We know that sample variance $\sum(x- \bar{x})^2/(n-1) = \sum(x^2- 2x\bar{x}+\bar{x}^2)/(n-1)$ is an unbiasedd estimator of variance
Is $\sum(x^2 - \bar{x}^2)/(n-1)$ also an unbiased estimator of variance?
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I realize that they are the same.
$\sum(x- \bar{x})^2 = \sum(x^2- 2x\bar{x}+\bar{x}^2) = \sum x^2- \sum 2x\bar{x}+\sum \bar{x}^2 = \sum x^2 - \sum \bar{x}^2$
The last step is because $\sum x\bar{x} = \bar{x} \sum x = \bar{x} \sum \bar{x} = \sum \bar{x}^2$
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$\begingroup$ It's not very intuitive algebraically for your last step for most readers, maybe you'd better add the summation index and show clearly for the above last step? $\endgroup$ Dec 15, 2022 at 0:35