Unconditional distribution of the length of a stick broken twice

Question

A stick lies on the interval $[0,1]$, and is broken at the point $X$~$U(0,1)$. The left part is then broken again at the point $Y$~$U(0,X)$, i.e the pdf of $Y|X$ is $f_{Y|X=x}(y)= \mathbb{I}_{[0,x]}1/x$. I'm trying to find the unconditional distribution of $Y$, this is what I did so far and I'm wondering what I am doing wrong since the last result is obviously unbounded: $$P(Y<y) = \int_{-\infty}^{\infty}P(Y<y|X=x)f_{X}(x)dx$$ $$= \int_0^1P(Y<y|X=x)\cdot 1 dx$$ $$= \int_0^1 1\int_0^y \frac{1}{x}dz dx= \int_0^1\frac{y}{x}dx$$ $$=y(ln(1)-ln(0))$$ Any help is appreciated

Answer 1

Nice question!

You are correct, under the conditions given $$P(Y<y)= \int_0^1P(Y<y|X=x)dx$$ The next question is to find $P(Y<y|X=x)$, and it is easy to show that it is equal to $y/x$ but (!) only if $y<x$; generally \begin{equation} P(Y<y|X=x)= \begin{cases} \frac{y}{x}, &\text{if 0<y<x}\\ 1, &\text{if y>x} \end{cases} \end{equation} Now,

$$P(Y<y)= \int_0^1P(Y<y|X=x)dx=\int_0^y1dx+\int_y^1\frac{y}{x}dx=y(1-\ln y)$$

Answer 2

first step should be $$ P(Y<y) = \int_{-\infty}^{\infty}P(Y<y, X=x)f_{X}(x)dx $$ except you cannot use $X =x$ for continuous variables, but instead use it's $\epsilon$-analog...