Distribution of $(XY)^Z$ if $(X,Y,Z)$ is i.i.d. uniform on $[0,1]$ $X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$
How is random variable $(XY)^Z$ distributed?
I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
 A: Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$\begin{align}P(w\geq W) &= \int_0^1\!dx\int_0^1\!dy\int_0^1\!dz\, \theta(w-(xy)^z)\\
&= \int_0^1\!dx\int_0^1\!dy \max\{1-\log_{xy} w,0\}\\
&=\int_0^1\!d\eta\int_\eta^1\!\frac{dx}{x}\max\{1-\log_{\eta} w,0\} \\
&=-\int_0^w\!d\eta \log \eta (1-\log_{\eta} w)\\
&=w.
\end{align}$$
with $\eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
A: Hints:


*

*The random variable $X$ is uniform on $(0,1)$ if and only if $-\log X$ is exponential with parameter $1$.

*If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $w\mapsto w\mathrm e^{-w}\mathbf 1_{w\gt0}$.

*If $W$  is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:


*

*If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.

A: Using the definition of weak convergence, it is so easy. First, for any positive integer $k\ge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(\cdot)$, we can find a polynomial function $f(\cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $W\sim U(0,1)$.
