# From Joint pdf to Conditional pdf and from Conditional pdf to Joint pdf

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?

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