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Calculate conditional expectation of exponential variables

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Let X and Y be independent exponential random variables with parameters a and b. Calculate E(X|X+Y).

    2. Relevant equations


    3. The attempt at a solution
    I'm pretty sure I have it, just want to make sure.

    Joint density for X and Y is abe^(-ax)e^(-by) for x,y>0. Let Z=X and W=X+Y so X=Z and Y=W-Z. The Jacobian is 1, so the joint density is

    abe^(-az)e^(-bw+bz)=abe^(+bz-az) e^(-bw)

    Marginal density for w is

    \int_0^ w abe^(bz-az) e^(-bw) dz=ab(e^(bw-aw)-1) /(b-a)

    So then conditional density is

    (b-a)e^(bz-az) e^(-bw)/(e^(bw-aw)-1)

    Then just integrate z from 0 to infinity to get
    e^(-bw)/(1-e^(bw-aw))

    Is this okay?
     
  2. jcsd
  3. Sep 27, 2014 #2

    Ray Vickson

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    No, it is not OK. Given ##X+Y = w##, ##X## has range ##[0,w]##; in your notation, ##0 \leq (Z|W=w) \leq w##. Therefore, integrating out to ##\infty## is wrong. Also, your "conditional density" ##f(z|w)## for ##(Z|W=w)## looks incorrect a bit. Go back to square one and proceed very carefully. The correct final answer is more complicated than yours.
     
  4. Sep 27, 2014 #3
    Thanks. Let me try once more.

    The joint density is
    ##abe^{-ax}e^{-by}##
    Then making the substitution ##X=Z## and ##Y=W-Z##.
    ##\rho_{Z.W}=abe^{-az}e^{-bw+bz}##
    ##\rho_{Z,W}=abe^{(b-a)z}e^{-bw}##

    Thus the marginal density is
    ##\rho_W=\int_0^w abe^{(b-a)z}e^{-bw} dz##
    ##\rho_W=abe^{-bw} \int_0^w e^{(b-a)z} dz##
    ##rho_W=abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)##

    And the conditional density is
    ##\rho_{Z|W}=abe^{(b-a)z}e^{-bw}\frac{1}{abe^{-bw} \frac{1}{b-a} (e^{(b-a)w}-1)}##
    ##\rho_{Z|W}=e^{(b-a)z}\frac{1}{\frac{1}{b-a} (e^{(b-a)w}-1)}##
    ##\rho_{Z|W}=e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)}##

    Then integrate from ##0## to ##w##

    ##\mathbb{E}(Z|W)=\int_0^w z e^{(b-a)z}\frac{b-a}{(e^{(b-a)w}-1)} dz##
    ##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \int_0^w z e^{(b-a)z} dz##
    ##\mathbb{E}(Z|W)=\frac{b-a}{(e^{(b-a)w}-1)} \frac{e^{(b-a)w}((b-a)w-1)+1}{(b-a)^2}##

    Is this okay? Thanks for your help.
     
  5. Sep 27, 2014 #4

    Ray Vickson

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    Yes, it's OK, but it should be simplified a bit. You can make it look nicer by writing
    [tex] E(X|X+Y=w) = \frac{e^{cw}(cw-1)+1}{c(e^{cw}-1)}, \: c = b-a[/tex]
     
  6. Sep 27, 2014 #5
    Awesome, thanks for all your help!
     
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