Variance of the sample mean

Question

You observe X1, \dots, Xn, independent draws with mean \mu and variance \sigma^2, and form the sample mean \bar{X} = \frac{1}{n}\sum{i=1}^n Xi. What are the mean and variance of \bar{X}? How many observations do you need to cut the standard error in half?

Accepted Answer

Mean. By linearity of expectation, \mathbb{E}[\bar{X}] = \frac{1}{n}\sum{i=1}^n \mathbb{E}[Xi] = \mu, so \bar{X} is an unbiased estimator of \mu. Note that linearity needs no independence at all. Variance. Because the Xi are independent, variances add: \operatorname{Var}(\bar{X}) = \frac{1}{n^2}\sum{i=1}^n \operatorname{Var}(Xi) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}. The standard error is therefore \sigma/\sqrt{n}: precision improves with the square root of the sample size. To halve the standard error you need 4	imes the data; to get one extra decimal digit of precision you need 100	…

Variance of the sample mean

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