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PDF of function of 3 continuous, uniform random variables?

  1. Mar 26, 2009 #1
    Hi. The question is:

    Given X, Y and Z are all continuous, independant 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 :)
  2. jcsd
  3. Mar 26, 2009 #2
    Fix x in (0,1). We could start with

    P(WZ ≤ x) = E[P(WZ ≤ x | W)].

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

    E[P(W^Z \le x \mid W)] &= E\left[{\left({1 - \frac{\ln(x)}{\ln(W)}}\right)1_{\{W\le x\}}}\right]\\
    &= \int_0^x \left({1 - \frac{\ln(x)}{\ln(w)}}\right)(-\ln(w))\,dw.

    Now do the integral and check that the result is x.
  4. Mar 26, 2009 #3
    We can do this the same way. If [tex]w\in(0,1)[/tex], then

    P(W\le w) &= P(XY \le w)\\
    &= E[P(XY\le w \mid Y)]\\
    &= E\left[{P\left({X\le\frac wY\mid Y}\right)}\right].

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

    P(W\le w) &= E\left[{1_{\{Y\le w\}} + \frac wY1_{\{Y > w\}}}\right]\\
    &= P(Y \le w) + \int_w^1 \frac wy\,dy\\
    &= w - w\ln(w).

    To get the density, we differentiate, which gives [tex]-\ln(w)[/tex].
  5. Mar 27, 2009 #4
    Thanks a lot, that's really very useful.

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