Understanding Uncertainty · Mini-Lesson · pairs with Class 04 & Class 10

Why the Sample Variance Undershoots

A geometric fact about the mean, turned into Bessel's correction

Understanding Uncertainty · in-class group work

A standalone mini-lesson in a small, growing set on estimator properties. A companion lesson shows that centering on the sample mean always beats centering on the true θ for the sum of squared deviations; this page turns that geometric fact into a statement about bias, and the fix for it.

A companion exploration built to accompany the sampling-distribution material in Class 10 — no external dataset, every widget generates its own die rolls in the browser.

New vocabulary is underlined like this — hover or tap any term for a one-line explanation that appears right below the line you are reading.

Think · Pair · Share · from "always smaller" to "biased"

Think · 2 minIf centering on \(\bar X_n\) always gives a smaller sum of squares than centering on the true \(\theta\), what does that predict about the average behavior of \(\frac1n\sum(x_i-\bar x)^2\), used again and again across many different samples, as an estimator of \(\sigma^2\)?

I've done the Think step — reveal Pair & Share

Pair · 3 minCompare predictions — should it hit \(\sigma^2\) exactly, land above it, or land below it, on average across many samples?
Share · 2 minAgree as a table, then check the direction (not yet the exact size) against the widget below.

For any sample, splitting the squared deviation from the true \(\theta\) into a within-sample part and a centering part always holds:

$$\sum_i (x_i-\theta)^2 = \sum_i(x_i-\bar x)^2 + n(\bar x - \theta)^2$$

The second term on the right is a square, so it is never negative — which means the within-sample sum of squares, centered on \(\bar x\), can never exceed the sum centered on the true \(\theta\). Take expectations of both sides (using \(\mathbb{V}[\bar X_n] = \sigma^2/n\) from Class 04) and the naive sample variance comes out biased low by a factor of \((n-1)/n\):

$$\mathbb{E}\left[\frac1n\sum_i(x_i-\bar x)^2\right] = \frac{n-1}{n}\,\sigma^2$$

Bessel's correction — dividing by \((n-1)\) instead of n — is the fix. Watch it happen across 500 freshly drawn samples at a time, rather than just one:

Each histogram bar is one sample's variance estimate. Toggle the correction and watch the whole histogram shift to recenter on the true σ².

The bias shrinks with n, but never on its own without the correction. The undershoot factor \((n-1)/n\) gets closer to 1 as n grows — at n = 30 it's a 3% shrink, barely visible in the histogram — but it is never exactly 1 at any finite n. Bessel's correction removes it exactly, at every n, not just approximately at large n.

FAQ

Does Bessel's correction fix bias for every parameter, not just variance?

No — the \((n-1)\) correction is specific to the sample variance, and it comes from exactly one source: using the estimated center \(\bar x\) instead of the true \(\theta\) when measuring spread. Other biased estimators (a sample maximum, for instance) need entirely different corrections, if a simple one exists at all.

More estimator-properties mini-lessons: Unbiased, but never consistent What the sample mean minimizes How bad is my estimator? One sample, one point One sample, one point (skewed) One sample, one point (variance)