Interesting limit with Poisson and Chi-squared Distribution

Question

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.

Answer 1

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. $\square$

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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 $3/2$.

Answer 2

Let $\xi_1,\xi_2,\ldots$ be i.i.d. $N(0,1)$. Let $N_1,N_2,\ldots$ be i.i.d., unit rate Poisson processes, constructed so that $\{N_j\}$ and $\{\xi_j\}$ are independent.

Define $Y_j=N_j(\xi_j^2)$. Then $Y_1,Y_2,\ldots$ are i.i.d. and the conditional distribution of $Y_j$ given $\xi=\{\xi_1,\xi_2,...\}$ is Poisson with parameter $\xi_j^2$.

Let $X_n=Y_1+\cdots+Y_n$ and $U_n=\xi_1^2+\cdots+\xi_n^2$. Then $U_n$ has a chi-squared distribution with $n$ degrees of freedom and the conditional distribution of $X_n$ given $\xi$ is Poisson with parameter $U_n$. We are interested in $\lim_{n\to\infty}EV_n$, where $V_n=(E[Z_n\mid U_n])^2$ and $$ Z_n = \sqrt{\frac{X_n}n + \frac12}. $$ Note that $V_n=(E[Z_n\mid\xi])^2$.

By the law of large numbers, $X_n/n\to EY_1=E\xi_1^2=1$ a.s. Thus, $Z_n\to\sqrt{3/2}$ a.s. Since $EZ_n^2=3/2$ for all $n$, it follows that $\{Z_n\}$ is uniformly integrable. Therefore, $Z_n\to\sqrt{3/2}$ in $L^1$, which implies that $E[Z_n\mid\xi]\to\sqrt{3/2}$ in $L^1$.

By passing to a subsequence, we may assume this convergence is almost sure. Then, along this subsequence, $V_n\to3/2$ a.s. Now, \begin{align} EV_n^2 &\le EZ_n^4\\ &= E\left|{\frac{X_n}n + \frac12}\right|^2\\ &= n^{-2}EX_n^2 + n^{-1}EX_n + \frac14\\ &= n^{-2}EX_n^2 + 1 + \frac14. \end{align} Since \begin{align} EX_n^2 &= E[E[X_n^2\mid U_n]]\\ &= E[U_n + U_n^2]\\ &= n + n(n+1)\\ &= n^2 + 2n, \end{align} we have $$ \sup_nEV_n^2 \le \sup_n\left({1 + \frac2n + 1 + \frac14}\right) < \infty. $$ Hence, $\{V_n\}$ is uniformly integrable, and so $EV_n\to3/2$ along the given subsequence.

This shows that every subsequence of $\{EV_n\}$ has a further subsequence converging to $3/2$. Therefore $EV_n\to3/2$.

EDIT:

Here is an alternative to the second half of the proof which uses a modification of your dominated convergence idea.

Follow the proof up until the point where we show that $V_n\to3/2$ a.s. Now, $$ V_n \le W_n := E\left[{ \frac{X_n}n + \frac12 \;\bigg|\; U_n }\right] = \frac{U_n}n + \frac12. $$ By the law of large numbers, $W_n\to3/2$ a.s. Also, $EW_n=3/2$ for all $n$. Thus, by the generalized dominated convergence theorem (see General Lebesgue Dominated Convergence Theorem, for example), we have $EV_n\to 3/2$.

EDIT 2:

In fact, by mimicking the proof of the generalized DCT (as given in the above link) and using Fatou's lemma for conditional expectation (Theorem 6.56 in these notes), you can prove the following:

Theorem. Suppose $X_n\to X$ a.s., $|X_n|\le Y_n$, $Y_n\to Y$ a.s., and $E[Y_n\mid\mathcal{G}]\to E[Y\mid\mathcal{G}]$. Then $E[X_n\mid\mathcal{G}]\to E[X\mid\mathcal{G}]$ a.s.

We can now use this theorem to give an alternative proof that $V_n\to3/2$ a.s. First, $Z_n\to\sqrt{3/2}$ a.s. by the law of large numbers. Next, since $x\ge1/2$ implies $\sqrt x\,\le\sqrt 2\,x$, we have $$ Z_n \le \zeta_n := \sqrt2\,\left({\frac{X_n}n + \frac12}\right). $$ Since $\zeta_n\to3/\sqrt2$ a.s. and $E\zeta_n=3/\sqrt2$ for all $n$, the above theorem implies $$ E[Z_n\mid U_n] = E[Z_n\mid\xi]\to\sqrt{3/2}\quad\text{a.s.} $$ If you use both these alternates, then you have a rigorous incarnation of what you originally envisioned, which was two applications of the dominated convergence theorem.