Chain rule for Bayesian Networks

Question

Suppose we have a simple Bayesian Network as follows: $X_1$ --> $X_3$ <-- $X_2$. Using the chain rule of Bayesian Networks, we can say the following:

$$ f(x_1,x_2,x_3) = f(x_1) f(x_2) f(x_3 | x_1, x_2) \ \ \ \ \ \ \ \ \ \ \ \ [1]$$

We know that

$$ f(x_3 | x_1, x_2) = \frac{f(x_3,x_1,x_2)}{f(x_1,x_2)} \ \ \ \ \ \ \ \ \ \ \ [2]$$

Thus, for Equation 1 above to be valid, $f(x_1,x_2)$ must be equal to $f(x_1) * f(x_2)$, implying that $X_1$ must be independent to $X_2$.

I want to confirm that the following logic for $X_1$ being independent to $X_2$ is correct:

In the Directed Acyclic Graph (DAG) above, $X_1$ is indeed independent to $X_2$ because of d-separation. $X_1$ and $X_2$ are d-separated as long as $X_3$ is not observed, which is the scenario when computing $f(x_1,x_2)$.

This seems somewhat subtle to me, I haven't seen this explicitly stated in any notes that I found on the Internet ...

Answer 1

Your logic sounds correct, however I would like to add a few things.

• The chain rule results in $f(x_1|x_2,x_3)f(x_2|x_3)f(x3)$, basically its the entire parameter space.

The DAG $X_1 \rightarrow X_3 \leftarrow X_2$ will fall under 1 of these 2 cases:

• if $X_3$ is not in the conditioning set $Z$, the path (that's all paths in this case) from $X_1$ to $X_2$ will be inactive, therefore we can say $X_1\!\perp\!\!\!\perp X_2 | X_3 \implies P(X_1,X_2,X_3) =P(X_1|X_3)*P(X_2|X3)$
• if $X_3$ or any of its' descendants are in the conditioning set $Z$, we have an active path, so we cannot guarantee independence as in the first case, note however that this does not imply the opposite.