CS 486/686 Lecture 7 — Independence and Bayesian Networks I

CS 486/686
Independence and
Bayesian Networks I

Yuntian Deng

Lecture 7

RN 12.4, 13.1–13.2 · PM 8.2–8.3

Reminders

Due tonight Tue Jun 2, 11:59 PM

Chat 3 (Probabilities) on Chrysalis.

Learning goals

Tell if two variables are independent, or conditionally independent given a third
Derive a compact representation of a joint distribution using independence
Describe the components of a Bayesian network
Compute joint probabilities from a Bayes net
Explain independence in the three key structures

Why Bayes nets? The \(2^n\) problem

A joint distribution over \(n\) Boolean variables needs \(2^n\) probabilities.

Variables	Probabilities
6 (Holmes)	64
10	1,024
20	~ 1 million
30	~ 1 billion

Independence lets us factor this huge joint into tiny pieces.

Unconditional independence

Independence. \(X\) and \(Y\) are (unconditionally) independent iff (any, and therefore all):

\(P(X | Y) = P(X)\)
\(P(Y | X) = P(Y)\)
\(P(X \wedge Y) = P(X)\, P(Y)\)

Learning \(Y\) does not change our belief about \(X\).

Two probabilities (the priors) suffice to recover four joint probabilities.

Conditional independence

Conditional independence. \(X\) and \(Y\) are conditionally independent given \(Z\) iff (any, and therefore all):

\(P(X | Y \wedge Z) = P(X | Z)\)
\(P(Y | X \wedge Z) = P(Y | Z)\)
\(P(X \wedge Y | Z) = P(X | Z)\, P(Y | Z)\)

Once we know \(Z\), learning \(Y\) doesn't change our belief about \(X\).

Important: independence does not imply conditional independence, and vice versa. We'll see both on later slides.

Quiz pack: compact representation

Three Boolean variables \(A, B, C\). How many probabilities specify the joint?

Q1. No independence assumed.

7. Chain rule: \(P(A) + P(B | A) + P(C | A \wedge B) = 1 + 2 + 4 = 7\).

Q2. \(A, B, C\) all mutually independent.

3. Just the priors: \(P(A), P(B), P(C) = 3\).

Q3. \(A, B\) conditionally independent given \(C\).

5. \(P(C) + P(A | C) + P(B | C) = 1 + 2 + 2 = 5\).

Bayes net = DAG + CPTs

A Bayesian network is a directed acyclic graph (DAG) in which:

each node is a random variable,
each edge \(X \to Y\) says "\(X\) directly affects \(Y\)" — \(X\) is a parent of \(Y\),
each node \(X_i\) carries a conditional probability table \(P(X_i \,|\, \mathrm{Parents}(X_i))\).

Examples of Bayes nets

Inheritance of handedness

Car diagnostic network

The Holmes Bayes net

P(E)= 0.0003

P(B)= 0.0001

P(R | ·)
E: 0.9 ¬E: 0.0002

P(A | ·)
B∧E: 0.96 B∧¬E: 0.95
¬B∧E: 0.20 ¬B∧¬E: 0.01

P(W | ·)
A: 0.80 ¬A: 0.40

P(G | ·)
A: 0.40 ¬A: 0.04

From 64 numbers to 12

Representation	Probabilities needed
Full joint over 6 Boolean variables	\(2^6 = 64\)
Holmes Bayes net: \(1 + 1 + 2 + 4 + 2 + 2\)	12

From those 12 numbers we can reconstruct every one of the 64 joint probabilities.

For larger networks, the savings are exponential.

Joint = product of conditionals

Bayes-net factorization. For a Bayes net over variables \(X_1, \ldots, X_n\):

\(P(X_1, \ldots, X_n) = \prod_{i=1}^{n} P(X_i \,|\, \mathrm{Parents}(X_i))\).

Each variable is conditionally independent of its non-descendants given its parents.

Worked example: a joint from the Holmes BN

Goal: \(P(\neg B \wedge \neg E \wedge A \wedge \neg R \wedge G \wedge W)\). All neighbours call but neither a burglary nor an earthquake actually happened.

Step 1. Factor by the BN structure (using parents):

\(= P(\neg B) \cdot P(\neg E) \cdot P(A | \neg B \wedge \neg E) \cdot P(\neg R | \neg E) \cdot P(G | A) \cdot P(W | A)\)

Step 2. Plug in numbers:

\(= 0.9999 \times 0.9997 \times 0.01 \times 0.9998 \times 0.40 \times 0.80\)

\(\approx \mathbf{3.2 \times 10^{-3}}\)

Q: a different all-negatives joint

Q. What is \(P(\neg B \wedge \neg E \wedge \neg A \wedge \neg R \wedge \neg G \wedge \neg W)\)?

0.5699
0.6699
0.7699
0.8699
0.9699

A — 0.5699. By the same factorization, each factor is now a "no event happened" version:
\(= 0.9999 \times 0.9997 \times (1 - 0.01) \times (1 - 0.0002) \times (1 - 0.04) \times (1 - 0.40) \approx 0.5699\).

Three key structures

Chain

\(A\) and \(C\) talk through \(B\).

Common cause

\(B\) and \(C\) share a hidden cause \(A\).

Common effect

\(A\) and \(B\) both cause \(C\) (a v-structure).

Each structure has its own independence pattern. Let's check.

Chain: \(B \to A \to W\)

Q1. Are \(B\) and \(W\) unconditionally independent?

No. Knowing \(B\) makes the alarm more likely, which makes \(W\) more likely. Information flows through \(A\).

Q2. Are \(B\) and \(W\) conditionally independent given \(A\)?

Yes. Once we know whether the alarm is going, \(W\) only depends on \(A\); \(B\) adds nothing more. The middle node "blocks" the path.

Common cause: \(W \leftarrow A \to G\)

Q1. Are \(W\) and \(G\) unconditionally independent?

No. If \(W\) is calling, the alarm is more likely going, so \(G\) is more likely too. The shared cause \(A\) couples them.

Q2. Are \(W\) and \(G\) conditionally independent given \(A\)?

Yes. Once \(A\) is known, the two callers are independent observers. The common cause "blocks" the path when conditioned.

Common effect: \(E \to A \leftarrow B\)

Q1. Are \(E\) and \(B\) unconditionally independent?

Yes. Earthquakes and burglaries don't influence each other — they only share the same downstream alarm.

Q2. Are \(E\) and \(B\) conditionally independent given \(A\)?

No — the opposite. Given the alarm, learning of an earthquake makes a burglary less likely: explaining away. Conditioning on the common effect creates a dependence.

The v-structure is the surprise: conditioning on the middle node opens a path instead of blocking it.

The three rules

Structure	Are endpoints independent?	Given the middle node?
Chain \(A \to B \to C\)	No	Yes blocked
Common cause \(B \leftarrow A \to C\)	No	Yes blocked
Common effect \(A \to C \leftarrow B\)	Yes	No opens up

In the chain and common-cause structures, knowing the middle node blocks information flow. In the v-structure, knowing the middle node opens it (explaining away).

Learning goals (recap)

✓ Tell if two variables are independent or conditionally independent
✓ Derive a compact representation using independence
✓ Describe the components of a Bayesian network
✓ Compute joint probabilities from a Bayes net
✓ Explain independence in the three key structures

Next: d-separation

The three structures cover paths of length 2. For arbitrary paths in a Bayes net, we generalize to d-separation — the rule for reading off all (conditional) independences from any DAG.