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Homework Help: Moment Generating Function Given pdf

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data
    X is distributed exponentially with λa=2. Y is distributed exponentially with λb = 3. X and Y are independent.
    Let W=max(X,Y), the time until both persons catch their first fish. Let k be a positive integer. Find E(W^k).

    Also, find P{(1/3)<X/(X+Y)<(1/2)}


    2. Relevant equations
    f(X) = λa e^(-λa x)
    f(Y) = λb e^(-λb y)
    f(X,Y) = f(X)f(Y)
    Mx(t) = E(e^t)
    Mx^k(0)=E(W^k)


    3. The attempt at a solution
    I found f(w)=3e^(-3w)+2e^(-2w)-5e^(-2w-3w) but not sure where to go from here
     
  2. jcsd
  3. Nov 30, 2012 #2

    haruspex

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    Having found the pdf of W, how do you calculate its moment generating function?
    (You quote that it's E[et], but it's E[etW].)
     
  4. Nov 30, 2012 #3
    That's correct, sorry, typo on my part. It's actually where to go from there that confuses me. I know the ∫0 to ∞((e^tw)f(w)dw gives me Mw(t), so I think ∫0 to ∞ ((w^k)f(w))dw should give me E(W^k), but I don't know how to actually find this or if it is the best way to approach it.

    Also, I got the second probability part on my own. I really just need help with the E(W^k).
     
  5. Nov 30, 2012 #4
    I have f(w) = 3e^(-3w)+2e^(-2w)-5e^(-5w), w>0 for the pdf, but I am not sure how to proceed to E(W^k) from here.
     
  6. Nov 30, 2012 #5

    Ray Vickson

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    In other words,
    [tex] f(w) = 2e^{-2w} + 3 e^{-3w} - 5e^{-5w} = f_2(w) +f_3(w) - f_5(w),[/tex] where f_r(w) is the exponential density for rate r. Do you know how to compute
    [tex] E_r(W^k) \equiv \int_0^{\infty} r e^{-rw} w^k \, dw ?[/tex]
     
  7. Nov 30, 2012 #6
    My solution: E(X^k) for an exponential is k!/(λ^k), so E(W^k) = k!/(3^k)+k!/(2^k)-k!/(5^k) = k!(10^k + 15^k - 6^k)/(30^k). Say λ=r, and this is exactly what you are saying, yes?
     
  8. Nov 30, 2012 #7
    If not, then I am not sure how to compute an integral involving (w^k)e^(-rw)...
     
  9. Nov 30, 2012 #8

    Ray Vickson

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    Just to be accurate: No, you are saying it, not me. However, your result is correct.
     
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