Understanding Uncertainty · Mini-Lesson · pairs with Class 04

Unbiased, but Never Consistent

Two die-roll estimators, same center, very different spread
Understanding Uncertainty · in-class group work

A fair six-sided die is the running example: \(\theta = \mathbb{E}[X] = 3.5\). This is a standalone mini-lesson in a small, growing set on estimator properties — see also what the sample mean minimizes and why the naive sample variance undershoots.

A companion exploration built to accompany the sample-mean material in Class 04 — no external dataset, the 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 · a deliberately bad estimator
I've done the Think step — reveal Pair & Share
  • Pair · 3 minCompare your reasoning about whether "unbiased" and "good" have to mean the same thing.
  • Share · 2 minAs a table, agree on one sentence describing what this estimator is missing that \(\bar X_n\) has — then check it against the widget below.

Unbiased, but never consistent

Compare two estimators of \(\theta\), both built from n die rolls. The sample mean \(\bar X_n\) uses all n rolls. The other, \(T_n\), uses exactly one: whatever the first roll happened to be, no matter how large n grows afterward. Both have expectation \(\theta\) — rolling one die and averaging n of them are both centered at 3.5. What differs is what happens as n grows.

Two distributions are stacked in the widget below — keep track of which is which. The die itself has one distribution: six faces, each with probability 1/6. But no dot in the plot below is a single die roll. Every dot is an entire sample's worth of rolls, collapsed into one number — an average of n rolls for \(\bar X_n\), or just the first of them for \(T_n\). Plotting 250 of those collapsed numbers side by side draws a second, different distribution: the sampling distribution of the estimator, one level removed from the die. What tightens (or doesn't) as n grows is the spread of that distribution — not the die's, which never changes at all.
Drag n from 1 to 200 and watch each estimator's cloud of 250 replicate values. Both stay centered on the true θ; only one of them tightens.
Same center, different spread. At every n, both \(\bar X_n\) and \(T_n\) average out to \(\theta = 3.5\) across replications — that is exactly what unbiased means, and it is true for both. But \(\bar X_n\)'s spread shrinks like \(1/\sqrt n\) while \(T_n\)'s spread never moves — it is always just the spread of one die roll, forever. \(\bar X_n\) is consistent; \(T_n\) is not. Unbiasedness is a finite-sample fact, checked at each n on its own; consistency is a claim about the limit, and nothing about being unbiased at every n gets you there for free.
Check your intuition

At n = 200, \(T_n\)'s replicate std is still about 1.71 — the same as at n = 1. Why doesn't collecting 199 more rolls help \(T_n\) at all?

Show answer

\(T_n\) is defined to depend only on the first roll; the formula literally never looks at rolls 2 through n. Collecting more data only helps an estimator if the estimator's formula actually uses that data — \(\bar X_n\) divides by n and folds every roll into the average, so more rolls mechanically shrinks its variance (\(\sigma^2/n\)). \(T_n\) has no n in its formula at all.

FAQ

Why does Tn even count as an estimator if it throws away almost all the data?

Nothing in the definition of "estimator" requires using all your data well — an estimator is any rule that turns a sample into a number meant to approximate a parameter. \(T_n\) is a legitimate, if deliberately bad, estimator. It exists here purely to separate two properties that are easy to conflate: being centered correctly on average, and getting more precise with more data.

More estimator-properties mini-lessons: What the sample mean minimizes Why the sample variance undershoots How bad is my estimator? One sample, one point One sample, one point (skewed) One sample, one point (variance)