Ratio of moment generating function with it's derivative

Question

Let $X$ be a random variable with distribution function $F$. Let $\phi(t) = E[e^{t X}]$ be the moment generating function. Let $x_0 = \sup\{x : F(x) < 1\}$ be a finite quantity. Prove that $\phi(t)$ exists for all $t > 0$, and $\frac{\phi'(t)}{\phi(t)} \to x_0$ as $t \to \infty$.

The first part of this problem is very easy. Note that $e^{tX} = e^{tX}(1_{\{X > x_0\}} + 1_{\{x_0 \geq X > 0\}} + 1_{\{X \leq 0\}})$, which is bounded by $e^{tx_0}1_{\{x_0 \geq X > 0\}} + 1$, and hence the expectation is bounded by $e^{tx_0} + 1$ for all $t$.

However, for the second part,I have found a few things, but I am not able to piece then together:

• $\phi'(t) = E[Xe^{tX}]$.
• $\frac{\phi'(t)}{\phi(t)}$ is an increasing function bounded by $x_0$.

Hence, it's sufficient if I can prove that $\sup_t {\frac{\phi'(t)}{\phi(t)}} = x_0$. However, I am somewhat stuck here.

Answer 1

The first part is even easier: the $X\leq 0$ doesn't need to be treated as a special case; you can just use $1_{X\leq x_0}.$

For the second part, one way you could start is to consider $a<x_0$ and show that $\mathbb E[1_{X\leq a}e^{tX}]/\mathbb E[e^{tX}]\to 0.$ This is because $\mathbb E[1_{X\leq a}e^{tX}]\leq e^{at},$ whereas for any $a<b<x_0$ we have $\mathbb E[e^{tX}]\geq \mathbb P[X\geq b]e^{bt}$ which dominates $e^{at}$ for large $t.$

Answer 2

Let $\gamma(t)=\ln(\phi(t))$ and $p_\epsilon=\mathsf{P}(X\ge x_0-\epsilon)$. $\gamma'$ is non-decreasing ($\because \gamma''\ge 0$) and bounded by $x_0$. For any $\epsilon>0$, $p_{\epsilon}>0$ and $\gamma(t)\ge t(x_0-\epsilon)+\ln(p_{\epsilon})$ so that $\lim_{t\to\infty}\gamma'(t)\ge x_0-\epsilon$.