题目
Let X∼Gamma(α,λ)X \sim \text{Gamma}(\alpha, \lambda)X∼Gamma(α,λ) with PDF f(x)=λαΓ(α)xα−1e−λxf(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x}f(x)=Γ(α)λαxα−1e−λx for x>0x > 0x>0.
(a) Derive E[X]E[X]E[X] by direct integration, using the fact that ∫0∞λβΓ(β)xβ−1e−λx dx=1\int_0^\infty \frac{\lambda^\beta}{\Gamma(\beta)} x^{\beta - 1} e^{-\lambda x}\,dx = 1∫0∞Γ(β)λβxβ−1e−λxdx=1 for any β>0\beta > 0β>0.
(b) Derive E[X2]E[X^2]E[X2] similarly and compute Var(X)\text{Var}(X)Var(X).
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你的答案
E[X]
Var(X)