Can you please help me with this. Let $X,Y,Z$ are independent random variables uniformly distributed over $[0,1]$. Show that $(XY)^Z$ is also uniformly distributed over $[0,1]$

I try to show that $-Z(\ln X+\ln Y)$ has an exponential distribution, but i am not sure. Thanks in advance.


So give -ln(X) $\sim$ exp(1) which is well established where X is $U(0,1)$, then similarly -ln(Y) $\sim exp(1)$ where Y is uniform.

Then sums of exponentials are gamma(2,1) indicates that -ln(X)-ln(Y) $\sim Gamma(2,1)$

$f_{W}(y)$= $ye^{-y}, y\geq 0$

then the joint distribution $f_{Z,W} = ye^{-y}, y \geq 0$ since Z $\sim U(0,1)$

now the joint distribution is established. if you solve for the transformation U=ZW, and V=$\frac{W}{Z}$, set up your jacobian and then find the marginal of U by integrating over all V and you should get your answer which is an exponential term.

  • $\begingroup$ It would be great if you could add details for this answer. $\endgroup$
    – graffe
    Nov 30 '18 at 22:09

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