graphical models for causal inference - uclaftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf ·...
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Graphical Models for Causal
Inference
Karthika Mohan and Judea Pearl
University of California, Los Angeles
August 14, 2012
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Introduction
Why do we need graphs?
Figure: Motivating Example
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 3: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/3.jpg)
Introduction
Figure: Motivating Example
Variables in the study:
I Season
I Sprinkler
I Rain
I Wetness of pavement(Wet)
I Slipperiness ofpavement(Slippery)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Introduction
Figure: Motivating Example
# Variables Table size5 326 647 1288 2569 51210 1, 02420 1, 048, 57630 1, 073, 741, 824
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Introduction
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG Representation
Conditional ProbabilityDistributions
I P (X1) : 2
I P (X3|X1) : 4
I P (X2|X1) : 4
I P (X4|X2, X3) : 8
I P (X5|X4) : 4
Total # of Table Entries = 22
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Graphs: Notations
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
I Adjacent Nodes
I Root and Leaf Nodes
I Skeleton
I Path
I Kinship Terminology
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Graphs: Notations
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Chain X1 → X3 → X4 → X5
X1 → X3 → X4 → X5
Fork X3 ← X1 → X2
Collider X3 → X4 ← X2
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 q Uy and Figure(b)& (c) express the assumption Ux 6 qUy
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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) =
∏j P (xj |x1, ..., xj−1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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) =
∏j P (xj |x1, ..., xj−1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 11: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/11.jpg)
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) =
∏j P (xj |x1, ..., xj−1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 12: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/12.jpg)
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) =
∏j P (xj |x1, ..., xj−1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 13: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/13.jpg)
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) =
∏j P (xj |x1, ..., xj−1)
Is it possible that conditional probability of some variable Xj isnot sensitive to all its predecessors?
Yes!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 14: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/14.jpg)
Markovian Parents
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Markovian ParentsX1 : φX2 : {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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markovian Parents
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: Bayesian Networkrepresenting dependencies
Markovian ParentsX1 : φX2 : {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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Compatibility
Let V = {x1, x2, ..., xn} be the set of observed nodes and pai bethe Markovian parents of xi. 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 G and P areMarkov compatible.
Example
X Y P (X,Y )
1 1 0.225
1 0 0.375
0 1 0.125
0 0 0.275
Markov Compatible DAGs:X → YX ← Y
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Compatibility
Let V = {x1, x2, ..., xn} be the set of observed nodes and pai bethe Markovian parents of xi. 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 G and P areMarkov compatible.
Example
X Y P (X,Y )
1 1 0.225
1 0 0.375
0 1 0.125
0 0 0.275
Markov Compatible DAGs:X → YX ← Y
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Compatibility
Let V = {x1, x2, ..., xn} be the set of observed nodes and pai bethe Markovian parents of xi. 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 G and P areMarkov compatible.
Example
X Y P (X,Y )
1 1 0.225
1 0 0.375
0 1 0.125
0 0 0.275
Markov Compatible DAGs:X → YX ← Y
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
![Page 19: Graphical Models for Causal Inference - UCLAftp.cs.ucla.edu/pub/stat_ser/uai12-mohan-pearl.pdf · Graphical Models for Causal Inference Karthika Mohan and Judea Pearl University of](https://reader033.vdocument.in/reader033/viewer/2022042114/5e917d37322ed1379620ff78/html5/thumbnails/19.jpg)
Testing Markov Compatibility
Given a DAG G and distribution P , how can you conclude thatP and G are compatible?
I Parents shielding testsI non-descendantsI predecessors
I d-separation
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Testing Markov Compatibility
Given a DAG G and distribution P , how can you conclude thatP and G are compatible?
I Parents shielding testsI non-descendantsI predecessors
I d-separation
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 q Y |Z)G) if and only if Z blocksevery path from a node in X to 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 i→ m→ j or a fork i← m→ j suchthat the middle node m is in Z, or(2) ...
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 q Y |Z)G) if and only if Z blocksevery path from a node in X to 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 i→ m→ j or a fork i← m→ j suchthat the middle node m is in Z, or(2) ...
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: d-Separation
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
I X1 qX4|X2, X3
I X3 qX5|X4
I X1 qX5|X4
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 q Y |Z)G) if and only if Z blocksevery path from a node in X to 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 i→ m→ j or a fork i← m→ j suchthat the middle node m is in Z, or(2) p contains an inverted fork (or collider) i→ m← such thatthe middle node m is not in Z and such that no descendant ofm is in Z.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: d-Separation
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
I X3 qX2|X1
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
Case (a): X q Y |φCase (b): X��qY
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
Case (a): X q Y |φ
Case (b): X��qY
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: d-Separation
Z2 Z3Z1X Y Z2 Z3Z1
(b)
X Y
(a)
Case (a): X q Y |φCase (b): X��qY
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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d-Separation
When is it impossible to d-separate 2 non-adjacent nodes Xand Y ?Do we need to test all sets for possible separation?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 Z3Z1X Y
(a) (b)
Note: There are no separators for X and Y .
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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 orgraphical form.
Identify Determine if Q is identifiable.
Estimate Estimate Q if it is identifiable; approximate it, if itis not.
Test If M has testable implications
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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A “Mini” Turing Test in Causal Conversation
Figure: Turing Test
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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...
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Q1: If the season is dry and the pavement is slippery, did itrain?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
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.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Q3: Do you mean that if we actually turn the sprinkler ON, therain will be less likely?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Q3: Do you mean that if we actually turn the sprinkler ON, therain 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
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Q4: Suppose we see that the sprinkler is ON and pavement iswet. What if the sprinkler were OFF?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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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
Q4: Suppose we see that the sprinkler is ON and pavement iswet. 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?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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InterventionsQuery: Would the pavement be slippery if we make sure thatthe the sprinkler is on?Compute: P (x5|do(x3))May be equivalently represented as:(a) P (x5|x3)(b) Px3(x5)
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Interventions
Compute: P (x5|do(x3))
X1
X23X
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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Interventions
Compute: P (x5|do(x3))
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23X
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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Interventions
Compute: P (x5|do(x3))
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23X
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 (x5|do(x3)) =
∑x1,x2,x4
P (x1)P (x2|x1)P (x4|x2, x3)P (x5|x4)Note: P (x5|do(x3)) 6= P (x5|x3) i.e. Doing 6= Seeing
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Examples
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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Examples
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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Examples
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)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Identifiability
DefinitionLet Q(M) be any computable quantity of a model M .
We saythat Q is identifiable in a class M of models if, for any pairs ofmodels M1 and M2 from M , Q(M1) = Q(M2) wheneverPM1(v) = PM2(v).
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Identifiability
DefinitionLet Q(M) be any computable quantity of a model M . We saythat Q is identifiable in a class M of models if, for any pairs ofmodels M1 and M2 from M , Q(M1) = Q(M2) wheneverPM1(v) = PM2(v).
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect
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,w P (yw|x, z)P (z) =∑
z P (y|x, z)P (z)
X Y
Z
X Y
Z
WW
Figure: DAGs before and after intervention
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Theorem (Adjustment for direct causes)
Let PAi denote the set of direct causes of Xi and let Y be anyset of variables disjoint of {Xi ∪ Pai}. The causal effect of Xi
on Y is given by:
P (y|xi) =∑
paiP (y|xi, pai)P (pai)
where P (y|xi, pai) and P (pai) represent pre-interventionalprobabilities.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: Adjustment for direct causes
Query: Would the pavement be slippery if we make sure thatthe the sprinkler is on?
P (x5|x3) =∑
x1P (x5|x3, x1)P (x1)
X1
X23X
X4
X5
SPRINKLER
SEASON
RAIN
WET
SLIPPERY
Figure: DAG before intervention
X1
X23X
X4
X5
RAIN
SLIPPERY
WET
SPRINKLER= ON
SEASON
Figure: DAG after intervention
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating Causal Effect
Compute: P (Xj |do(Xi))How can we find a set Z of concomitants that are sufficient foridentifying causal effect?
X1
Xi X6
X4X3
X2
X5
Xj
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Back-door Criterion for Identifiability
Definition (Pearl-1993)
A set of variables Z satisfies the back-door criterion relative toan ordered pair of variables (Xi, Xj) in a DAG G if:
(i) no node in Z is a descendant of Xi; and(ii) Z blocks every path between Xi and Xj that contains an
arrow into Xi.P (xj |do(xi)) =
∑z P (xj |xi, z)P (z)
X1
Xi X6
X4X3
X2
X5
Xj
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Can you adjust for direct cause?
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Can you adjust for direct cause? NO!
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Can you apply backdoor criterion?
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Can you apply backdoor criterion? NO!
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Is P (y|do(x)) identifiable?
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
I Is P (y|do(x)) identifiable? YES!
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
Given: P (y|x) =∑
z P (y|z)P (z|x)
ZX Y
U (Unobserved)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect: P (y|do(x))
Given: P (y|x) =∑
z P (y|z)P (z|x)P (z|x) = P (z|x)P (y|z) =
∑x′ P (y|x′, z)P (x′)
Therefore,P (y|x) =
∑z P (z|x)
∑x′ P (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 Criterion for Identifiability
Definition (Pearl-1995)
A set of variables Z satisfies the front-door criterion relative toan ordered pair of variables (Xi, Xj) in a DAG G if:
(i) Z intercepts all directed paths from X to Y ; and(ii) there is no unblocked back-door path from X to Z; and(iii) all back-door paths from Z to Y are blocked by X
1
3
2
Z
Z
Z
X Y
U (Unobserved)
Figure: Frontdoor criterion is satisfied by Z = {Z1, Z2, Z3}
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Front-door Adjustment
If Z satisfies the front door criterion relative to (X,Y ) and ifP (x, z) > 0, then the causal effect of X on Y is identifiable andis given by:
P (y|x) =∑
z P (z|x)∑
x′ P (y|x′, z)P (x′)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect P (y|do(x))
How can you syntactically derive claims about interventions?
I 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 of G used in the derivation of causal effects.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Estimating causal effect P (y|do(x))
How can you syntactically derive claims about interventions?
I 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 of G used in the derivation of causal effects.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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do-Calculus-[Pearl-1995]
Rule-1 Insertion or deletion of observationsP (y|x, z, w) = P (y|x, w) if (Y q 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
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do-Calculus-[Pearl-1995]
Rule-2 Action/Observation exchangeP (y|x, z, w) = P (y|x, z, w) if (Y q 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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do-Calculus-[Pearl-1995]
Rule-3 Insertion or deletion of actionsP (y|x, z, w) = P (y|x, w) if (Y q Z|X,W )G
X,Z(W )
where Z(W ) is the set of Z nodes that are not ancestors of anyW node in 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
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Deriving causal effect using 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
Compute: P (y|z)P (y|z) =
∑x P (y|x, z)P (x|z)
P (x|z) = P (x) since (Z qX)GZ
P (y|x, z) = P (y|x, z) since (Z q Y |X)GZ
P (y|z) =∑
x P (y|x, z)P (x)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Prove: P (y|x) =∑
z P (y|z)P (z|x)
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 q Z in GXZ
= P (y|z) since Y qX in GZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Prove: P (y|x) =∑
z P (y|z)P (z|x)
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 q Z in GXZ= P (y|z) since Y qX in GZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Prove: P (y|x) =∑
z P (y|z)P (z|x)
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 q Z in GXZ
= P (y|z) since Y qX in GZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Prove: P (y|x) =∑
z P (y|z)P (z|x)
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 q Z in GXZ
= P (y|z) since Y qX in GZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Graphical Models in which P (y|x) is Identifiable
X
Y
(a)
Y
ZX
(d)
X
Y
X
Y
Z
Z1
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
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Graphical Models in which P (y|x) is notIdentifiable
(e)
ZX
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
ZX
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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C-components and C-factor
Two variables are said to be in the same C-component if theyare connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W1
2Z
2W
X Y
T
S1 = {X,Y,W1,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
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C-components and C-factor
Two variables are said to be in the same C-component if theyare connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W1
2Z
2W
X Y
T
S1 = {X,Y,W1,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
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C-components and C-factor
Two variables are said to be in the same C-component if theyare connected by a path comprising of only bi-directionaledges[Tian & Pearl, 2002].
Z1
W1
2Z
2W
X Y
T
S1 = {X,Y,W1,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
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Identifiability of C-factor
Lemma (Tian & Pearl, 2002)
Let a topological order over V be V1 < V2 < ... < Vn and letV (i) = {V1, V2, ..., Vi}, i = 1, ..., n and V (0) = φ. For any set C,let GC denote the subgraph of G composed only of variables inC. Then:(i) Each C-factor Qj , j = 1, ..., k is identifiable and is given by:
Qj =∏{i:Vi∈Sj} P (vi|v(i−1))
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: Identifiability of C-factor
X1X2 X3 X4
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
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Necessary and Sufficient condition foridentifiability of Px(v)
Theorem (Tian & Pearl, 2002)
Let X be a singleton. Px(v) is identifiable if and only if there isno bi-directed path connecting X to any of its children.
WhenPx(v) is identifiable, it is given by:
Px(v) = P (v)QX
∑xQ
X ,
where QX is the c-factor corresponding to the c-component SX
that contains X.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Necessary and Sufficient condition foridentifiability of Px(v)
Theorem (Tian & Pearl, 2002)
Let X be a singleton. Px(v) is identifiable if and only if there isno bi-directed path connecting X to any of its children. WhenPx(v) is identifiable, it is given by:
Px(v) = P (v)QX
∑xQ
X ,
where QX is the c-factor corresponding to the c-component SX
that contains X.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Example: Necessary and Sufficient condition foridentifiability of Px(v)
X1X2 X3 X4
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
∑x1Q2
= P (x4|x1, x2, x3)P (x2|x1)∑x1P (y|x1, x2, x3, x4)P (x3|x1, x2)P (x1)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Causal Effect Identifiability
Identification of Px(y|z) where X ∩ Y ∩ Z = φ and X is notnecessarily a singleton, [Shpitser & Pearl,2006]
I Hedge Criterion
I IDC - Sound and Complete Algorithm
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Counterfactuals
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)
I Abduction
I Intervention
I Prediction
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Counterfactuals
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)
I Abduction
I Intervention
I Prediction
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Counterfactuals
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)
I Abduction
I Intervention
I Prediction
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Counterfactuals
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)
I Abduction
I Intervention
I Prediction
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Counterfactuals
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 MC = UA = CB = CD = A ∨BFacts: DConclusions:U,A,B,C,D
Model MqA
C = UqAB = CD = A ∨BFacts: UConclusions:U, qA,B,C,D
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Equivalence
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
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Markov Equivalence
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 2Z2WW1
X Y
T
Hard to enumerate all separation conditions.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Observational Equivalence
Theorem (Verma & Pearl 1990)
Two DAGs are observationally equivalent iff they have the samesets of edges and the same sets of v-structures, that is, twoconverging arrows whose tails are not connected by an arrow.
X1
X23X
X4
X5
X1
X23X
X4
X5
Figure: Observationally Equivalent DAGs
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Equivalence and ObservationalEquivalence
If two DAGs are Markov Equivalent, then they areObservationally Equivalent as well. True/False?
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Equivalence and ObservationalEquivalence
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.
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Markov Equivalence and ObservationalEquivalence
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.
X T Z YX T Z Y
Figure: DAGs that are Markov Equivalent but not ObservationallyEquivalent
How would you distinguish between the two?
I Verma Constraints (Refer slide:115)
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Ancestral Graphs
Definition (Ancestral Graphs)
A graph which may contain directed or bi-directed edges isancestral if:(i) there are no directed cycles(ii) whenever there is an edge X ←→ Y , then there is nodirected path from X to Y or from Y to X.
Z1 2Z2WW1
X Y
T
Figure: Ancestral graph
Z1
W1
2Z
2W
X Y
T
Figure: Not an Ancestral graph
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Maximal Ancestral Graphs (MAGs)
Definition (Spirtes & Richardson, 2002)
An ancestral graph is said to be maximal if, for every pair ofnon-adjacent nodes X, Y there exists a set Z such that X andY are d-separated conditional on Z.
Z1
W1
2Z
2W
X Y
T
Z1 2
Z
W1 2
W
X Y
T
Figure: DAG and its corresponding MAG
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Construction of a MAG
Given : DAG GStep-1: Construct a graph M comprising 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-2: For every bi-directed edge A↔ B in G,(i) add A→ B to M if A is an ancestor of B in G(ii) add A← B to M if B is an ancestor of A in G(iii) copy A↔ B to M if (i) and (ii) do not hold true
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,connected by an inducing path ,(i) add A→ B to M if A is an ancestor of B in G(ii) add A← B to M if B is an ancestor of A in G(iii) add A↔ B to M if (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: MAG M
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Markov Equivalence
TheoremTwo graphs G1 and G2 are said to be Markov equivalent if theirMAGs are Markov Equivalent
Are these MAGs Markov Equivalent?
Z1 2Z2WW1
X Y
T
Z1 2
Z
W1 2
W
X Y
T
Note: Markov Equivalence in MAGs are easier to checkComplete criterion for determining Markov Equivalence of 2MAGs: [Ali, Richardson and Spirtes, 2009]
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Reversing an edge in a MAG
Definition (Screened Edge)
[Tian,2005] An edge X → Y is a screened edge in a MAG ifPa(Y ) = Pa(X) ∪ {X} and Sp(Y ) = Sp(X)1.
X YZ
W
Figure: MAG with Screened Edge:X → Y
1Nodes X and Y are spouses, if they are connected by a bi-directed edge.Karthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Reversing an edge in a MAG
Theorem (Tian,2005)
Let M be a MAG with edge X → Y and M ′ be a graph withedge X ←− Y , otherwise identical to M . Then M ′ is a MAGthat is Markov Equivalent to M if and only if X → Y is ascreened edge in M .
X YZ
W
X YZ
W
Figure: Markov Equivalent MAGs
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Confounding Equivalence
Definition (Pearl and Paz,2009)
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) =∑
z P (y|x, z)P (z) ∀x, y
W1 2W
V1 2V
X Y
Examples:
I T = {W1, V2} ≈ Z = {W2, V1}I T = {W1, V1} ≈ Z = {W2, V2}I T = {W1,W2} ≈ Z = {W1}I T = {W1,W2} 6≈ Z = {W2}
Note: C-equivalence is testable
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Necessary and Sufficient Condition forC-Equivalence
Theorem (Pearl and Paz, 2009)
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:
I X q (Z ∪ T ) | (Z ∩ T ) or
I Z and T are G-admissible2
W1 2W
V1 2V
X Y
Examples:
I T = {W1, V2} ≈ Z = {W2, V1}I T = {W1, V1} ≈ Z = {W2, V2}I T = {W1,W2} ≈ Z = {W1}I T = {W1,W2} 6≈ Z = {W2}
2satisfies back-door criterionKarthika Mohan and Judea Pearl Graphical Models for Causal Inference
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Linear Models and Causal Diagrams
Assume all variables are normalized to have zero mean and unitvariance.
Z X Y
W
a
b
c
α β
γ
Z = e1W = e2X = aZ + e3Y = bW + cX + e4Cov(e1, e2) = α 6= 0Cov(e2, e3) = β 6= 0Cov(e3, e4) = γ 6= 0
Which parameters can be identified?
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Vanishing Regression Coefficient
DefinitionFor any linear model for a causal diagram D that may includecycles and bi-directed arcs, the partial correlation ρXY.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 I rTX.W1Z1 = 0
Find more
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Single Door Criterion for Direct Effects
TheoremLet G be any path diagram in which α is the path coefficientassociated with link X → Y and let Gα denote the diagram thatresults when X → Y is deleted from G. The coefficient α isidentifiable if there exists a set of variables Z such that :(i) Z contains no descendant of Y and(ii) Z d-separates X from Y in GαMoreover, if Z satisfies these two conditions, then α is equal tothe regression coefficient rY X.Z .
Gα
βZ YX
G
Z YXβ α
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Instrumental Variables (IV)DefinitionA variable Z is an instrument relative to a cause X and aneffect Y if:
I Z is independent of all error terms that have an influenceon Y when X is held constant, and
I Z is not independent of X.
In linear systems, Causal effect of X on Y = rZYrZX
UYUY
UWUYUY
Y
(b)
X
Z
Y
(c)
XZ
Y
(d)
X
Z
W
Y
(a)
X
Z
Figure: Z is an instrument in (a), (b) and (c) but not in (d)
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Conditional Instrumental Variable
Definition (Brito & Pearl, 2002)
Z is an instrumental variable if ∃ a set W such that:
I W contains only non-descendants of YI W d-separates Z from Y in the sub-graph Gα obtained by
removing the edge X → YI W does not d-separate Z from X in Gα
Z X Y
W
α
Z X Y
W
Figure: Graph G and corresponding subgraph Gα
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Conditional Instrumental Variable
I Z is a conditional instrumental variable. Hence,α = Causal effect of X on Y = rZY.W
rZX.WI W does not satisfy single-door criterion. So, α cannot be
identified using single-door.
Z X Y
W
α
Z X Y
W
Figure: Graph G and corresponding subgraph Gα
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Verma Constraints([Tian and Pearl, 2002])
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) =
∑b P (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) =
∑b P (d|a, b, c)P (b|a)∑
b P (d|a, b, c)P (b|a) is notindependent of a.
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Conclusions
Graphs are indispensable for:
I encoding causal knowledge
I identifying parameters and causal effects
I identifying testable implications
Go ahead and Exploit the Power of Graphs!
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Thank You!
Karthika Mohan and Judea Pearl Graphical Models for Causal Inference