We have two random variables $X \in R$ and $Y \in R$.
1.) If we know the joint pdf $f_{X,Y}(x,y)$, can we find the conditional pdf $f_{X \mid Y}(x \mid y)$? From the Bayes rule we have
$f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} $
We don't know the $f_X(x)$, but we can find it using
$f_Y(y)=\int\limits_{-\infty}^\infty f_{X,Y}(x,y)dx$.
Is this correct?
2.) If we know the conditional pdf $f_{X \mid Y}(x\mid y)$, can we find the joint pdf $f_{X,Y}(x,y)$? From the Bayes rule we have
$f_{X,Y}(x,y) = f_{X \mid Y}(x\mid y) f_Y(y)$
but we don't know the $f_Y(y)$.
Can we find it?
Best Answer
For 1),$$f_{X|Y}(x|y)=\frac{d}{dx}P\left(X\le x|Y=y\right)=\frac{f_{X,\,Y}(x,\,y)}{f_Y(y)}=\frac{f_{X,\,Y}(x,\,y)}{\int_{\Bbb R}f_{X,\,Y}(x^\prime,\,y)dx^\prime}.$$In answer to 2), we cannot in general obtain $f_{X,\,Y}(x,\,y)$ from $f_{X|Y}(x|y)$. For example, if $f_{X|Y}(x|y)$ is $y$-independent, it can be taken as the PDF of $X$, but the pdf of $Y$ can be chosen arbitrarily, viz. $f_{X,\,Y}(x|y)=f_{X}(x)f_Y(y)$ for an arbitrary PDF $f_Y$.