Converting joint probability to conditional: Is P(A and B | C) = P(A| B,C)P(B)?

Question

This question came up in the context of Bayesian networks. If I have a network where variable C is dependent on both A and B, and I want to find P(A,B|C) (probability of A and B are true given C is true), is it valid to say that it's equal to the probability of A being true given both B and C are true, multiplied by P(B)?

So: $P(A,B \ |\ C) = P(A|B,C)P(B)$ ?

I got this idea from the formula I already know to be true, $P(A,B) = P(A|B)P(B)$. I'm wondering if this can be extended to the case where A and B are conditioned on another variable, and why it is true or not true.

Thanks in advance!

Answer 1

No, this can't hold. Consider e.g. a situation where $A,B,C$ are all the same event. Then $P(A\land B\mid C)$ and $P(A\mid B\land C)$ are both $1$, and your proposed equality becomes $1=P(B)$.

The right relativization of $P(A\land B)=P(A\mid B)P(B)$ to $C$ would be $$P(A\land B\mid C)=P(A\mid B\land C)P(B\mid C)$$

Answer 2

No. Counterexample:

In this example, $\mathbb P(A \cap B \mid C) = 1/3$, $\mathbb P(A \mid B \cap C) = 1/2$, and $\mathbb P(B) = 1/2$.

Answer 3

Assume $P(A\cap B\cap C) > 0$.

Then \begin{align*} P(A,B|C) &=P\bigl((A\cap B)|C\bigr)\\[4pt] &=\frac{P\bigl((A\cap B)\cap C\bigr)}{P(C)}\\[4pt] &=\frac{P(A\cap B \cap C)}{P(C)}\\[4pt] \end{align*} and \begin{align*} P(A|B,C)P(B) &=P\bigl(A|(B\cap C)\bigr)P(B)\\[4pt] &=\frac{P\bigl(A\cap(B \cap C)\bigr)}{P(B\cap C)}P(B)\\[4pt] &=\frac{P(A\cap B \cap C)}{P(B\cap C)}P(B)\\[4pt] \end{align*} hence \begin{align*} &P(A,B|C)=P(A|B,C)P(B)\\[4pt] \iff\;& \frac{P(A\cap B \cap C)}{P(C)} = \frac{P(A\cap B \cap C)}{P(B\cap C)}P(B) \\[4pt] \iff\;&P(B\cap C)=P(B)P(C)\\[4pt] \end{align*} which is true if and only if $B,C$ are independent.

Answer 4

Normally $\mathsf P_{A,B\mid C}(a,b\mid c)=\mathsf P_{A\mid C}(a\mid b,c)~\mathsf P_{B\mid C}(b\mid c)$, unless some form of conditional independence applies (and such will be indicated by a Bayesian Directed Acyclic Graph).

If I have a network where variable C is dependent on both A and B, and I want to find P(A,B|C)

If your network is the following and the factorisation is $\mathsf P_{A,B,C}(a,b,c)=\mathsf P_A(a)~\mathsf P_B(b)~\mathsf P_{C\mid A,B}(c\mid a,b)$ $$\require{enclose}\def\circle{\enclose{circle}}\raise{2ex}{\circle A} \searrow\lower{2ex}{\circle C} \swarrow\raise{2ex}{ \circle B}$$

Then indeed the diagram states that A and B are conditionally independent when given C, and an alternate factorisation would be $\mathsf P_{A,B,C}(a,b,c)=\mathsf P_{A\mid C}(a\mid c)~\mathsf P_{B\mid C}(b\mid c)~\mathsf P_C(c)$. Thus...

$$\mathsf P_{A,B\mid C}(a,b\mid c)=\mathsf P_{A\mid C}(a\mid c)~\mathsf P_{B\mid C}(b\mid c)$$