PDF of function of 3 continuous, uniform random variables?by Phillips101 Tags: continuous, function, random, uniform, variables 

#1
Mar2609, 02:17 PM

P: 33

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
Mar2609, 10:24 PM

P: 5

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, [tex] \begin{align*} 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. \end{align*} [/tex] Now do the integral and check that the result is x. 



#3
Mar2609, 10:37 PM

P: 5

[tex]\begin{align*} 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]. \end{align*} [/tex] If [tex]Y\le w[/tex], then the probability is 1; otherwise, it is w/Y. Thus, [tex]\begin{align*} 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). \end{align*} [/tex] To get the density, we differentiate, which gives [tex]\ln(w)[/tex]. 



#4
Mar2709, 02:19 PM

P: 33

PDF of function of 3 continuous, uniform random variables?
Thanks a lot, that's really very useful.
James 


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