Markov's inequality

Markov's inequality gives an upper bound for the measure of the set (indicated in red) where $f(x)$ exceeds a given level $\varepsilon$. The bound combines the level $\varepsilon$ with the average value of $f$.

In probability theory, Markov's inequality gives an upper bound for the probability that a non-negative function of a random variable is greater than or equal to some positive constant. 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.

Statement

If X is a nonnegative random variable and a > 0, then the probability that X is at least a is at most the expectation of X divided by a:

$\operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}.$

Let $a={\tilde {a}}\cdot \operatorname {E} (X)$ (where ${\tilde {a}}>0$); then we can rewrite the previous inequality as

$\operatorname {P} (X\geq {\tilde {a}}\cdot \operatorname {E} (X))\leq {\frac {1}{\tilde {a}}}.$

In the language of measure theory, Markov's inequality states that if (X, Σ, μ) is a measure space, $f$ is a measurable extended real-valued function, and ε > 0, then

$\mu (\{x\in X:|f(x)|\geq \varepsilon \})\leq {\frac {1}{\varepsilon }}\int _{X}|f|\,d\mu .$

This measure-theoretic definition is sometimes referred to as Chebyshev's inequality.

Extended version for nondecreasing functions

If φ is a nondecreasing nonnegative function, X is a (not necessarily nonnegative) random variable, and φ(a) > 0, then

$\operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (\varphi (X))}{\varphi (a)}}.$

An immediate corollary, using higher moments of X supported on values larger than 0, is

$\operatorname {P} (|X|\geq a)\leq {\frac {\operatorname {E} (|X|^{n})}{a^{n}}}.$

Proofs

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.

Intuition

$\operatorname {E} (X)=\operatorname {P} (X<a)\cdot \operatorname {E} (X|X<a)+\operatorname {P} (X\geq a)\cdot \operatorname {E} (X|X\geq a)$ where $\operatorname {E} (X|X<a)$ is larger than or equal to 0 as the random variable $X$ is non-negative and $\operatorname {E} (X|X\geq a)$ is larger than or equal to $a$ because the conditional expectation only takes into account of values larger than or equal to $a$ which r.v. $X$ can take.

Hence intuitively $\operatorname {E} (X)\geq \operatorname {P} (X\geq a)\cdot \operatorname {E} (X|X\geq a)\geq a\cdot \operatorname {P} (X\geq a)$, which directly leads to $\operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}$.

Probability-theoretic proof

Method 1: From the definition of expectation:

$\operatorname {E} (X)=\int _{-\infty }^{\infty }xf(x)\,dx$

However, X is a non-negative random variable thus,

$\operatorname {E} (X)=\int _{-\infty }^{\infty }xf(x)\,dx=\int _{0}^{\infty }xf(x)\,dx$

From this we can derive,

$\operatorname {E} (X)=\int _{0}^{a}xf(x)\,dx+\int _{a}^{\infty }xf(x)\,dx\geq \int _{a}^{\infty }xf(x)\,dx\geq \int _{a}^{\infty }af(x)\,dx=a\int _{a}^{\infty }f(x)\,dx=a\operatorname {Pr} (X\geq a)$

From here, dividing through by $a$ allows us to see that

$\Pr(X\geq a)\leq \operatorname {E} (X)/a$

Method 2: For any event $E$, let $I_{E}$ be the indicator random variable of $E$, that is, $I_{E}=1$ if $E$ occurs and $I_{E}=0$ otherwise.

Using this notation, we have $I_{(X\geq a)}=1$ if the event $X\geq a$ occurs, and $I_{(X\geq a)}=0$ if $X<a$. Then, given $a>0$,

$aI_{(X\geq a)}\leq X$

which is clear if we consider the two possible values of $X\geq a$. If $X<a$, then $I_{(X\geq a)}=0$, and so $aI_{(X\geq a)}=0\leq X$. Otherwise, we have $X\geq a$, for which $I_{X\geq a}=1$ and so $aI_{X\geq a}=a\leq X$.

Since $\operatorname {E}$ is a monotonically increasing function, taking expectation of both sides of an inequality cannot reverse it. Therefore,

$\operatorname {E} (aI_{(X\geq a)})\leq \operatorname {E} (X).$

Now, using linearity of expectations, the left side of this inequality is the same as

$a\operatorname {E} (I_{(X\geq a)})=a(1\cdot \operatorname {P} (X\geq a)+0\cdot \operatorname {P} (X<a))=a\operatorname {P} (X\geq a).$

Thus we have

$a\operatorname {P} (X\geq a)\leq \operatorname {E} (X)$

and since a > 0, we can divide both sides by a.

Measure-theoretic proof

We may assume that the function $f$ is non-negative, since only its absolute value enters in the equation. Now, consider the real-valued function s on X given by

$s(x)={\begin{cases}\varepsilon ,&{\text{if }}f(x)\geq \varepsilon \\0,&{\text{if }}f(x)<\varepsilon \end{cases}}$

Then $0\leq s(x)\leq f(x)$. By the definition of the Lebesgue integral

$\int _{X}f(x)\,d\mu \geq \int _{X}s(x)\,d\mu =\varepsilon \mu (\{x\in X:\,f(x)\geq \varepsilon \})$

and since $\varepsilon >0$, both sides can be divided by $\varepsilon$, obtaining

$\mu (\{x\in X:\,f(x)\geq \varepsilon \})\leq {1 \over \varepsilon }\int _{X}f\,d\mu .$

Corollaries

Chebyshev's inequality

Chebyshev's inequality uses the variance to bound the probability that a random variable deviates far from the mean. Specifically,

$\operatorname {P} (|X-\operatorname {E} (X)|\geq a)\leq {\frac {\operatorname {Var} (X)}{a^{2}}},$

for any a > 0. Here Var(X) is the variance of X, defined as:

$\operatorname {Var} (X)=\operatorname {E} [(X-\operatorname {E} (X))^{2}].$

Chebyshev's inequality follows from Markov's inequality by considering the random variable

$(X-\operatorname {E} (X))^{2}$

and the constant $a^{2},$ for which Markov's inequality reads

$\operatorname {P} ((X-\operatorname {E} (X))^{2}\geq a^{2})\leq {\frac {\operatorname {Var} (X)}{a^{2}}}.$

This argument can be summarized (where "MI" indicates use of Markov's inequality):

$\operatorname {P} (|X-\operatorname {E} (X)|\geq a)=\operatorname {P} \left((X-\operatorname {E} (X))^{2}\geq a^{2}\right)\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} \left((X-\operatorname {E} (X))^{2}\right)}{a^{2}}}={\frac {\operatorname {Var} (X)}{a^{2}}}.$

Other corollaries

1. The "monotonic" result can be demonstrated by:

   $\operatorname {P} (|X|\geq a)=\operatorname {P} {\big (}\varphi (|X|)\geq \varphi (a){\big )}\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} (\varphi (|X|))}{\varphi (a)}}$

2. The result that, for a nonnegative random variable X, the quantile function of X satisfies:

   $Q_{X}(1-p)\leq {\frac {\operatorname {E} (X)}{p}},$

   the proof using

   $p\leq \operatorname {P} (X\geq Q_{X}(1-p))\,{\overset {\underset {\mathrm {MI} }{}}{\leq }}\,{\frac {\operatorname {E} (X)}{Q_{X}(1-p)}}.$

3. Let $M\succeq 0$ be a self-adjoint matrix-valued random variable and a > 0. Then

   $\operatorname {P} (M\npreceq a\cdot I)\leq {\frac {\operatorname {tr} \left(E(M)\right)}{na}}.$

   can be shown in a similar manner.

Examples

Assuming no income is negative, Markov's inequality shows that no more than 1/5 of the population can have more than 5 times the average income.

See also

• Paley–Zygmund inequality – a corresponding lower bound
• Concentration inequality – a summary of tail-bounds on random variables.

External links

• The formal proof of Markov's inequality in the Mizar system.