If \(X\) is a positive random variable, prove that its negative moments are given by, for \(r>0\) :

(a) \(\quad \mathbb{E}\left(X^{-r}\right)=\frac{1}{\Gamma(r)} \int_{0}^{\infty} t^{r-1} \mathbb{E}\left(e^{-t X}\right) d t\)

where \(\Gamma\) is the Gamma function and its positive moments are, for \(0

(b) \[\begin{equation*}\mathbb{E}\left(X^{r}\right)=\frac{r}{\Gamma(1-r)} \int_{0}^{\infty} \frac{1-\mathbb{E}\left(e^{-t X}\right)}{t^{r+1}} d t \tag{b}\end{equation*}\]

and for \(n

(c) \(\quad \mathbb{E}\left(X^{r}\right)=\frac{r-n}{\Gamma(n+1-r)} \int_{0}^{\infty}(-1)^{n} \frac{\phi^{(n)}(0)-\phi^{(n)}(t)}{t^{r+1-n}} d t\).

\[s^{r} \Gamma(1-r)=r \int_{0}^{\infty} \frac{1-e^{-s t}}{t^{r+1}} d t\]

For \(r=n\), one has \(\mathbb{E}\left(X^{n}\right)=(-1)^{n} \phi^{(n)}(0)\). See Schürger [774] for more results and applications.