Verified answer

1) Since the pdf for the gamma distribution is f(x) = (1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) for non-neg. x,

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.

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For the exponential distribution, simply let α = 1 in the results above.

I hope this helps!

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