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Puzzle with moment generating function for gamma function

  1. Nov 29, 2013 #1
    I am given that The kth moment of an exponential random variable with mean mu is
    E[Y^k] = k!*mu^k for nonnegative integer k.

    I found m^2 (0) = (-a)(-a-1)(-beta)^2. The answer I found is however mu^2+a*beta^2 which is different from the k! From the given formula.
    Could someone help me figure it out what happen?
     
    Last edited: Nov 29, 2013
  2. jcsd
  3. Nov 30, 2013 #2

    Ray Vickson

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    You need to show all the steps, in detail, because you are certainly obtaining the wrong answer.
     
  4. Nov 30, 2013 #3
    m(t) = (1-beta*t)^-a

    m'(t) = -a * (1-beta*t)^(-a-1)*(-beta)
    m'(0) = -a *(-beta)
    m''(t) = -a*(-a-1)* (1-beta*t)^(-a-1)*(-beta)^2
    m''(0) = -a*(-a-1)* (-beta)^2
    m'''(t) = -a*(-a-1)*(-a-2)*(1-beta*t)^(-a-1)*(-beta)^3
    m'''(0) = a*(-a-1)*(-a-2)*(-beta)^3
     
  5. Nov 30, 2013 #4

    Ray Vickson

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    You are supplying no useful information at all. I need details, right from the very beginning. In other words, I need to know: (1) What is your formula for the density function of Y? (2) What is the definition of the moment-generating function m(t)? (3) For your particular f, what is the integration you perform to get your m(t)?

    Only after I see all that can I possibly help you. I know the definition of m(t), but the question is whether YOU know it. Judging from your formula for m(t), I would guess the answer to be NO, you do not know it. I need to see that. I need details, details, details!
     
  6. Nov 30, 2013 #5
    f(y) = y^(a-1)*e^(-y/beta)/(beta^(a)*Γ(a))

    moment-generating function m(t) is alternative specification for its probabilit function, E(Y^k) = m^k (0)
     
  7. Nov 30, 2013 #6

    Ray Vickson

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    Your first message said that Y had an exponential distribution (or so it seemed so say). Is that not true? Does Y really have a (general) Gamma distribution? I will assume so, from what you have now just written. Do you see why your messages cause confusion?

    OK, so now tell me: what is the definition of m(t)? Don't just tell me "... m(t) is alternative specification for its probabilit function..."; tell me what it IS! Then I will be able to judge whether you have a correct formula or not.
     
  8. Nov 30, 2013 #7

    I think My first message is that gamma distribution while exponential distribution is (1/beta)*e^(-t/beta). What makes you think my message causes confusion?
    m(t) = E(e^(tY))
     
  9. Nov 30, 2013 #8

    Ray Vickson

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    Because it confused me. Here is EXACTLY what you said:

    "I am given that The kth moment of an exponential random variable with mean mu is E[Y^k] = k!*mu^k for nonnegative integer k.
    I found m^2 (0) = (-a)(-a-1)(-beta)^2. The answer I found is however mu^2+a*beta^2 which is different from the k! From the given formula.
    Could someone help me figure it out what happen?"

    You did not mention the Gamma distribution anywhere in that first message---only the exponential!

    Anyway, your computation of the mgf for Gamma-distributed Y is OK, and your subsequent computations of ##m^{(k)}(0)## are OK for k = 1 and 2, but your ##m'''(0)## has the wrong sign. You should always simplify carefully before substitution k=0, and that means you should always get rid of the negatives in factors such as (-a), etc---take out the '-' signs carefully.

    As to your original question: why would you assume the moments of the exponential distribution have anything to do with the moments of the Gamma distribution?
     
    Last edited: Nov 30, 2013
  10. Nov 30, 2013 #9
    Let holds this a little bit. I need to figure out what context I took this from
     
  11. Nov 30, 2013 #10

    Ray Vickson

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    Note that the exponential is a special case of the gamma, obtained by setting b = 1. What happens to b(b+1)(b+2) ... (b+k-1) when you put b = 1?
     
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