Is it true that $P(A|B) = P(A|C) \cdot P(C|B) $?

Question

I think that $$P(A|B) = P(A|C) \cdot P(C|B) $$ is True.

You are just transforming the information from $B$ through $C$.
Is this correct and if it is, what's the name for this property?

Answer 1

It is not true.

Let $P(C|B)=0$ and let $P(A|B) >0$ and we can see a counterexample.

We have

$$P(A|B) = P(A|B, C)P(C|B) + P(A|B, C^c)P(C^c|B)$$

Answer 2

It is not necessarily true. We have

$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$ $$P(A|C) = \frac{P(A\cap C)}{P(C)}$$ $$P(C|B) = \frac{P(B\cap C)}{P(B)}$$

And what you claimed is that $$\frac{P(A\cap B)}{P(B)} = \frac{P(A\cap C)}{P(C)}\cdot\frac{P(B\cap C)}{P(B)}$$ which can be false. For instance, if $A, B$ and $C$ are mutually independent events, then from this equality, we have $$P(A) = P(A)\cdot P(C) \implies P(C) = 1$$ but what happens if $P(C) < 1$?

Answer 3

Multiply both sides with $P(B)P(C)$:

• $P(A\mid B)P(B)P(C)=P(A\cap B)P(C)$

• $P(A\mid C)P(C\mid B)P(B)P(C)=P(A\cap B)P(C\cap B)$

Of course we do not have $P(C)=P(C\cap B)$ in general, so the statement is not true.