Given some normally distributed observations $x_1,x_2,...,x_n$

$\forall i\ x_i\sim\mathcal{N}(\mu, \sigma^2)$

the ML estimator decides that the variance that maximizes the likelihood function is (see here):


Now, I am trying to find the variance of this estimation:


If we note that: $\hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2) \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}\frac{1}{n}\sum_{i=1}^{n}x_i+\frac{1}{n}\sum_{i=1}^{n}\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}^2+\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-\bar{x}^2$

we have:


but I am stuck here since I think that $x_i$ and $\bar{x}$ are not independent in order to use the property that says that the variance of the sum is the sum of the variances.


1 Answer 1


Do you know the famous result that if $X_1, \ldots, X_n \text{ i.i.d. } \sim N(\mu, \sigma^2)$, then $$\frac{1}{\sigma^2}\sum_{i = 1}^n (X_i - \bar{X})^2 \sim \chi_{n - 1}^2?$$ It is also well-known that the variance of a $\chi_k^2$ random variable is $2k$.

An alternative treatment is to express the sample variance as a quadratic form of the random vector $\mathbf{X} = (X_1, \ldots, X_n)$ then apply the quadratic form variance formula. See this answer for a solution to a slightly more general case.

  • $\begingroup$ I know about the second part about the variance of chi-squared distribution you are referring to but where does the first result come from? I mean, how can I prove it? Finally, what is the distribution of $X_i-\bar{X}$? $\endgroup$
    – mgus
    Jul 14, 2015 at 18:54
  • $\begingroup$ @KonstantinosKonstantinidis For one possible justification, see stats.stackexchange.com/questions/121662/… $\endgroup$
    – Zhanxiong
    Jul 14, 2015 at 21:26

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