# 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 $\mu =\operatorname {E} [X]$ and covariance matrix
    $V=\operatorname {E} [(X-\mu )(X-\mu )^{T}].\,$
If $V$ is a positive-definite matrix, for any real number $t>0$:
    $\Pr \left({\sqrt {(X-\mu )^{T}V^{-1}(X-\mu )}}>t\right)\leq {\frac {N}{t^{2}}}$

## Proof

Since $V$ is positive-definite, so is $V^{-1}$. Define the random variable
    $y=(X-\mu )^{T}V^{-1}(X-\mu ).$
Since $y$ is positive, Markov's inequality holds:
    $\Pr \left({\sqrt {(X-\mu )^{T}V^{-1}(X-\mu )}}>t\right)=\Pr({\sqrt {y}}>t)=\Pr(y>t^{2})\leq {\frac {\operatorname {E} [y]}{t^{2}}}.$
Finally,
    ${\begin{aligned}\operatorname {E} [y]&=\operatorname {E} [(X-\mu )^{T}V^{-1}(X-\mu )]\\[6pt]&=\operatorname {E} [\operatorname {trace} (V^{-1}(X-\mu )(X-\mu )^{T})]\\[6pt]&=\operatorname {trace} (V^{-1}V)=N\end{aligned}}.$

## Infinite dimensions

There is a straightforward extension of the vector version of Chebyshev's inequality to infinite dimensional settings$^{[morerefs.needed]}$. Let X be a random variable which takes values in a Fréchet space ${\mathcal {X}}$ (equipped with seminorms || ⋅ ||$_α$). This includes most common settings of vector-valued random variables, e.g., when ${\mathcal {X}}$ 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
    $\operatorname {E} \left(\|X\|_{\alpha }^{2}\right)<\infty$
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 $\mu \in {\mathcal {X}}$ be the Pettis integral of X (i.e., the vector generalization of the mean), and let
    $\sigma _{a}:={\sqrt {\operatorname {E} \|X-\mu \|_{\alpha }^{2}}}$
be the standard deviation with respect to the seminorm || ⋅ ||$_α$. In this setting we can state the following:
    General version of Chebyshev's inequality. $\forall k>0:\quad \Pr \left(\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }\right)\leq {\frac {1}{k^{2}}}.$
Proof. The proof is straightforward, and essentially the same as the finitary version$^{[sourceneeded]}$. If σ$_α$ = 0, then X is constant (and equal to μ) almost surely, so the inequality is trivial.
If
    $\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }$
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 $1={\tfrac {\|X-\mu \|_{\alpha }^{2}}{\|X-\mu \|_{\alpha }^{2}}}$.
The following calculations complete the proof:
    ${\begin{aligned}\Pr \left(\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }\right)&=\int _{\Omega }\mathbf {1} _{\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }}\,\mathrm {d} \Pr \\&=\int _{\Omega }\left({\frac {\|X-\mu \|_{\alpha }^{2}}{\|X-\mu \|_{\alpha }^{2}}}\right)\cdot \mathbf {1} _{\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }}\,\mathrm {d} \Pr \\[6pt]&\leq \int _{\Omega }\left({\frac {\|X-\mu \|_{\alpha }^{2}}{(k\sigma _{\alpha })^{2}}}\right)\cdot \mathbf {1} _{\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }}\,\mathrm {d} \Pr \\[6pt]&\leq {\frac {1}{k^{2}\sigma _{\alpha }^{2}}}\int _{\Omega }\|X-\mu \|_{\alpha }^{2}\,\mathrm {d} \Pr &&\mathbf {1} _{\|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }}\leq 1\\[6pt]&={\frac {1}{k^{2}\sigma _{\alpha }^{2}}}\left(\operatorname {E} \|X-\mu \|_{\alpha }^{2}\right)\\[6pt]&={\frac {1}{k^{2}\sigma _{\alpha }^{2}}}\left(\sigma _{\alpha }^{2}\right)\\[6pt]&={\frac {1}{k^{2}}}\end{aligned}}$