In probability theory , the multidimensional Chebyshev's inequality [1] 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}}}}
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}}.}
[1] [2]
Infinite dimensions [ edit ]
There is a straightforward extension of the vector version of Chebyshev's inequality to infinite dimensional settings.[3] Let X be a random variable which takes values in a Fréchet space
X
{\displaystyle {\mathcal {X}}}
(equipped with seminorms || ⋅ ||α ). This includes most common settings of vector-valued random variables, e.g., when
X
{\displaystyle {\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
E
(
‖
X
‖
α
2
)
<
∞
{\displaystyle \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 .[4]
Let
μ
∈
X
{\displaystyle \mu \in {\mathcal {X}}}
be the Pettis integral of X (i.e., the vector generalization of the mean), and let
σ
a
:=
E
‖
X
−
μ
‖
α
2
{\displaystyle \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.
∀
k
>
0
:
Pr
(
‖
X
−
μ
‖
α
≥
k
σ
α
)
≤
1
k
2
.
{\displaystyle \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. If σα = 0 , then X is constant (and equal to μ ) almost surely, so the inequality is trivial.
If
‖
X
−
μ
‖
α
≥
k
σ
α
2
{\displaystyle \|X-\mu \|_{\alpha }\geq k\sigma _{\alpha }^{2}}
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
‖
α
2
‖
X
−
μ
‖
α
2
{\displaystyle 1={\tfrac {\|X-\mu \|_{\alpha }^{2}}{\|X-\mu \|_{\alpha }^{2}}}}
.
The following calculations complete the proof:
Pr
(
‖
X
−
μ
‖
α
≥
k
σ
α
)
=
∫
Ω
1
‖
X
−
μ
‖
α
≥
k
σ
α
d
Pr
=
∫
Ω
(
‖
X
−
μ
‖
α
2
‖
X
−
μ
‖
α
2
)
⋅
1
‖
X
−
μ
‖
α
≥
k
σ
α
d
Pr
≤
∫
Ω
(
‖
X
−
μ
‖
α
2
(
k
σ
α
)
2
)
⋅
1
‖
X
−
μ
‖
α
≥
k
σ
α
d
Pr
≤
1
k
2
σ
α
2
∫
Ω
‖
X
−
μ
‖
α
2
d
Pr
1
‖
X
−
μ
‖
α
≥
k
σ
α
≤
1
=
1
k
2
σ
α
2
(
E
‖
X
−
μ
‖
α
2
)
=
1
k
2
σ
α
2
(
σ
α
2
)
=
1
k
2
{\displaystyle {\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}}}
References [ edit ]
^ a b 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 .
^ Navarro, Jorge (2013-05-24), A simple proof for the multivariate Chebyshev inequality , doi :10.48550/arXiv.1305.5646 , retrieved 2024-05-24
^ Ait-Haddou, Rachid; Mazure, Marie-Laurence (2018-02-01). "The Fundamental Blossoming Inequality in Chebyshev Spaces—I: Applications to Schur Functions" . Foundations of Computational Mathematics . 18 (1): 135–158. doi :10.1007/s10208-016-9334-8 . ISSN 1615-3383 .
^ Vakhania, Nikolai Nikolaevich. Probability distributions on linear spaces. New York: North Holland, 1981.