Concept in probability theory
Markov's inequality gives an upper bound for the measure of the set (indicated in red) where
f
(
x
)
{\displaystyle f(x)}
ε
{\displaystyle \varepsilon }
ε
{\displaystyle \varepsilon }
f
{\displaystyle f}
In probability theory , Markov's inequality gives an upper bound on the probability that a non-negative random variable is greater than or equal to some positive constant . Markov's inequality is tight in the sense that for each chosen positive constant, there exists a random variable such that the inequality is in fact an equality.[ 1]
It is named after the Russian mathematician Andrey Markov , although it appeared earlier in the work of Pafnuty Chebyshev (Markov's teacher), and many sources, especially in analysis , refer to it as Chebyshev's inequality (sometimes, calling it the first Chebyshev inequality, while referring to Chebyshev's inequality as the second Chebyshev inequality) or Bienaymé 's inequality.
Markov's inequality (and other similar inequalities) relate probabilities to expectations , and provide (frequently loose but still useful) bounds for the cumulative distribution function of a random variable. Markov's inequality can also be used to upper bound the expectation of a non-negative random variable in terms of its distribution function.
If X is a nonnegative random variable and a > 0X is at least a is at most the expectation of X divided by a :[ 1]
P
(
X
≥
a
)
≤
E
(
X
)
a
.
{\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}.}
When
E
(
X
)
>
0
{\displaystyle \operatorname {E} (X)>0}
a
=
a
~
⋅
E
(
X
)
{\displaystyle a={\tilde {a}}\cdot \operatorname {E} (X)}
a
~
>
0
{\displaystyle {\tilde {a}}>0}
P
(
X
≥
a
~
⋅
E
(
X
)
)
≤
1
a
~
.
{\displaystyle \operatorname {P} (X\geq {\tilde {a}}\cdot \operatorname {E} (X))\leq {\frac {1}{\tilde {a}}}.}
In the language of measure theory , Markov's inequality states that if (X , Σ, μ ) is a measure space ,
f
{\displaystyle f}
measurable extended real -valued function, and ε > 0
μ
(
{
x
∈
X
:
|
f
(
x
)
|
≥
ε
}
)
≤
1
ε
∫
X
|
f
|
d
μ
.
{\displaystyle \mu (\{x\in X:|f(x)|\geq \varepsilon \})\leq {\frac {1}{\varepsilon }}\int _{X}|f|\,d\mu .}
This measure-theoretic definition is sometimes referred to as Chebyshev's inequality .[ 2]
Extended version for nondecreasing functions [ edit ] If φ is a nondecreasing nonnegative function, X is a (not necessarily nonnegative) random variable, and φ (a ) > 0[ 3]
P
(
X
≥
a
)
≤
E
(
φ
(
X
)
)
φ
(
a
)
.
{\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (\varphi (X))}{\varphi (a)}}.}
An immediate corollary, using higher moments of X supported on values larger than 0, is
P
(
|
X
|
≥
a
)
≤
E
(
|
X
|
n
)
a
n
.
{\displaystyle \operatorname {P} (|X|\geq a)\leq {\frac {\operatorname {E} (|X|^{n})}{a^{n}}}.}
If X is a nonnegative random variable and a > 0U is a uniformly distributed random variable on
[
0
,
1
]
{\displaystyle [0,1]}
X , then[ 4]
P
(
X
≥
U
a
)
≤
E
(
X
)
a
.
{\displaystyle \operatorname {P} (X\geq Ua)\leq {\frac {\operatorname {E} (X)}{a}}.}
Since U is almost surely smaller than one, this bound is strictly stronger than Markov's inequality. Remarkably, U cannot be replaced by any constant smaller than one, meaning that deterministic improvements to Markov's inequality cannot exist in general. While Markov's inequality holds with equality for distributions supported on
{
0
,
a
}
{\displaystyle \{0,a\}}
[
0
,
a
]
{\displaystyle [0,a]}
We separate the case in which the measure space is a probability space from the more general case because the probability case is more accessible for the general reader.
E
(
X
)
=
P
(
X
<
a
)
⋅
E
(
X
|
X
<
a
)
+
P
(
X
≥
a
)
⋅
E
(
X
|
X
≥
a
)
{\displaystyle \operatorname {E} (X)=\operatorname {P} (X<a)\cdot \operatorname {E} (X|X<a)+\operatorname {P} (X\geq a)\cdot \operatorname {E} (X|X\geq a)}
E
(
X
|
X
<
a
)
{\displaystyle \operatorname {E} (X|X<a)}
X
{\displaystyle X}
E
(
X
|
X
≥
a
)
{\displaystyle \operatorname {E} (X|X\geq a)}
a
{\displaystyle a}
a
{\displaystyle a}
X
{\displaystyle X}
Property 1:
P
(
X
<
a
)
⋅
E
(
X
∣
X
<
a
)
≥
0
{\displaystyle \operatorname {P} (X<a)\cdot \operatorname {E} (X\mid X<a)\geq 0}
Given a non-negative random variable
X
{\displaystyle X}
E
(
X
∣
X
<
a
)
≥
0
{\displaystyle \operatorname {E} (X\mid X<a)\geq 0}
X
≥
0
{\displaystyle X\geq 0}
P
(
X
<
a
)
≥
0
{\displaystyle \operatorname {P} (X<a)\geq 0}
P
(
X
<
a
)
⋅
E
(
X
∣
X
<
a
)
≥
0
{\displaystyle \operatorname {P} (X<a)\cdot \operatorname {E} (X\mid X<a)\geq 0}
This is intuitive since conditioning on
X
<
a
{\displaystyle X<a}
Property 2:
P
(
X
≥
a
)
⋅
E
(
X
∣
X
≥
a
)
≥
a
⋅
P
(
X
≥
a
)
{\displaystyle \operatorname {P} (X\geq a)\cdot \operatorname {E} (X\mid X\geq a)\geq a\cdot \operatorname {P} (X\geq a)}
For
X
≥
a
{\displaystyle X\geq a}
X
≥
a
{\displaystyle X\geq a}
a
.
E
(
X
∣
X
≥
a
)
≥
a
{\displaystyle a.\operatorname {E} (X\mid X\geq a)\geq a}
P
(
X
≥
a
)
{\displaystyle \operatorname {P} (X\geq a)}
P
(
X
≥
a
)
⋅
E
(
X
∣
X
≥
a
)
≥
a
⋅
P
(
X
≥
a
)
{\displaystyle \operatorname {P} (X\geq a)\cdot \operatorname {E} (X\mid X\geq a)\geq a\cdot \operatorname {P} (X\geq a)}
This is intuitive since all values considered are at least
a
{\displaystyle a}
a
{\displaystyle a}
Hence intuitively,
E
(
X
)
≥
P
(
X
≥
a
)
⋅
E
(
X
|
X
≥
a
)
≥
a
⋅
P
(
X
≥
a
)
{\displaystyle \operatorname {E} (X)\geq \operatorname {P} (X\geq a)\cdot \operatorname {E} (X|X\geq a)\geq a\cdot \operatorname {P} (X\geq a)}
P
(
X
≥
a
)
≤
E
(
X
)
a
{\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}}
Probability-theoretic proof [ edit ] Method 1:
From the definition of expectation:
E
(
X
)
=
∫
−
∞
∞
x
f
(
x
)
d
x
{\displaystyle \operatorname {E} (X)=\int _{-\infty }^{\infty }xf(x)\,dx}
However, X is a non-negative random variable thus,
E
(
X
)
=
∫
−
∞
∞
x
f
(
x
)
d
x
=
∫
0
∞
x
f
(
x
)
d
x
{\displaystyle \operatorname {E} (X)=\int _{-\infty }^{\infty }xf(x)\,dx=\int _{0}^{\infty }xf(x)\,dx}
From this we can derive,
E
(
X
)
=
∫
0
a
x
f
(
x
)
d
x
+
∫
a
∞
x
f
(
x
)
d
x
≥
∫
a
∞
x
f
(
x
)
d
x
≥
∫
a
∞
a
f
(
x
)
d
x
=
a
∫
a
∞
f
(
x
)
d
x
=
a
Pr
(
X
≥
a
)
{\displaystyle \operatorname {E} (X)=\int _{0}^{a}xf(x)\,dx+\int _{a}^{\infty }xf(x)\,dx\geq \int _{a}^{\infty }xf(x)\,dx\geq \int _{a}^{\infty }af(x)\,dx=a\int _{a}^{\infty }f(x)\,dx=a\operatorname {Pr} (X\geq a)}
From here, dividing through by
a
{\displaystyle a}
Pr
(
X
≥
a
)
≤
E
(
X
)
/
a
{\displaystyle \Pr(X\geq a)\leq \operatorname {E} (X)/a}
Method 2:
For any event
E
{\displaystyle E}
I
E
{\displaystyle I_{E}}
E
{\displaystyle E}
I
E
=
1
{\displaystyle I_{E}=1}
E
{\displaystyle E}
I
E
=
0
{\displaystyle I_{E}=0}
Using this notation, we have
I
(
X
≥
a
)
=
1
{\displaystyle I_{(X\geq a)}=1}
X
≥
a
{\displaystyle X\geq a}
I
(
X
≥
a
)
=
0
{\displaystyle I_{(X\geq a)}=0}
X
<
a
{\displaystyle X<a}
a
>
0
{\displaystyle a>0}
a
I
(
X
≥
a
)
≤
X
{\displaystyle aI_{(X\geq a)}\leq X}
which is clear if we consider the two possible values of
X
≥
a
{\displaystyle X\geq a}
X
<
a
{\displaystyle X<a}
I
(
X
≥
a
)
=
0
{\displaystyle I_{(X\geq a)}=0}
a
I
(
X
≥
a
)
=
0
≤
X
{\displaystyle aI_{(X\geq a)}=0\leq X}
X
≥
a
{\displaystyle X\geq a}
I
X
≥
a
=
1
{\displaystyle I_{X\geq a}=1}
a
I
X
≥
a
=
a
≤
X
{\displaystyle aI_{X\geq a}=a\leq X}
Since
E
{\displaystyle \operatorname {E} }
E
(
a
I
(
X
≥
a
)
)
≤
E
(
X
)
.
{\displaystyle \operatorname {E} (aI_{(X\geq a)})\leq \operatorname {E} (X).}
Now, using linearity of expectations, the left side of this inequality is the same as
a
E
(
I
(
X
≥
a
)
)
=
a
(
1
⋅
P
(
X
≥
a
)
+
0
⋅
P
(
X
<
a
)
)
=
a
P
(
X
≥
a
)
.
{\displaystyle a\operatorname {E} (I_{(X\geq a)})=a(1\cdot \operatorname {P} (X\geq a)+0\cdot \operatorname {P} (X<a))=a\operatorname {P} (X\geq a).}
Thus we have
a
P
(
X
≥
a
)
≤
E
(
X
)
{\displaystyle a\operatorname {P} (X\geq a)\leq \operatorname {E} (X)}
and since a > 0, we can divide both sides by a .
Measure-theoretic proof [ edit ] We may assume that the function
f
{\displaystyle f}
s on X given by
s
(
x
)
=
{
ε
,
if
f
(
x
)
≥
ε
0
,
if
f
(
x
)
<
ε
{\displaystyle s(x)={\begin{cases}\varepsilon ,&{\text{if }}f(x)\geq \varepsilon \\0,&{\text{if }}f(x)<\varepsilon \end{cases}}}
Then
0
≤
s
(
x
)
≤
f
(
x
)
{\displaystyle 0\leq s(x)\leq f(x)}
Lebesgue integral
∫
X
f
(
x
)
d
μ
≥
∫
X
s
(
x
)
d
μ
=
ε
μ
(
{
x
∈
X
:
f
(
x
)
≥
ε
}
)
{\displaystyle \int _{X}f(x)\,d\mu \geq \int _{X}s(x)\,d\mu =\varepsilon \mu (\{x\in X:\,f(x)\geq \varepsilon \})}
and since
ε
>
0
{\displaystyle \varepsilon >0}
ε
{\displaystyle \varepsilon }
μ
(
{
x
∈
X
:
f
(
x
)
≥
ε
}
)
≤
1
ε
∫
X
f
d
μ
.
{\displaystyle \mu (\{x\in X:\,f(x)\geq \varepsilon \})\leq {1 \over \varepsilon }\int _{X}f\,d\mu .}
We now provide a proof for the special case when
X
{\displaystyle X}
Let
a
{\displaystyle a}
a
Pr
(
X
>
a
)
{\displaystyle a\operatorname {Pr} (X>a)}
=
a
Pr
(
X
=
a
+
1
)
+
a
Pr
(
X
=
a
+
2
)
+
a
Pr
(
X
=
a
+
3
)
+
.
.
.
{\displaystyle =a\operatorname {Pr} (X=a+1)+a\operatorname {Pr} (X=a+2)+a\operatorname {Pr} (X=a+3)+...}
≤
a
Pr
(
X
=
a
)
+
(
a
+
1
)
Pr
(
X
=
a
+
1
)
+
(
a
+
2
)
Pr
(
X
=
a
+
2
)
+
.
.
.
{\displaystyle \leq a\operatorname {Pr} (X=a)+(a+1)\operatorname {Pr} (X=a+1)+(a+2)\operatorname {Pr} (X=a+2)+...}
≤
Pr
(
X
=
1
)
+
2
Pr
(
X
=
2
)
+
3
Pr
(
X
=
3
)
+
.
.
.
{\displaystyle \leq \operatorname {Pr} (X=1)+2\operatorname {Pr} (X=2)+3\operatorname {Pr} (X=3)+...}
+
a
Pr
(
X
=
a
)
+
(
a
+
1
)
Pr
(
X
=
a
+
1
)
+
(
a
+
2
)
Pr
(
X
=
a
+
2
)
+
.
.
.
{\displaystyle +a\operatorname {Pr} (X=a)+(a+1)\operatorname {Pr} (X=a+1)+(a+2)\operatorname {Pr} (X=a+2)+...}
=
E
(
X
)
{\displaystyle =\operatorname {E} (X)}
Dividing by
a
{\displaystyle a}
[ edit ] Chebyshev's inequality uses the variance to bound the probability that a random variable deviates far from the mean. Specifically,
P
(
|
X
−
E
(
X
)
|
≥
a
)
≤
Var
(
X
)
a
2
,
{\displaystyle \operatorname {P} (|X-\operatorname {E} (X)|\geq a)\leq {\frac {\operatorname {Var} (X)}{a^{2}}},}
for any a > 0[ 3] Var(X ) is the variance of X, defined as:
Var
(
X
)
=
E
[
(
X
−
E
(
X
)
)
2
]
.
{\displaystyle \operatorname {Var} (X)=\operatorname {E} [(X-\operatorname {E} (X))^{2}].}
Chebyshev's inequality follows from Markov's inequality by considering the random variable
(
X
−
E
(
X
)
)
2
{\displaystyle (X-\operatorname {E} (X))^{2}}
and the constant
a
2
,
{\displaystyle a^{2},}
P
(
(
X
−
E
(
X
)
)
2
≥
a
2
)
≤
Var
(
X
)
a
2
.
{\displaystyle \operatorname {P} ((X-\operatorname {E} (X))^{2}\geq a^{2})\leq {\frac {\operatorname {Var} (X)}{a^{2}}}.}
This argument can be summarized (where "MI" indicates use of Markov's inequality):
P
(
|
X
−
E
(
X
)
|
≥
a
)
=
P
(
(
X
−
E
(
X
)
)
2
≥
a
2
)
≤
M
I
E
(
(
X
−
E
(
X
)
)
2
)
a
2
=
Var
(
X
)
a
2
.
{\displaystyle \operatorname {P} (|X-\operatorname {E} (X)|\geq a)=\operatorname {P} \left((X-\operatorname {E} (X))^{2}\geq a^{2}\right)\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} \left((X-\operatorname {E} (X))^{2}\right)}{a^{2}}}={\frac {\operatorname {Var} (X)}{a^{2}}}.}
The "monotonic" result can be demonstrated by:
P
(
|
X
|
≥
a
)
=
P
(
φ
(
|
X
|
)
≥
φ
(
a
)
)
≤
M
I
E
(
φ
(
|
X
|
)
)
φ
(
a
)
{\displaystyle \operatorname {P} (|X|\geq a)=\operatorname {P} {\big (}\varphi (|X|)\geq \varphi (a){\big )}\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} (\varphi (|X|))}{\varphi (a)}}}
The result that, for a nonnegative random variable X , the quantile function of X satisfies:
Q
X
(
1
−
p
)
≤
E
(
X
)
p
,
{\displaystyle Q_{X}(1-p)\leq {\frac {\operatorname {E} (X)}{p}},}
the proof using
p
≤
P
(
X
≥
Q
X
(
1
−
p
)
)
≤
M
I
E
(
X
)
Q
X
(
1
−
p
)
.
{\displaystyle p\leq \operatorname {P} (X\geq Q_{X}(1-p))\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} (X)}{Q_{X}(1-p)}}.}
Let
M
⪰
0
{\displaystyle M\succeq 0}
A
≻
0
{\displaystyle A\succ 0}
P
(
M
⋠
A
)
≤
tr
(
E
(
X
)
A
−
1
)
{\displaystyle \operatorname {P} (M\npreceq A)\leq \operatorname {tr} (\operatorname {E} (X)A^{-1})}
which can be proved similarly.[ 5] Assuming no income is negative, Markov's inequality shows that no more than 10% (1/10) of the population can have more than 10 times the average income.[ 6]
Another simple example is as follows: Andrew makes 4 mistakes on average on his Statistics course tests. The best upper bound on the probability that Andrew will do at least 10 mistakes is 0.4 as
P
(
X
≥
10
)
≤
E
(
X
)
α
=
4
10
.
{\displaystyle \operatorname {P} (X\geq 10)\leq {\frac {\operatorname {E} (X)}{\alpha }}={\frac {4}{10}}.}
^ a b Huber, Mark (2019-11-26). "Halving the Bounds for the Markov, Chebyshev, and Chernoff Inequalities Using Smoothing" . The American Mathematical Monthly . 126 (10): 915–927. arXiv :1803.06361 doi :10.1080/00029890.2019.1656484 . ISSN 0002-9890 . ^ Stein, E. M. ; Shakarchi, R. (2005), Real Analysis , Princeton Lectures in Analysis , vol. 3 (1st ed.), p. 91^ a b Lin, Zhengyan (2010). Probability inequalities . Springer. p. 52. ^ Ramdas, Aaditya; Manole, Tudor (2023), Randomized and Exchangeable Improvements of Markov's, Chebyshev's and Chernoff's Inequalities , arXiv :2304.02611 ^ Tu, Stephen (2017-11-04). "Markov's Inequality for Matrices" . Retrieved May 27, 2024 . ^ Ross, Kevin. 5.4 Probability inequalitlies | An Introduction to Probability and Simulation
![{\displaystyle [0,1]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
![{\displaystyle [0,a]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/63d050b0ffe6cc6f635808b9a013366a60e6d0c0)
![{\displaystyle \operatorname {Var} (X)=\operatorname {E} [(X-\operatorname {E} (X))^{2}].}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/71c7a116967cab98cb1eb56e626497e77ce354a2)
