4

Is the following statement true? How? I'm having trouble seeing whether not it is true or false.

$$P(A\mid B) = P(A\mid B \cap C)P(C\mid B) + P(A\mid B \cap C')P(C'\mid B)$$

Michael Hardy
  • 1
britt
  • 79

4 Answers4

3

It is true, generally speaking:

\begin{align} P(A \mid B) & = \frac{P(A, B)}{P(B)} \\ & = \frac{P(A, B, C)+P(A, B, C')}{P(B)} \\ & = \frac{P(A, B, C)}{P(B)}+\frac{P(A, B, C')}{P(B)} \\ & = \frac{P(A, B, C)}{P(B, C)} \cdot \frac{P(B, C)}{P(B)} + \frac{P(A, B, C')}{P(B, C')} \cdot \frac{P(B, C')}{P(B)} \\ & = P(A \mid B, C) \cdot P(C \mid B) + P(A \mid B, C') \cdot P(C' \mid B) \end{align}

Note any potential gotchas in the denominators.

Brian Tung
  • 34,160
1

$$P(A|B\cap C)P(C|B)=\frac{P(A\cap B\cap C)}{P(B\cap C)}\cdot{\frac{P(B\cap C)}{P(B)}}=\frac{P(A\cap B\cap C)}{P(B)} \\ P(A|B\cap C')P(C'|B)=\frac{P(A\cap B\cap C')}{P(B\cap C')}\cdot\frac{P(B\cap C')}{P(B)}=\frac{P(A\cap B\cap C')}{P(B)}$$Hence RHS is $$\frac{P(A\cap B\cap C)}{P(B)}+\frac{P(A\cap B\cap C')}{P(B)}=\frac{P(A\cap B)}{P(B)}=P(A|B)$$

Galc127
  • 4,451
0

Here's a handy device: Write (1) $P_B(A)$ in place of $P(A\mid B)$ and (2) $P_B(A\mid C)$ in place of $P(A\mid B\cap C)$. Then the identity in question becomes clearer: $$ P_B(A) = P_B(A\mid C) P_B(C) + P_B(A\mid C') P_B(C'), $$ which is true because $P_B$ is a probability.

You can think of moves (1) and (2) as mentally erasing the conditioning on $B$. (Note that move (2) is justified by (1) and the definition of conditional probability.) Once you remember this device it becomes a lot easier to prove, or derive, similar expressions from other identities that you know to be true.

grand_chat
  • 38,951
0

To literally see this, I drew a diagram. First, notice that $$P(A|BC)P(C|B) = \frac{P(ABC)}{P(BC)}\frac{P(BC)}{P(B)} = \frac{P(ABC)}{P(B)} = \frac{P(AC|B)P(B)}{P(B)} = P(AC|B).$$ This is saying the probability of $A$ and $C$ given $B$. Similarly $P(A|B C')P(C'|B) = P(AC'|B)$, which says the probability of $A$ and $C'$ given $B$. Then we are just partitioning $A|B$ into the part that includes $C$ and the part that excludes $C$. ABC $$P(A|B) = P(AC|B)+P(AC'|B).$$

David
  • 401