Naive Bayes Derivation I was going over the derivation of Naive Bayes, and the following 3 lines were given:
Suppose $X = \left < X_1, X_2 \right>$
\begin{align}
P(X|Y) &= P(X_1, X_2 | Y)  \\[2pt]
       &= P(X_1 | X_2, Y)P(X_2 | Y)  \\[2pt]
       &= P(X_1 | Y)P(X_2 | Y)
\end{align}
So the third line comes from the fact that we have made the assumption that each $X_i$ is conditionally independent given Y, but how was line 2 derived? 
 A: I've figured it out.
The second line comes from the "chain rule" of probability which states that:
$$P(A, B, \dots, Z) = P(A |B,\dots,Z) \times P(B|C,\dots,Z) \times \dots \times P(Y|Z) \times P(Z)$$
In fact, this can be proven by expanding each term in the equality:
$$P(A, B, ..., Z) = \frac{P(A |B,\dots,Z)}{P(B,\dots,Z)} \times \frac{P(B |C,\dots,Z)}{P(C,\dots,Z)} \times \dots \times \frac{P(Y|Z)}{P(Z)} \times P(Z)$$
and we can see that the terms cancel out.
My main confusion stemmed from the fact that I didn't know that the commas were in fact a notation for the intersection. This post clarifies it: 

Now, by common convention, a list of events is interpreted as the intersection (AND) of those events, such that $P(E_1,E_2) = P(E_1 \cap E_2)$ or, using logical connectives instead of set-theory ones, $P(E_1 \land E_2)$.  However, that convention is by no means universal, so if you want to be sure to avoid ambiguity, you should explicitly use $\cap$ (or $\land$) to denote the intersection of events.

