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We have the following Random variable:

$Y=\frac{\log Z}{\log Z+\log (1-W)}$ where both $W$ and $Z$ are $U(0,1)$. The question asks to calculate the variance of the random variable $Y$.

I think the random variable $Y$ will follow the exponential distribution with parameter $2$. How I obtained this result is as follows:

First, after a little simplification, the random variable $Y$ turned out to be $\log ZW$ which can further be written as $\log Z+ \log W$. Each of which follows exponential distribution. Hence, the variance of the random variable$Y$ can be obtained as $\frac{1}{4}$. What is wrong in this approach? Because at the back, the answer is $\frac{1}{12}$ as they are saying that $Y$ will follow uniform distribution.

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    $\begingroup$ How you turned $Y$ to be $\log ZW$? Note $\frac{logA}{logB}\ne log (A-B)$ $\endgroup$
    – Deep North
    Jan 9, 2018 at 6:07
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    $\begingroup$ Ok, I got your point. I used wrong log property. Then, how $Y$ follows uniform distribution. I am not able to understand that. Could you please explain it to me? $\endgroup$
    – userNoOne
    Jan 9, 2018 at 6:59

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When $U$ is Uniform $\mathcal{U}(0,1)$, $-\log(U)$ is distributed as an Exponential $\mathcal{E}(1)$ (just invert the cdf). The same is true for $\log(1-U)$. Thus $$Y=\dfrac{\log Z}{\log Z+\log (1-W)}=\dfrac{-\log Z}{-\log Z-\log (1-W)}=\dfrac{X_1}{X_1+X_2}$$ is the ratio of one exponential over the sum of two exponentials, which is a Beta variable. More exactly a Beta $\mathcal{B}(1,1)$ variable aka a Uniform variable.

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  • $\begingroup$ I got it Xi'an. Thank you so much. You explained nicely. $\endgroup$
    – userNoOne
    Jan 9, 2018 at 12:20

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