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Using the basic definition of Sample Variance

$$\ S^{2} = \frac{\sum_{i=1}^n (X_i- \bar{X})^2} {n-1} $$ I obtained expectation of $S^2$ as $ \frac{n \sigma^2}{n-1} $

How do I remove the constant factor? I saw similar questions to this on the forum but the answers listed in those are simply too complicated and use an entirely different approach than directly calculating expectation of the expression for sample mean

Edit: Steps involved in derivation

$$ E(S^2)= E \left(\frac{\sum_{i=1}^n (X_i- \bar{X})^2}{n-1} \right) \\= \frac{\sum_{i=1}^n E(X_i- \bar{X})^2}{n-1}\\=\frac{\sum_{i=1}^n E(X_i^2)-2E(X_i)\bar{X}+(E(\bar{X}))^2}{n-1} \\=\frac{\sum_{i=1}^n \sigma^2} {n-1}\\=\frac{n\sigma^2}{n-1} $$

Hari07
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1 Answers1

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The main problem with any "simple" evaluation of the expectation of the sample variance is that the $X_i$ and the $\bar{X}$ aren't independent. To fix this problem, we can note the following: $$X_i - \bar{X} = X_i - \frac{1}{n}\sum_{k=1}^n X_k = \frac{n-1}{n}\left(X_i - \frac{1}{n-1}\sum_{k=1\\k\neq i}^nX_k\right) := \frac{n-1}{n}(X_i - Y_i)$$ where now $X_i - Y_i$ is the difference of two independent random variables with $E(X_i) = E(Y_i)$ and $\text{var}(Y_i) = \frac{\sigma^2}{n-1}$

Then we can show directly that $$E(S^2) = \frac{n-1}{n}E(X_1-Y_1)^2 = \frac{n-1}{n}\text{var}(X_1-Y_1) = \frac{n-1}{n}\left(\sigma^2 + \frac{\sigma^2}{n-1}\right) = \sigma^2$$

Brian Moehring
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