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 ...