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$X$ has a Gamma distribution with parameters $\lambda$ and $\alpha$. I must find $E(X^r)$ and $r$ is a positive integer.

How can I do this?

I am guessing I have to use the Gamma function but I don't know how to do this?

savinq
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  • From the gamma function, the integral of $x^{\alpha-1}e^{-x}$ between $0$ and infinity, I have derived this to equal: $(\alpha-1)\Gamma(\alpha-1)$. I don't even know if this is correct or whether anything can come from this, basically I don't know what I am doing! – savinq Nov 27 '13 at 18:56

1 Answers1

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You leave it to be inferred by us based only on conventions that $\alpha$, rather than $\lambda$ is the shape parameter. There's also the question of whether $\lambda$ is supposed to be the intensity parameter, so that the distribution is $$ \frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx)\text{ for }x>0 $$ or its reciprocal, the scale parameter, so that the distribution is $$ \frac{1}{\Gamma(\alpha)}\cdot (x/\lambda)^{\alpha-1} e^{-x/\lambda}\,(dx/\lambda)\text{ for }x>0. $$

I'm going to guess that you mean the first of these two alternatives.

Once you understand the integral that defines the Gamma function, you're almost done. You have $$ \begin{align} \mathbb E(X^r) & = \int_0^\infty x^r \frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt] & = \frac{1}{\Gamma(\alpha)} \cdot\frac{1}{\lambda^r} \int_0^\infty (\lambda x)^{r+\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt] & = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r} \int_0^\infty u^{r+\alpha-1} e^{-u}\,du \\[12pt] & = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r}\cdot\Gamma(r+\alpha). \tag1 \end{align} $$

This bears simplification. We have $$ \begin{align} \Gamma(r+\alpha) & = (r-1)\Gamma(r-1+\alpha)= (r-1)(r-2)\Gamma(r-2+\alpha)=\cdots \\[12pt] & \cdots=(r+\alpha-1)(r+\alpha-2)(r+\alpha-3)\cdots(r+\alpha-r)\Gamma(\alpha). \end{align} $$ Then you can reduce the fraction in $(1)$.

Michael Hardy
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  • Quick question, in the integral, should it be $({\lambda}x)^{\alpha-1}$ or ${\lambda}^{\alpha}x^{\alpha-1}$ – savinq Nov 27 '13 at 19:17
  • The former, because $(\lambda,dx)$ appears. – Michael Hardy Nov 27 '13 at 19:22
  • In textbooks one often sees $\displaystyle\frac{1}{\Gamma(\alpha)}\lambda^\alpha x^{\alpha-1} e^{-\lambda x},dx$. That is the same thing as $\displaystyle\frac{1}{\Gamma(\alpha)}(\lambda x)^{\alpha-1} e^{-\lambda x},(\lambda,dx)$, which is how I wrote it above. I find the latter form more convenient because it makes the substitution $u=\lambda x$ clearer and simpler. – Michael Hardy Nov 27 '13 at 19:25
  • Ah right I got you, missed the $\lambda$ at the end there. – savinq Nov 27 '13 at 19:25