Reasoning under Uncertainty

• Variable elimination is an exact method. It multiplies the relevant conditional probability tables and sums out the hidden variables one at a time, so A is false.
• What it returns is a conditional probability such as \(P(Q \mid e)\) for a query \(Q\) given evidence \(e\), so C is true.
• Gibbs sampling and rejection sampling both estimate probabilities from random samples. Their answers are approximate and improve as more samples are drawn, so B is false and D is true.

Answer: C, D

The factorisation is the Bayesian network \(U \rightarrow W \leftarrow V\), \(W \rightarrow X\), \(W \rightarrow Y\).

• A and B: \(U\) and \(V\) reach \(X\) and \(Y\) only through chains such as \(V \rightarrow W \rightarrow Y\), and observing \(W\) blocks a chain. Both are true.
• D: \(X \leftarrow W \rightarrow Y\) has \(W\) as a common cause, so \(X\) and \(Y\) are independent once \(W\) is known. True.
• C: \(U\) and \(V\) are independent on their own (the factor \(P(U)P(V)\)), but \(W\) is their common effect. Observing a common effect couples its causes (explaining away): once \(W\) is known, learning \(U\) changes what \(V\) is likely to be. So C does not hold in general.

Answer: C

A Bayesian network factorises the joint probability into one factor per node, each conditioned on that node's parents. Here U has no parents, V and W each have parent U, and Z has parents V and W:

\[ P(U, V, W, Z) = P(U)\,P(V|U)\,P(W|U)\,P(Z|V, W) \]

Reading the tables for \(U = 1, V = 1, W = 1, Z = 1\):

• \(P(U = 1) = 0.5\)
• \(P(V = 1|U = 1) = 0.5\)
• \(P(W = 1|U = 1) = 1\)
• \(P(Z = 1|V = 1, W = 1) = 0.5\)

So the joint probability is \(0.5 \times 0.5 \times 1 \times 0.5 = 0.125\).

Answer: 0.125