Convergence in distribution for this estimator based on the second moment I am having trouble proving the following:
Claim: $\sqrt{n}(\hat\theta_2 - \theta)$ converges in distribution and give an asymptotic confidence interval centered for $\theta$ in $\hat\theta_2$, where:


*

*$\hat\theta_2 = \frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2$, $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ and the $X_i$'s are i.i.d ~ $Poi(\theta), \theta > 0$.

*I know that $\hat\theta_2 = \frac{1}{n}\sum_{i=1}^n (X_i - \theta)^2 - (\theta - \bar{X}_n)^2$.

*I also known that $Var[(X_i - \theta)^2] = 2\theta^2 + \theta$
Until now, I used the transformation of the second bullet point to consider 
$\sqrt{n}\big(\frac{1}{n}\sum_{i=1}^n (X_i - \theta)^2 - (\theta - \bar{X}_n)^2 - \theta\big)$. 
Then, taking each summand inside the brackets separately, I determined that $(\theta - \bar{X}_n)^2 \xrightarrow{n \to \infty} 0$ a.s. using the LLN and continuous mapping theorem.
Applying the same method to $\frac{1}{n}\sum_{i=1}^n (X_i - \theta)^2$ gives:
$\frac{1}{n}\sum_{i=1}^n (X_i - \theta)^2 \xrightarrow{n \to \infty} E[(X_i - \theta)^2] = Var[(X_i - \theta)^2] + E[(X_i - \theta)]^2 = 2\theta^2 + \theta$. 
Overall, this means that I am left with $\sqrt{n}(2\theta^2)$, and do not know how to proceed further to show convergence and build the confidence interval.
I am grateful for any tip and piece of advice.
 A: The convergence in distribution can be shown using Cramer's theorem on the asymptotic normality of functions of sample moments, studied through a Taylor series expansion to one term.


Reference: A Course in Large Sample Theory (first edition) by Thomas Ferguson.

A direct proof can also be given as shown in the accepted answer here.
Suppose $X_1,\ldots,X_n$ are i.i.d with $E(X_i)=\mu,\operatorname{Var}(X_i)=\sigma^2$ and $\operatorname{Var}(X_i^2)=\tau^2<\infty$. 
Let $S^2=\frac1n\sum\limits_{i=1}^n (X_i-\overline X)^2$. Then $$\sqrt n(S^2-\sigma^2)=\sqrt n\left(\frac1n\sum_{i=1}^n X_i^2-\overline X^2-\sigma^2\right)$$
Since the distribution of $S^2$ does not depend on $\mu$, we can take $\mu=0$ without loss of generality. 
Then by the CLT, $$\sqrt n\left(\frac1n\sum_{i=1}^n X_i^2-\sigma^2\right)\stackrel{L}\longrightarrow N(0,\tau^2)\tag{2}$$
It can be argued (see accepted answer in linked post) that $$\sqrt n\overline X^2\stackrel{P}\longrightarrow 0\tag{3}$$
From $(2)$ and $(3)$ it follows that $$\sqrt n(S^2-\sigma^2)\stackrel{L}\longrightarrow N(0,\tau^2)$$
Or equivalently with $\mu_4=E(X_i-\mu)^4$,

$$\sqrt n(S^2-\sigma^2)\stackrel{L}\longrightarrow N(0,\mu_4-\sigma^4)\,,$$

Now you can apply this result to your specific question and construct a pivot for a Wald-type asymptotic confidence interval.
