I have a random sample drawn from a $N(\theta,\sigma^2)$ distribution with $\sigma^2$ known. I am trying to estimate $\theta$.
I need to calculate the efficiency of the unbiased estimator, $\bar{X}^2-\frac{\sigma^2}{n}$. To do so, I have been trying to find the pdf of the estimator by using the MGF. However, I have been having some difficulty in doing so.
My thought process thus far:
$M_{\frac{1}{n^2} \sum_{i=1}^{n} X_{i}^{2} -\frac{\sigma^2}{n} (t)} =e^{\frac{-\sigma^2}{n}t}M_{\sum_{i=1}^{n}X_{i}^2}(\frac{1}{n^2}t)$
Can we say that $X_{i}^2=\chi_{1}^2$? If so:
$e^{-\frac{\sigma^2}{n}t}[M_{X_{i}^2}(\frac{1}{n^2}t)]^n =e^{-\frac{\sigma^2}{n}t}[(\frac{1}{1-2(\frac{1}{n^2}t)})^{\frac{1}{2}}]^n =e^{-\frac{\sigma^2}{n}t}[\frac{1}{1-\frac{2}{n^2}}t]^{\frac{n}{2}}$
Do we instead need to take into account that $X\sim N(\theta,\sigma^2)$? Does that mean we need to make $\frac{(X_{i}-\theta)^2}{\sigma^2}$ appear in our equation? If so, how would I do that? Does the following make any sense?
$\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}X_{i}^2-\frac{\sigma^2}{n} =\sigma^2 \frac{1}{\sigma^2}\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}X_{i}^2-\frac{\sigma^2}{n} =\sigma^2\frac{1}{\sigma^2}\frac{1}{n^2}\displaystyle \sum_{i=1}^nX_{i}^2-2\theta X_i+\theta^2+2\theta X_i-\theta^2-\frac{\sigma^2}{n} =\sigma^2\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}\frac{(X_i-\theta)^2}{\sigma^2}+\frac{1}{n^2}\displaystyle \sum_{i=1}^n 2\theta X_i-\frac{1}{n^2}\displaystyle \sum_{i=1}^n \theta^2-\frac{\sigma^2}{n} =\frac{\sigma^2}{n^2}\chi^2_n+\frac{2\theta}{n^2}\displaystyle \sum_{i=1}^{n}X_i-\frac{\theta^2}{n}-\frac{\sigma^2}{n}$
*We'd then take the MGF of the above expression.
Are we now dealing with the sum of a chi-squared and normal distribution? Am I overthinking this?
I'd appreciate any guidance! As you can see, I am quite confused as to what to do. Thank you in advance!