Is conditional probability $P(A\mid B)$ proportional to $P(B\mid A)$?

Question

It feels a bit odd but since $$P(A\mid B) = \frac{P(A,B)}{\sum_A P(A,B)} \propto P(A,B)\text{ and }P(B\mid A) = \frac{P(A,B)}{\sum_B P(A,B)} \propto P(A,B)$$

can we say that $P(A\mid B) \propto P(B\mid A)$? Is this correct?

Answer 1

"Proportional to" is a phrase that should never be uttered without specifying some context.

$f(x,y)$ may be proportional to $g(x,y)$ as a function of one of the two variables and not of the other. If one has $f(x,y) = h(y) g(x,y)$ for all $x,y$, then $f(x,y)\propto g(x,y)$ as a function of $x$ but not as a function of $y$.

It makes no sense to say one number is proportional to another. One function of a variable may be proportional to another.

If one has \begin{align} \Pr( A_i\mid B) & = \frac{\Pr(B\mid A_i)\Pr(A_i)}{\Pr(B\mid A_1)\Pr(A_1) + \cdots + \Pr(B\mid A_n)\Pr(A_n)} \\[10pt] & = \frac{\Pr(B\cap A_i)}{\Pr(B\cap A_1)+\cdots+\Pr(B\cap A_n)} = \frac{\Pr(B\cap A_i)}{\Pr(B)} \end{align} and thus $$ \Pr( A_i\mid B) = \frac{\Pr(B\cap A_i)}{\Pr(B)}. $$ then one can say that $\Pr(A_i\mid B)$ is proportional to $\Pr(B\cap A_i)$ as a function of $A_i$ or as a function of $i$. The thing that makes the denominator "constant" is that it doesn't change as $i$ goes from $1$ to $n$, or as $A_i$ runs through the list $A_1,\ldots,A_n$.

See if you can re-write your argument for proportionality between $\Pr(A\mid B)$ and $\Pr(B\mid A)$ in a way that bears all this in mind, and you will learn that you are not just trying to prove something that is false, but rather something that doesn't even say anything. As a famous physicist (was it Wigner?) once said, it isn't even wrong.