Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume there are $n$ independent random variables $X_1,X_2,\ldots,X_n$ and i wonder why the sample variance is $S^2=\frac{\sum\limits_{i=1}^n \ (X_i-X)^2}{n-1}$ where $X$=$\frac{X_1+X_2+\cdots+X_n}{n}$ instead of $S^2=\frac{\sum\limits_{i=1}^{n}\ (X_i-X)^2}{n}$.

share|cite|improve this question
i think there should be some derivation right? – johnny Jan 2 '12 at 7:57
Did you try to compute the expectation of $S^2$ in both cases? (And please add some squares in your formulas.) – Did Jan 2 '12 at 8:16
up vote 1 down vote accepted

You have said that $X_1, X_2, \ldots, X_n$ are independent, but not added, as most people do, that they are also identically distributed (or at the very least, have the same mean and variance). With this added condition, we have $$E\left[\sum_{i=1}^n (X_i-X)^2\right] = (n-1)\sigma^2$$ where $\sigma^2$ is the common variance of the $n$ random variables. Thus, defining the sample variance as $$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i-X)^2$$ has the nice property that $E[S^2] = \sigma^2$. As Didier Piau has already suggested to you in a comment, you should work out the expectation calculation for yourself.

You might also want to try and work out what happens when the $X_i$ have the same mean $\mu$ but different variances $\sigma_i^2$, and also when they have different means $\mu_i$ and variances $\sigma_i^2$.

share|cite|improve this answer
$$E\left[\sum_{i=1}^n (X_i-X)^2\right] = (n)\sigma^2$$ there are n terms so $$E\left[\sum_{i=1}^n (X_i-X)^2\right]=\sum_{i=1}^{n} E\left[(X_i-X)\right] = (n)\sigma^2$$(since all of them have the same varience) but i cant figure out why the answer is (n-1) $\sigma^2$ – johnny Jan 3 '12 at 3:26
No, since $E[X_i]=\mu=E[X]$, $X_i-X$ is a zero-mean random variable, and so $\text{var}(X_i-X)=E[(X_i-X)^2]$ can be computed as $$\begin{align*}\text{var}(X_i-X)&=\text{var}(X_i)+\text{var}(X)-2\text{cov}(X_i‌​,X)\\&=\sigma^2+\frac{\sigma^2}{n}-2\text{cov}(X_i,X)\\&=\frac{n-1}{n}\sigma^2.\end‌​{align*}$$ I will leave it to you to work out why $\text{cov}(X_i,X)=\frac{\sigma^2}{n}$. Putting it all together, $$E\left[\sum_{i=1}^n (X_i-X)^2\right]=\sum_{i=1}^{n} E\left[(X_i-X)^2\right] = (n-1)\sigma^2.$$ Note that your assertion is missing a square in the middle sum. – Dilip Sarwate Jan 3 '12 at 12:49

There is a Wikipedia article about this: Bessel's correction. (I wrote some of it myself, but in this case others did most of it.)

Bessels correction eliminates bias. Eliminating bias is sometimes a very bad idea, as I explained in this paper: "An Illuminating Counterexample, American Mathematical Monthly, volume 110, number 3 (March, 2003), pages 234-238.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.