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If $X_i$ iid with variance $\sigma$ then I want to prove that $S_n^2 = \frac{1}{n-1}\sum_{i=1}^{n}(X_i -\bar X_n )^2$ is an unbiased estimate of the variance $\sigma$. So here I go:

\begin{equation} \begin{aligned} \mathbb{E}(S_n ^2)&= \frac{1}{n-1}\sum_{i=1} ^{n} \mathbb{E}(X_i -\bar X_n )^2\\ &= \frac{1}{n-1}\sum_{i=1} ^{n} \mathbb{E}(X_i^2 -2X_i\bar X_n + \bar X_n^2) \\ &= \frac{1}{n-1}\sum_{i=1} ^{n} \mathbb{E}(X_i^2 -\frac{2}{n}X_i ^2 -\frac{2}{n}\sum_{j\neq i} X_i X_j + \bar X_n^2)\\ &= \frac{1}{n-1}\left\{ (n-2)\mathbb{E}(X_1 ^2) -\frac{2}{n}\sum_{i=1} ^{n}\sum_{j\neq i}\mathbb{E}(X_i)\mathbb{E}(X_j) + \sigma ^2 + \mathbb{E}(X_1)^2 \right\} \end{aligned} \end{equation} where I used the fact that for $X_i$, $X_j$ independent we have $\mathbb{E}(X_i X_j) = \mathbb{E}(X_i)\mathbb{E}X_j)$ and that $\mathbb{E}(\bar X_n ^2) = \frac{\sigma^2 + \mathbb{E}(X_1)^2}{n}$. Finally, after rearranging the first and last terms:

\begin{equation} \begin{aligned} \mathbb{E}(S_n ^2) &= \frac{1}{n-1}\left\{ (n-1)\mathbb{E}(X_1 ^2) -\frac{2}{n}n(n-1)\mathbb{E}(X_1)^2 \right\}\\ &= \mathbb{E}(X_1 ^2) -2\mathbb{E}(X_1)^2\\ &\neq \mathbb{E}(X_1 ^2) -\mathbb{E}(X_1)^2 \end{aligned} \end{equation}

I'm off by a factor $2$. Can someone help me point out my mistake?

Michael Hardy
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fricadelle
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2 Answers2

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Where you have $\displaystyle\sum_{i=1}^n \sum_{j\ne i}$ should you be summing over all ordered pairs $(i,j)$ with $j\ne i$, or over all unordered pairs? Whether the $2$ belongs there depends on how you answer that. You can do it one way with the $2$ and the other way without the $2$.

Michael Hardy
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  • I added an extra step: $-2X_i\bar X_n = -\frac{2}{n}X_i ^2 -\frac{2}{n}\sum_{j\neq i} X_i X_j$ which is completely unnecessary. But if you go this way and use Landon Carter's comment then you end up with the correct answer. – fricadelle Jan 02 '17 at 11:29
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Probably the easiest way to show this is to use Fisher's Lemma: $\frac{S_{n}^2 (n-1)}{\sigma^2} \sim \chi^2_{n-1}$.

$\mathbb{E}(\chi^2_{n}) = n \Rightarrow \mathbb{E} \left( \frac{S_{n}^2 (n-1)}{\sigma^2} \right) = \frac{n-1}{\sigma^2} \cdot \mathbb{E}S^2_{n} = \mathbb{E}(\chi^2_{n-1})=(n-1) \Rightarrow \mathbb{E}(S^2_{n}) = \frac{n-1}{n-1} \cdot \sigma^2$

Joitandr
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