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Tony Jebara, Columbia University
Machine Learning 4771
Instructor: Tony Jebara
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Tony Jebara, Columbia University
Topic 15 • Graphical Models
• Maximum Likelihood for Graphical Models • Testing for
Conditional Independence & D-Separation • Bayes Ball
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Tony Jebara, Columbia University
Learning Fully Observed Models • Easiest scenario: we have
observed all the nodes • Want to learn the probability tables from
data… • Have N iid patients:
• Simplest case: least general graph handle each dim
individually as Bernoulli/Multinomial
• 2nd Simplest case: most general, count each entry in pdf
• What about learning graphs in between?
PATIENT FLU FEVER SINUS TEMP SWELL HEAD
1 Y Y N L Y Y
2 N N N M N Y
3 Y N Y H Y N
4 Y N Y M N N
Divide by total count Since FLU
TEMP SINUS
+1
+1 ! p x( )
x6
∑x1
∑ = 1
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Tony Jebara, Columbia University
Maximum Likelihood CPTs • Each conditional probability table θi
part of our parameters • Given table, have pdf
• Have M variables:
• Have N x M dataset:
• Maximum likelihood:
θ4
p X
U| θ( ) = p xi | πi, θi( )i=1
M∏
X
U= x
1,…,x
M{ }
D = X
U ,1,…,X
U ,N{ }
θ* = arg maxθlog p D | θ( )
= arg maxθ
log p XU ,n
| θ( )n=1N∑
= arg maxθ
log p xi,n
| πi,nθ
i( )i=1M∏n=1
N∑= arg max
θlog p x
i,n| π
i,nθ
i( )i=1M∑n=1
N∑
each θi appears independently, can do ML for each CPT alone!
efficient storage & efficient learning
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Tony Jebara, Columbia University
Maximum Likelihood CPTs
δ XU ,n
,XU ,m( ) =
1 if XU ,n
= XU ,m
0 otherwise
⎧⎨⎪⎪⎪
⎩⎪⎪⎪
m x
i( ) = δ xi,xi,n( )n=1N∑
l θ( ) = log p XU ,n | θ( )n=1N∑
= log p XU
| θ( )XU∏δ XU ,XU ,n( )
n=1
N∑= δ X
U,X
U ,n( ) log p XU | θ( )XU∑n=1N∑
= m XU( ) log p XU | θ( )XU∑ = m XU( ) log p xi | πi, θi(
)i=1
M∏XU∑= m X
U( ) log p xi | πi, θi( )i=1M∑XU∑
m X
U( ) = δ XU ,XU ,n( )n=1N∑
m X
C( ) = m XU( )XU\C∑
N = m x
1( )x1∑ = m x1,x2( )x2∑( )x1∑ = m x1,x2,x3( )x3∑( )x2∑( )x1∑
Counts: # of times what’s in the bracket appeared in data, for
example:
• So…
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Tony Jebara, Columbia University
• Continuing:
• Define: Constraint: • Now have above with Lagrange
multipliers:
• Plug constraint:
• Final solution (trivial!):
Maximum Likelihood CPTs
l θ( ) = m XU( ) log p xi | πi, θi( )i=1M∑XU∑
= m XU( ) log p xi | πi, θi( )XU\xi \πi∑xi ,πi∑i=1
M∑= m x
i,π
i( ) log p xi | πi, θi( )xi ,πi∑i=1M∑
θ x
i,π
i( ) = p xi | πi, θi( ) θ xi,πi( )xi∑ = 1
l θ( ) = m xi,πi( ) log θ xi,πi( )πi∑xi∑i=1
M∑ − λπi θ xi,πi( )xi∑ −1( )πi∑i=1M∑
∂l θ( )∂θ x
i,π
i( )=
m xi,π
i( )θ x
i,π
i( )−λ
πi= 0
→ θ xi,π
i( ) =m x
i,π
i( )λπi
m xi,π
i( )λπi
xi∑ = 1 → λπi = m xi,πi( )xi∑ = m πi( )
θ xi,π
i( ) =m x
i,π
i( )m π
i( )
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Tony Jebara, Columbia University
• Continuing:
• Define: Constraint: • Now have above with Lagrange
multipliers:
• Plug constraint:
• Final solution (trivial!): MAP VERSION
Maximum Likelihood CPTs
l θ( ) = m XU( ) log p xi | πi, θi( )i=1M∑XU∑
= m XU( ) log p xi | πi, θi( )XU\xi \πi∑xi ,πi∑i=1
M∑= m x
i,π
i( ) log p xi | πi, θi( )xi ,πi∑i=1M∑
θ x
i,π
i( ) = p xi | πi, θi( ) θ xi,πi( )xi∑ = 1
l θ( ) = m xi,πi( ) log θ xi,πi( )πi∑xi∑i=1
M∑ − λπi θ xi,πi( )xi∑ −1( )πi∑i=1M∑
∂l θ( )∂θ x
i,π
i( )=
m xi,π
i( )θ x
i,π
i( )−λ
πi= 0
→ θ xi,π
i( ) =m x
i,π
i( )λπi
m xi,π
i( )λπi
xi∑ = 1 → λπi = m xi,πi( )xi∑ = m πi( )
θ xi,π
i( ) =m x
i,π
i( ) + εm π
i( ) + ε xi
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Tony Jebara, Columbia University
Maximum Likelihood CPTs • Let’s try an example: • Compute the
cpt
• Using the formula:
θ xi,π
i( ) =m x
i,π
i( )m π
i( )
PATIENT FLU FEVER SINUS TEMP SWELL HEAD
1 Y Y N L Y Y
2 N N N M N Y
3 Y N Y H Y N
4 Y N Y M N N p x
3| x
1( )
1 1
0 2
Note, here 0/0 = prior constant
1 3
1 1/3
0 2/3
m x
3,x
1( )
m x
1( )
p x
3| x
1( )
x1 = 0 x
1= 1
x3 = 0
x3 = 1
Efficient, only count over subset of variables in p(XB|XA) Not
all p(x1,…,xM)
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Tony Jebara, Columbia University
Conditional Dependence Tests • Another thing we would like to do
with a graphical model: Check conditional independencies… “Is
Temperature Indep. of Flu Given Fever?” “Is Temperature Indep. of
Sinus Infection Given Fever?” • Try computing & simplify
marginals of p(x)
• In this case it was easy, what if checking: • Hard to compute
want efficient algorithm…
p X( ) = p x1( )p x2 | x1( )p x3 | x1( )p x4 | x2( )p x5 | x3(
)p x6 | x2,x5( )
p x4| x
1,x
2,x
3( ) =p x
1,x
2,x
3,x
4( )p x
1,x
2,x
3( )=
p X( )x6∑x5∑p X( )x6∑x5∑x4∑
=p x
1( )p x2 | x1( )p x3 | x1( )p x4 | x2( )p x
1( )p x2 | x1( )p x3 | x1( )= p x
4| x
2( ) x4 ! x1,x3 | x2 x1 ! x6 | x2,x3
p x
1| x
2,x
3,x
6( )
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Tony Jebara, Columbia University
D-Separation & Bayes Ball • There is a graph algorithm for
checking independence • Intuition: separation or blocking of some
nodes by others • Example: if nodes “block” path from to we might
say that
• This is not exact for directed graphs (true for Undirected)
• We need more than just simple Separation • Need D-Separation
(directed separation) • D-Separation is computed via the Bayes Ball
algorithm • Use to prove general statements over subsets of
vars:
XA ! XB | XC
x2,x3
x1 x
6
x1 ! x6 | x2,x3
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Tony Jebara, Columbia University
Bayes Ball Algorithm • The algorithm: 1) Shade nodes XC 2) Place
a ball at each node in XA 3) Bounce balls around graph according to
some rules 4) If no balls reach XB, then is true (else false)
Balls can travel along/against arrows Pick any incoming &
outgoing path Test each to see if ball goes through or bounces
back
Look at canonical sub-graphs & leaf cases for rules…
XA ! XB | XC
XA ! XB | XC
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Tony Jebara, Columbia University
Bayes Ball Algorithm 1) Markov Chain:
Rule depends only on shading of middle node Ball stops Go
Through
2) Two Effects:
Rule depends only on shading of middle node Ball stops Go
Through
3) Two Causes (V): Rule depends only on shading of middle node x
Go Through Ball stops
x ! z | y x ! z
x ! z | y
x ! z | y x ! zx
x
x ! zx
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Tony Jebara, Columbia University
Bayes Ball Algorithm • Also need to look at special ‘leaf’
cases:
• Example:
• Example: x1 ! x5 | x3
∴ x
1! x
6| x
2,x
3{ }
X1,X2,X4 Stopped X1,X2,X6 Stopped X1,X3,X5 Stopped
Flu is independent of headache given fever & sinus
infection!
Ball is stopped
Bounces back
Bounces back
Ball is stopped
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Tony Jebara, Columbia University
Bayes Ball Algorithm • Example:
• Example:
X2,X6,X5 Goes Through Because of V-structure
∴ x
2! x
3| x
1,x
6{ }x
aliens ! watch aliens ! watch | reportx Ball bounces back from
report leaf and goes to right if report is shaded. Bob is waiting
for Alice but can’t know if she is late. Instead a security guard
says if she is. She can be late if aliens abduct her or Bob’s watch
is ahead (daylight savings time). Guard reports she is late. If
watch is ahead, p(alien=true) goes down, they are dependent.
