# Can you derive the distribution of the Gaussian variance estimator without using moment generating functions?

Given a dataset of Gaussian i.i.d. $X_i's$, we can show that the maximum likelihood estimates for this Gaussian distribution's mean and variance are given as: \begin{align} \hat{\mu} &= \frac{1}{n}\sum_{i = 1}^{n} X_i \\[10pt] \widehat{\sigma^2} &= \frac{1}{n}\sum_{i = 1}^{n} (X_i - \hat{\mu})^2 \end{align} We can easily tell that $\hat{\mu}$ will be Gaussian distributed as it is a linear function of Gaussian random variables. But the same cannot be said for $\widehat{\sigma^2}$ as it is a non-linear function of $X_i's$. My question is, what is the best way to determine the variance estimator's distribution for arbitrary $n$?

For the case where $n = 2$ we have: \begin{align} \hat{\sigma} &= \frac{1}{2}\left( \big(X_1 - \frac{1}{2} \left( X_1 + X_2 \right)^2 \big) + \big(X_2 - \frac{1}{2} \left( X_1 + X_2 \right)^2 \big)\right) \\[10pt] \widehat{\sigma^2} &= \frac{1}{4} \left( X_1^2 + X_2^2 - 2X_1X_2 \right) \end{align} Even for this case it is not exactly clear to me how to go about determining the estimator's distribution analytically.

• When the $X$'s are Gaussian, the variance will be distributed as chi-squared. Cf: Why is chi square used when creating a confidence interval for the variance? – gung - Reinstate Monica Nov 22 '15 at 18:32
• So working with moment generating functions or characteristic functions would be the best way to solve for the distribution? Is there a treatment of going about it without using MGFs? Although I imagine it's unwieldy, I'd be interested in seeing how one might attempt it. – Falimond Nov 22 '15 at 18:53

It is possible to prove that $$S^2=\sum_{i=1}^n (X_i-\bar{X})^2\sim \chi^2_{n-1}$$without the mgf. This comes from the property that, if $$Z$$ is a $$n$$-dimension Normal random vector, $$Z\sim\text{N}(0,I_n)$$ and if $$A$$ is an $$n\times n$$ symmetric and idempotent matrix, then $$\xi^2=Z^\text{T} A Z \sim \chi^2_{\text{tr}(A)}$$ where $$\text{tr}(A)$$ denotes the trace of $$A$$. (The proof of this result does not necessarily involves the mgf.) Since $$S^2=\sum_{i=1}^n (X_i-\bar{X})^2=Z^\text{T} A Z$$when $$A=I_n-\frac{1}{n} \mathbf{1}\mathbf{1}^\text{T}$$and $$A$$ is idempotent with $$n-1$$ eigenvalues equal to $$1$$ and one equal to $$0$$ (with eigenvector $$\mathbf{1}$$), the above lemma applies and$$S^2\sim\chi^2_{n-1}$$