Why does $P(A,B | C) = P(A|C) \cdot P(B|C)$

Question

I'm in an NLP course learning about naive bayes statistics. We briefly went over joint and conditional probabilities.

Why does $P(A,B | C) = P(A|C)\cdot P(B|C)$

Answer 1

That's not true in general. For example, take the three following events from rolling a six-sided die:

• A = "Is odd"
• B = "Is prime"
• C = "Is at least 3"

Then $P(A|C) = 1/2$ and $P(B|C) = 1/2$, but $P(A,B|C) = 1/2$ as well.

The equation $P(A|C) P(B|C) = P(A,B|C)$ is the definition of what it means for $A$ and $B$ to be independent events when conditioned on $C$.