Let X be a variable uniformly distributed between 0 to 1
Let Y be a variable uniformly distributed between X to 1
What is the PDF of Y?
The following is what I understand:
PDF of X is 1 between 0 to 1
CDF of X is x between 0 to 1
I believe that the PDF of Y is $\frac{1}{1-X}$
However, I don't know how to eliminate X from the PDF of Y
EDIT:
From @NCh's answer: $$ f_Y(y) = \int_0^y \frac{1}{1-x} \, dx $$ Further simplification: $$ f_Y(y) = -ln(1-y) $$
You are given the conditional distribution: given $X=x$, $Y$ is uniformly distributed between $x$ and $1$. That is, for any $x\in(0,\,1)$ and any $y\in(x,\,1)$, the conditional density function of $Y$ is given as $$f_{Y|X}(y\mid x)=\frac{1}{1-x}$$ To find unconditional probability density function $f_Y(y)$, you need to find joint PDF and then integrate it over $x$: $$ f_{X,Y}(x,y) = f_X(x)\,f_{Y|X}(y\mid x) = {\mathbb 1}_{\{0<x<1\}}\frac{1}{1-x}{\mathbb 1}_{\{x<y<1\}} $$ And $$ f_Y(y) = \int f_{X,Y}(x,y) \,dx = \int_0^y \frac{1}{1-x} \, dx $$ The upper limit in the integral is $y$ since given $y$, $x$ can vary from $0$ to $y$ only.