Chebyshev's Inequality — Mathematics Knowledge System

Overview

Chebyshev's inequality provides a universal bound on how far a random variable can stray from its mean. Given only the mean μ and variance σ² of a random variable, it guarantees that the probability of the variable being more than k standard deviations from the mean is at most 1/k². Unlike many probabilistic bounds, it makes no assumption about the shape of the distribution — it applies to any distribution with finite mean and variance. It is a direct consequence of Markov's inequality applied to the squared deviation.

Intuition

Chebyshev's inequality says that no matter what distribution you have (as long as mean and variance exist), at least 75% of the probability mass lies within 2 standard deviations of the mean, and at least 89% lies within 3 standard deviations. These bounds are much weaker than what holds for the normal distribution (95% and 99.7%), but they apply universally. The key idea: if the variance is small, the distribution cannot be spread out far from its mean.

Formal Definition

Definition

Let X be a random variable with mean μ = E[X] and finite variance σ² = Var(X). For any k > 0:

P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}

Chebyshev's Inequality (standard form)

P(|X - \mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}

Chebyshev's Inequality (epsilon form)

P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \quad \Leftarrow \quad P\!\left((X-\mu)^2 \geq k^2\sigma^2\right) \leq \frac{E[(X-\mu)^2]}{k^2\sigma^2} = \frac{\sigma^2}{k^2\sigma^2}

Derivation via Markov's inequality

Worked Examples

1
Write 15 in terms of standard deviations: 15 = k × 5, so k = 3.
$k = \frac{15}{5} = 3$
2
Apply Chebyshev's inequality.
$P(|X - 50| \geq 15) \leq \frac{1}{k^2} = \frac{1}{9} \approx 0.111$

✓ Answer

P(|X − 50| ≥ 15) ≤ 1/9 ≈ 11.1%.

Practice Problems

Easyfree response

A random variable X has mean 20 and variance 9. Give an upper bound for P(|X − 20| ≥ 6).

Mediumfree response

Using Chebyshev's inequality, what is a lower bound on P(|X − μ| < 4σ)?

Quiz

Summary

Chebyshev's inequality bounds the probability that a random variable deviates from its mean by more than k standard deviations: P(|X − μ| ≥ kσ) ≤ 1/k².
It requires only finite mean and variance — it is distribution-free.
At least 75% of probability mass lies within 2σ, and at least 89% within 3σ, for any distribution.
It is proved by applying Markov's inequality to the squared deviation (X − μ)².

References

WebsiteWikipedia — Chebyshev's inequality