Cramer Rao bound

Question

Let $X_1\dots X_n$ be a random sample with distribution $N(\mu, \sigma^2)$, with known variance $\sigma^2$.

Show that the estimator $\delta(X) = \bar{X}^2 - \frac{\sigma^2}{n}$ of $\mu^2$ doesn't reach the Cramer Rao bound.

Here $\bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_i$.

Proof. First, the estimator $\delta$ is an unbiased estimator for $\mu^2$. The Fisher information is $I(\mu) = \frac{n}{\sigma^2}$. So, the Cramer-Rao bound gives us

$$\operatorname{var}(\delta) \geq \frac{g'(\mu)^2}{I(\mu)}=\frac{4\mu^2 \sigma^2}{n}$$ with $g(\mu) = \mu^2$. Now, I want to find out what is

$$\operatorname{var}(\bar{X}^2 - \frac{\sigma^2}{n})=\operatorname{var}(\bar{X}^2).$$

I've found a paper with the expression for $\operatorname{var}(\bar{X}^2)$, but uses a lot of things. Maybe there is a simplest way.

Answer 1

Recall that $ \bar{X}_n \sim N(\mu, \sigma ^2/n) $ and $ n ( \bar{X}_n - \mu ) ^ 2 / \sigma^2 \sim \chi^2 (1)$, hence \begin{align} \operatorname{var}(\bar{X}^2) &=\operatorname{var}( (\bar{X} - \mu + \mu )^2)\\ &= \operatorname{var}( (\bar{X} - \mu ) ^2 + 2\mu( \bar{X} - \mu) + \mu ^2)\\ &=\frac{\sigma^4} {n^2} \operatorname{var}(\chi^2(1)) + 4\mu^2 \frac{\sigma^2}{n} \\ & = \frac{2\sigma^4}{n^2} + \frac{4\mu^2\sigma^2} n \\ & > \frac{4\mu^2\sigma^2} n. \end{align}