Multidimensional Chebyshev's inequality

In probability theory, the multidimensional Chebyshev's inequality is a generalization of Chebyshev's inequality, which puts a bound on the probability of the event that a random variable differs from its expected value by more than a specified amount.

Let X be an N-dimensional random vector with expected value μ=E⁡[X] and covariance matrix

V=E⁡[(X−μ)(X−μ)T].

If V is a positive-definite matrix, for any real number t>0:

Pr⁡((X−μ)TV−1(X−μ)>t)≤Nt2

Proof

Since V is positive-definite and symmetric, so is V−1. Define the random variable

y=(X−μ)TV−1(X−μ).

Since y is positive, Markov's inequality holds:

Pr⁡((X−μ)TV−1(X−μ)>t)=Pr⁡(y>t)=Pr⁡(y>t2)≤E⁡[y]t2.

Finally,

E⁡[y]=E⁡[(X−μ)TV−1(X−μ)]=E⁡[trace⁡(V−1(X−μ)(X−μ)T)]=trace⁡(V−1V)=N.

Infinite dimensions

There is a straightforward extension of the vector version of Chebyshev's inequality to infinite dimensional settings[more refs. needed]. Let X be a random variable which takes values in a Fréchet space 𝒳 (equipped with seminorms || ⋅ ||α). This includes most common settings of vector-valued random variables, e.g., when 𝒳 is a Banach space (equipped with a single norm), a Hilbert space, or the finite-dimensional setting as described above.

Suppose that X is of "strong order two", meaning that

E⁡(‖X‖α2)<∞

for every seminorm || ⋅ ||α. This is a generalization of the requirement that X have finite variance, and is necessary for this strong form of Chebyshev's inequality in infinite dimensions. The terminology "strong order two" is due to Vakhania.

Let μ∈𝒳 be the Pettis integral of X (i.e., the vector generalization of the mean), and let

σa:=E⁡‖X−μ‖α2

be the standard deviation with respect to the seminorm || ⋅ ||α. In this setting we can state the following:

General version of Chebyshev's inequality. ∀k>0:Pr⁡(‖X−μ‖α≥kσα)≤1k2.

Proof. The proof is straightforward, and essentially the same as the finitary version[source needed]. If σα = 0, then X is constant (and equal to μ) almost surely, so the inequality is trivial.

If

‖X−μ‖α≥kσα

then ||X − μ||α > 0, so we may safely divide by ||X − μ||α. The crucial trick in Chebyshev's inequality is to recognize that 1=‖X−μ‖α2‖X−μ‖α2.

The following calculations complete the proof:

Pr⁡(‖X−μ‖α≥kσα)=∫Ω𝟏‖X−μ‖α≥kσαdPr=∫Ω(‖X−μ‖α2‖X−μ‖α2)⋅𝟏‖X−μ‖α≥kσαdPr≤∫Ω(‖X−μ‖α2(kσα)2)⋅𝟏‖X−μ‖α≥kσαdPr≤1k2σα2∫Ω‖X−μ‖α2dPr𝟏‖X−μ‖α≥kσα≤1=1k2σα2(E⁡‖X−μ‖α2)=1k2σα2(σα2)=1k2

References

  1. 1 2 Marshall, Albert W.; Olkin, Ingram (December 1960). "Multivariate Chebyshev Inequalities". The Annals of Mathematical Statistics. 31 (4): 1001–1014. doi:10.1214/aoms/1177705673. ISSN 0003-4851.
  2. ↑ Navarro, Jorge (2013-05-24). "A simple proof for the multivariate Chebyshev inequality". arXiv:1305.5646 [math.ST].
  3. ↑ Altomare, Francesco; Campiti, Michele (1994). De Gruyter (ed.). Korovkin-type Approximation Theory and Its Applications. p. 313. doi:10.1515/9783110884586. ISBN 978-3-11-014178-8.
  4. ↑ Vakhania, Nikolai Nikolaevich. Probability distributions on linear spaces. New York: North Holland, 1981.