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In [[probability theory]], the '''multidimensional Chebyshev's inequality'''<ref name=awm>{{Cite journal |last1=Marshall |first1=Albert W. |last2=Olkin |first2=Ingram |date=December 1960 |title=Multivariate Chebyshev Inequalities |url=https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-31/issue-4/Multivariate-Chebyshev-Inequalities/10.1214/aoms/1177705673.full |journal=The Annals of Mathematical Statistics |volume=31 |issue=4 |pages=1001–1014 |doi=10.1214/aoms/1177705673 |issn=0003-4851}}</ref> 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 <math>X</math> be an <math>N</math>-dimensional [[random vector]] with [[expected value]] <math>\mu=\operatorname{E}[X] </math> and [[covariance matrix]]

: <math>V=\operatorname{E} [(X - \mu) (X - \mu)^T]. \, </math>

If <math>V</math> is a [[positive-definite matrix]], for any [[real number]] <math>t>0</math>:
:<math>
\Pr \left( \sqrt{( X-\mu)^T  V^{-1} (X-\mu) } > t\right) \le \frac N {t^2}
</math>

==Proof==
Since <math>V</math> is positive-definite, so is <math>V^{-1}</math>. Define the random variable

:<math>
y = (X-\mu)^T V^{-1} (X-\mu).
</math>

Since <math>y</math> is positive, [[Markov's inequality]] holds:

:<math>
\Pr\left( \sqrt{(X-\mu)^T V^{-1} (X-\mu) } > t\right) = \Pr( \sqrt{y} > t) = \Pr(y > t^2)
\le \frac{\operatorname{E}[y]}{t^2}.
</math>

Finally,

:<math>\begin{align}
\operatorname{E}[y] &= \operatorname{E}[(X-\mu)^T V^{-1} (X-\mu)]\\[6pt]
&=\operatorname{E}[ \operatorname{trace} (  V^{-1} (X-\mu) (X-\mu)^T )]\\[6pt]
&= \operatorname{trace} (  V^{-1} V ) = N
\end{align}.</math><ref name=awm/><ref>{{cite arXiv|last=Navarro |first=Jorge |title=A simple proof for the multivariate Chebyshev inequality |date=2013-05-24 |class=math.ST |eprint=1305.5646 }}</ref>

==Infinite dimensions==
There is a straightforward extension of the vector version of Chebyshev's inequality to infinite dimensional settings<sup>[more refs. needed]</sup>.<ref>{{Cite book |last1=Altomare |first1=Francesco |last2=Campiti |first2=Michele |date=1994 |title=Korovkin-type Approximation Theory and Its Applications |url=https://doi.org/10.1515/9783110884586 |editor=De Gruyter |language=en  |pages=313 |doi=10.1515/9783110884586}}</ref> Let {{mvar|X}} be a random variable which takes values in a [[Fréchet space]] <math>\mathcal X</math> (equipped with seminorms {{math|{{!!}} ⋅ {{!!}}<sub>''α''</sub>}}). This includes most common settings of vector-valued random variables, e.g., when <math>\mathcal X</math> is a [[Banach space]] (equipped with a single norm), a [[Hilbert space]], or the finite-dimensional setting as described above.

Suppose that {{mvar|X}} is of "[[strong order two]]", meaning that

: <math> \operatorname{E}\left(\| X\|_\alpha^2 \right) < \infty </math>

for every seminorm {{math|{{!!}} ⋅ {{!!}}<sub>''α''</sub>}}. This is a generalization of the requirement that {{mvar|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]].<ref>Vakhania, Nikolai Nikolaevich. Probability distributions on linear spaces. New York: North Holland, 1981.</ref>

Let <math>\mu \in \mathcal X</math> be the [[Pettis integral]] of {{mvar|X}} (i.e., the vector generalization of the mean), and let

:<math>\sigma_a := \sqrt{\operatorname{E}\|X - \mu\|_\alpha^2}</math>

be the standard deviation with respect to the seminorm {{math|{{!!}} ⋅ {{!!}}<sub>''α''</sub>}}. In this setting we can state the following:

:'''General version of Chebyshev's inequality.''' <math>\forall k > 0: \quad \Pr\left( \|X - \mu\|_\alpha \ge k \sigma_\alpha \right) \le \frac{1}{ k^2 }.</math>

'''Proof.''' The proof is straightforward, and essentially the same as the finitary version<sup>[source needed]</sup>. If {{math|''σ<sub>α</sub>'' {{=}} 0}}, then {{mvar|X}} is constant (and equal to {{mvar|μ}}) almost surely, so the inequality is trivial.

If

:<math>\|X - \mu\|_\alpha \ge k \sigma_\alpha</math>

then {{math|{{!!}}''X'' − ''μ''{{!!}}<sub>''α''</sub> > 0}}, so we may safely divide by {{math|{{!!}}''X'' − ''μ''{{!!}}<sub>''α''</sub>}}. The crucial trick in Chebyshev's inequality is to recognize that <math>1 = \tfrac{\|X - \mu\|_\alpha^2}{\|X - \mu\|_\alpha^2}</math>.

The following calculations complete the proof:

:<math>\begin{align}
\Pr\left( \|X - \mu\|_\alpha \ge k \sigma_\alpha \right) &= \int_\Omega \mathbf{1}_{\|X - \mu\|_\alpha \ge k \sigma_\alpha} \, \mathrm d\Pr \\
& = \int_\Omega \left ( \frac{\|X - \mu\|_\alpha^2}{\|X - \mu\|_\alpha^2} \right ) \cdot \mathbf{1}_{\|X - \mu\|_\alpha \ge k \sigma_\alpha} \, \mathrm d\Pr \\[6pt]
&\le \int_\Omega \left (\frac{\|X - \mu\|_\alpha^2}{(k\sigma_\alpha)^2} \right ) \cdot \mathbf{1}_{\|X - \mu\|_\alpha \ge k \sigma_\alpha} \, \mathrm d\Pr \\[6pt]
&\le \frac{1}{k^2 \sigma_\alpha^2} \int_\Omega \|X - \mu\|_\alpha^2 \, \mathrm d\Pr && \mathbf{1}_{\|X - \mu\|_\alpha \ge k \sigma_\alpha} \le 1\\[6pt]
&= \frac{1}{k^2 \sigma_\alpha^2} \left (\operatorname{E}\|X - \mu\|_\alpha^2 \right )\\[6pt]
&= \frac{1}{k^2 \sigma_\alpha^2} \left (\sigma_\alpha^2 \right )\\[6pt]
&= \frac{1}{k^2}
\end{align}</math>

==References==
{{Reflist}}

[[Category:Probabilistic inequalities]]
[[Category:Statistical inequalities]]