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Show that $E(\varepsilon)={\alpha}m$ when gamma distribution function is given as

$$f(\varepsilon)=\frac{1}{{\alpha^m\Gamma(m)}}\varepsilon^{m-1}\text{exp}[-\frac{\varepsilon}{\alpha}]$$

I am so stuck with calculations of this. I am using the formula of $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ but I can't get the required result. I am stuck while counting the integral at the part where I would already plug in the limits. I get something like this before plugging in the limits: $\frac{m\Gamma(m)e^\frac{1}{\alpha}}{\alpha\Gamma(m)}$.

lovetimberland
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    You are computing $\frac{1}{\alpha^m \Gamma(m)} \int_0^\infty x^m \exp(-x/\alpha) dx$. Change variables to $u=x/\alpha$ and pull constants out. You should get a constant times $\Gamma(m+1)$. Then $\frac{\Gamma(m+1)}{\Gamma(m)}=m$ finishes the job. – Ian Oct 13 '16 at 15:03
  • I think I have nailed it, thank you. – lovetimberland Oct 13 '16 at 22:33

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