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Let $X$, $Y$ and $Z$ be i.i.d. uniform (0,1) random variables. What is the distribution of $(XY)^Z$? I've tried to solve it via mgfs, and what I've found is:

$$E\left(e^{(XY)^Z}\right)=E\left(E\left(e^{(XY)^Z}|Z\right)\right)$$ $$=E\left(E\left(e^{(XY)}\right)^Z\right)$$ $$=E\left(2\left(\frac{e^t-1}{t}\right)^Z\right)$$ $$=2\int_0^1\left(\frac{e^t-1}{t}\right)^zdz$$ $$=2\int_0^1\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)dz$$ $$=2\left[\frac{\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)}{\log\left(\frac{e^t-1}{t}\right)}\right]_0^1$$ $$=2\frac{{\frac{e^t-1}{t}}-1}{\log\left(\frac{e^t-1}{t}\right)}$$

But I don't see where to go from here. It doesn't look like the mgf of any distribution that I know. How can I determine the distribution from this, or is there another way?

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    $\begingroup$ Why do you think that $$E\left(e^{(XY)^Z}\mid Z\right)=E\left(e^{(XY)}\right)^Z\ ?$$ $\endgroup$
    – Did
    Jan 13 '17 at 22:51
  • $\begingroup$ @Did You're right - I was trying to think it through as in this problem: math.stackexchange.com/questions/912869/…. However, this is a slightly different case, and so it's not clear that this is allowed. $\endgroup$
    – Number 34
    Jan 14 '17 at 8:55
  • $\begingroup$ Actually this step is rather clearly wrong, already for positive real numbers $(x,y,z)$, $e^{(xy)^z}$ and $(e^{xy})^z=e^{xyz}$ do not coincide. $\endgroup$
    – Did
    Jan 14 '17 at 12:03
  • $\begingroup$ Oops, you're right. That means that I'm even further from solving this one. $\endgroup$
    – Number 34
    Jan 15 '17 at 19:06
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Elementary but slightly boring: Let $T$ be $(0,1)$ valued and independent of $Z$, then, for every $t$ in $(0,1)$, $$[T^Z\leqslant t]=\left[T\leqslant t,Z\geqslant\frac{\ln t}{\ln T}\right]$$ hence $$P(T^Z\leqslant t)=E\left(1-\frac{\ln t}{\ln T};T\leqslant t\right)$$ Apply this to $T=XY$, then, for every $t$ in $(0,1)$, $$P(T\leqslant t)=P\left(X\leqslant\frac{t}Y\right)=E\left(\min\left(1,\frac{t}Y\right)\right)=P(Y\leqslant t)+E\left(\frac{t}Y;Y>t\right)$$ that is, $$P(T\leqslant t)=t+t \int_t^1\frac{dy}y=t-t\ln t$$ hence $XY$ has PDF $f$ given by $$f(t)=\frac{d}{dt}\left(t-t\ln t\right)=-\ln t$$ In particular, $$P((XY)^Z\leqslant t)=\int_0^t\left(1-\frac{\ln t}{\ln u}\right)f(u)du=\int_0^t\ln\left(\frac{t}u\right)du=\int_0^1\ln\left(\frac1v\right)(tdv)$$ that is, $$P((XY)^Z\leqslant t)=t\int_0^1\ln\left(\frac1v\right)dv=t$$ hence $(XY)^Z$ is uniform on $(0,1)$.

More elaborate: One has $X=e^{-U}$ and $Y=e^{-V}$ where $(U,V)$ are independent of $Z$ and i.i.d. with standard exponential distributions. Thus, $$(XY)^Z=e^{-(U+V)Z}$$ From here, it should be well known that $W=U+V$ is gamma $(2,1)$ with PDF $we^{-w}$ on $w>0$, and it might be well known that for such $W$, $ZW$ is again standard exponential, hence $e^{-ZW}$ is uniform on $(0,1)$.

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