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The time when I studied probability theory was about three years ago, so I have forgot many things and want to refresh them.

I want to prove that if $X,Y,Z$ are independent and uniformly distributed on $[0,1]$ random variables then $(XY)^Z$ is also uniformly distributed on $[0,1]$.

As I know uniformly distributed on $[0,1]$ variable has density $\rho (t)= \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}$.

Denote $\xi = XY$. Since $X$ and $Y$ are independent $\rho_\xi (t) = \rho_X (t) \rho_Y (t) = \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}$, so $\xi$ is uniformly distributed on $[0,1]$.

Denote $\zeta = (XY)^Z$. Since $\xi$ and $Z$ are independent, we obtain in the same way that $\rho_\zeta (t) = \rho_\xi (t) \rho_Z (t) = \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}$ and we are done.

It seems to me that I'm wrong. Can anyone explain me why it is incorrect (I believe that it is) and show the right solution?

Hasek
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  • You can't multiply pdfs like that. If X and Y are independent, then the pdf of the vector (X,Y) is $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. – Blaza Feb 05 '17 at 21:37

1 Answers1

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Let me answer my own question, maybe it will be helpful for someone in the future.

We may find cumulative distribution function of $XY$ from geometric considerations (draw a square $[0,1]\times[0,1]$ and calculate the corresponding area under the graph): $$F_{XY}(t) = P(xy<t) = t + \int\limits_t^1 dx \int\limits_0^{\frac{t}{x}}dy = t + \int\limits_{t}^1 \frac{t}{x}dx = t(1 - \ln t)$$

Similarly for $(XY)^Z = u^Z$: $$F_{u^Z}(t) = P(u^Z<t) = \int\limits_0^t (-\ln u)du \int\limits_{\frac{\ln t}{\ln u}}^1 dy = \int\limits_0^t (- \ln u)(1 - \frac{\ln t}{\ln u})du = t$$

We also know that the derivative from cumulative distribution function is a probability density function, which gives us that this random variable is uniformly distributed.

Hasek
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