Conditional Density

Question

Question:

X follows uniform [0, 1] and Y|X follows uniform[0,X]. What is the distribution of X|Y?

My Try: $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{1}{x}, \, with\, f_X(x)=1$$ $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$

I am not sure how to get $f_Y(y)$, any hints or help are appreciated!

Just use Bayes' rule $f(x|y)=\frac{f(y|x)f(x)}{\int f(y|x)f(x)\mathrm{d}x}$ and the fact that $X|Y=y$ is supported on $[y,1]$ — Matthew H., Sep 26 '21 at 23:33
Hi Matthew, Thanks for your help. One thing I am still not sure is that what are boundaries for the integral on the denominator — quant_to_be, Sep 28 '21 at 00:32

score 1 · Accepted Answer · answered Sep 28 '21 at 02:16

Suppose $X\sim \mathcal{U}(0,1)$ and $Y|X\sim \mathcal{U}(0,X)$.

First notice $f_{X,Y}(x,y)=f_{Y|X}(y|x)f_{X}(x)=\frac{1}{x}\cdot 1_{\{0 < y < x,0< x < 1\}}$ so for any $x\in (y,1)$ we have $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}=\frac{f_{X,Y}(x,y)}{\int_{y}^1f_{X,Y}(x,y)\mathrm{d}x}=-\frac{1}{x \ln (y)}$$

Conditional Density

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