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Thread: Finding a PDF with Two Variables Changing to One

  1. #1
    Super Member Aryth's Avatar
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    Finding a PDF with Two Variables Changing to One

    I have been stumped with this problem as well. I have tried several methods, but cannot seem to get the right answer.

    I have two variables $X_1$ and $X_2$ which have the joint PDF $f(x_1,x_2) = e^{-x_1}e^{-x_2}$ if $x_1,x_2 > 0$ and $0$ otherwise. Now, we let $W = w_1X_1 + w_2X_2$ with $w_1,w_2 > 0$, and we need to confirm that the PDF is

    $f(w) = \frac{1}{w_1-w_2} \left(e^{-\frac{w}{w_1}} - e^{-\frac{w}{w_2}}\right)$ when $w>0$ and $0$ otherwise.

    Any help is appreciated! I can only seem to get one of the exponential terms and not both.
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    Re: Finding a PDF with Two Variables Changing to One

    toying with this I found the easiest way to reproduce this answer is via first finding the distributions for the weighted variables $w_1 X_1$ and $w_2 X_2$ and then just doing the convolution.

    It's well known that given two random variables the distribution of the sum is the convolution of the individual distributions.

    it's quickly seen that

    $w_1 X_1 \sim \dfrac 1 {w_1} f_{X_1}\left(\dfrac{x}{w_1}\right)$

    and so the convolution integral we are after is

    $\displaystyle{\int_0^x}\dfrac{e^{-t/w_1}}{w_1}~\dfrac{e^{-(x-t)/w_2}}{w_2}~dt$

    this is readily evaluated to

    $f_W(w) = \dfrac{e^{-\frac{w}{w_1}}-e^{-\frac{w}{w_2}}}{w_1-w_2}$
    Thanks from Aryth
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    Super Member Aryth's Avatar
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    Re: Finding a PDF with Two Variables Changing to One

    I used the convolution method but I had the wrong limits for my integral. Thanks for your help!
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