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Gamma distribution proving

  1. May 10, 2012 #1
    prove that μ=αθ and σ^2=αθ^2




    μ = E(X)
    ...= ∫(x = 0 to ∞) x *(1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
    ...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^α e^(-x/θ) dx
    ...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^α e^(-t) * (θ dt), letting t = x/θ
    ...= (θ/Γ(α)) ∫(t = 0 to ∞) t^α e^(-t) dt
    ...= (θ/Γ(α)) Γ(α+1), by definition of Gamma function
    ...= (θ/Γ(α)) * α Γ(α), by Gamma recurrence
    ...= αθ.

    E(X^2) = ∫(x = 0 to ∞) x^2 * (1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
    ...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^(α+1) e^(-x/θ) dx
    ...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^(α+1) e^(-t)* (θ dt), letting t = x/θ
    ...= (θ^2/Γ(α)) ∫(t = 0 to ∞) t^(α+1) e^(-t) dt
    ...= (θ^2/Γ(α)) Γ(α+2), by definition of Gamma function
    ...= (θ^2/Γ(α)) * α(α+1) Γ(α), by Gamma recurrence (used twice)
    ...= α(α+1)θ^2.

    So, σ^2 = E(X^2) - (E(X))^2 = α(α+1)θ^2 - (αθ)^2 = αθ^2




    i would like to know if it is correct and tnx :biggrin:
     
  2. jcsd
  3. May 10, 2012 #2

    Ray Vickson

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    Science Advisor
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    It looks OK.

    RGV
     
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