Variance of the $\hat{\sigma^2}$ of a Maximum Likelihood estimator - Cross Validated most recent 30 from stats.stackexchange.com 2019-10-21T16:07:54Z https://stats.stackexchange.com/feeds/question/161448 https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/161448 1 Variance of the $\hat{\sigma^2}$ of a Maximum Likelihood estimator mgus https://stats.stackexchange.com/users/81385 2015-07-14T18:11:10Z 2015-07-14T22:05:56Z <p>Given some normally distributed observations $x_1,x_2,...,x_n$ </p> <p>$\forall i\ x_i\sim\mathcal{N}(\mu, \sigma^2)$</p> <p>the ML estimator decides that the variance that maximizes the likelihood function is (see <a href="https://en.wikipedia.org/wiki/Maximum_likelihood#Iterative_procedures" rel="nofollow">here</a>):</p> <p>$\hat{\sigma^2}=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x}^2)$</p> <p>Now, I am trying to find the variance of this estimation:</p> <p>$\sigma^2_{\hat{\sigma^2}}=Var[\hat{\sigma^2}]=Var[\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x}^2)]$ </p> <p>If we note that: $\hat{\sigma^2}=\frac{1}{n}\sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2) \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}\frac{1}{n}\sum_{i=1}^{n}x_i+\frac{1}{n}\sum_{i=1}^{n}\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}^2+\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-\bar{x}^2$</p> <p>we have:</p> <p>$\sigma^2_{\hat{\sigma^2}}=Var[\frac{1}{n}\sum_{i=1}^{n}x_i^2-\bar{x}^2]$</p> <p>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.</p> https://stats.stackexchange.com/questions/161448/variance-of-the-hat-sigma2-of-a-maximum-likelihood-estimator/161452#161452 2 Answer by Zhanxiong for Variance of the $\hat{\sigma^2}$ of a Maximum Likelihood estimator Zhanxiong https://stats.stackexchange.com/users/20519 2015-07-14T18:19:06Z 2015-07-14T22:05:56Z <p>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 $2k$.</p>