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For a Gamma distribution where $Y\sim \mathrm{Gamma}(\alpha, \beta)$ we have

$$f(y)=\frac{y^{\alpha -1}e^{-y/\beta}}{\beta^{\alpha} \Gamma(\alpha)}$$

The question is, given that random variable $Y \sim \textrm{Expo}(\beta)$, derive a closed-form expression for $E[Y^a].$

An exponential distribution with parameter $\beta$ is simply a Gamma distribution with $\alpha = 1$.

The answer is $\beta^a\Gamma(a+1)$. But how may I reach this answer?

Thank you very much.

The solution is as follows.

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Update: Thanks Robin. I think I forgot about the following fact. The problem should be solved now. Thank you very much!

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Michael Hardy
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Sam Shen
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  • I know that the probability density function would be $$f(y)=\frac{e^{-y/\beta}}{\beta}$$ by the definition of a gamma distribution with $\alpha=1, \beta=\beta$. – Sam Shen Nov 02 '14 at 23:47
  • And also that, for a Gamma distribution, $$E(Y)=\alpha\beta$$ So for our distribution, we should have $$E(Y)=\beta$$ (Since $\alpha=1$) – Sam Shen Nov 02 '14 at 23:49
  • @RobinGoodfellow I get it now. Thank you very much! – Sam Shen Nov 02 '14 at 23:59

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