Limiting variance of normal mean

Question

In Casella's Statistical Inference，in Example 10.1.8 on page 470, it says that the limiting variance of normal mean $\bar X_n$, is $\lim_{n\to\infty}\sqrt n\text{Var}\bar X_n=\sigma^2$. However, since $\bar X_n$ is iid normal mean, we have $\text{Var}\bar X_n=\sigma^2/n$, hence the limiting variance is $\lim_{n\to\infty}n\text{Var}\bar X_n=\sigma^2$.

So, is it a typo in the book or I misunderstood something?

Thanks

Answer 1

Think about a general estimator $T_n$ of a parameter $\theta$. Consider $ n^{1/2} (T_n - \theta) / \sigma \rightarrow_d N(0,1)$ which is the same as $ n^{1/2} (T_n - \theta) \rightarrow_d N(0,\sigma^2)$. Here $\sigma^2$ is referred to as the asymptotic variance of $T_n$.

From CLT, $ \bar{X}_n \dot \sim N(\mu, \sigma^2/n)$. This means that $$n^{1/2} (\bar{X}_n - \mu) / \sigma \rightarrow_d N(0,1)$$ or $$n^{1/2} (\bar{X}_n - \mu) \rightarrow_d N(0,\sigma^2)$$ which implies also that $ \bar{X}_n - \mu = O_p (n^{-1/2})$. The limiting variance is $$\lim_{n\rightarrow \infty} V \{n^{1/2} (\bar{X}_n - \mu )\} = \lim_{n\rightarrow \infty} V (n^{1/2} \bar{X}_n) = \lim_{n\rightarrow \infty} n V(\bar{X}_n) = \lim_{n\rightarrow \infty} n \frac{\sigma^2}{n} = \sigma^2$$