# Denominator of variance estimator

I'm reading a peer-reviewed article from a highly respected member of my field. In it the author defines equations for the mean and variance of the mean.

I'm trying to understand why the denominator for variance has both n and n-1 in it. I thought the basic equation for variance from a simple random sample was simply

$\frac{1}{n}$ $\sum^{n}_{i=1} (x_{i} - \bar x ) ^ 2$

and it seems like this author has added an additional component to the denominator- which changes the value of the variance substantially. Can someone enlighten me?

The estimator you have written is not unbiased! The unbiased estimator of the population variance that is most oftenly used is

$$S^2 =\frac{1}{n-1} \sum_{i=1}^n \left(x_i - \bar{x} \right)^2$$

Your article then refers to the variance of the sample mean and provides an estimator for it. To make sense of that estimator recall that if we denote by $\sigma^2$ the population variance of a single observation, then the (population) variance of the sample mean is

$$Var\left(\frac{1}{n} \sum_{i=1}^n x_i \right) = \frac{\sigma^2}{n}$$

but since the population variance, $\sigma^2$, is unknown we replace it by its unbiased estimator, namely the $S^2$. Putting that in the place of $\sigma^2$, you will obtain equation (5).

Here is a useful - and exciting - read on the matter.

Howard Wainer, Picturing the Uncertain World, The Most Dangerous Equation.

• so the article provides a formula for the variance of the sample mean rather than the variance of a sample? When is the variance of the sample mean of interest?
– Nan
Jan 6, 2016 at 21:06
• @Nan I cannot think of a situation where it is not of interest! The sample mean alone is half the story because it does not tell you anything about the variation around it. If the variation is large, the sample mean cannot really be trusted as roughly speaking there is a large margin of error. There is much more that can be said about this fantastic equation, so in my answer I have included a very interesting paper if you want to take a look. Jan 6, 2016 at 21:18
• Perhaps I misunderstood your answer? doesn't $s^2$ provide us with a measure of the variation around the sample mean? When do we calculate Var($\bar x$) rather than var(x) or $s^2$?
– Nan
Jan 6, 2016 at 21:34
• @Nan We need the variance of the sample mean when we use it as an estimator of location for the data. It tells us how reliable the sample mean really is. You don't have to use it but it's good to understand it. Jan 6, 2016 at 21:41