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?
Ok, let $a\in U([0,1])$ and put $u = \log a$. We know that CDF of $a$ is $$ F_a(t) = \mathsf P(a\leq t) = t\cdot1_{[0,1)}(t)+1_{[1,\infty)}(t). $$ Let us find CDF of $u$: $$ F_u(t) = \mathsf P(u\leq t) = \mathsf P(a\leq e^t) = F_a(e^t) = e^t 1_{(-\infty,0)}(t)+1_{(0,\infty)}(t). $$ So the PDF of $u$ is $f_u(t) = e^t1_{(-\infty,0)}(t)$. The similar holds for $v$, i.e. $f_v(s) = e^s1_{(-\infty,0)}(s)$.
Finally, for the CDF of the variable $z = u+v$ we have $$ \begin{align} \mathsf P(u+v\leq r) &= \int\limits_{-\infty}^\infty \mathsf P(u\leq r-s)f_v(s)\mathrm ds \\ &=\int\limits_{-\infty}^0 F_u(r-s)e^{s}\mathrm ds = \int\limits_0^\infty F_u(r+s)e^{-s}\mathrm ds \end{align} $$ where the latter integral clearly converges. If you need a help to compute it, please tell me - I'll extend the answer.