Let $U$, $V$ be independent random variables following a continuous uniform distribution on $(0,1).$ Let $X=U.V$ then find $f_X$ and $f_V|_X$

Question

Let $U$, $V$ be independent random variables following a continuous uniform distribution on $(0,1).$ Let $X=U.V$ then find $f_X$ and $f_V|_X$

My Working

for $f_X$ we have:

$P(X\leq x)=P(UV\leq x)=\int_{0}^{1} F_V(\frac{x}{u}) \,du=\int_{0}^{x} F_V(\frac{x}{u}) \,du+\int_{x}^{1} F_V(\frac{x}{u}) \,du=\int_{0}^{x} 1\,du+\int_{x}^{1} \frac{x}{u}\,du=-\log{x}$

Is my working for $f_X$ correct? Also how do find $f_V|_X$. Kindly help me

Answer 1

Yes your reasoning is correct but there is a method by which you can solve both question together. Consider the following system

$$ \begin{cases} x=uv\\ y=v \end{cases}\rightarrow\begin{cases} u=\frac{x}{y}\\ v=y \end{cases}$$

calculate the jacobian obtaining $|J|=\frac{1}{y}$ and thus

$$f_{XY}(x,y)=\frac{1}{y}$$

Now observe that $0<\frac{x}{y}<1$ that is $0<x<y<1$ (a triangle) thus, integrating, immediately you find

$$f_X(x)=\int_x^1\frac{1}{y}dy=-\log x$$

and again immediately you get

$$f_{Y|X}(y|x)=\frac{f_{XY}(x,y)}{f_X(x)}=\frac{1}{-y\log x}$$