Finding the density of $X$ selected from $[Y,1]$, where $Y$ is first selected in $[0,1]$

Question

Based on the problem statement:

$f_{X|Y}(x|y) = \frac{1}{1-y}$ for $0<y<x<1$.

$f_X(x) = \int_{-\infty}^{\infty} f_{X|Y}(x|y)f_Y(y)dy = \int_{-\infty}^{\infty}= \frac{1}{1-y}dy$

My question is on the limits of integration, since $f_{X|Y}(x|y)$ is only non-zero in $0<y<x$ (and $f_Y(y)$ in $0<y<1$), the limit of integration should be from $0$ to $x$ and $f_X(x)=-ln(1-x)$ for $0<x<1$. However, I have trouble seeing why this makes sense geometrically. Is there another way to derive the limits of integration?

Answer 1

Your calculation is correct. Because if $x$ is fixed then the integrand is $0$ if $y$ is not in in $(0,x)$. So your integral will be

$$f_X(x) = \int_{-\infty}^{\infty} f_{X|Y}(x|y)f_Y(y)dy = \int_{0}^{x} \frac{1}{1-y}dy=-\ln(1-x)$$

if $x\in (0,1)$ and $0$ outside of the same.

I don't understand what is wrong with $f_X(x)=-\ln(1-x)$ for $0<x<1$.

This is a nice positive function whose integral is $1$ over $(0,1)$. The shape of this function is as shown below: