I am asked to show that the unbiased estimator $\hat{\sigma^2}=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X})^2$ is not efficient.
So far I was able to show that the Rao-Cramer Lower Bound is $\frac{2\sigma^4}{n}$ so I know that I need to find $Var[\hat{\sigma^2}]>RCLB$.
What I have tried here is to simply plug it in using the identity
$$\sum_{i=1}^n (X_i-\bar{X})^2 = \sum_{i=1}^n X_i^2 - n\bar{X}^2$$
What makes me uncomfortable is that unlike expectation, I need to know whether these two terms are independent or not, and I do not know how to find $Var[\bar{X}^2]$ even if they were.
I also tried to express the sum of square as
$$\sum_{i=1}^n (X_i-\mu +\mu-\bar{X})^2 = \sigma^2 \left[ \sum_{i=1}^n\left(\frac{X_i-\bar{X}}{\sigma}\right)^2 - 2\left(\frac{X_i-\bar{X}}{\sigma}\right)\sum_{i=1}^n\left( \frac{X_i-\mu}{\sigma} \right) + n\left( \frac{\bar{X}-\mu}{\sigma} \right)^2 \right]$$
hoping that $\chi_{n}^2$ and $N(0,1)$ could come into play but I face the same issue again.
I would appreciate your input.
Because: $$\hat{\sigma^2} \sim \frac{\chi_{n-1}^2 \sigma^2}{n-1}$$ And since the variance of the Chi-squared distribution is $2k$ where $k$ is the number of the degrees of freedom we have: $$Var[\hat{\sigma^2}]=\frac{2(n-1)\sigma^4}{(n-1)^2}=\frac{2\sigma^4}{n-1}>\frac{2\sigma^4}{n}$$