Dominating function for derivative of moment generating function

Question

Let $X$ be a random variable and the moment generating function

$$\psi_X:(-\varepsilon,\varepsilon)\rightarrow \mathbb{R}_+,\quad \psi_X(t):=E[e^{tX}]$$ be defined, such that $\psi_X(t)<\infty$ for all $t\in(-\varepsilon,\varepsilon)$. According my textbook we have

$$\psi^{(n)}_X(0)=\frac{d^n}{dt^n}E[e^{tX}]\bigg|_{t=0}=E[\frac{d^n}{dt^n}e^{tX}\bigg|_{t=0}]=E[X^n],$$

where differentiation and integration can be interchanged by using dominated convergence.

However what is the dominating function here? It is clear that for all $t\in(-\varepsilon,\varepsilon)$, we have

$$\frac{d^n}{dt^n}e^{tX}=X^n\ e^{tX}$$

So I need to find a integrable dominating function $h:\Omega\rightarrow\mathbb{R}_+$, such that

$$\forall t\in(-\varepsilon,\varepsilon): |X^n\ e^{tX}|\le h$$ Does someone has a hint on this one?

Thanks a lot in advance!

Answer 1

There's no need for differentiation under the integral sign. First note that for $|t|<\varepsilon$, $$\frac{|t|^n|X|^n}{n!}\leq e^{|tX|}\leq e^{tX}+e^{-tX}$$ So $X$ has moments at any order.

Next, $E(e^{tX})=E\left(\sum_{k=0}^\infty\frac{(tX)^k}{k!}\right)$ and note that for all $n$, $$\left| \sum_{k=0}^n\frac{(tX)^k}{k!} \right|\leq \sum_{k=0}^n\frac{|tX|^k}{k!} \leq e^{|tX|}\leq e^{tX}+e^{-tX}$$

The DCT applies and yields $\Psi_X(t)=E\left(\sum_{k=0}^\infty\frac{(tX)^k}{k!}\right) = \sum_{k=0}^\infty\frac{t^k}{k!}E(X^k)$.

Hence $\Psi_X$ coincides with a power series over $(-\varepsilon, \varepsilon)$. It is therefore infinitely differentiable at $0$ with $\psi^{(k)}_X(0)=E(X^k)$.

Answer 2

The short answer to your question is that you can always find a $\delta >0$ s.t. for all $t<\epsilon - \delta$, $|x^n e^{t|x|}| \leq |x|^n e^{(\epsilon-\delta)|x|}$, and $|x|^n e^{-\delta |x|}$ is bounded for all $n$ and $\delta$, while $e^{\epsilon |x|}$ is integrable by assumption.

First, consider the case $n=1$.Then, $$\psi'(t) = \lim_{h\rightarrow 0} \frac{\psi(t+h)-\psi(t)}{h} = \lim_{h\rightarrow 0} E[\frac{e^{(t+h)X}-e^{tX}}{h}].$$ Now, choose $\delta>0$ s.t. $0<\delta<\epsilon$. Assume $t$ and $h$ are small enough s.t. $t\pm h \in (-\epsilon + \delta, \epsilon - \delta)$. By the Mean Value Theorem, for any x, there exists a $t' \in [t,t+h]$ s.t. $$ \frac{e^{(t+h)x}-e^{tx}}{h} = x e^{t'x}. $$

Since $t' \leq \epsilon - \delta$, \begin{equation} |xe^{t'x}| \leq |x|e^{|t'x|} \leq |x|e^{(\epsilon -\delta)|x|} \leq C e^{\epsilon|x|}, \end{equation} where $C=\delta^{-1}e^{-1}$, the maximum of $f(x)=|x|e^{-\delta|x|}$. Next, observe that $e^{\epsilon|x|}\leq e^{\epsilon x}+e^{-\epsilon x}$, which is integrable by assumption. This allows to use dominated convergence to deduct the claim.

The case $n>1$ follows by induction and using a similar bound, i.e., $$|x^{n}e^{t'x}| \leq |x^{n}| e^{(\epsilon-\delta)|x|} \leq Ce^{\epsilon|x|},$$ where $C = \max_x f(x) = max_x\{|x|^n e^{-\delta|x|}\} \leq (n/\delta)^ne^{-n}$, which, again, is integrable by the same logic as in the base case $n=1$.