Chebyshev's Inequality

Definition of Chebyshev’s Inequality

Statement of Chebyshev’s Inequality

Chebyshev's Inequality

Let the expectation of a random variable XX be μ\mu and its variance be σ2\sigma^2. Then for any positive number ε\varepsilon: P(∣X−μ∣≥ε)≤σ2ε2P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}

Proof

Proof for a discrete random variable: P(∣X−μ∣≥ε)=∑∣xi−μ∣≥εP(X=xi)P(|X - \mu| \geq \varepsilon) = \sum_{|x_i - \mu| \geq \varepsilon} P(X = x_i) ≤∑∣xi−μ∣≥ε(xi−μ)2ε2P(X=xi)\leq \sum_{|x_i - \mu| \geq \varepsilon} \frac{(x_i - \mu)^2}{\varepsilon^2} P(X = x_i) ≤1ε2∑i=1∞(xi−μ)2P(X=xi)=σ2ε2\leq \frac{1}{\varepsilon^2} \sum_{i=1}^{\infty} (x_i - \mu)^2 P(X = x_i) = \frac{\sigma^2}{\varepsilon^2}

Proof for a continuous random variable: P(∣X−μ∣≥ε)=∫∣x−μ∣≥εf(x)dxP(|X - \mu| \geq \varepsilon) = \int_{|x - \mu| \geq \varepsilon} f(x) dx ≤∫∣x−μ∣≥ε(x−μ)2ε2f(x)dx\leq \int_{|x - \mu| \geq \varepsilon} \frac{(x - \mu)^2}{\varepsilon^2} f(x) dx ≤1ε2∫−∞+∞(x−μ)2f(x)dx=σ2ε2\leq \frac{1}{\varepsilon^2} \int_{-\infty}^{+\infty} (x - \mu)^2 f(x) dx = \frac{\sigma^2}{\varepsilon^2}

Equivalent Forms

Form 1: P(∣X−μ∣<ε)≥1−σ2ε2P(|X - \mu| < \varepsilon) \geq 1 - \frac{\sigma^2}{\varepsilon^2}

Form 2: P(∣X−μ∣≥kσ)≤1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}, where k>0k > 0

Intuitive Understanding of Chebyshev’s Inequality

Understanding: Chebyshev’s inequality gives an upper bound on the probability that a random variable deviates from its expected value. It tells us that the farther the values of a random variable deviate from the expectation, the smaller the probability of such a deviation.

Key Idea Behind the Proof

Key idea:

  1. In the region ∣X−μ∣≥ε|X - \mu| \geq \varepsilon, we have (X−μ)2≥ε2(X - \mu)^2 \geq \varepsilon^2
  2. Therefore (X−μ)2ε2≥1\frac{(X - \mu)^2}{\varepsilon^2} \geq 1
  3. This inequality is used to bound the probability

Properties of Chebyshev’s Inequality

Generality

Property 1: Chebyshev’s inequality applies to random variables with any distribution, as long as their expectation and variance exist.

Conservativeness

Property 2: the upper bound given by Chebyshev’s inequality is usually conservative, that is, the actual probability may be far smaller than this upper bound.

Distribution-Free Character

Property 3: Chebyshev’s inequality places no special requirement on the distribution of the random variable and depends only on the expectation and the variance.

Applications of Chebyshev’s Inequality

Applications in Probability Estimation

Application 1: estimating the probability that a random variable deviates from its expectation

Example 1: let the expectation of a random variable XX be 00 and its variance be 44. Estimate an upper bound for P(∣X∣≥3)P(|X| \geq 3).

Solution: P(∣X∣≥3)≤49≈0.444P(|X| \geq 3) \leq \frac{4}{9} \approx 0.444

Applications in Statistical Inference

Application 2: proving the law of large numbers

Application 3: estimating the precision of the sample mean

Applications in Quality Control

Application 4: estimating the range of fluctuation of product quality

Application 5: setting quality control standards

Limitations of Chebyshev’s Inequality

Conservativeness

Limitation 1: the upper bound given by Chebyshev’s inequality is usually conservative; for specific distributions there may be more precise estimates.

Incomplete Use of Information

Limitation 2: Chebyshev’s inequality uses only the information of the expectation and the variance, and does not use other features of the distribution.

Restricted Applicability

Limitation 3: for certain special distributions, there may be better inequalities.

Relations to Other Inequalities

Relation to Markov’s Inequality

Markov’s inequality: let XX be a non-negative random variable. Then for any a>0a > 0: P(X≥a)≤E(X)aP(X \geq a) \leq \frac{E(X)}{a}

Relation: Chebyshev’s inequality is a special case of Markov’s inequality, obtained by replacing XX with (X−μ)2(X - \mu)^2.

Relation to Hoeffding’s Inequality

Hoeffding’s inequality: for bounded random variables, Hoeffding’s inequality gives a more precise estimate.

Generalizations of Chebyshev’s Inequality

Multidimensional Chebyshev’s Inequality

Generalization 1: for multidimensional random variables, there is a corresponding Chebyshev’s inequality.

Conditional Chebyshev’s Inequality

Generalization 2: under given conditions, there is a corresponding Chebyshev’s inequality.

Exercises

Exercise 1

Use Chebyshev’s inequality to estimate an upper bound for P(∣X∣≥3)P(|X|\geq3) when E(X)=0,D(X)=4E(X)=0, D(X)=4.

Chebyshev's inequalityprobability estimationinequality for concentration

Approach: Use Chebyshev’s inequality.

Steps:

  1. Chebyshev’s inequality: P(∣X−μ∣≥ε)≤σ2ε2P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}
  2. Here μ=0\mu = 0, σ2=4\sigma^2 = 4, and ε=3\varepsilon = 3
  3. P(∣X∣≥3)≤49≈0.444P(|X| \geq 3) \leq \frac{4}{9} \approx 0.444

Answer: 49\frac{4}{9}

Exercise 2

Let the expectation of a random variable XX be 55 and its variance be 99. Use Chebyshev’s inequality to estimate a lower bound for P(2<X<8)P(2 < X < 8).

Chebyshev's inequalitycomplementary eventprobability estimation

Approach: First find an upper bound for P(∣X−5∣≥3)P(|X - 5| \geq 3), then use the relation of complementary events.

Steps:

  1. P(∣X−5∣≥3)≤99=1P(|X - 5| \geq 3) \leq \frac{9}{9} = 1
  2. P(∣X−5∣<3)≥1−1=0P(|X - 5| < 3) \geq 1 - 1 = 0
  3. Since 2<X<82 < X < 8 is equivalent to ∣X−5∣<3|X - 5| < 3, we get P(2<X<8)≥0P(2 < X < 8) \geq 0

Answer: 00

Exercise 3

Let the expectation of a random variable XX be 1010 and its variance be 1616. Use Chebyshev’s inequality to estimate an upper bound for P(∣X−10∣≥8)P(|X - 10| \geq 8).

Chebyshev's inequalityupper bound on probabilityinequality for concentration

Approach: Apply Chebyshev’s inequality directly.

Steps:

  1. P(∣X−10∣≥8)≤1664=14P(|X - 10| \geq 8) \leq \frac{16}{64} = \frac{1}{4}
  2. Substituting into the general form of Chebyshev’s inequality P(∣X−μ∣≥kσ)≤1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}, with kσ=8k\sigma = 8 and k2=4k^2 = 4, the upper bound is 14\frac{1}{4}.

Answer: P(∣X−10∣≥8)≤14P(|X - 10| \geq 8) \leq \frac{1}{4}, that is, the probability of this event occurring is at most one quarter.

Exercise 4

Prove that for any random variable XX, if E(X)=μE(X) = \mu and D(X)=σ2D(X) = \sigma^2, then for any k>0k > 0: P(∣X−μ∣≥kσ)≤1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}

Chebyshev's inequalityproofinequality for concentration

Approach: Use Chebyshev’s inequality with ε=kσ\varepsilon = k\sigma.

Steps:

  1. Chebyshev’s inequality: P(∣X−μ∣≥ε)≤σ2ε2P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}
  2. Setting ε=kσ\varepsilon = k\sigma gives: P(∣X−μ∣≥kσ)≤σ2(kσ)2=1k2P(|X - \mu| \geq k\sigma) \leq \frac{\sigma^2}{(k\sigma)^2} = \frac{1}{k^2}

Answer: the proof is complete

Exercise 5

Let a random variable XX follow the normal distribution N(μ,σ2)N(\mu, \sigma^2). Compare the upper bound given by Chebyshev’s inequality with the actual probability.

Chebyshev's inequalitynormal distributiontightness of the bound

Approach: Compute the actual probability of the normal distribution and compare it with the upper bound given by Chebyshev’s inequality.

Steps:

  1. For the normal distribution, P(∣X−μ∣≥2σ)=2(1−Φ(2))≈0.0455P(|X - \mu| \geq 2\sigma) = 2(1 - \Phi(2)) \approx 0.0455
  2. Chebyshev’s inequality gives: P(∣X−μ∣≥2σ)≤14=0.25P(|X - \mu| \geq 2\sigma) \leq \frac{1}{4} = 0.25
  3. The actual probability 0.04550.0455 is far smaller than the upper bound 0.250.25, showing that Chebyshev’s inequality is fairly conservative

Answer: the upper bound given by Chebyshev’s inequality is much larger than the actual probability, showing that it is fairly conservative


Summary

Symbols Used in This Article

Greek LetterPronunciationDescription
ε\varepsilonEpsilon (epsilon)denotes an arbitrary positive number; in Chebyshev’s inequality it represents the degree of deviation
μ\muMu (mu)denotes the population mean or the expectation
σ\sigmaSigma (sigma)denotes the population standard deviation
Φ\PhiPhi (phi)denotes the distribution function of the standard normal distribution

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