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 {\displaystyle X} be an N {\displaystyle N} -dimensional random vector with expected value μ = E [ X ] {\displaystyle \mu =\operatorname {E} [X]} and covariance matrix
V = E [ ( X − μ ) ( X − μ ) T ] . {\displaystyle V=\operatorname {E} [(X-\mu )(X-\mu )^{T}].\,}
If V {\displaystyle V} is a positive-definite matrix, for any real number t > 0 {\displaystyle t>0} :
Pr ( ( X − μ ) T V − 1 ( X − μ ) > t ) ≤ N t 2 {\displaystyle \Pr \left({\sqrt {(X-\mu )^{T}V^{-1}(X-\mu )}}>t\right)\leq {\frac {N}{t^{2}}}}
Proof Since V {\displaystyle V} is positive-definite, so is V − 1 {\displaystyle V^{-1}} . Define the random variable
y = ( X − μ ) T V − 1 ( X − μ ) . {\displaystyle y=(X-\mu )^{T}V^{-1}(X-\mu ).}
Since y {\displaystyle y} is positive, Markov's inequality holds:
Pr ( ( X − μ ) T V − 1 ( X − μ ) > t ) = Pr ( y > t ) = Pr ( y > t 2 ) ≤ E [ y ] t 2 . {\displaystyle \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,
E [ y ] = E [ ( X − μ ) T V − 1 ( X − μ ) ] = E [ trace ( V − 1 ( X − μ ) ( X − μ ) T ) ] = trace ( V − 1 V ) = N . {\displaystyle {\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}}.}
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