# A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

Let $$C$$, $$B$$, and $$A$$ be events in the same probability space, such that $$A$$ and $$B$$ are independent and $$P(A \cap C) > 0$$, $$P(B \cap C) > 0$$.

Prove or disprove:

$$P(A \cap B|C) = P(A|C)P(B|C).$$

• Very relevant: Wikipedia on Conditional independence. Your question can be rephrased as "If A and B are independent, then are A and B also conditionally independent given C?"
– Stef
Feb 13 at 10:26
• This is relevant in graphical models too
– qwr
Feb 14 at 18:11
• Duplicate (after some algebra): stats.stackexchange.com/questions/630951/… Feb 14 at 22:03

No this is not in general true, as you can see from a simple counter example:

Toss two independent coins.

Event $$A$$ is coin 1 head. $$P(A)=0.5$$

Event $$B$$ is coin 2 head. $$P(B)=0.5$$

Event $$C$$ is either coin heads. So $$C=A\cup B$$.

$$P(A \cap B \mid C) = 1/3$$

$$P(A \mid C) = 2/3$$

$$P(B \mid C) = 2/3$$

So $$P(A \cap B \mid C) \neq P(A\mid C) \times P(B\mid C)$$

This example is related to the idea of Berkson's bias. If you have two independent events, then conditioning on a third event that depends on both can induce a correlation between them. (See Berkson's paradox.)

Here's a drawing if it makes it easier to follow. $$A$$ is the blue area, $$B$$ is the red area, $$C$$ is the total shaded area. $$A$$ and $$B$$ are independent, but if we condition on $$C$$ (being in the shaded area), then $$A$$ and $$B$$ are no longer independent because if we know $$A$$ is false then $$B$$ must be true.

A related question is If $X, Y$ are independent of $Z$, is $P(X|Y, Z) = P(X|Y)$?

The example there, $$C = XOR(A,B)$$, is a simple counter example to this question as well

A B C   probability
0 1 1   1/4
1 0 1   1/4
0 0 0   1/4
1 1 0   1/4


Then

$$P(A=1 \text{ and } B=1|C=1) = 0$$

whereas

$$P(A=1|C=1) \cdot P(B=1|C=1) = 0.5 \cdot 0.5 = 0.25$$

Also related: