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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.2012-05-15
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    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.2012-05-15

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