Let $(X_1,...,X_n)$ be a random sample from a normal distribution $N(\mu,1)$.
$T=\frac{1}{n}\sum_{i=1}^{n}X_{i}^2-1$ is an unbiased estimator for $\mu^2$ since : $$ E(T)=E(\frac{1}{n}\sum_{i=1}^{n}X_{i}^2-1)$$ $$= \frac{1}{n}\sum_{i=1}^{n}E(X_{i}^2)-1$$ $$=\frac{1}{n}\sum_{i=1}^{n}[Var(X_{i}+E(X_{i})^2]-1$$ $$=\frac{1}{n}\sum_{i=1}^{n}[1+\mu^2]-1 = \mu^2$$
My question is , can we prove that $T$ is not the MVUE for $\mu^2$ based only on comparing its variance with the CRLB for $\mu^2$?. In other words does the MVUE attain the CRLB (or at least in this case where $\sigma^2$ is known ) ? .
I have tried to find the variance of $T$ and compare it with CRLB , but there is something wrong . Here is my attempt :
The CRLB for $q(\theta)= \mu^2$ is $\frac{4\mu^2}{n}$
$$Var(T)=Var(\frac{1}{n}\sum_{i=1}^{n}X_{i}^2-1)=\frac{1}{n}\sum_{i=1}^{n}X_{i}^2$$ $$=\frac{1}{n^2}\sum_{i=1}^{n} Var(X_{i}^2)$$
Now $Var(X_{i}^2)$ could be calculated as follows :
$$Var(X_{i}^2) = E[(X_{i}^2 - E(X_{i}^2))^2]= E[(X_{i}^2 - (1+\mu^2))^2]$$ $$=\int_{-\infty}^{+\infty}(X_{i}^2 - (1+\mu^2))^2f(x_{i},\mu)dx_{i}$$ $$=\int_{-\infty}^{+\infty}x_{i}^4f(x_{i},\mu)dx_{i} -2 (1+\mu^2)\int_{-\infty}^{+\infty}x_{i}^2f(x_{i},\mu)dx_{i}+(1+\mu^2)^2\int_{-\infty}^{+\infty}f(x_{i},\mu)dx_{i}$$ $$=\int_{-\infty}^{+\infty}x_{i}^4f(x_{i},\mu)dx_{i}-(1+\mu^2)^2$$ If we standardize the first term we will get $ Var(X_{i}^2) = 3-(1+\mu^2)^2$ then $Var(T)=\frac{3-(1+\mu^2)^2}{n}$ which could be negative ! any help please.
