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  1. #1
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    gamma distribution

    If X is a gamma(2,lambda) distribution

    lambda(^2)xe^(-lambda x) x >0
    fX(x)=
    0 otherwise

    How do i obtain the mean and variance of X?
    thanks.
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  2. #2
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    Quote Originally Posted by math_lete View Post
    If X is a gamma(2,lambda) distribution

    lambda(^2)xe^(-lambda x) x >0
    fX(x)=
    0 otherwise

    How do i obtain the mean and variance of X?
    thanks.
    f(x) = \lambda^2 \, x \, e^{-\lambda x} for x > 0 and zero elsewhere.

    E(X) = \lambda^2 \int_{0}^{+\infty} x^2 \, e^{-\lambda x} \, dx.

    Var(X) = E(X^2) - [E(X)]^2.

    E(X^2) = \lambda^2 \int_{0}^{+\infty} x^3 \, e^{-\lambda x} \, dx.

    The integrals can be evaluated by repeated application of integration by parts.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    f(x) = \lambda^2 \, x \, e^{-\lambda x} for x > 0 and zero elsewhere.

    E(X) = \lambda^2 \int_{0}^{+\infty} x^2 \, e^{-\lambda x} \, dx.

    Var(X) = E(X^2) - [E(X)]^2.

    E(X^2) = \lambda^2 \int_{0}^{+\infty} x^3 \, e^{-\lambda x} \, dx.

    The integrals can be evaluated by repeated application of integration by parts.
    Or by doing it in a way where integration by parts is not required

    I'll quickly do E(X^2). A similar thing can be done with E(X).

    Let u=\lambda x\implies x=\frac{u}{\lambda}. Thus, \,dx=\frac{\,du}{\lambda}

    The integral transforms into \lambda^2\int_0^{\infty}\frac{1}{\lambda}\left(\fr  ac{u}{\lambda}\right)^3e^{-u}\,du\implies\frac{1}{\lambda^2}\int_0^{\infty}u^  3e^{-u}\,du.

    But note that \int_0^{\infty}e^{-u}u^3\,du=\Gamma(4).

    So we have \frac{1}{\lambda^2}\int_0^{\infty}u^3e^{-u}\,du=\frac{\Gamma(4)}{\lambda^2}.

    Since \Gamma(4)=3!=6, we now see that \color{red}\boxed{E(X^2)=\frac{6}{\lambda^2}}

    --Chris
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