"T is a random variable that follows an exponential law. Show that (for q a whole number)
$$ E[T^q] = (q-1)!E[S](µ_T)^{q-1} $$
With $ µ = E[T] $ and $ S ∼ Γ (1/(µ_T),q) $ "
Anyway I know that it simplifies to $ E[T^q] = q!(µ_T)^q $... My question is simply how could this be? Shouldn't the answer simply be $ E[T^q] = (E[T])^q = (µ_T)^q $? Or have I severely misunderstood something?
If $T$ is exponentially distributed with mean $\mu$, then $T$ has density $\frac1\mu e^{-\frac t\mu}\mathsf 1_{(0,\infty)}(t)$. Hence $$ \mathbb E[T] = \int_0^\infty \frac t\mu e^{-\frac t\mu}\ \mathsf dt = te^{-\frac t\mu}|_0^\infty -\int_0^\infty e^{-\frac t\mu}\ \mathsf dt = \frac1\mu e^{-\frac t\mu}|_\infty^0 = \frac1\mu. $$ Assume now that $\mathbb E[T^q] = q!\mu^q$ for some positive integer $q$. Then \begin{align} \mathbb E[T^{q+1}] &= \int_0^\infty \frac{t^{q+1}}\mu e^{-\frac t\mu}\ \mathsf dt\\ &= (q+1)t^{q+1}e^{-\frac t\mu}|_0^\infty - \mu(q+1)\int_0^\infty \frac{t^q}\mu e^{-\frac t\mu}\ \mathsf dt\\ &= \mu(q+1)\mathbb E[T^q]\\ &= \mu(q+1)q!\mu^q\\ &= (q+1)!\mu^{q+1}, \end{align} so by induction we conclude that $\mathbb E[T^q] = q!\mu^q$ for all positive integers $q$.