I have computed $\frac{g(y\mid x)f(x)}{g(y)}$, where $g(y\mid x)$, $f(x)$, and $g(y)$ are density functions (i.e. they integrate to 1 and the functions are always $\geq0$).
What assumptions must I make to conclude that this fraction is equal to $f(x\mid y)$? Since the resulting distribution doesn't integrate to 1 (not even close), it seems as if I'm missing some unknown assumption....
Does the fact that $g(y)$ is only positive when $y\subseteq[0,1]$ have any special relevance? (Note $f(x)>0$ for all $x$).