Slide 1
Directed GraphicalProbabilistic Models
William W. Cohen
Machine Learning 10-601
Feb 2008
Slide 2
Some practical problems
• I have 3 standard d20 dice, 1 loaded die.
• Experiment: (1) pick a d20 uniformly at random then (2)roll it. Let A=d20 picked is fair and B=roll 19 or 20 withthat die. What is P(B)?
P(B) = P(B|A) P(A) + P(B|~A) P(~A)
= 0.1*0.75 + 0.5*0.25 = 0.2
•What is P(A=fair|B=criticalHit)?
“Loaded” means P(19 or 20)=0.5
Slide 3
Definition of Conditional Probability
P(A ^ B)
P(A|B) = -----------
P(B)
Corollaries:
P(B|A) = P(A|B) P(B) / P(A)
P(A ^ B) = P(A|B) P(B)
Chain rule
Bayes rule
Slide 4
The (highly practical) Monty Hall problem
•You’re in a game show.Behind one door is a prize.Behind the others, goats.
•You pick one of threedoors, say #1
•The host, Monty Hall,opens one door,revealing…a goat!
3
You now can either
• stick with your guess
• always change doors
• flip a coin and pick a new doorrandomly according to the coin
Slide 5
Some practical problems
• I have 1 standard d6 die, 2 loaded d6 die.
• Loaded high: P(X=6)=0.50 Loaded low: P(X=1)=0.50
• Experiment: pick two d6 uniformly at random (A) and roll them. Whatis more likely – rolling a seven or rolling doubles?
Three combinations: HL, HF, FL
P(D) = P(D ^ A=HL) + P(D ^ A=HF) + P(D ^ A=FL)
= P(D | A=HL)*P(A=HL) + P(D|A=HF)*P(A=HF) + P(A|A=FL)*P(A=FL)
Slide 6
A brute-force solution
A
Roll 1
Roll 2
P
FL
1
1
1/3 * 1/6 * ½
FL
1
2
1/3 * 1/6 * 1/10
FL
1
…
…
…
1
6
FL
2
1
FL
2
…
…
…
…
FL
6
6
HL
1
1
HL
1
2
…
…
…
HF
1
1
…
Comment
doubles
seven
doubles
doubles
A joint probability table shows P(X1=x1 and … andXk=xk) for every possible combination of valuesx1,x2,…., xk
With this you can compute any P(A) where A is anyboolean combination of the primitive events(Xi=Xk), e.g.
• P(doubles)
• P(seven or eleven)
• P(total is higher than 5)
• ….
Slide 7
The Joint Distribution
Recipe for making a joint distributionof M variables:
Example: Booleanvariables A, B, C
Slide 8
The Joint Distribution
Recipe for making a joint distributionof M variables:
1.Make a truth table listing allcombinations of values of yourvariables (if there are M Booleanvariables then the table will have2M rows).
Example: Booleanvariables A, B, C
A
B
C
0
0
0
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
Slide 9
The Joint Distribution
Recipe for making a joint distributionof M variables:
1.Make a truth table listing allcombinations of values of yourvariables (if there are M Booleanvariables then the table will have2M rows).
2.For each combination of values,say how probable it is.
Example: Booleanvariables A, B, C
A
B
C
Prob
0
0
0
0.30
0
0
1
0.05
0
1
0
0.10
0
1
1
0.05
1
0
0
0.05
1
0
1
0.10
1
1
0
0.25
1
1
1
0.10
Slide 10
The Joint Distribution
Recipe for making a joint distributionof M variables:
1.Make a truth table listing allcombinations of values of yourvariables (if there are M Booleanvariables then the table will have2M rows).
2.For each combination of values,say how probable it is.
3.If you subscribe to the axioms ofprobability, those numbers mustsum to 1.
Example: Booleanvariables A, B, C
A
B
C
Prob
0
0
0
0.30
0
0
1
0.05
0
1
0
0.10
0
1
1
0.05
1
0
0
0.05
1
0
1
0.10
1
1
0
0.25
1
1
1
0.10
A
B
C
0.05
0.25
0.10
0.05
0.05
0.10
0.10
0.30
Slide 11
Using theJoint
One you have the JD you canask for the probability of anylogical expression involvingyour attribute
Slide 12
Using theJoint
P(Poor Male) = 0.4654
Slide 13
Using theJoint
P(Poor) = 0.7604
Slide 14
Inferencewith theJoint
Slide 15
Inferencewith theJoint
P(Male | Poor) = 0.4654 / 0.7604 = 0.612
Slide 16
Inference is a big deal
•I’ve got this evidence. What’s the chance that thisconclusion is true?
•I’ve got a sore neck: how likely am I to have meningitis?
•I see my lights are out and it’s 9pm. What’s the chance my spouseis already asleep?
•…
•Lots and lots of important problems can be solvedalgorithmically using joint inference.
•How do you get from the statement of the problem to the jointprobability table? (What is a “statement of the problem”?)
•The joint probability table is usually too big to represent explicitly,so how do you represent it compactly?
Slide 17
Problem 1: A description of the experiment
• I have 3 standard d20 dice, 1 loaded die.
• Experiment: (1) pick a d20 uniformly at random then (2)roll it. Let A=d20 picked is fair and B=roll 19 or 20 withthat die. What is P(B)?
A
B
P(A=fair)=0.75
P(A=loaded)=0.25
P(B=critical | A=fair=0.1
P(B=noncritical | A=fair)=0.9
P(B=critical | A=loaded)=0.5
P(B=noncritical | A=loaded)=0.5
A
P(A)
Fair
0.75
Loaded
0.25
A
B
P(B|A)
Fair
Critical
0.1
Fair
Noncritical
0.9
Loaded
Critical
0.5
Loaded
Noncritical
0.5
Slide 18
Problem 1: A description of the experiment
A
B
A
P(A)
Fair
0.75
Loaded
0.25
A
B
P(B|A)
Fair
Critical
0.1
Fair
Noncritical
0.9
Loaded
Critical
0.5
Loaded
Noncritical
0.5
This is
• an “influence diagram” (informal: what influences what)
• a Bayes net / belief network / … (formal: more later!)
• a description of the experiment (i.e., how data could be generated)
• everything you need to reconstruct the joint distribution
Chain rule:
P(A,B)=P(B|A) P(A)
problem with probs:Bayes net::word problem:algebra
Slide 19
Problem 2: a description
• I have 1 standard d6 die, 2 loaded d6 die.
• Loaded high: P(X=6)=0.50 Loaded low: P(X=1)=0.50
• Experiment: pick two d6 without replacement (D1,D2) and roll them.What is more likely – rolling a seven or rolling doubles?
D1
D2
D1
P(D1)
Fair
0.33
LoadedLo
0.33
LoadedHi
0.33
D1
D2
P(D2|D1)
Fair
LoadedLo
0.5
Fair
LoadedHi
0.5
LoadedLo
Fair
0.5
LoadedLo
LoadedHi
0.5
LoadedHi
Fair
0.5
LoadedHi
LoadedLo
0.5
R
D1
D2
R
P(R|D1,D2)
Fair
Lo
1,1
1/6 * 1/2
Fair
Lo
1,2
1/6 * 1/10
..
…
…
…
Slide 20
Problem 2: a description
D1
D2
D1
P(D2)
Fair
0.33
LoadedLo
0.33
LoadedHi
0.33
D1
D2
P(D2|D1)
Fair
LoadedLo
0.5
Fair
LoadedHi
0.5
LoadedLo
Fair
0.5
LoadedLo
LoadedHi
0.5
LoadedHi
Fair
0.5
LoadedHi
LoadedLo
0.5
R
D1
D2
R
P(R|D1,D2)
Fair
Lo
1,1
1/6 * 1/2
Fair
Lo
1,2
1/6 * 1/10
..
…
…
…
P(A ^ B) = P(A|B) P(B)
Chain rule
P(R,D1,D2) = P(R|D1,D2) P(D1,D2)
P(R,D1,D2) = P(R|D1,D2) P(D2|D1) P(D1)
Slide 21
Problem 2: another description
D1
D2
R1
R2
Is7
R2=1,2,…,6
R1=1,2,…,6
Is7=yes,no
Slide 22
The (highly practical) Monty Hall problem
•You’re in a game show. Behind onedoor is a prize. Behind the others,goats.
•You pick one of three doors, say #1
•The host, Monty Hall, opens onedoor, revealing…a goat!
•You now can either stick with yourguess or change doors
A
B
First guess
The money
C
The goat
D
Stick, orswap?
E
Second guess
A
P(A)
1
0.33
2
0.33
3
0.33
B
P(B)
1
0.33
2
0.33
3
0.33
D
P(D)
Stick
0.5
Swap
0.5
A
B
C
P(C|A,B)
1
1
2
0.5
1
1
3
0.5
1
2
3
1.0
1
3
2
1.0
…
…
…
…
Slide 23
The (highly practical) Monty Hall problem
A
B
First guess
The money
C
The goat
D
Stick orswap?
E
Second guess
A
P(A)
1
0.33
2
0.33
3
0.33
B
P(B)
1
0.33
2
0.33
3
0.33
A
B
C
P(C|A,B)
1
1
2
0.5
1
1
3
0.5
1
2
3
1.0
1
3
2
1.0
…
…
…
…
A
C
D
P(E|A,C,D)
…
…
…
…
Slide 24
The (highly practical) Monty Hall problem
A
B
First guess
The money
C
The goat
D
Stick orswap?
E
Second guess
A
P(A)
1
0.33
2
0.33
3
0.33
B
P(B)
1
0.33
2
0.33
3
0.33
A
B
C
P(C|A,B)
1
1
2
0.5
1
1
3
0.5
1
2
3
1.0
1
3
2
1.0
…
…
…
…
A
C
D
P(E|A,C,D)
…
…
…
…
…again by the chain rule:
P(A,B,C,D,E) =
P(E|A,C,D) *
P(D) *
P(C | A,B ) *
P(B ) *
P(A)
We could construct thejoint and computeP(E=B|D=swap)
Slide 25
The (highly practical) Monty Hall problem
A
B
First guess
The money
C
The goat
D
Stick orswap?
E
Second guess
A
P(A)
1
0.33
2
0.33
3
0.33
B
P(B)
1
0.33
2
0.33
3
0.33
A
B
C
P(C|A,B)
1
1
2
0.5
1
1
3
0.5
1
2
3
1.0
1
3
2
1.0
…
…
…
…
A
C
D
P(E|A,C,D)
…
…
…
…
The joint table has…?
3*3*3*2*3 = 162 rows
The conditionalprobability tables (CPTs)shown have … ?
3 + 3 + 3*3*3 + 2*3*3= 51 rows < 162 rows
Big questions:
• why are the CPTs smaller?
•how much smaller are theCPTs than the joint?
• can we compute the answersto queries like P(E=B|d) withoutbuilding the joint probabilitytables, just using the CPTs?
Slide 26
The (highly practical) Monty Hall problem
A
B
First guess
The money
C
The goat
D
Stick orswap?
E
Second guess
A
P(A)
1
0.33
2
0.33
3
0.33
B
P(B)
1
0.33
2
0.33
3
0.33
A
B
C
P(C|A,B)
1
1
2
0.5
1
1
3
0.5
1
2
3
1.0
1
3
2
1.0
…
…
…
…
A
C
D
P(E|A,C,D)
…
…
…
…
Why is the CPTsrepresentation smaller?Follow the money! (B)
E is conditionallyindependent of Bgiven A,D,C
Slide 27
Conditional Independence formalized
Definition: R and L are conditionally independent given M if
for all x,y,z in {T,F}:
P(R=x M=y ^ L=z) = P(R=x M=y)
More generally:
Let S1 and S2 and S3 be sets of variables.
Set-of-variables S1 and set-of-variables S2 areconditionally independent given S3 if for allassignments of values to the variables in the sets,
P(S1’s assignments S2’s assignments & S3’s assignments)=
P(S1’s assignments S3’s assignments)
Slide 28
The (highly practical) Monty Hall problem
A
B
First guess
The money
C
The goat
D
Stick orswap?
E
Second guess
What are the conditional indepencies?
• I<A, {B}, C> ?
• I<A, {C}, B> ?
• I<E, {A,C}, B> ?
• I<D, {E}, B> ?
• …
Slide 29
Bayes Nets Formalized
A Bayes net (also called a belief network) is an augmenteddirected acyclic graph, represented by the pair V , E where:
•V is a set of vertices.
•E is a set of directed edges joining vertices. No loops ofany length are allowed.
Each vertex in V contains the following information:
•The name of a random variable
•A probability distribution table indicating how theprobability of this variable’s values depends on allpossible combinations of parental values.
Slide 30
Building a Bayes Net
1.Choose a set of relevant variables.
2.Choose an ordering for them
3.Assume they’re called X1 .. Xm (where X1 is the first in theordering, X1 is the second, etc)
4.For i = 1 to m:
1.Add the Xi node to the network
2.Set Parents(Xi ) to be a minimal subset of {X1…Xi-1}such that we have conditional independence of Xi andall other members of {X1…Xi-1} given Parents(Xi )
3.Define the probability table of
P(Xi =k Assignments of Parents(Xi ) ).
Slide 31
The general case
P(X1=x1 ^ X2=x2 ^ ….Xn-1=xn-1 ^ Xn=xn) =
P(Xn=xn ^ Xn-1=xn-1 ^ ….X2=x2 ^ X1=x1) =
P(Xn=xn Xn-1=xn-1 ^ ….X2=x2 ^ X1=x1) * P(Xn-1=xn-1 ^…. X2=x2 ^ X1=x1) =
P(Xn=xn Xn-1=xn-1 ^ ….X2=x2 ^ X1=x1) * P(Xn-1=xn-1 …. X2=x2 ^ X1=x1) *
P(Xn-2=xn-2 ^…. X2=x2 ^ X1=x1) =
:
:
=
So any entry in joint pdf table can be computed. And so anyconditional probability can be computed.
Slide 32
Building a Bayes Net: Example 1
1.Choose a set of relevant variables.
2.Choose an ordering for them
3.Assume they’re called X1 .. Xm(where X1 is the first in theordering, X1 is the second, etc)
4.For i = 1 to m:
•Add the Xi node to the network
•Set Parents(Xi ) to be a minimalsubset of {X1…Xi-1} such that wehave conditional independenceof Xi and all other members of{X1…Xi-1} given Parents(Xi )
•Define the probability table of
P(Xi =k values of Parents(Xi ) )
D1 – first die; D2 – second; R – roll, Is7
D1, D2, R, Is7
D1
D2
R
Is7
D1
P(D2)
Fair
0.33
Lo
0.33
Hi
0.33
D1
D2
P(D2|D1)
Fair
Lo
0.5
Fair
Hi
0.5
Lo
Fair
0.5
Lo
Hi
0.5
Hi
Fair
0.5
Hi
Lo
0.5
D1
D2
R
P(R|D1,D2)
Fair
Lo
1,1
1/6 * 1/2
Fair
Lo
1,2
1/6 * 1/10
..
…
…
…
R
Is7
P(D2|D1)
1,1
No
1.0
1,2
No
1.0
…
…
…
1,6
Yes
1.0
2,1
No
…
…
No
…
P(D1,D2,R,Is7)=
P(Is7|R)*P(R|D1,D2)*P(D2|D1)*P(D1)
Slide 33
Another construction
D1
D2
R1
R2
Is7
R2=1,2,…,6
R1=1,2,…,6
Is7=yes,no
Pick the order D1, D2, R1, R2, Is7
Network will be:
What if I pick otherorders? WhataboutR2,R1,D2,D1,Is7?
What about thefirst network AB?suppose I pickedthe order B,A ?
Slide 34
Another simple example
•Bigram language model:
•Pick word w0 from a multinomial 1 P(w0)
•For t=1 to 100
•Pick word wt from multinomial 2 P(wt|wt-1)
w0
w1
w2
…
w100
w0
P(w0)
aardvark
0.0002
…
…
the
0.179
…
…
zymurgy
0.097
Wt-1
Wt
P(w0)
aardvark
aardvark
0.00000001
aardvark
…
…
aardvark
tastes
0.1275
…
…
…
aardvark
zymurgy
0.0000076
Slide 35
A simple example
•Bigram languagemodel:
•Pick word w0 from amultinomial 1 P(w0)
•For t=1 to 100
•Pick word wt frommultinomial 2P(wt|wt-1)
w0
w1
w2
…
w100
w0
P(w0)
aardvark
0.0002
…
…
the
0.179
…
…
zymurgy
0.097
Wt-1
Wt
P(wt|wt-1)
aardvark
aardvark
0.00000001
aardvark
…
…
aardvark
tastes
0.1275
…
…
…
aardvark
zymurgy
0.0000076
Size of CPTs: |V| + |V|2
Size of joint table: |V| ^ 101
|V|=20 for amino acids (proteins)
Slide 36
A simple example
•Trigram language model:
•Pick word w0 from a multinomial M1 P(w0)
•Pick word w1 from a multinomial M2 P(w1|w0)
•For t=2 to 100
•Pick word wt from multinomial M3 P(wt|wt-1,wt-2)
w0
w1
w2
…
w99
Size of CPTs: |V| + |V|2 + |V|3
Size of joint table: |V| ^ 101
|V|=20 for amino acids (proteins)
w100
Slide 37
What Independencies does a Bayes Net Model?
•In order for a Bayesian network to model a probabilitydistribution, the following must be true by definition:
Each variable is conditionally independent of all its non-descendants in the graph given the value of all its parents.
•This implies
•But what else does it imply?
Slide 38
What Independencies does a Bayes Net Model?
•Example:
Z
Y
X
Given Y, does learning the value of Z tell us
nothing new about X?
I.e., is P(X|Y, Z) equal to P(X | Y)?
Yes. Since we know the value of all of X’sparents (namely, Y), and Z is not adescendant of X, X is conditionallyindependent of Z.
Also, since independence is symmetric,
P(Z|Y, X) = P(Z|Y).
Slide 39
Quick proof that independence is symmetric
•Assume: P(X|Y, Z) = P(X|Y)
•Then:
(Bayes’s Rule)
(Chain Rule)
(By Assumption)
(Bayes’s Rule)
Slide 40
What Independencies does a Bayes Net Model?
•Let I<X,Y,Z> represent X and Z being conditionally independentgiven Y.
•I<X,Y,Z>? Yes, just as in previous example: All X’s parents given,and Z is not a descendant.
Y
X
Z
Slide 41
What Independencies does a Bayes Net Model?
•I<X,{U},Z>? No.
•I<X,{U,V},Z>? Yes.
•Maybe I<X, S, Z> iff S acts a cutsetbetween X and Z in an undirectedversion of the graph…?
Z
V
U
X
Slide 42
Things get a little more confusing
•X has no parents, so we know all its parents’values trivially
•Z is not a descendant of X
•So, I<X,{},Z>, even though there’s aundirected path from X to Z through anunknown variable Y.
•What if we do know the value of Y, though?Or one of its descendants?
Z
X
Y
Slide 43
The “Burglar Alarm” example
•Your house has a twitchy burglaralarm that is also sometimestriggered by earthquakes.
•Earth arguably doesn’t care whetheryour house is currently being burgled
•While you are on vacation, one ofyour neighbors calls and tells youyour home’s burglar alarm is ringing.Uh oh!
Burglar
Earthquake
Alarm
Phone Call
Slide 44
Things get a lot more confusing
•But now suppose you learnthat there was a medium-sizedearthquake in yourneighborhood. Oh, whew!Probably not a burglar after all.
•Earthquake “explains away” thehypothetical burglar.
•But then it must not be thecase that
I<Burglar,{Phone Call},Earthquake>, even though
I<Burglar,{}, Earthquake>!
Burglar
Earthquake
Alarm
Phone Call
Slide 45
d-separation to the rescue
•Fortunately, there is a relatively simple algorithm fordetermining whether two variables in a Bayesian networkare conditionally independent: d-separation.
•Definition: X and Z are d-separated by a set of evidencevariables E iff every undirected path from X to Z is“blocked”, where a path is “blocked” iff one or more of thefollowing conditions is true: ...
ie. X and Z are dependent iff there exists an unblocked path
Slide 46
A path is “blocked” when...
•There exists a variable V on the path such that
•it is in the evidence set E
•the arcs putting Y in the path are “tail-to-tail”
•Or, there exists a variable V on the path such that
•it is in the evidence set E
•the arcs putting Y in the path are “tail-to-head”
•Or, ...
Y
Y
unknown“commoncauses” of Xand Z imposedependency
unknown“causalchains”connecting Xan Z imposedependency
Slide 47
A path is “blocked” when… (the funky case)
•… Or, there exists a variable Y on the path such that
•it is NOT in the evidence set E
•neither are any of its descendants
•the arcs putting Y on the path are “head-to-head”
Y
Known “commonsymptoms” of Xand Z imposedependencies… Xmay “explainaway” Z
Slide 48
d-separation to the rescue, cont’d
•Theorem [Verma & Pearl, 1998]:
•If a set of evidence variables E d-separates X and Z in a Bayesiannetwork’s graph, then I<X, E, Z>.
•d-separation can be computed in linear time using a depth-first-search-like algorithm.
•Great! We now have a fast algorithm for automatically inferringwhether learning the value of one variable might give us any additionalhints about some other variable, given what we already know.
•“Might”: Variables may actually be independent when they’re not d-separated, depending on the actual probabilities involved