parents
X1:
X2:{X1}X3:{X1}X4:{X2, X3}X5:{X4}
P(x1, x2, x3, x4, x5) =P(x1)P(x2|x1)P(x3|x1)P(x4|x2, x3)P(x5|x4)
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Markov Compatibility
Let V ={x1, x2,...,xn}be the set of observed nodes and pai bethe Markovian parents ofxi. Then,P(v) =P(x1, x2,..,xn) =
i P(xi|pai).
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Markov Compatibility
Let V ={x1, x2,...,xn}be the set of observed nodes and pai bethe Markovian parents ofxi. Then,P(v) =P(x1, x2,..,xn) =
i P(xi|pai).
Definition (Markov Compatibility)
If a probability distribution P admits Markovian factorizationof observed nodes relative to DAG G, we say that Gand P areMarkov compatible.
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Markov Compatibility
Let V ={x1, x2,...,xn}be the set of observed nodes and pai bethe Markovian parents ofxi. Then,P(v) =P(x1, x2,..,xn) =
i P(xi|pai).
Definition (Markov Compatibility)
If a probability distribution P admits Markovian factorizationof observed nodes relative to DAG G, we say that Gand P areMarkov compatible.
Example
X Y P(X, Y)
1 1 0.2251 0 0.375
0 1 0.125
0 0 0.275
Markov Compatible DAGs:XYXY
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Testing Markov Compatibility
Given a DAG Gand distribution P, how can you conclude that
P and Gare compatible?
Parents shielding tests non-descendants predecessors
d-separation
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d-Separation
DefinitionLet X, Y and Z be disjoint sets in DAG G. X and Y ared-separated by Z (written (X Y|Z)G) if and only ifZ blocksevery path from a node in Xto a node in Y.
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d-Separation
DefinitionLet X, Y and Z be disjoint sets in DAG G. X and Y ared-separated by Z (written (X Y|Z)G) if and only ifZ blocksevery path from a node in Xto a node in Y.
A path p is said to be d-separated(or blocked) by a set of nodesZ if and only if :(1) p contains a chain im j or a fork i m j suchthat the middle node m is in Z, or(2) ...
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Example: d-Separation
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
X1
X4|
X2
, X3
X3 X5|X4 X1 X5|X4
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d-Separation
DefinitionLet X, Y and Z be disjoint sets in DAG G. X and Y ared-separated by Z (written (X Y|Z)G) if and only ifZ blocksevery path from a node in Xto a node in Y.
A path p is said to be d-separated(or blocked) by a set of nodesZ if and only if :(1) p contains a chain im j or a fork i m j suchthat the middle node m is in Z, or
(2) p contains an inverted fork (or collider) im such thatthe middle node m is not in Zand such that no descendant ofm is in Z.
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Example: d-Separation
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
X3 X2|X1
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
Case (a): X Y|
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
Case (a): X Y|Case (b): XY
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d-Separation
When is it impossible to d-separate 2 non-adjacent nodes X
and Y?Do we need to test all sets for possible separation?
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I d i h
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Inducing path
DefinitionPath between 2 nodes X and Y is termed inducing if everynon-terminal node on the path:(i) is a collider and(ii) an ancestor of either X or Y (or both)
X Y
Z1
Z2 Z2 Z3Z1
X Y
(a) (b)
Note: There are no separators for X and Y.
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Th Fi N St f C l A l i
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The Five Necessary Steps of Causal Analysis
Define Express the target quantity Q as property of themodel M.
Assume Express causal assumptions in structural or
graphical form.Identify Determine ifQ is identifiable.
Estimate Estimate Q if it is identifiable; approximate it, if itis not.
Test IfMhas testable implications
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A Mi i T i T t i C l C ti
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
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A Mi i T i T t i C l C ti
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
Input: Story
Question: What if? What is? Why?
Answers: I believe that...
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A Mi i T i Te t i C l C e ti
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q1: If the season is dry and the pavement is slippery, did itrain?
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q1: If the season is dry and the pavement is slippery, did itrain?A1: Unlikely, it is more likely that the sprinkler was ON with avery slight possibility that it is not even wet.
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q2: But what if we see that the sprinkler is OFF?
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q2: But what if we see that the sprinkler is OFF?A2: Then it is more likely that it rained.Without graphs,# of Table Entries = 32
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q3: Do you mean that if we actually turn the sprinkler ON, therain will be less likely?
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q3: Do you mean that if we actually turn the sprinkler ON, the
rain will be less likely?A3: No, the likelihood of rain would remain the same but thepavement would surely get wet.Without graphs,# of Table Entries = 32 * 32
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q4: Suppose we see that the sprinkler is ON and pavement iswet. What if the sprinkler were OFF?
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A Mini Turing Test in Causal Conversation
Figure: Turing Test
The Story
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Q4: Suppose we see that the sprinkler is ON and pavement is
wet. What if the sprinkler were OFF?A4: The pavement would be dry because the season is likely tobe dryWithout graphs, what would be the # of table entries?
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Interventions
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InterventionsQuery: Would the pavement be slippery if we make sure thatthethe sprinkler is on?
Compute: P(x5|do(x3))May be equivalently represented as:(a) P(x5|x3)(b) Px3 (x5)
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
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Interventions
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Compute: P(x5|do(x3))
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
P(v) =P(x1)P(x2|x1)P(x3|x1)P(x4|x2, x3)P(x5|x4)
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Interventions
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Compute: P(x5|do(x3))
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23
X
X4
X5
RAIN
SLIPPERY
WET
SPRINKLER
= ON
SEASON
Figure: DAG after intervention
P(v) =P(x1)P(x2|x1) P(x3|x1) P(x4|x2, x3)P(x5|x4)P(x1, x2, x4, x5|do(x3)) =P(x1)P(x2|x1)P(x4|x2, x3)P(x5|x4)
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Interventions
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Compute: P(x5|do(x3))
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23
X
X4
X5
RAIN
SLIPPERY
WET
SPRINKLER
= ON
SEASON
Figure: DAG after interventionP(v) =P(x1)P(x2|x1) P(x3|x1) P(x4|x2, x3)P(x5|x4)P(x5|do(x3)) =
x1,x2,x4
P(x1)P(x2|x1)P(x4|x2, x3)P(x5|x4)Note: P(x5|do(x3))=P(x5|x3) i.e. Doing = Seeing
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Examples
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p
X YX Y
U
Question : Can you estimate P(y|do(x)), given P(x, y)?
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Examples
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X YX Y
U
Question : Can you estimate P(y|do(x)), given P(x, y)?
NO!
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Examples
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X YX Y
U
Question : Can you estimate P(y|do(x)), given P(x, y)?
NO!
P(x, y) =
u P(x,y,u) =
u P(y|x, u)P(x|u)P(u)
P(y|do(x)) =
u P(y|x, u)P(u)
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Identifiability
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DefinitionLet Q(M) be any computable quantity of a model M.
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Identifiability
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DefinitionLet Q(M) be any computable quantity of a model M. We say
that Q is identifiable in a class Mof models if, for any pairs ofmodels M1 and M2 from M, Q(M1) =Q(M2) wheneverPM1 (v) =PM2 (v).
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Estimating causal effect
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Adjustment for direct causesCompute: P(y|x)
P(x , y , z, w) =P(y|x, w)P(x|z)P(w|z)P(z)P(y , z , w|do(x)) =P(y|x, w)P(w|z)P(z) P(x|z)
P(x|z)
P(y , z , w|do(x)) = P(x,y,z,w)P(x|z)
P(y|do(x)) =
z,wP(yw|x, z)P(z) =
zP(y|x, z)P(z)
X Y
Z
X Y
Z
WW
Figure: DAGs before and after intervention
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Theorem (Adjustment for direct causes)LetP Ai denote the set of direct causes ofXi and letY be anyset of variables disjoint of{Xi P ai}. The causal effect ofXionY is given by:
P(y|xi) =
pai P(y|xi, pai)P(pai)
whereP(y|xi, pai) andP(pai) represent pre-interventionalprobabilities.
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Example: Adjustment for direct causes
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Query: Would the pavement be slippery if we make sure thatthethe sprinkler is on?
P(x5|x3) =
x1P(x5|x3, x1)P(x1)
X1
X23
X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23
X
X4
X5
RAIN
SLIPPERY
WET
SPRINKLER
= ON
SEASON
Figure: DAG after intervention
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Estimating Causal Effect
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Compute: P(Xj |do(Xi))How can we find a set Zof concomitants that are sufficient foridentifying causal effect?
X1
Xi X6
X4X3
X2
X5
Xj
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Back-door Criterion for Identifiability
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Definition (Pearl-1993)
A set of variables Zsatisfies the back-door criterion relative toan ordered pair of variables (Xi, Xj) in a DAG G if:
(i) no node in Zis a descendant ofXi; and(ii) Zblocks every path between Xi and Xj that contains an
arrow into Xi.
P(xj|do(xi)) =
zP(xj |xi, z)P(z)
X1
Xi X6
X4X3
X2
X5
Xj
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Estimating causal effect: P(y|do(x))
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Can you adjust for direct cause?
ZX Y
U (Unobserved)
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Can you adjust for direct cause? NO!
ZX Y
U (Unobserved)
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Can you apply backdoor criterion?
ZX Y
U (Unobserved)
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Can you apply backdoor criterion? NO!
ZX Y
U (Unobserved)
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Estimating causal effect: P(y|do(x))
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Is P(y|do(x)) identifiable?
ZX Y
U (Unobserved)
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Estimating causal effect: P(y|do(x))
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Is P(y|do(x)) identifiable? YES!
ZX Y
U (Unobserved)
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Estimating causal effect: P(y|do(x))
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Given: P(y|x) =
zP(y|z)P(z|x)
ZX Y
U (Unobserved)
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Estimating causal effect: P(y|do(x))
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Given: P(y|x) =
zP(y|z)P(z|x)
P(z|x) =P(z|x)P(y|z) =
xP(y|x, z)P(x)
Therefore,
P(y|x) =
zP(z|x)
xP(y|x, z)P(x)
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Front-door Adjustment
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IfZsatisfies the front door criterion relative to (X, Y) and ifP(x, z)>0, then the causal effect ofX on Y is identifiable andis given by:
P(y|x) =
zP(z|x)
xP(y|x, z)P(x)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Estimating causal effect P(y|do(x))
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How can you syntactically derive claims about interventions?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Estimating causal effect P(y|do(x))
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How can you syntactically derive claims about interventions?
do-calculus
X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
Figure: Subgraphs ofG used in the derivation of causal effects.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
do-Calculus-[Pearl-1995]
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Rule-1Insertion or deletion of observations
P(y|x , z, w) =P(y|x, w) if (Y Z|X, W)GX
X Z Y X Z YX Z Y
X Z Y
U (Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
do-Calculus-[Pearl-1995]
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Rule-2Action/Observation exchange
P(y|x, z, w) =P(y|x , z, w) if (Y Z|X, W)GXZ
X Z Y X Z YX Z Y
X Z Y
U (Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Deriving causal effect using do-calculus
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X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
Compute: P(y|z)
P(y|z) =
x P(y|x, z)P(x|z)P(x|z) =P(x) since (Z X)GZ
P(y|x, z) =P(y|x, z) since (Z Y|X)GZP(y|z) =
x P(y|x, z)P(x)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Prove: P(y|x) =
zP(y|z)P(z|x)
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X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Prove: P(y|x) =
zP(y|z)P(z|x)
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X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
P(y|x) =
z
P(yz|x) =
z
P(y|xz)P(z|x)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Prove: P(y|x) =
zP(y|z)P(z|x)
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X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
P(y|x) =
z
P(yz|x) =
z
P(y|xz)P(z|x)P(y|zx) =P(y|zx) since Y Z in GXZ
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Prove: P(y|x) =
zP(y|z)P(z|x)
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X Z Y X Z YX Z Y
X Z Y
U(Unobserved)
X Z Y
Z X
Z XZXZG G G
G = GG
P(y|x) =
z
P(yz|x) =
z
P(y|xz)P(z|x)P(y|zx) =P(y|zx) since Y Z in GXZ
=P(y|z) since Y X in GZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Graphical Models in which P(y|x) is Identifiable
ZZ
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X
Y
(a)
Y
X
(d)
X
Y
X
Y
Z
1
Z2
Y
(g)
Z3
XX
Y
Z
(e)
X
Y
Z
(b) (c)
Z1
Z2
(f)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Graphical Models in which P(y|x) is notIdentifiable
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(e)
Z
X
Y
Z1Z2
X
Y
(g)(f)
Y
Z
X
Y
(h)
Z
W
X
X
Y
(a) (b)
X
Y
Z
X
Y
Z
(c) (d)Y
Z
X
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
C-components and C-factor
bl d b h C f h
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Two variables are said to be in the same C-component if they
are connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W
1
2Z
2
W
X Y
T
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
C-components and C-factor
T i bl id b i h C if h
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Two variables are said to be in the same C-component if they
are connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W
1
2Z
2
W
X Y
T
S1={X , Y , W 1, W2}S2={Z1}S3={Z2}S4={T}
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
C-components and C-factor
T i bl id t b i th C t if th
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Two variables are said to be in the same C-component if they
are connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W
1
2Z
2
W
X Y
T
S1={X , Y , W 1, W2}S2={Z1}S3={Z2}S4={T}
C-factor: Q[Si](v) =Pv\si(si)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Identifiability of C-factor
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Lemma (Tian & Pearl, 2002)
Let a topological order overV beV1 < V2 < ... < Vn and letV(i) ={V1, V2,...,Vi}, i= 1,...,nandV
(0) =. For any setC,letGCdenote the subgraph ofG composed only of variables in
C. Then:(i) Each C-factorQj , j = 1,...,k is identifiable and is given by:
Qj =
{i:ViSj}P(vi|v
(i1))
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Example: Identifiability of C-factor
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X1
X2 X3 X
4
Y
Admissible order: X1 < X2 < X3 < X4 < YQ1=P(x4|x1, x2, x3)P(x2|x1)Q2=P(y|x1, x2, x3, x4)P(x3|x1, x2)P(x1)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Necessary and Sufficient condition foridentifiability ofPx(v)
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Theorem (Tian & Pearl, 2002)
LetXbe a singleton. Px(v) is identifiable if and only if there isno bi-directed path connectingX to any of its children.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Necessary and Sufficient condition foridentifiability ofPx(v)
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Theorem (Tian & Pearl, 2002)
LetXbe a singleton. Px(v) is identifiable if and only if there isno bi-directed path connectingX to any of its children. When
Px(v) is identifiable, it is given by:
Px(v) = P(v)QX
x Q
X,
whereQX is the c-factor corresponding to the c-componentSX
that containsX.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Example: Necessary and Sufficient condition foridentifiability ofPx(v)
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X1
X2 X3 X
4
Y
Admissible order: X1 < X2 < X3 < X4 < YQ1=P(x4|x1, x2, x3)P(x2|x1)Q2=P(y|x1, x2, x3, x4)P(x3|x1, x2)P(x1)Px1 (x2, x3, x4, y) =Q1
x1
Q2
=P(x4|x1, x2, x3)P(x2|x1)x1
P(y|x1, x2, x3, x4)P(x3|x1, x2)P(x1)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Causal Effect Identifiability
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Identification ofPx(y|z) where X Y Z= and X is notnecessarily a singleton, [Shpitser & Pearl,2006]
Hedge Criterion IDC - Sound and Complete Algorithm
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Counterfactuals
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Query: Would the prisoner be dead had rifleman A not shot
him, given that the prisoner is dead and rifleman A shot him?
C (Captain)
U (Court order)
B (Riflemen)A
D (Death)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Counterfactuals
Q ld h b d d h d fl A h
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Query: Would the prisoner be dead had rifleman A not shot
him, given that the prisoner is dead and rifleman A shot him?
C (Captain)
U (Court order)
B (Riflemen)A
D (Death)
Abduction
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Counterfactuals
Q W ld h i b d d h d ifl A h
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Query: Would the prisoner be dead had rifleman A not shot
him, given that the prisoner is dead and rifleman A shot him?
C (Captain)
U (Court order)
B (Riflemen)A
D (Death)
Abduction Intervention
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Counterfactuals
Q W ld th i b d d h d ifl A t h t
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Query: Would the prisoner be dead had rifleman A not shot
him, given that the prisoner is dead and rifleman A shot him?
C (Captain)
U (Court order)
B (Riflemen)A
D (Death)
Abduction Intervention
Prediction
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Counterfactuals
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Query: Would the prisoner be dead had rifleman A not shothim, given that the prisoner is dead and rifleman A shot him?
C (Captain)
U(Court order)
B (Riflemen)A
D (Death)
Model M
C=UA= C
B =CD=A BFacts: D
Conclusions:
U,A,B,C,D
Model MAC=UA
B=CD=A BFacts: UConclusions:
U, A, B, C, D
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Markov Equivalence
Given 2 models is there a test that would tell them apart?
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Given 2 models, is there a test that would tell them apart?
DefinitionTwo graphs G1 and G2 are said to be Markov equivalent ifevery d-separation condition in one also holds in the other. .
A
B
Z
1G( )
A
B
Z
G( )2
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Markov Equivalence
Given 2 models is there a test that would tell them apart?
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Given 2 models, is there a test that would tell them apart?
DefinitionTwo graphs G1 and G2 are said to be Markov equivalent ifevery d-separation condition in one also holds in the other. .
Are these DAGs Markov Equivalent?
Z1
W1
2Z
2W
X Y
T
Z1 2
Z2
WW1
X Y
T
Hard to enumerate all separation conditions.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Observational Equivalence
Theorem (Verma & Pearl 1990)
Two DAGs are observationally equivalent iff they have the same
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y q ff y
sets of edges and the same sets of v-structures, that is, twoconverging arrows whose tails are not connected by an arrow.
X1
X23
X
X4
X5
X1
X23
X
X4
X5
Figure: Observationally Equivalent DAGs
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Markov Equivalence and ObservationalEquivalence
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If two DAGs are Markov Equivalent, then they areObservationally Equivalent as well. True/False?
True if all variables are observed(i.e. no bi-directed edges) andFalse otherwise.
K thik M h d J d P l G hi l M d ls f C s l I f
Markov Equivalence and ObservationalEquivalence
If two DAGs are Markov Equivalent, then they are
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q , y
Observationally Equivalent as well. True/False?True if all variables are observed(i.e. no bi-directed edges) andFalse otherwise.
X T Z YX T Z Y
Figure: DAGs that are Markov Equivalent but not ObservationallyEquivalent
How would you distinguish between the two?
Verma Constraints (Refer slide:115)
K thik M h d J d P l G hi l M d l f C l I f
Ancestral Graphs
Definition (Ancestral Graphs)
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A graph which may contain directed or bi-directed edges isancestral if:(i) there are no directed cycles(ii) whenever there is an edge XY, then there is nodirected path from X to Y or from Y to X.
Z1 2
Z2
WW1
X Y
T
Figure: Ancestral graph
Z1
W1
2Z
2W
X Y
T
Figure: Notan Ancestral graph
K thik M h d J d P l G hi l M d l f C l I f
Maximal Ancestral Graphs (MAGs)
Definition (Spirtes & Richardson, 2002)
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Definition (Spirtes & Richardson, 2002)
An ancestral graph is said to be maximal if, for every pair ofnon-adjacent nodes X, Ythere exists a set Zsuch that X and
Yare d-separated conditional on Z.
Z1
W1
2Z
2W
X Y
T
Z1 2
Z
W1 2W
X Y
T
Figure: DAG and its corresponding MAG
K thik M h d J d P l G hi l M d l f C l I f
Construction of a MAG
Given : DAG G
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Given: DAG G
Step-1: Construct a graph Mcomprising of:(i) all nodes in G(ii) all uni-directional edges in G
Z1
W1
2Z
2W
X Y
T
Figure: DAG G
Z1
W1
2Z
2W
X Y
T
Figure: Graph M
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Construction of a MAG
Step-3: For every pair of non-adjacent nodes A and B in G,
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connected by an inducing path ,(i) add A B to M ifA is an ancestor ofB in G(ii) add AB to M ifB is an ancestor ofAin G(iii) add AB to Mif (i) and (ii) do not hold true
Z1
W1
2Z
2W
X Y
T
Figure: DAG G
Z1 2
Z
W1 2
W
X Y
T
Figure: MAGM
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Markov Equivalence
Theorem
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Two graphsG1 andG2 are said to be Markov equivalent if theirMAGs are Markov Equivalent
Are these MAGs Markov Equivalent?
Z1 2
Z2
WW1
X Y
T
Z1 2
Z
W1 2
W
X Y
T
Note: Markov Equivalence in MAGs are easier to check
Complete criterion for determining Markov Equivalence of 2MAGs: [Ali, Richardson and Spirtes, 2009]
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Reversing an edge in a MAG
Definition (Screened Edge)
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Definition (Screened Edge)
[Tian,2005] An edge XY is a screened edge in a MAG ifP a(Y) =P a(X) {X} and Sp(Y) =Sp(X)1.
X YZ
W
Figure: MAG with Screened Edge:XY
1Nodes X and Yare spouses, if they are connected by a bi-directed edge.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Reversing an edge in a MAG
Theorem (Tian,2005)
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LetMbe a MAG with edgeXY andM
be a graph withedgeXY, otherwise identical to M. ThenM is a MAGthat is Markov Equivalent to M if and only ifXY is ascreened edge inM.
X YZ
W
X YZ
W
Figure: Markov Equivalent MAGs
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Confounding Equivalence
Definition (Pearl and Paz,2009)
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Define two sets, T and Z as c-equivalent (relative to X and Y),written T Z, if the following equality holds for every x and y:
t P(y|x, t)P(t) =
zP(y|x, z)P(z) x, y
W1 2
W
V1 2
V
X Y
Examples:
T ={W1, V2} Z={W2, V1}
T ={W1, V1} Z={W2, V2}
T ={W1, W2} Z={W1}
T ={W1, W2} Z={W2}
Note: C-equivalence is testable
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Necessary and Sufficient Condition forC-Equivalence
Theorem (Pearl and Paz, 2009)
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Let Z and T be two sets of variables containing no descendantof X. A necessary and sufficient condition for Z and T to bec-equivalent is that at least one of the following conditions hold:
X (Z T) | (Z T) or
Z and T are G-admissible2
W1 2
W
V1 2
V
X Y
Examples:
T ={W1, V2} Z={W2, V1}
T ={W1, V1} Z={W2, V2} T ={W1, W2} Z={W1}
T ={W1, W2} Z={W2}
2satisfies back-door criterion
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Linear Models and Causal Diagrams
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Assume all variables are normalized to have zero mean and unitvariance.
ZX
Y
W
a
b
c
Z=e1W =e2X=aZ+ e3
Y =bW+ cX+ e4Cov(e1, e2) = = 0Cov(e2, e3) == 0Cov(e3, e4) == 0
Which parameters can be identified?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Vanishing Regression Coefficient
Definition
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DefinitionFor any linear model for a causal diagram D that may includecycles and bi-directed arcs, the partial correlation XY.Z mustvanish if and only if node X is d-separated from node Y by thevariables of Z in D [Spirtes et al., 1997b].
Z1
W1
2Z
2W
X Y
T rTX.W1Z1 = 0
Find more
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Single Door Criterion for Direct Effects
TheoremLetG be any path diagram in which is the path coefficient
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associated with linkXY and letG denote the diagram thatresults whenXY is deleted fromG. The coefficient isidentifiable if there exists a set of variablesZsuch that :(i) Zcontains no descendant ofY and(ii) Z d-separatesX fromY inG
Moreover, ifZ satisfies these two conditions, then is equal tothe regression coefficientrY X.Z.
G
Z YX
G
Z YX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Instrumental Variables (IV)DefinitionA variable Zis an instrument relative to a cause X and aneffect Y if:
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Z is independent of all error terms that have an influenceon Y when X is held constant, and
Z is notindependent of X.
In linear systems, Causal effect ofX on Y = rZYrZX
UY
UY
UW
UYUY
Y
(b)
X
Z
Y
(c)
X
Z
Y
(d)
X
Z
W
Y
(a)
X
Z
Figure: Z is an instrument in (a), (b) and (c) but not in (d)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Conditional Instrumental Variable
Definition (Brito & Pearl, 2002)
Zis an instrumental variable ifa set Wsuch that:
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Wcontains only non-descendants ofY W d-separates Z from Y in the sub-graph G obtained by
removing the edge XY W does not d-separate Z from X in G
Z X Y
W
Z X Y
W
Figure: Graph Gand corresponding subgraph G
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Conditional Instrumental Variable
Z is a conditional instrumental variable. Hence,
= Causal effect ofX on Y = rZY.WrZX.W
Wdoes not satisfy single-door criterion. So, cannot be
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y g ,identified using single-door.
Z X Y
W
Z X Y
W
Figure: Graph Gand corresponding subgraph G
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Verma Constraints([Tian and Pearl, 2002])
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A B C D
(a) (b)
A B C D
Q[{B, D}] =
u P(b|a, u)P(d|c, u)P(u)Pv\d(d) =
u P(d|c, u)P(u)
Also,Q[{B, D}] =P(d|a,b,c)P(b|a)Pv\d
(d) =
bP(d|a,b,c)P(b|a)
b P(d|a,b,c)P(b|a) isindependent of a.
Q[{B, D}] =
u P(b|a, u)P(d|a,c,u)P(u)Pv\d(d) =
u P(d|a,u,c)P(u)
Also,Q[{B, D}] =P(d|a,b,c)P(b|a)Pv\d
(d) =
bP(d|a,b,c)P(b|a)
b P(d|a,b,c)P(b|a) isnotindependent of a.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
Conclusions
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Graphs are indispensable for:
encoding causal knowledge
identifying parameters and causal effects
identifying testable implications
Go ahead and Exploit the Power of Graphs!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Thank You!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference