GATE 2024 DA Q.24: solution and answer

Official answer: CSource: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. Read the factorisation as a Bayesian network. Each factor $P(\text{child}\mid \text{parents})$ gives edges from parent to child, so the edges are $U\to W$, $V\to W$, $W\to X$ and $W\to Y$. The statement that is FALSE is the one that d-separation (or direct algebra) does not support.

2. (A) The only path from $V$ to $Y$ is $V\to W\to Y$. Here $W$ is a chain (pass-through) node, so conditioning on $W$ blocks the path. Hence $Y\perp V\mid W$ holds, and (A) is true.

3. (B) The only path from $U$ to $X$ is $U\to W\to X$. Again $W$ is a chain node, so conditioning on $W$ blocks it. Hence $X\perp U\mid W$ holds, and (B) is true.

4. (D) The only path from $X$ to $Y$ is $X\leftarrow W\to Y$, where $W$ is a common cause (fork). Conditioning on $W$ blocks it. Algebraically, $P(X,Y\mid W)=\frac{{\sum }_{U,V}P(U)P(V)P(W\mid U,V)P(X\mid W)P(Y\mid W)}{P(W)}=P(X\mid W)P(Y\mid W)$, because the factors $P(X\mid W)$ and $P(Y\mid W)$ do not involve $U$ or $V$. So (D) is true.

5. (C) The path $U\to W\leftarrow V$ has $W$ as a collider. Marginally $P(U,V)=P(U)P(V)$, so $U$ and $V$ are independent. Conditioning on a collider opens the path, so $U$ and $V$ are in general dependent given $W$.

6. Check with a counterexample. Let $U,V$ be independent fair bits and let $W=U\oplus V$, which is allowed because $P(W\mid U,V)$ is an arbitrary conditional distribution. Given $W=0$, we have $U=V$, so $P(U=1,V=1\mid W=0)=\frac{1}{2}$. But $P(U=1\mid W=0)P(V=1\mid W=0)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$. These differ, so $U/\perp V\mid W$.

7. In general $P(U,V\mid W)=\frac{P(U)P(V)P(W\mid U,V)}{P(W)}$, which factors into a function of $U$ times a function of $V$ only if $P(W\mid U,V)$ itself factors. The network does not guarantee this, so statement (C) is the false one.