TITLE: tail limit of Laplace transform of a bounded random variable
QUESTION [0 upvotes]: Suppose that $X$ is a variable such that $0<X<m$. I would like to know some information on the behavior of the function
$$\phi(p)=\frac{1}p \log E e^{pX} $$
when $p\to\infty$. 
Here are some information that I know. Because $X$ is bounded by $m$, we have an obvious bound $\phi(p)\le m$ for all $p>0$. In addition, by Jensen inequality, we can also estimate $\phi(p)$ from below
$$\phi(p)\ge\frac1p \log e^{pEX}=EX$$
So to summarize, my questions are: can one say anything more about the limit $\lim_{p\to\infty}\phi(p)$? If this is not possible, can one say anything about the limits $\limsup_{p\to\infty}\phi(p)$ and $\liminf_{p\to\infty}\phi(p)$?

REPLY [1 votes]: The limit is the supremum $\xi$ of the support of $X$, that is, $$\xi=\sup\{x\mid P(X\geqslant x)\ne0\}=\inf\{x\mid P(X\leqslant x)=1\}.$$
To prove this, note that for every positive $\varepsilon$:

$X\leqslant\xi$ almost surely hence $E(e^{pX})\leqslant e^{p\xi}$
$q=P(X\geqslant\xi-\varepsilon)$ is positive and $E(e^{pX})\geqslant qe^{p(\xi-\varepsilon)}$

Thus, $\xi-\varepsilon+\frac1p\log q\leqslant\phi(p)\leqslant\xi$ for every $p$, which is enough to conclude.
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