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Let $\mu_n, \mu$ be probability measures on $\mathbb{R}$ such that $\mu_n$ converges in distribution to $\mu$. Let $M_n(s) = \mathbb{E}(e^{sX_n})$ and $M(s) = \mathbb{E}(e^{sX})$ be the respective moment generating functions. Assume that $M_n(s)$ is finite in a common interval $[-s_0, s_0]$, $s_0 > 0$. Does it follow that $M_n(s) \rightarrow M(s)$ in this interval?

I want to know if this holds in order to prove a version of the continuity theorem for MGFs. The issue I am having is that the function $e^{sx}$ is continuous but not bounded, unlike the case for characteristic functions.

Nikhil Ghosh
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1 Answers1

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If uniform integrability of $\{\exp\left(sX_n\right),n\geqslant 1\}$ takes place, then usingthe fact that uniform integrability and convergence in distribution implies convergence of the expectations.

This holds if we assume that for all $s\in (-s_0.s_0)$, $\sup_{n\geqslant 1}\mathbb E\left[\exp\left(sX_n\right)\right]$ is finite.

However, the convergence may not hold at $s_0$. For example, if $s_0=1$, take a sequence of positive random variables $(Y_n)$ which converges in distribution say to one and such that $\mathbb EY_n$ does not converge to one ($Y_n=n^2$ and $1$ with respective probabilitites $1/n$ and $1-1/n$) and let $X_n=\ln(Y_n)$. In this example, $M_n(s)$ is finite for all $s$ but converges only for $s\leqslant 0$.

Davide Giraudo
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  • Thanks @Davide, but why does it hold that $\exp(sX_n)$ is uniformly integrable for $s \in (-s_0, s_0)$? – Nikhil Ghosh Dec 30 '19 at 20:48
  • @NikhilGhosh Actually, it seems that uniform integrability may not hold without further assumptions. – Davide Giraudo Dec 30 '19 at 20:55
  • Ok I think I see now. Uniform integrability now follows from the fact that $\mathbb{E}(\exp(sX_n) ; \exp(sX_n) \geq \alpha) \leq \frac{1}{\alpha^{\delta}} \int \exp(s(1+\delta)X_n) dP$ – Nikhil Ghosh Dec 30 '19 at 21:16
  • @DavideGiraudo For the counterexample given, does it show that the continuity theorem does not hold for moment generating function? But the tricky thing is I believe I have seen the continuity of MGF several times... – Zorualyh Oct 12 '20 at 15:06