How do I find the variance of $\overline{X}^2 - \frac{1}{n}$ where $X_i\sim N(\theta,1)$?

Question

I have a problem here. The question asks to compare the UMVUE and MVBE of $\theta^2$ considering a random sample $X_1,...,X_n$ from a $N(\theta,1)$ distribution.

I found out the Cramer Rao Lower Bound for the variance of an unbiased estimator for estimating $\theta^2$ to be $\frac{4\theta^2}{n}$. Since $\overline{X}$ came out to be a sufficient and complete statistic, I used this to get the UMVUE which is $T = \overline{X}^2 - \frac{1}{n}$. Now I have to find the variance of $T$.

Now,$Var[T]=Var[\overline{X}^2]$. I see that $Z=n(\overline{X} - \theta)^2$ has a $\chi_1^2$ distribution. So, $Var[Z] = 2$ $\implies Var[(\overline{X} - \theta)^2]= \frac{2}{n^2}$ $\implies Var[\overline{X}^2 +\theta^2 - 2\overline{X}\theta] = \frac{2}{n^2}$ $\implies Var[\overline{X}^2] + Var[2\overline{X}\theta] = \frac{2}{n^2}$ $\implies Var[\overline{X}^2] = \frac{2}{n^2} - Var[2\overline{X}\theta]$ $\implies Var[\overline{X}^2] = \frac{2}{n^2} - \frac{4\theta^2}{n}=Var[T]$

The answer says that $Var[T] = \frac{2}{n^2} + \frac{4\theta^2}{n}$. So where am I going wrong?

Edit: My teacher had used this method which I wasn’t able to understand in two places. If someone could clarify this for me,would be great.

$Var[T] = Var[\overline{X}^2] = E[\overline{X}^4] - E^2[\overline{X}^2]$ $=E[\overline{X}^4] - (\mu^2 +\frac{1}{n})^2$ Now, $E[\overline{X}^4]=E[\overline{X} - \mu + \mu]^4$ $=E[\overline{X} - \mu]^4 + 4E[\overline{X}-\mu]^3\mu + 6E[\overline{X}-\mu]^2\mu^2 +4E[\overline{X}-\mu]\mu^3 +\mu^4$ $=3\frac{1}{n^2} + 0 + 6\frac{1}{n}\mu^2 + 0 + \mu^4$.

I can’t understand how the first two terms came about in the last step.

Answer 1

If $X_1,\dots,X_n$ are iid normal rvs with mean $\mu$ and variance $1$ then: $$Z:=\sqrt{n}\left(\overline{X}-\mu\right)=\frac{\sqrt{n}}{n}\sum_{i=1}^{n}\left(X_{i}-\mu\right)$$ has standard normal distribution so that $\mathbb{E}Z^{4}=3$ (see here).

That leads to $\mathbb{E}\left(\left(\overline{X}-\mu\right)^{4}\right)=\mathbb{E}\frac{1}{n^{2}}Z^{4}=\frac{3}{n^{2}}$ and $\mathbb{E}\left(\left(\overline{X}-\mu\right)^{3}\right)=\mathbb{E}\frac{1}{n\sqrt{n}}Z^{3}=0$