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Homework Help: Proofing Varience in Gamma Distribution

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Fy(y) = 1/2 λ^3 y^2 e^(-λy)

    Assume E(1/y) = λ/2
    Find E(1/y^2) and prove Var(1/y) = λ^2 / 4


    2. Relevant equations
    E(y) = k/λ
    Var (y) = k/ λ^2

    3. The attempt at a solution
    Hi guys,

    From the equation given from gamma distribution, I understands clearly that to get Var(y) what I have to do is to square y. i.e. 1/y^2 = (λ/2)^2 = λ^2 / 4. Which is equivalent to E(1/y^2)

    but I'm no clue on how to set up a proof to show this is correct. I needed help on this, as I have no idea at all on how to proof this.
     
  2. jcsd
  3. Feb 3, 2012 #2

    lanedance

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    the problem doesn't ask fior var(y) it asks for var(1/y)

    the expectation of any function of y, say g(y), can be written
    [tex] E(g(y)) = \int g(y) f_Y(y)dy[/tex]

    so for E(1/y^2) this gives
    [tex] E(\frac{1}{y^2}) = \int \frac{1}{y^2} f_Y(y)dy[/tex]
     
  4. Feb 3, 2012 #3
    Ok I've retried to do E(X). Can you please check if I have done it correctly before I moved on trying to work out the variance.

    E(1/y^2) = ∫1/y^2 fy(y) dy
    = ∫ 1/4 * λ^2 * 1/2 * λ^3 * y^2 * e^-λy dy
    =∫1/8 * λ^5 * y^2 * e^-λy dy
    =1/8 * 1/3 * λ^5 * y^3 * e^-λy - 1/λ * λ^6 * y^2 * e^-λy
    =1/8 * 1/3 * λ^5 * y^3 * e^-λy - λe^-λy * λ^5 * y^2
    =1/24 * λ^5 * y^3 * e^-λy - λ^6 * y^2 * e^-λy
    =1/24 * λ^5 * y^2 * e^-λy (y-λ)

    Up to this point I knew I did something wrong. But I run over my calculation again, but doesn't seem to be able to pick out where I did wrong.

    It would be nice if you can pick out where I did wrong. thanks
     
  5. Feb 4, 2012 #4

    lanedance

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    what happened to the 1/y^2 in the 2nd line?
     
  6. Feb 4, 2012 #5
    ar! I see where I get wrong now. I misunderstood it as (EX)^2 and substitute in the its value as 1/y^2. I got it now. should be

    ∫1/y^2 * fx(x) dy
    ∫1/y^2 * 1/2 λ^3 * y^2 * e(-λy) dy
    ∫1/2 * λ^3 * e(-λy) dy

    = 1/2 λ^3 - 1/λ e(-λy)
     
  7. Feb 5, 2012 #6

    lanedance

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    bit hard to read or follow, but if you mean times (-1/gamma) you're on the right track
     
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