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?

  • $\begingroup$ You can well write in (1) that $f_{X}(x) = \int_{-\infty}^{\infty} f_{X, Y}(x, y) \, dy$. $\endgroup$ – Rohan Oct 10 '19 at 11:13
  • $\begingroup$ Thank you @Rohan. Corrected. $\endgroup$ – tgeorgiop Oct 10 '19 at 11:18

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 $f_Y$.


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