<s>[INST] Interesting limit with Poisson and Chi-squared Distribution I am stuck with computing the following limit: 
\begin{align}
 \lim_{ n \to \infty} E \left[ \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U  \right] \right)^2 \right].
\end{align} 
In the above expression,  $X$ given $U$ follows Poisson with parameter $U$  and where $U$ is a Chi-square of degree $n$. 
Here is what I tried: 
Suppose we let
\begin{align}
V_n =\left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2
\end{align} 
Then, using Jensen's inequality
\begin{align}
V_n  \le E \left[  \frac{X}{n} + \frac{1}{2} \,  \Big | U \,  \right] =  \frac{U}{n}+\frac{1}{2}
\end{align} 
Moreover, we have that $E \left[  \frac{U}{n}+\frac{1}{2}   \right]=1+\frac{1}{2}$. Therefore, by the dominated convergence theorem we have that 
\begin{align}
\lim_{n \to \infty}  E[  V_n ]=   E[   \lim_{n \to \infty}  V_n ]
\end{align}
Therefore, assuming everything up to here is correct, to compute the limit we have to  find 
\begin{align}
 \lim_{n \to \infty}  V_n&=  \lim_{n \to \infty}   \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2\\
&=  \left(  \lim_{n \to \infty}  E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2
\end{align} 
This is the place where I am stuck.  Is it simply another applications dominated convergence theorem?  If so, then I think the answer is
\begin{align}
 \lim_{ n \to \infty} E \left[ \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U  \right] \right)^2 \right]=\frac{1}{2}.
\end{align} 
What I mean by another application of dominating convergence theorem is that for every $u>0$
\begin{align}
E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right] &\le  \sqrt{ E \left[   \frac{X}{n} + \frac{1}{2} \, \Big  | \, U=u  \right]}\\
&=   \sqrt{  \frac{u}{n} + \frac{1}{2} }\\
&= \sqrt{  u + \frac{1}{2}}
\end{align}
Therefore, 
\begin{align}
E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right]=  E \left[  \lim_{n \to \infty}  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right]=  \frac{1}{2}.
\end{align}
Is this a correct sequence of steps?  I feel a bit uneasy about the second application of the dominated convergence theorem. 
 [/INST] proof: Instead of the heuristic argument I made below you can simply prove it by finding a lower bound, since you already found an upper bound of $3/2$ by Jensen's inequality. Using
$$\sqrt{1+2x}\geq \sqrt{3} \left(1+\frac{x-1}{3}\right) - (x-1)^2$$
$\forall x \geq 0$, which then becomes with $x=\frac{X}{n}$
$$\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right] \geq \mathbb{E}\left[ \sqrt{3} \left(1+\frac{X/n-1}{3}\right) - (X/n-1)^2 \right] \\
=-{\frac {{U}^{2}}{{n}^{2}}}+ \left( -\frac{1}{n^2} + \frac{1}{\sqrt{3}\,n} + \frac{2}{n} \right) U + \frac{2}{\sqrt{3}}-1$$
and so since
$$\mathbb{E}\left(U\right) = n \\ \mathbb{E}\left(U^2\right) = n(n+2)$$
then
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right] \geq \sqrt{3}$$
which means
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]^2\right] \geq \lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right]^2 \geq \sqrt{3}^2 = 3 \tag{FKG}$$
because for any $u$ from the domain of the chi-squared distribution and since $\sqrt{1+2x}$ is increasing we have
$$\frac{{\rm d}}{{\rm d}u} \sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\sqrt{1+2k/n}=\sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\left(\sqrt{1+2(k+1)/n}-\sqrt{1+2k/n}\right)\geq 0$$ and so $\mathbb{E}\left[\sqrt{1+\frac{2X}{n}}\right]$ is an increasing function in $U=u$ and the reqiurements for the FKG inequality are fulfilled. $\square$

Originally I made a too long of a comment for a heuristic argument. Though I know I'm completely ignoring convergence here, when you simply manipulate the corresponding series expressions you obtain the Jensen upper bound result of $3/2$. Starting with
$$
\sqrt{1+\frac{2X}{n}}=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2X}{n}\right)^m
$$
and then calculating ${\mathbb{E}(\cdot)}$ with respect to Poisson one obtains
$$
\mathbb{E}\left(\sqrt{1+\frac{2X}{n}}\right)=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \mathbb{E}\left(X^m\right) \\
=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \sum_{i=0}^m {m \brace i} \, U^i \\
=\sum_{i=0}^\infty U^i \sum_{m=i}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m {m \brace i} \\
\stackrel{m=i+k}{=}\sum_{i=0}^\infty \left(\frac{2U}{n}\right)^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i} \, .
$$
The $i$-th moment with respect to the chi-squared distribution of degree $n$ is $$\mathbb{E}\left(U^i\right) = \frac{2^i \Gamma(i+n/2)}{\Gamma(n/2)} \sim n^i \left(1+\frac{i(i-1)}{n} + {\cal O}(1/n^2) \right)$$
which is merely an asymptotic series, but taking only the first term, squaring the previous equation and taking $\mathbb{E}(\cdot)$ we obtain
$$
\left(\sum_{i=0}^\infty 2^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i}\right)^2 \, .
$$
The terms $k>0$ vanish for $n\rightarrow \infty$, so we are left with
$$
\left( \sum_{i=0}^\infty \binom{1/2}{i} \, 2^i \right)^2 = \sqrt{1+2}^2 = 3
$$
where the series again doesn't converge, but only in the sense of analyticity. Gathering the left out overall factor $1/2$ the final result yields $3/2$.
</s>