So I was trying to prove the mean result of gamma distribution which is $\frac{\alpha}{\lambda}$.
My attempt,
$E(X)=\int_{0}^{\infty }x f(x)dx$
$=\int_{0}^{\infty } \frac{\lambda^{\alpha}}{\Gamma (\alpha)}x^{\alpha}e^{-\lambda x}dx$
After integrating it, I got the result $$\frac{\lambda^{\alpha}}{\Gamma (\alpha)} \cdot\frac{\alpha}{\lambda}(\int_{0}^{\infty } x^{\alpha-1}e^{-\lambda x}dx)$$. I'm stuck here. Could anyone continue it for me and explain? Thanks a lot.
$\endgroup$ 51 Answer
$\begingroup$Let us consider a random variable with Gamma distribution $X\sim \text{Gamma}(\alpha,\lambda)$. Its expected value is \begin{equation} E(X) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha} \, e^{-\lambda x} \, dx \, . \end{equation} Making the change of variable $y=\lambda x$ in the integral, one has \begin{aligned} E(X) &= \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty \left(\frac{y}{\lambda}\right)^{\alpha} \, e^{-y} \, \frac{dy}{\lambda} \\ &= \frac{1}{\lambda\Gamma(\alpha)} \int_0^\infty y^{\alpha} \, e^{-y} \, dy\\ &= \frac{\Gamma(\alpha+1)}{\lambda\Gamma(\alpha)} \, . \end{aligned} Due to the relationship $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$, one obtains $E(X) = \alpha/\lambda$.
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