The distribution of the sum of two independent exponential distributions

I am trying to calculate the distribution of the sum of two independent log-uniform distributions but something doesn't add up.

Suppose $a \sim \mathrm{uni}(0,1)$ and $b \sim \mathrm{uni}(0,1)$. Thus, $u=\log(a)$ has an exponential distribution of the form $e^u$, which is defined for values for which $u<0$ (the same applies to $v=\log(b)$ ).

Now, define a new r.v $z=u+v$. I have tried to compute the new distribution via the convolution formula, but I get a non-converging integral. Can anyone help?

2
2022-07-25 20:43:22
Source Share
Answers: 0