Moment generating function. Let \(X\) be a positive random variable on the probability space \((\Omega, \mathscr{A}, \mathbb{P})\). The function \(\phi_{X}(t):=\int e^{-t X} d \mathbb{P}\) is called moment generating function. Show that \(\phi_{X}\) is \(m\)-times differentiable at \(t=0+\) if the absolute \(m\) th moment \(\int|X|^{m} d \mathbb{P}\) exists.

If this is the case, the following formulae hold.
(i) \(M_{k}:=\int X^{k} d \mathbb{P}=\left.(-1)^{k} \frac{d^{k}}{d t^{k}} \phi_{X}(t)ight|_{t=0+} \quad\) for all \(0 \leqslant k \leqslant m\).
(ii) \(\phi_{X}(t)=\sum_{k=0}^{m} \frac{M_{k}}{k !}(-1)^{k} t^{k}+o\left(t^{m}ight) . \quad\left(f(t)=o\left(t^{m}ight)ight.\) means \(\left.\lim _{t ightarrow 0} f(t) / t^{m}=0.ight)\)
(iii) \(\left|\phi_{X}(t)-\sum_{k=0}^{m-1} \frac{M_{k}}{k !}(-1)^{k} t^{k}ight| \leqslant \frac{|t|^{m}}{m !} \int|X|^{m} d \mathbb{P}\).
(iv) If \(\int|X|^{k} d \mathbb{P}<\infty\) for all \(k \in \mathbb{N}\), then
\[
\phi_{X}(t)=\sum_{k=0}^{\infty} \frac{M_{k}}{k !}(-1)^{k} t^{k}
\] for all \(t\) within the convergence radius of the series.