Why n − 1 in Sample Variance?

Suppose our population consists of one million people who voted for Trump in the 2024 election. We want to know: roughly how old are they on average? And is their age distribution fairly concentrated, or is there a wide mix of younger and older voters?

In statistical terms, we are interested in the population mean \(\mu\) and variance \(\sigma^2\). But we can’t possibly ask all one million people, so here’s what we do:

We use the numbers calculated from our sample—say, \(\bar x=45\) and \(s^2=10\)—to estimate the mean \(\mu\) and variance \(\sigma^2\) of this population of one million people. Now here comes the question:

Why is the denominator in \(S^2\) equal to \(5-1\)? Or more generally, why define

\[ S^2=\frac{1}{\color{red}{\boldsymbol{n-1}}}\sum_{i=1}^{n}(X_i-\bar X)^2? \]

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To make the picture easier to draw, let’s represent the population with 15 points (we’re not drawing a million dots!). We sample five people, as shown below:

The population’s average age \(\mu\) is 39, and the ages are quite spread out. Now that we have sampled \((X_1,\ldots,X_5)\), we can naturally look at how far each person in our sample is from the true population mean:

\[ \begin{pmatrix} X_1-\mu\\ X_2-\mu\\ \vdots\\ X_5-\mu \end{pmatrix} \xrightarrow{\text{square, then average}} \frac{(X_1-\mu)^2+\cdots+(X_5-\mu)^2}{5}, \]

and use this to estimate the population variance \(\sigma^2\). This makes perfect sense: if we knew \(\mu\), dividing by \(n\) (five here) would be absolutely fine.

But here’s the trouble: we don’t know \(\mu\)! It’s the mean of all one million people, after all. So a natural thought is: we’re already using \(\bar X\) to estimate \(\mu\), so why not replace \(\mu\) with \(\bar X\) here too? That gives us our first step:

\[ \frac{1}{n}\sum_{i=1}^{n}(X_i-\mu)^2 \;\xrightarrow[\text{replace }\mu\text{ with }\bar X]{\text{Revision}}\; \underset{\star}{\underline{\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2}}. \]

A small problem has quietly slipped in. We wanted to measure “how far the sample is from the population’s average age \(\mu\).” But \(\bar X\) isn’t the population’s centre: it is chosen after seeing our sample \(X_1,\ldots,X_5\). In our example, \(\bar X=45\). Look back at the picture: it’s pretty close to the five sampled people, isn’t it? It makes us think, “Hmm, these people—the solid blue dots ●—don’t seem all that spread out.” Yet in this example, the population, including the hollow grey dots ○, is much more spread out than our sample.

So think of it this way: the true mean \(\mu\) stays put, while the sample mean \(\bar X\) “chases after our sample.” If we happen to sample a younger group today, \(\bar X\) moves towards the younger ages. If we happen to sample an older group tomorrow, \(\bar X\) moves towards the older ages instead. So using \(\underline{\star}\) to estimate the population’s spread tends to underestimate it, on average. This brings us to the next step:

\[ \underset{\star}{\underline{\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2}} \;\xrightarrow[\text{divide by }n-1\text{ instead}]{\text{Revision}}\; \frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar X)^2. \]

For our five-person sample, we go from dividing by five to dividing by four. This makes the estimate a little bigger, to offset the underestimation caused by using \(\bar X\)—and that answers our opening question.

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A small remark. How much does \(\underline{\star}\) underestimate the variance, exactly? We can describe this using the mathematical language of ST2131: expectation. With independent, identically distributed observations of finite variance, a little calculation gives:

\[ \begin{aligned} E\!\left[\underline{\star}\right] &=E\!\left[\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2\right]\\ &=\frac{1}{n}\sum_{i=1}^{n}E\!\left[(X_i-\bar X)^2\right]\\ &=\frac{1}{n}\sum_{i=1}^{n}E\!\left[\big[(X_i-\mu)+(\mu-\bar X)\big]^2\right]\\ &=\cdots=\frac{n-1}{n}\times\sigma^2. \end{aligned} \]

Don’t worry—try expanding the square, treating each expression in parentheses as one piece!

This means that if we repeat the “sample just five people” experiment many times, \(\underline{\star}\) averages out to \(4/5=80\%\) of the true variance. If we repeat the “sample 100 people” experiment, it averages out to \(99/100=99\%\). So as the sample size gets larger, dividing by \(n\) or \(n-1\) makes less and less difference.

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Let’s try sampling!

This time, let’s sample from \(N(0,10^2)\), so the true variance is 100. Choose a sample size, draw ten samples, and see where the two estimates end up on average.