Essential supremum via cumulant

Question

Let $p(t)=\log \mathbb{E}[\exp (tX)]$ for $X$ real valued random variable. Now it holds (assuming that $p$ is smooth and finite on $\mathbb{R}$) that $p'(\infty)=\text{ess}\sup X$. How can I prove that? For step functions it is easy, but how can I (by simple methods) show that it holds for all random variables?

Answer 1

Hmm. Say $X$ is uniformly distributed on $[-2,1]$. I'd say the essential supremuum of $X$ was $2$; from the way the question's turning out I gather you'd say it was $1$?

Assume first that $X$ is essentially bounded above.

Now $p'(t)=\mathbb E[Xe^{tX}]/\mathbb E[e^{tX}]$.

Say the essential sup of your $X$ is $S$. Then $X\le S$ almost surely; since exponentials are positive this says $\mathbb E[Xe^{tX}]\le S\mathbb E[e^{tX}]$, hence $p'(t)\le S$ for all $t$.

Now say $\delta>0$. Let $A=(X>S-\delta)$ and $B=(X>S-\delta/2)$. Note that $A$ and $B$ both have positive probability. Writing integrals instead of $\mathbb E$: $$\frac{\int_Ae^{tX}}{\int_{\Omega\setminus A}e^{tX}}\ge \frac{\int_Be^{tX}}{\int_{\Omega\setminus A}e^{tX}}\ge\frac{P(B)e^{t(S-\delta/2)}}{P(\Omega\setminus A)e^{t(S-\delta)}}\to\infty\quad(t\to\infty).$$

Let $\epsilon>0$. The above shows that if $t$ is large enough then $$\int_\Omega e^{tX}\le(1+\epsilon)\int_Ae^{tX}.$$So if $t$ is large enough then $$ \frac{\int Xe^{tX}}{\int e^{tX}}\ge \frac{\int_A Xe^{tX}}{(1+\epsilon)\int_A e^{tX}}\ge\frac{S-\delta}{1+\epsilon}$$ So for every $\delta,\epsilon>0$ we have $$\liminf_{t\to\infty}p'(t)\ge\frac{S-\delta}{1+\epsilon}.$$Hence $\liminf p'(t)\ge S$.

If $X$ is not essentially bounded above a similar but I believe somewhat simpler argument suffices to show $p'(\infty)=\infty$.

Answer 2

If $\operatorname{ess\,sup}(X)\gt0$, set $0\lt\lambda\lt\operatorname{ess\,sup}(X)$, $t\gt0$, and $0\lt\epsilon\lt\lambda$. Then $$ \begin{align} \int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x) &\ge\int_{x\gt\lambda}xe^{tx}\,\mathrm{d}\mu(x)\\ &\ge\mu\{x:x\gt\lambda\}\lambda e^{t\lambda}\tag{1} \end{align} $$ and $$ \begin{align} \int_{0\le x\le\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x) &\le\mu\{x:0\le x\le\lambda-\epsilon\}(\lambda-\epsilon)e^{t(\lambda-\epsilon)}\\ &\le\mu\{x:x\le\lambda\}(\lambda-\epsilon)e^{t(\lambda-\epsilon)}\tag{2} \end{align} $$ and since $xe^{tx}\ge-\frac1{et}$, $$ \int_{x\lt0}xe^{tx}\,\mathrm{d}\mu(x)\ge-\frac1{et}\tag{3} $$ If $\operatorname{ess\,sup}(X)\le0$, set $\lambda\lt\operatorname{ess\,sup}(X)$, $t\gt\frac{-1}{\lambda}$, and $\epsilon\gt0$. Then $$ \begin{align} \int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x) &\le\int_{x\gt\lambda}xe^{tx}\,\mathrm{d}\mu(x)\\ &\le\mu\{x:x\gt\lambda\}\lambda e^{t\lambda}\tag{4} \end{align} $$ and $$ \begin{align} \int_{x\le\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x) &\ge\mu\{x:x\le\lambda-\epsilon\}(\lambda-\epsilon)e^{t(\lambda-\epsilon)}\\ &\ge\mu\{x:x\le\lambda\}(\lambda-\epsilon)e^{t(\lambda-\epsilon)}\tag{5} \end{align} $$ In either case, using Iverson Brackets, $$ \begin{align} \lim_{t\to\infty}\left|\frac{\displaystyle\int_{x\le\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x)}\right| &\le\lim_{t\to\infty}\frac{\mu\{x:x\le\lambda\}(\lambda-\epsilon)e^{t(\lambda-\epsilon)}+\frac1{et}[\lambda\gt0]}{\mu\{x:x\gt\lambda\}\lambda e^{t\lambda}}\\[15pt] &=0\tag{6} \end{align} $$ Adding $1$ and taking reciprocals, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{t\to\infty}\frac{\displaystyle\int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{\mathbb{R}}xe^{tx}\,\mathrm{d}\mu(x)} =1}\tag{7} $$ Since $$ \int_{x\gt\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x) \ge\mu\{x:x\gt\lambda\}e^{t\lambda}\tag{8} $$ and $$ \int_{x\le\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x) \le\mu\{x:x\le\lambda\}e^{t(\lambda-\epsilon)}\tag{9} $$ we have $$ \begin{align} \lim_{t\to\infty}\frac{\displaystyle\int_{x\le\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{x\gt\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x)} &\le\lim_{t\to\infty}\frac{\mu\{x:x\le\lambda\}e^{t(\lambda-\epsilon)}}{\mu\{x:x\gt\lambda\}e^{t\lambda}}\\ &=0\tag{10} \end{align} $$ Adding $1$ and taking reciprocals, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{t\to\infty}\frac{\displaystyle\int_{x\gt\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{\mathbb{R}}e^{tx}\,\mathrm{d}\mu(x)} =1}\tag{11} $$ Therefore, $$ \begin{align} \lim_{t\to\infty}\frac{\mathrm{d}}{\mathrm{d}t}\log\left(\int_{\mathbb{R}}e^{tx}\,\mathrm{d}\mu(x)\right) &=\lim_{t\to\infty}\frac{\displaystyle\int_{\mathbb{R}}xe^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{\mathbb{R}}e^{tx}\,\mathrm{d}\mu(x)}\\ &=\lim_{t\to\infty}\frac{\displaystyle\int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x)}{\displaystyle\int_{x\gt\lambda-\epsilon}e^{tx}\,\mathrm{d}\mu(x)}\\[15pt] &\ge\lambda-\epsilon\tag{12} \end{align} $$ Sending $\lambda$ to $\operatorname{ess\,sup}(X)$ and $\epsilon$ to $0$, we get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{t\to\infty}\frac{\mathrm{d}}{\mathrm{d}t}\log\left(\int_{\mathbb{R}}e^{tx}\,\mathrm{d}\mu(x)\right) =\operatorname{ess\,sup}(X)}\tag{13} $$