Chebyshev’s theorem
Chebyshev’s inequality gives a guarantee that holds for any distribution with a finite mean and standard deviation, whatever its shape: skewed, bimodal or heavy-tailed. The share of values within k standard deviations of the mean is at least:
1 − 1 ÷ k², for any k > 1.
| k | At least within μ ± kσ | Normal distribution |
|---|---|---|
| 1.5 | 55.6% | 86.6% |
| 2 | 75% | 95.4% |
| 3 | 88.9% | 99.7% |
| 4 | 93.75% | 99.99% |
Example
A data set has a mean of 100 and a standard deviation of 15, with an unknown shape:
- with k = 2, at least 75% of values lie between 70 and 130, and at most 25% lie outside;
- the one-sided Cantelli bound says that at most 20% lie above 130 (and at most 20% below 70);
- if the data were normal, 95.45% would lie between 70 and 130;
- to guarantee at least 90% of values, you need k = √10 ≈ 3.16, so the interval 52.6 to 147.4.
Why the bound is so loose
Chebyshev’s bound must hold for every possible distribution, including the most extreme. It is much weaker than what applies to a normal or any bell-shaped distribution. The bound is reached exactly only by a distribution that puts all its mass at three points. For unimodal distributions, the tighter Vysochanskij–Petunin inequality gives at least 1 − 4 ÷ (9k²) for k > 1.63, which is 88.9% within 2 standard deviations.
Uses
- Robust guarantees: risk limits or tolerances when the distribution is unknown.
- Theory: Chebyshev’s inequality proves the weak law of large numbers. The mean of n observations has a standard deviation of σ ÷ √n, so it concentrates around μ as n grows.
- Quick checks: no more than 1/9 of any data set can lie more than 3 standard deviations from the mean. If you find more, check for errors in the mean or SD.
Markov’s inequality
Chebyshev’s inequality follows from Markov’s inequality: for a non-negative variable, P(X ≥ a) ≤ E[X] ÷ a. With an average wait of 10 minutes, at most 20% of waits exceed 50 minutes, whatever the distribution.