# Marginalizing the product of conditional probabilities

When learning about Bayesian networks, I come across a statement where one of the term $$X_2$$ was marginalized away:

$$\sum_{X_2} P(X_3|X_2)P(X_2|X_1) = P(X_3|X_1)$$

It is not clear to me why this is so. From the definition of conditional probability, I get: $$\sum_{X_2} P(X_3|X_2)P(X_2|X_1) = \sum_{X_2} \frac{P(X_3,X_2)}{P(X_2)}\frac{P(X_2,X_1)}{P(X_1)}$$

I know the equation for marginalizing for a joint probability, $$\sum_{X_2}P(X_1,X_2) = P(X_1)$$

But I don't know how to combine these together and handle the product between probabilities to prove the 1st statement

First check that the kwown marginalization formula $$\sum_{X_2}P(X_2, X_3) = P(X_3) \tag1$$
$$\sum_{X_2}P(X_2,X_3 | X_1) = P(X_3| X_1) \tag2$$
Then notice that $$P(X_2,X_3 | X_1) = P(X_3 | X_2, X_1) P(X_2 | X_1)$$ ( again, this is the same as the known formula $$P(X_2,X_3) = P(X_3 | X_2) P(X_2)$$, only that everything is aditionally conditioned on $$X_1$$)
Hence the formula you post is not true in general, it's true only if $$P(X_3 | X_2, X_1)=P(X_3 | X_2)$$ , that is, if $$X_1 \to X_2 \to X_3$$ (Markov property).