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?
Best Answer
Yes, you forgot to set $d(\lambda x)$ inside your integral. In other words "one" lambda has to be shifted in the differential
$$\Gamma(\beta+q)=\int_0^{\infty}(\lambda x)^{\beta+q-1}\cdot e^{-\lambda x} d(\lambda x)=\int_0^{\infty} \theta^{\beta+q-1}\cdot e^{-\theta}d\theta$$