# What is $P(A | B,C)$ when $B$, $C$ are both independent

Why is it true that $P(A|B,C) \neq P(A|B) P(A|C)$ when $B$ and $C$ are independent?

Intuitively I can understand that the LHS probability should be greater than the RHS and the above equation doesn't hold that. But is there any proof to convince me.

Another question: say if $A$ and $C$ are independent then can I tell $P(A|BC) = P(A|B)$?

• Do you intend that "/" means a conditional probability and "BC" refers to the conjunction of events "B" and "C"? If so, have you considered the validity of this first equality when "A" and "B" are the same event?
– whuber
Commented Dec 13, 2017 at 19:15
• @whuber - thanks and I have corrected it now. Commented Dec 14, 2017 at 3:19

I.

$$P(A|B,C) = \frac{P(A,B,C)}{P(B,C)} = \frac{P(A,B,C)}{P(B)P(C)}$$ (1)

$$P(A|B)P(A|C) = \frac{P(A,B)P(A,C)}{P(B)P(C)}$$ (2)

Equality depends on $$P(A,B,C) = P(A,B)P(A,C)$$, which usually does not hold.

II. still not true

Example: throw 4-sided dice

A = {get 1 or 2}

C = {get 1 or 3}

B = {get 1, 2 or 3}

you can check by definition that A,C are independent, but

P(A|B,C) = 1/2

P(A|B) = 2/3

A general comment on such statements of conditional probability: sometimes they may seem true, but if you cannot prove it from definition, they are usually not.

Edit: Minor fix replacing $$p(C)$$ with $$P(C)$$ in the first equation.

The whole concept of independence gets complicated once you move beyond 2 cases. The best example I have seen of this came from a textbook (I think it was by Feller):

Consider that your population is the following $(A,B,C)$ triplets:

$$(1,2,3); (1,3,2); (2,1,3); (2,3,1); (3,1,2); (3,2,1); (1,1,1); (2,2,2); (3,3,3)$$

So if you select the first triplet, then $A=1, B=2, C=3$.

It is simple to verify that $A$ and $B$ are independent (pairwise) since $p(A=a|B) = 1/3 = p(A=a)$, etc. Also $A$ and $C$ are pairwise independent and $B$ and $C$ are pairwise independent, but given any 2, you know exactly what the 3rd one will be (complete dependence).