Given events $A$ and $B$, if $P(A\mid B)$ and $P(B\mid A)$ are known, can $P(A \cap B)$ be found? I tried the following approach and came to an answer, but doubt its veracity. Here's my attempt:
I tried thinking of $P(A\mid B)$ as $P(B \implies A)$. This is logically equivalent to $(1-P(A))+P(B)-(1-P(A))P(B)$, right? This comes from $w \implies z = \neg w \lor z$ for Boolean variables. Then it treats $X$ and $Y$ as independent variables, and uses the common formula for the probability of event one or event two happening, $P(A\cup B) = P(A)+P(A)-P(A)P(B)$. In addition, it uses $P(\neg A)=1-P(A)$.
Taking that approach, if you assign $P(A\mid B)=a$ and $P(B\mid A)=b$, this system of equations results:
$$ 1-P(A)+P(B)-(1-P(A))P(B) = a $$ $$ 1-P(B)+P(A)-(1-P(B))P(A) = b $$
Well, this can be solved uniquely for $P(A)$ and $P(B)$. Given that solving this system with substitution involves solving a quadratic, the result is
$$ P(A\cap B) = \frac{3}{2} a + \frac{3}{2} b - 2 + \frac{1}{2}(a-b)^2 $$
OR
$$ P(A\cap B) = -\frac{1}{2} a - \frac{1}{2} b + 1 - \frac{1}{2}(a-b)^2 $$
Which looks dubious, since the answer here seems like it should be unique. Moreover, one but not both of these formulas give negative probability for certain combinations of $a$ and $b$.
Well, I believe applying Bayes' theorem is the right way to solve this problem, but it requires knowing either $P(A)$ or $P(B)$ beforehand. Is there a way to obtain $P(A\cap B)$ without the individual probabilities? In addition, what's wrong with the approach above? I assume it has to do with assuming independence between $A$ and $B$.
It's not possible to find $P(X\cap Y)$ given $P(X\mid Y)$ and $P(Y\mid X)$. To see this, look what happens when $X=Y$ - then, as long as $P(X)\neq 0$, we have $P(X\mid Y)=P(Y\mid X) = 1$. From this you can't deduce what $P(X\cap Y)=P(X)$ is.