## Sample Variance Intuition

This blog post will focus on the intuition behind the $n-1$ divisor of the standard variance. No proofs, but lots of context and motivation.

And you’ll be pleased to know that this post is self contained. You don’t need to read either of my previous two posts on sample variance.

For a population $X_1, X_2, \ldots , X_N$ the population variance is defined by

$\sigma^2 = \frac{1}{N} \sum\limits_{i=1}^{N} (X_i - \mu)^2$ where $\mu = \frac{1}{N} \sum\limits_{i=1}^{N} X_i$

and for a sample $x_1, x_2, \ldots , x_n$ the sample variance is defined by

$s^2 = \frac{1}{n-1} \sum\limits_{i=1}^{n} (x_i - \overline{x})^2$ where $\overline{x} = \frac{1}{n} \sum\limits_{i=1}^{n} x_i$

And the question is “why doesn’t the sample variance have the same formula as the population variance?” i.e why isn’t the sample variance given by the direct analogue of the population variance, which is known as the “uncorrected sample variance”:

${s_n}^2 = \frac{1}{n} \sum\limits_{i=1}^{n} (x_i - \overline{x})^2$

For the purpose of building intuition, there’s two facts that I think are extremely important to keep in mind.

## Rational and Irrational Values of Powers

Here’s a cute little existence argument that I was exposed to as an undergraduate and have never forgotten. It shows that there must be irrational positive reals $\alpha$ and $\beta$ for which $\alpha^\beta$ is rational. Furthermore, it’s done by showing that one of two very specific pairs of numbers (either $\alpha=\sqrt{2}, \beta=\sqrt{2}$ or $\alpha={\sqrt{2}}^{\sqrt{2}}, \beta=\sqrt{2}$) satisfies the condition, but without establishing which of the two pairs satisfies the condition.

The argument is elementary. First look at the case $\alpha=\sqrt{2}, \beta=\sqrt{2}$ If $\alpha^\beta = {\sqrt{2}}^{\sqrt{2}}$ is rational, we have an example, and we’re done. If not then ${\sqrt{2}}^{\sqrt{2}}$ must be irrational, and so $\alpha^\beta$ where $\alpha = {\sqrt{2}}^{\sqrt{2}}$ (assumed to be irrational) and $\beta=\sqrt{2}$ is an example of an irrational raised to an irrational. And it’s rational. In fact, it’s $2$, because $\alpha^\beta =({{\sqrt{2}}^{\sqrt{2}}})^{\sqrt{2}} = {{\sqrt{2}}^{{\sqrt{2}}{\sqrt{2}}}} = {\sqrt{2}}^2 = 2$

It’s a great little elementary existence argument, and I think it’s worth being exposed to it. But, it’s also worth knowing that with (much) more advanced techniques we can actually say which of the two choices satisfies the condition Continue reading “Rational and Irrational Values of Powers”

The construction is simple: consider an infinite sequence of sentences $S_1, S_2, S_3,\ldots$ where $S_i$ is
$S_j$ is false for all $j \textgreater i$