# Asymptotic Distributions of form: $\sqrt{n}(\hat{\mu} - \mu, \hat{\sigma}^2 - \sigma^2)$

Suppose $$X_1, \dots, X_n$$ iid normals $$N(\mu, \sigma^2)$$, and $$\hat{\mu}$$ and $$\hat{\sigma}^2$$ are the MLE. How would one go about finding

$$\sqrt{n}(\hat{\mu} - \mu, \hat{\sigma}^2 - \sigma^2).$$

When discussing the distributions of the forms $$\sqrt{n}(\hat{\mu} - \mu)$$ and $$\sqrt{n}(\hat{\sigma}^2 - \sigma^2)$$ it is quite clear to me using the regular tools and theorems:

The MLE are

$$\hat{\mu} = \bar{X} ~~~~~\text{and}~~~~~\hat{\sigma}^2 = \frac{1}{n}\sum_{i}(X_i - \bar{X})^2$$

and the asymptotic distribution are

\begin{align*} \sqrt{n}(\hat{\mu} - \mu) & \overset{d}{\Rightarrow} N(0, \sigma^2) \\ \sqrt{n}(\hat{\sigma}^2 - \sigma^2) & \overset{d}{\Rightarrow} N(0, 1). \end{align*}

But the joint $$\sqrt{n}(\hat{\mu} - \mu, \hat{\sigma}^2 - \sigma^2)$$ is not so obvious. How would one calculate this?

• $\sqrt{n}(\hat\sigma^2-\sigma^2) \to N(0, 1)$ should be $\sqrt{n}(\hat\sigma^2-\sigma^2) \to N(0, 2\sigma^4)$. The covariance of sample mean and sample variance for normal distribution is actually zero. You can crank out the (inverse) information matrix to show this. Mar 8, 2020 at 4:48