A broken stick problem: find distribution

[psie]

Homework Statement

A stick lies on the interval $[0,1]$, and is broken at the point $X\in U(0,1)$. The left part is then broken again at the point $Y\in U(0,X)$, i.e the conditional pdf of $Y$ given $X$ is $f_{Y|X=x}(y)= \mathbf{1}_{(0,x)}(y)\frac1x$, $0<y<x$, $0<x<1$. Find the unconditional distribution of $Y$.

Relevant Equations

We have $f_{Y|X=x}(y)=\frac{f_{X,Y}(x,y)}{f_X(x)}$. Since $X$ is uniformly distributed on $(0,1)$, $f_X(x)=1$ and thus $f_{Y|X=x}(y)=f_{X,Y}(x,y)$.

So I'd like to "integrate out" the $x$-variable, like $$f_Y(y)=\int_0^1 \mathbf{1}_{(0,x)}(y)\frac1x\,dx.$$ I am a bit hesitant on how to proceed, since I feel like I will get an unbounded density. Something's fishy, but I don't know what.

 

[Orodruin]

$\mathbf 1_{(0,x)}(y)$ is a function of both x and y. For what values of y is it non-zero?

Edit: I get something very well defined for the marginal distribution that very nicely integrates to 1.

 

[psie]

Ok, I get that $f_Y(y)$ is $0$ when $x<y<1$ because then $\mathbf{1}_{(0,x)}(y)=0$. For $0<y<x$, we have that $\mathbf{1}_{(0,x)}(y)=1$ and $y<x<1$, so $$f_Y(y)=\int_y^1 \frac1x\,dx=-\log(y).$$And integrating this over $0<y<1$ does indeed give $1$. Thanks!