I NFERENCE IN B AYESIAN N ETWORKS. A GENDA Reading off independence assumptions Efficient inference...

INFERENCE IN BAYESIAN NETWORKS

AGENDA

Reading off independence assumptions Efficient inference in Bayesian Networks

Top-down inference Variable elimination Monte-Carlo methods

SOME APPLICATIONS OF BN

Medical diagnosis Troubleshooting of hardware/software

systems Fraud/uncollectible debt detection Data mining Analysis of genetic sequences Data interpretation, computer vision, image

understanding

MORE COMPLICATED SINGLY-CONNECTED BELIEF NET

Radio

Battery

SparkPlugs

Starts

Gas

Moves

Region = {Sky, Tree, Grass, Rock}

R2

R4R3

R1

Above

BN to evaluate insurance risks

BN FROM LAST LECTURE

Burglary Earthquake

Alarm

MaryCallsJohnCalls

causes

effects

Directed acyclic graph

Intuitive meaning of arc from x to y:

“x has direct influence on y”

ARCS DO NOT NECESSARILY ENCODE CAUSALITY!

A

B

C

C

B

A

2 BN’s that can encode the same joint probability distribution

READING OFF INDEPENDENCE RELATIONSHIPS

Given B, does the value of A affect the probability of C? P(C|B,A) = P(C|B)?

No! C parent’s (B) are

given, and so it is independent of its non-descendents (A)

Independence is symmetric:C A | B => A C | B

A

B

C

WHAT DOES THE BN ENCODE?

Burglary EarthquakeJohnCalls MaryCalls | AlarmJohnCalls Burglary | AlarmJohnCalls Earthquake | AlarmMaryCalls Burglary | AlarmMaryCalls Earthquake | Alarm

Burglary Earthquake

Alarm

MaryCallsJohnCalls

A node is independent of its non-descendents, given its parents


How about Burglary Earthquake | Alarm ? No! Why?

Burglary Earthquake

Alarm

MaryCallsJohnCalls


How about Burglary Earthquake | JohnCalls? No! Why? Knowing JohnCalls affects the probability of

Alarm, which makes Burglary and Earthquake dependent

Burglary Earthquake

Alarm

MaryCallsJohnCalls

INDEPENDENCE RELATIONSHIPS

Rough intuition (this holds for tree-like graphs, polytrees): Evidence on the (directed) road between two

variables makes them independent Evidence on an “A” node makes descendants

independent Evidence on a “V” node, or below the V, makes

the ancestors of the variables dependent (otherwise they are independent)

Formal property in general case : D-separation independence (see R&N)

BENEFITS OF SPARSE MODELS

Modeling Fewer relationships need to be encoded (either

through understanding or statistics) Large networks can be built up from smaller ones

Intuition Dependencies/independencies between variables

can be inferred through network structures Tractable inference

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCESuppose we want to compute P(Alarm)

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCESuppose we want to compute P(Alarm)1. P(Alarm) = Σb,e P(A,b,e)2. P(Alarm) = Σb,e P(A|b,e)P(b)P(e)

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCESuppose we want to compute P(Alarm)1. P(A) = Σb,e P(A,b,e)2. P(A) = Σb,e P(A|b,e)P(b)P(e)3. P(A) = P(A|B,E)P(B)P(E) +

P(A|B, E)P(B)P(E) +P(A|B,E)P(B)P(E) +P(A|B,E)P(B)P(E)

4. P(A) = 0.95*0.001*0.002 +0.94*0.001*0.998 +0.29*0.999*0.002 +0.001*0.999*0.998= 0.00252

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCENow, suppose we want to compute P(MaryCalls)

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCENow, suppose we want to compute P(MaryCalls)1. P(M) = P(M|A)P(A) + P(M| A) P(A)

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCENow, suppose we want to compute P(MaryCalls)1. P(M) = P(M|A)P(A) + P(M| A) P(A)2. P(M) = 0.70*0.00252 + 0.01*(1-0.0252)

= 0.0117

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

TOP-DOWN INFERENCE WITH EVIDENCE

Suppose we want to compute P(Alarm|Earthquake)

TOP-DOWN INFERENCE

Only works if the graph of ancestors of a variable is a polytree

Evidence given on ancestor(s) of the query variable

Efficient: O(d 2k) time, where d is the number of ancestors

of a variable, with k a bound on # of parents Evidence on an ancestor cuts off influence of

portion of graph above evidence node

QUERYING THE BN

The BN gives P(T|C) What about P(C|T)?

Cavity

Toothache

P(C)

0.1

C P(T|C)

TF

0.40.01111

BAYES’ RULE

P(AB) = P(A|B) P(B)= P(B|A) P(A)

So… P(A|B) = P(B|A) P(A) / P(B)

APPLYING BAYES’ RULE Let A be a cause, B be an effect, and let’s say we

know P(B|A) and P(A) (conditional probability tables)

What’s P(B)?



What’s P(B)? P(B) = Sa P(B,A=a) [marginalization]

P(B,A=a) = P(B|A=a)P(A=a) [conditional probability]

So, P(B) = Sa P(B | A=a) P(A=a)



What’s P(A|B)?

HOW DO WE READ THIS?

P(A|B) = P(B|A)P(A) / [Sa P(B | A=a) P(A=a)] [An equation that holds for all values A can take on,

and all values B can take on] P(A=a|B=b) =



and all values B can take on] P(A=a|B=b) = P(B=b|A=a)P(A=a) /

[Sa P(B=b | A=a) P(A=a)]

Are these the same a?




[Sa P(B=b | A=a) P(A=a)]

Are these the same a?

NO!




[Sa’ P(B=b | A=a’) P(A=a’)]

Be careful about indices!

QUERYING THE BN The BN gives P(T|C) What about P(C|T)? P(Cavity|Toothache) =

P(Toothache|Cavity) P(Cavity)

P(Toothache)

[Bayes’ rule]

Querying a BN is just applying Bayes’ rule on a larger scale…

Cavity

Toothache

P(C)

0.1

C P(T|C)

TF

0.40.01111 Denominator computed by

summing out numerator over Cavity and Cavity

PERFORMING INFERENCE

Variables X Have evidence set E=e, query variable Q Want to compute the posterior probability

distribution over Q, given E=e Let the non-evidence variables be Y (= X \ E) Straight forward method:

1. Compute joint P(YE=e)2. Marginalize to get P(Q,E=e)3. Divide by P(E=e) to get P(Q|E=e)

INFERENCE IN THE ALARM EXAMPLE

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(J|M) = ??

Query Q

Evidence E=e


B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(J|MaryCalls) = ??


2. P(J,MaryCalls) =Sa,b,e P(J,A=a,B=b,E=e,MaryCalls)

2 entries:one for JohnCalls,the other for JohnCalls


B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(J|MaryCalls) = ??


2. P(J,MaryCalls) =Sa,b,e P(J,A=a,B=b,E=e,MaryCalls)

3. P(J|MaryCalls) = P(J,MaryCalls)/P(MaryCalls)= P(J,MaryCalls)/(SjP(j,MaryCalls))

HOW EXPENSIVE?

P(X) = P(x1x2…xn) = Pi=1,…,n P(xi|parents(Xi))

Straightforward method:1. Use above to compute P(Y,E=e)2. P(Q,E=e) = Sy1 … Syk P(Y,E=e)

3. P(E=e) = Sq P(Q,E=e) Step 1: O( 2n-|E| ) entries!

Normalization factor – no big deal once we have P(Q,E=e)

Can we do better?

VARIABLE ELIMINATION

Consider linear network X1X2X3

P(X) = P(X1) P(X2|X1) P(X3|X2)

P(X3) = Σx1 Σx2 P(x1) P(x2|x1) P(X3|x2)

VE IN ALARM EXAMPLE

P(E|j,m)=P(E,j,m)/P(j,m) P(E,j,m) = ΣaΣb P(E) P(b) P(a|E,b) P(j|a) P(m|a)

VE IN ALARM EXAMPLE


= P(E) Σb P(b) Σa P(a|E,b) P(j|a) P(m|a)

VE IN ALARM EXAMPLE



= P(E) Σb P(b) P(j,m|E,b) Compute for all values of E,b

VE IN ALARM EXAMPLE



= P(E) Σb P(b) P(j,m|E,b)

= P(E) P(j,m|E) Compute for all values of E

WHAT ORDER TO PERFORM VE?

For tree-like BNs (polytrees), order so parents come before children # of variables in each intermediate probability

table is 2^(# of parents of a node) If the number of parents of a node is

bounded, then VE is linear time!

Other networks: intermediate factors may become large

NON-POLYTREE NETWORKS

P(D) = Σa Σb Σc P(A)P(B|A)P(C|A)P(D|B,C) = Σb Σc P(D|B,C) Σa P(A)P(B|A)P(C|A)

A

B C

D

No more simplifications…

APPROXIMATE INFERENCE TECHNIQUES

Based on the idea of Monte Carlo simulation Basic idea:

To estimate the probability of a coin flipping heads, I can flip it a huge number of times and count the fraction of heads observed

Conditional simulation: To estimate the probability P(H) that a coin picked

out of bucket B flips heads, I can:1. Pick a coin C out of B (occurs with probability P(C))2. Flip C and observe whether it flips heads (occurs

with probability P(H|C))3. Put C back and repeat from step 1 many times4. Return the fraction of heads observed (estimate of

P(H))

APPROXIMATE INFERENCE: MONTE-CARLO SIMULATION

Sample from the joint distribution

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=0A=0J=1M=0


As more samples are generated, the distribution of the samples approaches the joint distribution!

B=0E=0A=0J=1M=0

B=0E=0A=0J=0M=0

B=0E=0A=0J=0M=0

B=1E=0A=1J=1M=0


Inference: given evidence E=e (e.g., J=1) Remove the samples that conflict

B=0E=0A=0J=1M=0

B=0E=0A=0J=0M=0

B=0E=0A=0J=0M=0

B=1E=0A=1J=1M=0

Distribution of remaining samples approximates the conditional distribution!

HOW MANY SAMPLES?

Error of estimate, for n samples, is on average

Variance-reduction techniques

RARE EVENT PROBLEM:

What if some events are really rare (e.g., burglary & earthquake ?)

# of samples must be huge to get a reasonable estimate

Solution: likelihood weighting Enforce that each sample agrees with evidence While generating a sample, keep track of the

ratio of(how likely the sampled value is to occur in the real world)

(how likely you were to generate the sampled value)

LIKELIHOOD WEIGHTING

Suppose evidence Alarm & MaryCalls Sample B,E with P=0.5

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

w=1



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=1

w=0.008



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=1A=1

w=0.0023

A=1 is enforced, and the weight updated to reflect the likelihood that this occurs



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=1A=1M=1J=1

w=0.0016



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=0

w=3.988



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=0A=1

w=0.004



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=0E=0A=1M=1J=1

w=0.0028



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=1E=0A=1

w=0.00375



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=1E=0A=1M=1J=1

w=0.0026



B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

B=1E=1A=1M=1J=1

w=5e-7



N=4 gives P(B|A,M)~=0.371 Exact inference gives P(B|A,M) = 0.375

B=0E=1A=1M=1J=1

w=0.0016

B=0E=0A=1M=1J=1

w=0.0028

B=1E=0A=1M=1J=1

w=0.0026

B=1E=1A=1M=1J=1

w~=0

RECAP

Efficient inference in BNs Variable elimination Approximate methods: Monte-Carlo sampling

NEXT LECTURE

Statistical learning: from data to distributions R&N 20.1-2

I NFERENCE IN B AYESIAN N ETWORKS. A GENDA Reading off independence assumptions Efficient inference...

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