Find function $h$ so that $h(U,V)$ equals density of $f(a) da$ for $f(a)=\frac{1}{2}e^{-\small|a|}$, $a \in \mathbb R$

Question

Let $f(a)=\frac{1}{2}e^{-\small|a|}$, $a \in \mathbb R$ and

let $U,V$ be independant and uniform distributed on [0,1].

Now I want to find a function $h$ so that $h(U,V)$ is equal to the density $f(a) da$.

I struggled for this task for quite some time but I just can't figure it out, there probably is a simple solution out there but I can't find it.

Your help is appreciated!

Edit: No one is able to help me? Did I miss to give some information about the task or what is wrong?

Answer 1

Yes, you might want to check your local library for the solution to this task.

My local library is THE place to be and I really think it could help you right now to check your local library for the solution to this task. Have a nice day

Answer 2

Hint: The density $f(a)=\frac12e^{-|a|}$ belongs to the Laplace distribution. The Laplace distribution arises when you subtract two iid exponential(1) variables.

Second hint: If $U$ has uniform distribution on $[0,1]$ then $X:=-\ln (U)$ has exponential(1) distribution.

Putting it all together: If $U$ and $V$ are independent uniform [0,1] variables, then [from hint 2] $X:=-\ln(U)$ and $Y:=-\ln(V)$ are independent exponential(1) variables. Therefore [from hint 1], $X-Y$ has Laplace distribution. Since $X-Y=-\ln U - (-\ln V) = \ln(V/U)$, we conclude that $\ln(V/U)$ has the desired distribution.