- #1
mcguiry03
- 8
- 0
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
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