Interchanging expectation and limsup using reverse Fatou 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?
 A: 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.
