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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?

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Should work,you are missing a negative sign, also exponentials are gammas, and the sum of two independent gammas with the same value of the exponential parameter is gamma. – mike May 15 '12 at 11:11
It is hard to point out where you might be making a mistake when you get a non-converging integral since you have not shown us what you have been doing, but here is an alternative strategy. $-\log(a)$ and $-\log(b)$ are independent exponential random variables with mean $1$, and their sum is a gamma random variable with parameters $(2,1)$. Have you tried working this convolution integral which is even found in many textbooks? If so, all you need is the observation that the density function of $-X$ is $f_X(-x)$ and you are done. – Dilip Sarwate May 15 '12 at 11:12

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.

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Thanks for all the replies and comments. Thanks for the thorough reply Ilya. Maybe it is the time of day here (noon), but I can't see how the latter integral converges. – Daniel May 15 '12 at 11:34
@Daniel: You're welcome. W.r.t. convergence, the CDF $F_u$ is a bounded function, hence the integral converges because of $e^{-s}$ term. – Ilya May 15 '12 at 11:36
Thanks. Now I see my error. I should have been more careful with the bounds. Thanks again :) – Daniel May 15 '12 at 11:55
@Daniel: you're welcome – Ilya May 15 '12 at 12:06

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