Regression to the Mean

Everyone has a steady true skill, and every score adds some luck to it. Choose how much of the spread is skill and how much is luck, select a group on one score, and see how it does on the next. The page works out the exact expected change and simulates a population to show it. The table below the chart does the arithmetic.

Worked example — Regression to the Mean

Reliability	0.50
Spread on each measurement	14.14
Cut-off in standard units	1.2816
Cut-off score	118.1
Density at the cut-off	0.1755
Inverse Mills ratio	1.7550
Group mean where selected x̄1	124.8
Expected group mean on the other x̄2	112.4

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Regression to the Mean

When a measurement is true skill plus independent luck, two measurements of the same people correlate at r = σ_S² ÷ (σ_S² + σ_L²), the share of the variance that is skill. A group selected for extreme scores on one measurement is expected to score μ + r(x̄ − μ) on the other: its distance from the mean multiplied by r. This is regression to the mean. It needs no cause, runs in both directions, and does not need the population's spread to shrink: here it stays the same.

What each symbol means

r: the correlation between two measurements, the share of the variance that is skill. 1 means no luck, 0 means all luck
σ_S: the spread of true skill in the population
σ_L: the spread of luck on a single measurement
μ: the population mean, the point the group regresses towards
λ: the inverse Mills ratio: the selected group's mean sits λ standard deviations from μ, so it is μ + σλ for a top group

The best do worse next time, and nobody got worse 🖖

Press Top 10%, r = 0.5. The people who scored highest the first time average 124.8, and the second time they are expected to average 112.4. Nobody lost any skill. The top 10% were, on average, both better and luckier than everyone else, and on the second sitting the skill stays while the luck is drawn afresh. The expected second score is μ + r(x̄ − μ): with r = 0.5, half the distance from the mean survives. Press Perfect reliability and there is no luck to lose, so nothing moves; press Pure luck and there is no skill to keep, so the group falls all the way back to 100.

A treatment that does nothing works on the worst 🖖

Press Worst 10% treated. The lowest scorers get a treatment whose effect is set to exactly zero, and at the next sitting they have improved by 12.4 points. Any tutor, remedy or new manager applied to whoever did worst last time will look better than it is, by about this much, unless there is a comparison group. Daniel Kahneman tells of flight instructors convinced that shouting at a cadet after a bad landing worked and praise after a good one backfired: both were the scores returning towards the mean. Press With a control group. Half the worst are treated and half are not, both gain about 12 points, and the only honest measure of the treatment, the difference between the two, is expected to be zero.

It runs backwards in time, and nobody is converging 🖖

Press Parent and child heights: parents in the tallest 10% average 187.4 cm and their children are expected to average 181.2. Francis Galton published this in 1886, from the heights of parents and their grown children, and called it regression towards mediocrity. Now press Heights, backwards: select the tallest children instead, and their parents are expected to be shorter than them, in this model, where both generations are equally spread, by the same amount. So it is not a force pulling people towards the average over time. And the spread does not shrink: the table gives almost the same spread on both measurements, 7.09 and 7.10 cm. Extreme people are replaced by other extreme people each time, because different people are lucky.

Problem solved in full

How far the top tenth falls 5 steps
Scores on a test have mean 100. The spread of true skill is 10 and the spread of luck on one sitting is also 10. The top 10% on the first sitting are tested again. Find their average on the first sitting and their expected average on the second. Press Top 10%, r = 0.5 to check.

Open this problem in the tool
1. The correlation between sittings is the share of the variance that is skill, and the total spread follows:
  
  $r = \frac{σ _{S}^{2}}{σ _{S}^{2} + σ _{L}^{2}} = \frac{100}{200} = 0.5, σ = 200 = 14.14$
2. The top 10% start at the 90th percentile, 1.2816 standard deviations above the mean:
  
  $z_{c} = Φ^{- 1} (0.90) = 1.2816, x_{c} = 100 + 14.14 \cdot 1.2816 = 118.1$
3. The mean of everyone above that cut-off, from the normal curve's height and tail there:
  
  $λ = \frac{φ ( z _{c} )}{1 - Φ ( z _{c} )} = \frac{0 . 1755}{0 . 10} = 1.7550, \overset{x}{ˉ}_{1} = 100 + 14.14 \cdot 1.7550 = 124.8$
4. On the second sitting the expected mean keeps only the fraction r of the distance from μ:
  
  $\overset{x}{ˉ}_{2} = μ + r (\overset{x}{ˉ}_{1} - μ) = 100 + 0.5 \cdot 24.8 = 112.4$
5. So the group is expected to fall by:
  
  $\overset{x}{ˉ}_{2} - \overset{x}{ˉ}_{1} = 112.4 - 124.8 = - 12.4$
Answer

The top tenth average 124.8 the first time and are expected to average 112.4 the second, a fall of 12.4 points with no change in anyone's skill. Their mean true skill is expected to be 112.4 all along: the second sitting is not biased by the luck that selected them, because that luck is not selected again.

$\overset{x}{ˉ}_{1} = 124.8, \overset{x}{ˉ}_{2} = 112.4$

By the 42Math team · Reviewed against published sources. Corrections welcome.