Interchanging expectation and limsup using reverse Fatou

Question

I'm trying to interchange expectation and limsup using reverse Fatou Lemma to conclude $$ \limsup_{n\rightarrow\infty}\mathbb{E}\left[\frac{X^{n}}{n}\right]\leq\mathbb{E}\left[\limsup_{n\rightarrow\infty}\frac{X^{n}}{n}\right], $$ where $\{X^n\}$ is a sequence of nonnegative random variables. For this, I need the sequence to be bounded by some nonnegtaive random variable with finite expectation.

I can bound $X^{n}$ by a Poisson r.v., i.e., $X^{n}\leq N^{n}$ almost surely where $N^{n}$ is Poisson with rate $\lambda^{n}$ such that $\lambda^{n}/n\rightarrow\lambda<\infty$ as $n\rightarrow\infty$.

I am not however sure this is enough since I can not bound $X^{n}$ for all $n$ by a random variable that does not depend on $n$.

Question: Is there a weaker version of Fatou's Lemma that could possibly apply to this case?

Answer 1

Firstly, if $\{Y_n,n\ge 1\}$ is uniformly integrable(u.i., cf. S.I. Resnick, A Probability Path, Springer,2014. p.182, 6.5.1), that is $$ \lim_{k\to\infty}\sup_{n\ge 1}\mathsf{E}[|Y_n|1_{(|Y_n|\ge k)}]=0,$$ then Fatou lemma is hold as following: $$ \mathsf{E}[\varliminf_{n\to\infty} Y_n]\le \varliminf_{n\to\infty}\mathsf{E}[Y_n]\le \varlimsup_{n\to\infty}\mathsf{E}[Y_n] \le \mathsf{E}[\varlimsup_{n\to\infty} Y_n]. \tag{1} $$ Proof of the right side: Since $$ \mathsf{E}[Y_n]=\mathsf{E}[Y_n1_{\{Y_n\le k\}}]+\mathsf{E}[Y_n1_{\{Y_n>k\}}] \le \mathsf{E}[Y_n\wedge k]+ \sup_{n\ge 1}\mathsf{E}[|Y_n|1_{\{|Y_n|>k\}}] $$ Hence, \begin{align} \varlimsup_{n\to\infty}\mathsf{E}[Y_n]&\le \varlimsup_{n\to\infty}\mathsf{E}[Y_n\wedge k]+ \sup_{n\ge 1}\mathsf{E}[|Y_n|1_{\{|Y_n|>k\}}]\\ \text{($Y_n\wedge k\le k$, by Fatou Lemma)}\qquad&\le \mathsf{E}[\varlimsup_{n\to\infty}(Y_n\wedge k)]+ \sup_{n\ge 1}\mathsf{E}[|Y_n|1_{\{|Y_n|>k\}}]\\ &\le \mathsf{E}[\varlimsup_{n\to\infty}Y_n]+ \sup_{n\ge 1}\mathsf{E}[|Y_n|1_{\{|Y_n|>k\}}] \end{align} Now letting $k\to+\infty$ and using the u.i. of $\{Y_n\}$ we get the right half of (1). The proof of left half of (1) is similar.

Secondly, $\{\frac{X_n}n,n\ge 1\}$ is u.i.: Since \begin{align} \mathsf{E}\Bigl[\frac{|X_n|}n 1_{(\frac{|X_n|}n\ge k)}\Bigr] &\le \frac{\mathsf{E}[X_n^2]}{kn^2}\le \frac{\mathsf{E}[N_n^2]}{kn^2}=\frac{\lambda_n^2+\lambda_n}{kn^2}\\ &\le\frac1k\sup_{n\ge1}\Bigl(\frac{\lambda_n^2+\lambda_n}{n^2}\Bigr) =\frac{C}k,\qquad \text{where $C\stackrel{\text{def}}{=}\sup_{n\ge 1}\Bigl(\frac{\lambda_n^2+\lambda_n}{n^2}\Bigr)<\infty$}. \end{align} Therefore, $\{\frac{X_n}n,n\ge 1\}$ is u.i. and under the conditions what you supposed, you could get what you want.