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$X,Y$ 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.

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  • $\begingroup$ Why do you need to know? Also, does $(XY)^{\frac{1}{2}}$ mean $\sqrt{XY}$ or $-\sqrt{XY}$. $\endgroup$ – Dilip Sarwate Dec 18 '12 at 23:45
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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 $(0,1)$.
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  • $\begingroup$ +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven. $\endgroup$ – Fabian Dec 19 '12 at 0:02
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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).

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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 $W\sim U(0,1)$.

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  • $\begingroup$ You can format your answer using latex. $\endgroup$ – user60610 Apr 18 '13 at 14:17
  • $\begingroup$ Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)? $\endgroup$ – Did Nov 27 '18 at 19:00

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