Let $\{X_i\}_{i=1}^{n}$ be $n \ $ i.i.d random variables such that $X_i \sim N(\theta,\theta)$ where $\theta > 0$.
Now, I am trying to simplify $$\text{var}\bigg(\sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}\bigg).$$ I got this expression in some calculation that I am doing.
Only, the square root is creating problems. I can't get my head around this. Please help.
Thanks!
EDIT : I have to show that $$\text{var}\bigg(\sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}\bigg) \to 0$$ as $n \to \infty.$
EDIT 2 : MLE of $\theta$ is $$\hat{\theta}_{MLE} = \frac{-1 + \sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}}{2}$$.
I am trying to prove the consistency of $\hat{\theta}_{MLE}$.
I do not know your purpose but surely you can approximate
$$\mathbb{V}\Bigg[\sqrt{1+4\overline{X}_n}\Bigg]$$
using Delta Method.
EDIT: your estimator is
$$T=\sqrt{1+4\frac{\Sigma_i X_i^2}{n}}$$
After usual calculations,
$$\hat{\theta}_{ML}=\frac{T-1}{2}$$
The asymptotic distribution of MLE is known...easy solved problem (as an approximation, obviously)
EDIT: at the end, the problem was different.
Reading the complete original text
After calculating $\hat{\theta}_{ML}$ that you did correctly let's go back to the Score that is
$$l^*(\theta)=-\frac{n}{2\theta}-\frac{n}{2}+\frac{\Sigma_i X_i^2}{2\theta^2}$$
The derivative of the score is
$$l^{**}(\theta)=\frac{n}{2\theta^2}-\frac{\Sigma_i X_i^2}{\theta^3}$$
Thus the Fischer information is
$$I_n(\theta)=-\mathbb{E}[l^{**}(\theta)]=-\frac{n}{2\theta^2}+\frac{(\theta+\theta^2)n}{\theta^3}=\frac{(1+2\theta)n}{2\theta^2}$$
Concluding:
The asymptotic distribution is the following
$$\hat{\theta}_{ML}\dot{\sim} N\Bigg[\theta;\frac{2\theta^2}{(1+2\theta)n}\Bigg]$$
This means that
$$\lim\limits_{n \to\infty}\mathbb{E}[\hat{\theta}_{ML}]=\theta$$
and
$$\lim\limits_{n \to\infty}\mathbb{V}[\hat{\theta}_{ML}]=0$$
Which is a necessary and sufficient condition for consistency