Does P(A and B|C)=P(A|C) * P(B|C) hold true if A and B are independent?

Question

Common sense tells me it's true, but can't rely just on that.

I have the classic problem with a disease (d) and 4 symptoms ($s_1,s_2,s_3,s_4$). I know the probability of each symptom manifesting given that you have the disease, the probability that you have the disease and probabilities that you have each symptom given that you don't have the disease.

The question is what is the probability that you have the disease, given that you present all 4 symptoms? So $P(d|s_1 \cap s_2 \cap s_3 \cap s_4)$=?

Started with

$P(d|s_1 \cap s_2 \cap s_3 \cap s_4)=\frac{P(s_1 \cap s_2 \cap s_3 \cap s_4|d)P(d)}{P(s_1 \cap s_2 \cap s_3 \cap s_4)}$

I found $P(s_1 \cap s_2 \cap s_3 \cap s_4)$, but I'm stuck at finding $P(s_1 \cap s_2 \cap s_3 \cap s_4|d)$ and I was thinking if I can just use:

$P(s_1 \cap s_2 \cap s_3 \cap s_4|d)=P(s_1|d)P(s_2|d)P(s_3|d)P(s_4|d)$

Answer 1

It appears that the original question has changed a lot since this answer was written.

Let's assume that $C$ has nonzero probability. Then, we may observe that $$ P(A\cap B|C)=\frac{P(A\cap B\cap C)}{P(C)}. $$ On the other hand, $$ P(A|C)P(B|C)=\frac{P(A\cap C)P(B\cap C)}{P(C)^2}. $$ Therefore, the two expressions that you write are equal if and only if $$ P(A\cap B\cap C)P(C)\stackrel{?}{=}P(A\cap C)P(B\cap C). $$ This equation looks a bit suspect. In particular, it looks like one might need $C$ to be independent of $A\cap B$ or perhaps that all the events are mutually independent. So, we try to find a counterexample.

Let $X$ and $Y$ be independent Bernoulli random variables which take values $\pm 1$ with probability $\frac{1}{2}$ (in other words, coin flips). Let \begin{align} A&=\text{The event where }X=1\\ B&=\text{The event where }Y=1\\ C&=\text{The event where }XY=-1. \end{align} These are pairwise independent, but knowing any two allows you to know the third. In our case, \begin{align} P(A\cap B\cap C)&=0\\ P(C)&=\frac{1}{2}\\ P(A\cap C)&=\frac{1}{4}\\ P(B\cap C)&=\frac{1}{4}. \end{align}

In this case, the necessary equality doesn't hold and the two expressions are not equal.

You can also go back and check the original expressions: \begin{align} P(A\cap B|C)&=0\\ P(A|C)&=\frac{1}{2}\\ P(B|C)&=\frac{1}{2}, \end{align} but $0\not=\frac{1}{4}$.

Answer 2

$P(a)p(a|c)=P(a|c)P(b|c)$ when $a$ and $b$ are independent events. $P(a|b)=P(a)$ and $P(b|a)=P(b)$. So:

• question $P(a)P(a|c)=P(a|c)P(b|c)$
• answer $P(a)P(a)=P(a)P(b)$

So it is proved that they are equal.