PDF of function of 3 continuous, uniform random variables?

[Phillips101]

Hi. The question is:

Given X, Y and Z are all continuous, independent random variables uniformly distributed on (0,1), prove that (XY)^Z is also uniformly distributed on (0,1).

I worked out the pdf of XY=W. I think it's -ln(w). I have no idea at all how to show that W^Z is U(0,1).

What do I integrate, how do I know how to combine the pdfs, how do I know what the limits are, what substitutions should I make if I need to make one? Etc, really. I just don't know how to tackle this sort of problem at all. The pdf I have for W came from a picture, not any real understanding of what I was doing.

Thanks for any help :)

 

[jason1995]

Fix x in (0,1). We could start with

P(W^Z ≤ x) = E[P(W^Z ≤ x | W)].

Since Z and W are independent, we can calculate P(W^Z ≤ x | W) by treating W as a constant. In this case, if W > x, then the probability is 0. Otherwise, W^Z ≤ x iff Z ≥ ln(x)/ln(W), which has probability 1 - ln(x)/ln(W). Hence,

$$\begin{align*}<br /> E[P(W^Z \le x \mid W)] &= E\left[{\left({1 - \frac{\ln(x)}{\ln(W)}}\right)1_{\{W\le x\}}}\right]\\<br /> &= \int_0^x \left({1 - \frac{\ln(x)}{\ln(w)}}\right)(-\ln(w))\,dw.<br /> \end{align*}$$

Now do the integral and check that the result is x.

 

[jason1995]

The pdf I have for W came from a picture, not any real understanding of what I was doing.

We can do this the same way. If $$w\in(0,1)$$, then

$$\begin{align*}<br /> P(W\le w) &= P(XY \le w)\\<br /> &= E[P(XY\le w \mid Y)]\\<br /> &= E\left[{P\left({X\le\frac wY\mid Y}\right)}\right].<br /> \end{align*}$$

If $$Y\le w$$, then the probability is 1; otherwise, it is w/Y. Thus,

$$\begin{align*}<br /> P(W\le w) &= E\left[{1_{\{Y\le w\}} + \frac wY1_{\{Y > w\}}}\right]\\<br /> &= P(Y \le w) + \int_w^1 \frac wy\,dy\\<br /> &= w - w\ln(w).<br /> \end{align*}$$

To get the density, we differentiate, which gives $$-\ln(w)$$.

 

[Phillips101]

Thanks a lot, that's really very useful.

James