I am trying to prove that the sample variance formula is unbiased.

Firts, let $\mathbb{E}(X_i) = \mu, \text{Var}(X_i)=\sigma^2=\mathbb{E}(X_i^2)-\mathbb{E}(X_i)^2, \bar{X}=\frac{\sum X_i}{n}$.

$\mathbb{E}( \frac{\sum(X_i-\bar{X})^2}{n-1}) =\sigma^2 $ can be transformed as follows.

$\mathbb{E}( \sum(X_i-\bar{X})^2/(n-1))= \frac{\sum\mathbb{E}(X_i-\bar{X})^2}{n-1}\cdot\dots\cdot (1)$

$ \sum\mathbb{E}(X_i-\bar{X})^2= \sum\mathbb{E}(X_i- \mu +\mu-\bar{X})^2 \cdot\dots\cdot (2)$

$ \mathbb{E}(X_i- \mu +\mu-\bar{X})^2 = \mathbb{E}( \underbrace{(X_i- \mu)^2}_{a} +\underbrace{(\mu-\bar{X})^2}_{b} +2\underbrace{(X_i- \mu)(\mu-\bar{X})}_{c}) $

$a: \mathbb{E}((X_i- \mu)^2 = \sigma^2 $

$b: \mathbb{E}((\mu-\bar{X})^2 = \frac{\sigma^2}{n} $

$c$ is arranged as follows.

$ \mathbb{E}(X_i- \mu)(\mu-\bar{X}) = \mathbb{E}(X_i\mu-X_i\bar{X} -\mu^2+\mu\bar{X})$

$ \mathbb{E}(X_i\mu-X_i\bar{X} -\mu^2+\mu\bar{X})=\mathbb{E}(-X_i\bar{X} +\mu\bar{X})$

$ \because \mathbb{E}(X_i\mu) = \mu^2$

$ \mathbb{E}(-X_i\bar{X}) =- \mathbb{E}(X_i\sum{X_i}/n)=-\frac{\sum\mathbb{E}(X_i^2)}{n} = -\mathbb{E}(X_i^2)$

$ \mathbb{E}(\mu\bar{X})= \mu\mathbb{E}(\sum{X_i}/n)= \mu\frac{\sum{\mathbb{E}(X_i)}}{n}=\mathbb{E}(X_i)^2$


$ \mathbb{E}(X_i\mu-X_i\bar{X} -\mu^2+\mu\bar{X})= -\mathbb{E}(X_i^2)+\mathbb{E}(X_i)^2=-\sigma^2$

Then, $(2)$ becomes $ \sigma^2+ \frac{\sigma^2}{n} -2\sigma^2=-\sigma^2+ \frac{\sigma^2}{n}=-\frac{n-1}{n}\sigma^2$, and plugging this into (1) leads to $-\sigma^2$.

Which stage is incorrect?

  • $\begingroup$ When the $X_i$ are independent, $$E[X_i\bar X]=\frac{1}{n}E[\sum_{j=1}^n X_iX_j]=\frac{1}{n}E[X_i^2]+\frac{1}{n}\sum_{j\ne i}E[X_iX_j]=\frac{1}{n}(\sigma^2+\mu^2)+\frac{n-1}{n}\mu^2=\frac{\sigma^2}{n}+\mu^2.$$ $\endgroup$
    – whuber
    Oct 10, 2022 at 17:20

2 Answers 2


While stats_model pointed out the possible error, problems like this should involve the following approach for smooth calculations:

\begin{align}\frac{\sum(X_i-\bar{X})^2}{n-1}&= \frac{1}{n-1}\left[\sum\left \{(X_i-\mu)-\left(\bar X -\mu\right)\right\}^2\right]\\ &= \frac{1}{n-1}\left[\sum(X_i-\mu)^2+n\left(\bar X -\mu\right)^2-2\left(\bar X -\mu\right)\underbrace{\sum(X_i-\mu)}_{n\left(\bar X-\mu\right)}\right]\\ &= \frac{1}{n-1}\left[\sum(X_i-\mu)^2-n\left(\bar X -\mu\right)^2\right].\tag 1\label 1\end{align}

Taking expectation of $\eqref 1, $ leads to

$$ \frac{1}{n-1} \left[\sum \operatorname{Var}(X_i) -n\operatorname{Var}\left(\bar X\right) \right].\tag 2$$

The rest follows.


Something that you have not explicitly stated in your question, but which I assume that you are implicitly assuming, is that your $X_i$'s are independent of one another (or at least uncorrelated). In that case, You have made an error in your calculation of $\mathbb E[-X_i\bar X]$. I believe the error is due to a bad notational choice when expanding out $\bar X$. Specifically, the correct way to make this calculation is: $$\mathbb E[-X_i\bar X] = \mathbb E\left[-X_i \frac{\sum_{j=1}^n X_j}{n}\right] = -\frac1n\mathbb E[X_i^2] - \mathbb E\left[\frac{\sum_{j\neq i} X_iX_j}{n}\right] = - \frac1n \mathbb E[X_i^2] - \frac{(n-1)\mu^2}{n}$$ where in the last line, I use the fact that $X_i$ and $X_j$ are uncorrelated, so $\mathbb E[X_i X_j] = \mathbb E[X_i]\mathbb E[X_j] = \mu^2$. In that case, we have $c = \mu^2 - \frac{1}{n}\mathbb E[X_i^2] - \frac{(n-1)\mu^2}{n} = -\frac1n\left[\mathbb E[X_i^2] - \mu^2\right] = -\frac1n \sigma^2$. Thus, (2) becomes $\sigma^2 + \frac{\sigma^2}{n} - 2 \frac{\sigma^2}{n} = \frac{(n-1)\sigma^2}{n}$ as desired.

  • $\begingroup$ Thank you so much. $\endgroup$
    – kurtkim
    Oct 11, 2022 at 12:10

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