Chebyshev’s inequality

Topic of Stochastic processes, part of Stochastic processes · Also in Functions of random variables and limit theorems, Estimation

Chebyshev’s inequality

Type Topic
Taught in Stochastic processes · Functions of random variables and limit theorems · Estimation
Part Stochastic processes · Probability · Statistical inference
Questions using it 2

For a random variable with finite variance and \varepsilon>0, \mathbb{P}(|X-\mathbb{E}X|\ge\varepsilon)\le\operatorname{Var}(X)/\varepsilon^2. It is Markov’s inequality applied to (X-\mathbb{E}X)^2, and it proves the weak law of large numbers and the consistency of an estimator whose mean squared error tends to zero.

In Stochastic processes

If X has a finite variance, apply the same bound to (X-\mathbb{E} X)^2 at level \varepsilon^2. That is Chebyshev’s inequality:

\mathbb{P}\bigl(|X-\mathbb{E} X|\ge\varepsilon\bigr)\le\frac{\operatorname{Var}(X)}{\varepsilon^2}.

In Functions of random variables and limit theorems

The weak law under a finite variance follows from Chebyshev’s inequality: \mathbb{P}(|S_n/n-\mu|>\varepsilon)\le \operatorname{Var}(S_n/n)/\varepsilon^2=\operatorname{Var}(X_1)/(n\varepsilon^2), which tends to zero.

In Estimation

From Estimation

For a large sample the leading question is whether the estimator settles on \theta. If the mean squared error tends to zero, the estimator is consistent, by Chebyshev’s inequality: \mathbb{P}(|T_n-\theta|>\varepsilon)\le\operatorname{MSE}(T_n)/\varepsilon^2. The law of large numbers says that the sample mean is consistent for \mu, and the same limit passes to continuous functions of a consistent estimator, which is why moment estimators are usually consistent. Both estimators in each example above are consistent. Maximum likelihood estimators are consistent in the regular models used here.

Used in

See also

Markov’s inequality · Jensen’s inequality · Cauchy–Schwarz inequality · Measurable random variable · Variance · Law of large numbers · Estimator · Mean squared error

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