Is $Pr(A | B , C) = Pr(A | B) \cdot Pr(B | C)$ a valid thing to write?

Question

The question heading actually highlights my point of confusion.

Answer 1

Suppose $X_1,X_2$ are independent coin throws, and take the events

$A: X_1=X_2$

$B: X_1\text{ is head }$

$C: X_2\text{ is head }$

Then $P(A|BC)=1$, and $P(A|B)=P(B|C)=\frac{1}{2}$

Answer 2

Definitely not.

Consider a simple example involving two independent coin flips (of a fair coin). If we set $A = (\text{Flip 1 is Tails})$ , $B = (\text{Flip 1 is Tails})$, and $C = (\text{Flip 2 is Tails})$, then we find $P(A|B\text{ and } C) = 1$, $P(A|B) = 1$, and $P(B|C) = 1/2$. In this scenario, we have $$ P(A|B\text{ and } C) = 1 $$ but $$ P(A|B)P(B|C) = 1\cdot 1/2 = 1/2\ne 1. $$

Answer 3

To complement the counterexamples in the other answers, I would also suggest writing out the definitions to see why you shouldn't expect this to hold in general. Recall $$P(D|E):=\frac{P(D\cap E)}{P(E)}.$$

So assuming $B,C$ have positive probability, you can write $$P(A|B,C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}=\frac{P(A\cap B|C)}{P(B|C)}=\frac{P(A\cap C|B)}{P(C|B)}.$$

If you assume $A,C$ are conditionally independent (conditional on $B$), then this simplifies to $P(A|B)$. That's about the closest you can get to something that looks like your expression.