Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$X,Y$ 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.

share|improve this question
    
Why do you need to know? Also, does $(XY)^{\frac{1}{2}}$ mean $\sqrt{XY}$ or $-\sqrt{XY}$. –  Dilip Sarwate Dec 18 '12 at 23:45
add comment

3 Answers 3

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 $(0,1)$.
share|improve this answer
    
+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. –  Fabian Dec 19 '12 at 0:02
add comment

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).

share|improve this answer
add comment

Using the definition of weak convergence, it is so easy. First, for any positive integer k>=0, 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(.), we can find a polynomial function f(.) such that f can approximate g uniformly by Weierstrass's theorem. Thus, Eg(W)=Eg(U). So W~U(0, 1).

share|improve this answer
    
You can format your answer using latex. –  user60610 Apr 18 '13 at 14:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.