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Let $X \sim \Gamma(\beta,\lambda)$ where $\beta>0$ is the rate paramter and $\lambda>0$ is the shape parameter. When I want to compute the $q$'th moment I get that \begin{align*} \mathbb{E}[X^q] &= \int_{0}^\infty x^q P_X(dx) \\ &= \int_0^\infty x^q \frac{\lambda^\beta x^{\beta-1}\exp(-\lambda x)}{\Gamma(\beta)} \ dx \\ &= \frac{\lambda^{-q+1}}{\Gamma(\beta) }\int_0^\infty \lambda^{q+\beta-1} x^{q+\beta-1}\exp(-\lambda x) \ dx \\ &= \frac{\lambda^{-q+1}}{\Gamma(\beta) }\Gamma(q+\beta). \end{align*} which is different ($\lambda^{-q}$ instead of $\lambda^{-q+1}$) from what my textbook says as well as https://en.wikipedia.org/wiki/Gamma_distribution#Higher_Moments.

Did I make a mistake somewhere? Or is there something I haven't thought of?

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1 Answer 1

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Did I make a mistake somewhere?

Yes, you forgot to set $d(\lambda x)$ inside your integral. In other words "one" lambda has to be shifted in the differential

$$\Gamma(\beta+q)=\int_0^{\infty}(\lambda x)^{\beta+q-1}\cdot e^{-\lambda x} d(\lambda x)=\int_0^{\infty} \theta^{\beta+q-1}\cdot e^{-\theta}d\theta$$

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  • $\begingroup$ Ahhh that makes sense, thanks! $\endgroup$
    – Jacobiman
    Commented Sep 28, 2021 at 14:12

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