1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    bit hard to read or follow, but if you mean times (-1/gamma) you're on the right track
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proofing Varience in Gamma Distribution
  1. Gamma distribution (Replies: 1)

Loading...