# Finding the Variance of the MLE Variance of a Joint Normal Distribution

I have a random sampling of $$Z_1,...Z_n$$ from a normal distribution $$N(\mu,\sigma^{2})$$. I am considering them within a joint likelihood function.

I know that the MLE ($$\hat\sigma^{2}$$) of $$\sigma^{2}$$ is $$\frac{1}{n}\Sigma_{i=1}^{n} (x_{i}-\bar{x})^{2}$$.

I want to find the variance of $$\hat\sigma^{2}$$.

Here is my reasoning so far:

I know that

$$\Sigma_{i=1}^{n} \frac{(x_{i}-\bar{x})^{2}}{\sigma^{2}} \sim \chi^{2}_{n-1}\\ \implies \frac{n\hat\sigma^{2}}{\sigma^{2}} \sim \chi^{2}_{n-1} \\ \therefore \text{Var}(\frac{n\hat\sigma^{2}}{\sigma^{2}}) = 2(n-1)$$
Here, I guessed that I need to make the assumption that $$\sigma^{2}$$ is a constant. $$\frac{n^{2}}{\sigma^{4}}\text{Var}(\hat\sigma^{2}) = 2(n-1) \\ \therefore \text{Var}(\hat\sigma^{2}) = \frac{2(n-1)\sigma^{4}}{n^{2}}$$

• What do you mean when you say "I need to make the assymption that $\sigma^2$ is constant" ? Are you considering it to be a random variable ? If you are not doing Bayesian statistics, $\sigma^2$ is usually constant (just unknown). Commented Aug 14, 2023 at 7:22

This problem can be solved much more generally for any distribution whose moments exist. Let $$(X_1, ..., X_n)$$ denote a random sample of size $$n$$ drawn from a population random variable $$X$$. By definition, your MLE estimator is the $$2^\text{nd}$$ sample central moment:

$$m_2 = \frac1n \sum _{i=1}^n \left(X_i-\bar{X}\right){}^2$$

The problem is to find: $$\text{Var}(m_2)$$ .

Background

The problem is just a special case of the much more general problem of 'moments of moments' which are usually defined in terms of power sum notation:

$$s_r=\sum _{i=1}^n X_i^r$$

The first step is to express the sample central moment $$m_2$$ in terms of power sums $$s_i$$. This is a well-known result, but for generality is done here by the mathStatica function SampleCentralToPowerSum[r]:

We seek $$\text{Var}(m_2)$$ ... i.e. the $$2^\text{nd}$$ CentralMoment of $$m_2$$. The solution can be obtained with the mathStatica function:

where the solutions is expressed in terms of central moments of the population $$\mu_i$$, assuming they exist of course.

Special case of Normality

The above is a completely general solution for any distribution whose moments exist. In the special case of a $$N(\mu, \sigma^2)$$ parent, we have that $$\mu_2 = \sigma^2$$ and $$\mu_4 = 3\sigma^4$$, which simplifies to: