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Did I correctly find the probability density function?

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A random variable x has a probability density function given by

    fX(x) = e-x , x ≥ 0

    and an independent random variable Y has a probability density function of

    fY(y) = ey , y ≤ 0

    using the characterisic functions, find the probability density function of Z = X + Y

    2. Relevant equations

    [itex]\Phi[/itex]X([itex]\omega[/itex]) = E[ejωx] = ∫fX(x)ejωxdx

    [itex]\Phi[/itex]Z(ω) = [itex]\Phi[/itex]X([itex]\omega[/itex]) * [itex]\Phi[/itex]Y([itex]\omega[/itex])

    fZ(z) = 1/ (2[itex]\pi[/itex]) ∫ [itex]\Phi[/itex]Z(ω) * e-jωz

    the integrals within section 2 are from negative infinity to positive infinity . . .
    also it's 1 over 2 pi . . didn't know how to make that into a fraction :confused:

    3. The attempt at a solution

    My attempt was scanned and uploaded via photobucket.. here's the scan!
    http://i359.photobucket.com/albums/oo39/r19ecua/scan0356.jpg

    I'm wondering if Cauchy's was a good way to handle this..
     
    Last edited: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2

    Ray Vickson

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    Since ##Z## is a real-valued random variable its pdf ##f_Z(z)## cannot be complex for real ##z .## So your solution is incorrect.
     
  4. Mar 28, 2013 #3
    I just realized that I did not replace the 1's with j's which led to my grand error! The j's are meant to cancel.. my answer should be:

    ez / 2 .. .. Is that correct?
     
  5. Mar 29, 2013 #4

    Ray Vickson

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    No, obviously not. Does e^z have a finite integral when you integrate z from -∞ to +∞?
     
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