# Thread:

1. ## A Funtion of 3 uniform, continuous random variables

The question is:

Given X,Y,Z are uniformly distributed on (0,1), and are independant, prove that (XY)^Z is also uniform on (0,1)

I know the PDF of XY. My main question is, how do I work out the PDF of W^Z, where W has pdf -ln(t) and Z has pdf of 1? (Or, even better, how do I do it with general functions?).

Thanks for any help.

2. I'd probably do a 2-2 transformation and then integrate out the dummy variable.

3. But... how? How do I know what I'm integrating, how do I know how to combine the PDFs in the right way, how do I know what the limits are?

4. I've been busy with grading exams, but here's a thought.
Try the CDF of your animal.
Now, I didn't verify your density of W nor did I draw what I'm proposing.
But here it is...

$P(W^Z\le a)=P(W\le a^{1/z} )=\int_0^1\int_0^{a^{1/z}} f(w,z)dwdz$

$=\int_0^1\int_0^{a^{1/z}} f(w)f(z)dwdz =\int_0^1\int_0^{a^{1/z}} f(w)dwdz$.

BUT you should draw this and make sure the bounds of integration are correct.

Back to my grading...

5. I've just tried that. I consistently get as part of the answer the integral of e^(1/x). I even tried the question a different way using the same approach (Taking the ln of (xy)^z and considering zlnx as a random variable, lnx being exponentially distributed) and I still got the integral of e^(1/x) in my answer.

We have most definitely never been taught anything about that function in our course, and nor have we ever integrated an infinite series. Something is either fundamentally wrong with what I'm doing or it's one of -those- questions which we're not actually supposed to be able to answer.

Either way, I don't think I'll get anywhere with it, so I'll leave it. I now know how to attempt these types of questions though - so thanks for that

6. I think I did it. I was worried about a negative sign, but that's not a problem.

I also have the density of W=XY as $-\ln w$ on (0,1).

I then let $A=W^Z$ and $B=W$.

Hence $Z={\ln A\over \ln B}$ and $W=B$.

My Jacobian is $\biggl|{1\over A\ln B}\biggr|$.

BUT 0<B<1, SO $\biggl| {1\over A\ln B} \biggr|= {-1\over A\ln B}$ or if you wish ${1\over A\ln B^{-1}}$.

So, the $\ln B$'s cancel and I get ${1\over A}$ for the joint density.

NOW, our region is 0<B<1 and $0<{\ln A\over \ln B}<1$.

That translates to 0<B<A<1, since both numbers are in (0,1).

Thus the density of A is $f_A(a)=\int_0^A {dB\over A} =1$ on 0<A<1.

7. That helps a lot - I have the question done now, thanks.
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