Can we simplify chain rule in probability?

Question

I have:

$ P(A,B,C) = P(A|B,C)*P(B|C) * P(C) $

According to chain rule we isolate first random variable and condition it to the rest (joint combinations). etc. etc. But then I was thinking, what if I would joint first two variables and condition it to the last in this manner:

$ P(A,B,C) = P(A,B|C) * P(C) $

This would be legit if

$P(A|B,C)*P(B|C) * P(C) = P(A,B|C) * P(C)$

which infact is. My question is, so this is a legit way of using chain rule or I was just lucky?

Answer 1

Short version: yes, this is legitimate.

Long version:

The main identity here is $P(A \cap B)=P(A \mid B) P(B)$. This identity also holds if all of these $P$'s are conditioned on some fixed event. So you can write $P(A \cap B \cap C)$ as $P(D \cap C)$ where $D=A \cap B$, and then this same identity gives $P(D \cap C)=P(D \mid C)P(C)$ which is what you wrote as $P(A,B \mid C)P(C)$. You can then do it again to get $P(A \cap B \mid C)=P(A \mid B \cap C) P(B \mid C)$, which when substituted in gives your first equation.

Answer 2

It's correct. If we define $E = A \cap B$, then your equation states that $P( E \cap C) = P( E \mid C ) P(C)$, which you already know is correct.