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The sum of two gamma distributions

  1. Nov 22, 2011 #1
    let X~gamma(x,λ), Y~gamma(y,λ)
    then Z = X+Y is gamma (x+y, λ)

    I'm trying to prove this. Is using the moment generating functions the only way to do it.

    and in such case, can I assume that MZ(t)= MX(t)*MY(t)
     
  2. jcsd
  3. Nov 22, 2011 #2

    chiro

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    Hint: Do you know what Moment Generating Functions are? Do you know the consequence of equating a MGF to a particular distribution?
     
  4. Nov 22, 2011 #3
    I know MX(t) = ∫X E[et*x] dx

    and MY(t) = ∫X E[et*y] dy

    hence I get MZ(t) = ∫X E[et*x] dx * ∫YE[et*y] dy
     
  5. Nov 22, 2011 #4

    chiro

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    Let Z = X + Y and calculate E[e^(tZ)]. Also you can use the fact that since X and Y are independent then E[e^(tZ)] = E[e^(t[X+Y])] = E[e^(tX + tY)] = E[e^(tX)] x E[e^(tY)].
     
  6. Nov 22, 2011 #5
    I know how to get to

    MX(t) = (λ/ λ-t)x

    and MY(t) = (λ/ λ-t)y

    hence since MZ(t) = MX(t)*MY(t)

    this implies MZ(t) = (λ/ λ-t)x+y

    which implies Z~gamma(x+y, λ)
     
  7. Nov 22, 2011 #6

    chiro

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    Yep thats it. As long as you have the assumption that X and Y are independent, you have your result which is correct.
     
  8. Nov 22, 2011 #7
    thanks
     
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