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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?

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    $\begingroup$ $\sqrt{n}(\hat\sigma^2-\sigma^2) \to N(0, 1)$ 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. $\endgroup$ Mar 8, 2020 at 4:48

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Either compute the information matrix or recall that the normal distribution is symmetric, meaning that the off-diagonal entries in the information matrix are equal to...

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