What the Sample Mean Minimizes
A standalone mini-lesson in a small, growing set on estimator properties. See also an unbiased estimator that never gets more precise; this page builds the geometric fact used to explain why the naive sample variance undershoots.
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- Think · 2 minYou roll a die five times and get {2, 6, 4, 3, 5}. Pick any single number c you like, and add up the five squared distances from c to each roll. By hand, try c = 3.5 (the true θ) and c = 4 (the sample average). Which gives a smaller total?
I've done the Think step — reveal Pair & Share
- Pair · 3 minCompare your two totals. Try a third value of c neither of you picked — does it ever beat 4?
- Share · 2 minAgree as a table on which single value of c minimizes the total, then check it against the widget below.
What the sample mean minimizes
Take the pinned sample {2, 6, 4, 3, 5} — five valid die rolls, sample mean \(\bar X = 4\), true die mean \(\theta = 3.5\). For any candidate center c, define the sum of squared deviations:
$$SS(c) = \sum_{i=1}^{5} (x_i - c)^2$$
This is a parabola in c, and it has exactly one minimum. Drag c across the slider below and watch where the marker bottoms out.
Confirm \(SS(3.5) = 11.25\) and \(SS(4) = 10\) directly: \((2-3.5)^2+(6-3.5)^2+(4-3.5)^2+(3-3.5)^2+(5-3.5)^2\) versus the same sum centered at 4.
Show the arithmetic
At c = 3.5: \(2.25+6.25+0.25+0.25+2.25 = 11.25\). At c = 4: \(4+4+0+1+1 = 10\). The gap, 1.25, is exactly \(n(\bar X - \theta)^2 = 5\times(0.5)^2 = 1.25\) — no coincidence, that identity holds for every sample, and it's exactly what the sample-variance mini-lesson uses to explain the bias.