denote by $X$ and $X_n$, $n\in \mathbb{N}$, random variables and $r\in\mathbb{R}_+$ with $E=\mathbb{E}\left[ e^{rX} \right] < \infty$ and $E_n=\mathbb{E}\left[ e^{rX_n} \right] < \infty$ for all $n$. Assume that the $X_n$ converge to $X$ for $n\rightarrow\infty$, say almost surely. Apart from $X_n \nearrow X$ (monotone convergence) or $X_n \leq 0$ (dominated convergence), are there any other criteria for the convergence $E_n \rightarrow E$ ?



1 Answer 1


The concept of uniform integrability gives a generalization of the conditions of the dominated convergence theorem. Uniform integrability of a sequence $\left(Y_n\right)_{n\geqslant 1}$ means that $$\lim_{x\to +\infty}\sup_{n\geqslant 1}\mathbb E\left[\left|Y_n\right|\mathbf 1\left\{|Y_n|\gt x\right\}\right]=0.$$ We can prove that a uniformly integrable sequence $\left(Y_n\right)_{n\geqslant 1}$ converges in probability to a random variables $Y$ if and only if it converges in $\mathbb L^1$ to $Y$. Therefore, the result holds if $\left(e^{rX_n}\right)_{n\geqslant 1}$ is uniformly integrable.

Conversely, we can prove that if $\left(Y_n\right)_{n\geqslant 1}$ is a sequence of non-negative random variables which converges in distribution to $Y$ and such that $\mathbb E[Y_n]\to \mathbb E[Y]$, then $\left(Y_n\right)_{n\geqslant 1}$ is uniformly integrable.

  • $\begingroup$ Thank you very much for this answer! In connection with this question, are there conditions that guarantee that the convergence $\mathbb{E}[e^{r X_n}] \to \mathbb{E}[e^{r X}]$ is uniform in $r$? $\endgroup$ May 11, 2023 at 9:56

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