If $X\ge0$, $e^{sX} $ is integrable for every $s\le a$ and $X^k$ is integrable for every $k$, then $e^{sX}X^k$ is integrable for every $s<a$

Question

I encounter the following problem in homework:

$X$ is nonnegative random variable. Show if $\mathbb{E}[e^{sX}] < \infty$ for $s\in(-\infty, a]$ and $\mathbb{E}[X^k] < \infty$ for all $k$, then $\mathbb{E}[e^{sX}X^k] < \infty$ for $s<a$. I tried using cauchy schwarz inequality and bounding $e^{sX}X^k$ by some other functions that have finite expectations, but they doesn't seem working. Also don't use differentiation inside the integral since this is actually part of the problem in proving differentiation under the integration.

Any help would be appreciated!!

Answer 1

Assume that $X$ is a $\mathbb{R}$-valued random variable such that $\mathbb{E}[e^{sX}] < \infty$ for all $s \in (-\infty, a)$. Then it follows that $\mathbb{E} \left| X^k e^{sX} \right| < \infty$ for all $k \geq 0$ and $s \in (-\infty, a)$.

This follows from $|x|^k \leq M(e^{\epsilon x} + e^{-\epsilon x})$, where $M = M(k,\epsilon) > 0$ is a constant depending only on $k$ and $\epsilon$. So if $s < a$ and we pick $\epsilon > 0$ so that $s + \epsilon < a$,

\begin{align*} \mathbb{E}[ |X|^k e^{sX}] \leq M \left( \mathbb{E}[ e^{(s+\epsilon)X}] + \mathbb{E}[ e^{(s-\epsilon)X}] \right) < \infty. \end{align*}

Answer 2

Let $X$ have density function proportional to $f(x)=\exp(-x)/(1+x^2)$ for $x\ge0$ and $f(x)=0$ for $x<0.$ Then obviously $E|X|^k<\infty$ for all $k>0$, and $E\exp(sX)<\infty$ for all $s\le 1$. But equally clearly, $E\exp (X) X^k = \infty$ for all $k\ge 1$.