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:^[1]

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

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

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

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

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

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, then^[3]

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

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

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

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 edit

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

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

Probability-theoretic proof edit

Method 1: From the definition of expectation:

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

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

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

From this we can derive,

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

From here, dividing through by ${\displaystyle a}$  allows us to see that

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

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

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

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

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

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

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

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

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

Thus we have

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

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

Measure-theoretic proof edit

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

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

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

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

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

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

Discrete case edit

We now provide a proof for the special case when ${\displaystyle X}$  is a discrete random variable which only takes on non-negative integer values.

Let ${\displaystyle a}$  be a positive integer. By definition ${\displaystyle a\operatorname {Pr} (X>a)}$  ${\displaystyle =a\operatorname {Pr} (X=a+1)+a\operatorname {Pr} (X=a+2)+a\operatorname {Pr} (X=a+3)+...}$  ${\displaystyle \leq a\operatorname {Pr} (X=a)+(a+1)\operatorname {Pr} (X=a+1)+(a+2)\operatorname {Pr} (X=a+2)+...}$  ${\displaystyle \leq \operatorname {Pr} (X=1)+2\operatorname {Pr} (X=2)+3\operatorname {Pr} (X=3)+...}$  ${\displaystyle +a\operatorname {Pr} (X=a)+(a+1)\operatorname {Pr} (X=a+1)+(a+2)\operatorname {Pr} (X=a+2)+...}$  ${\displaystyle =\operatorname {E} (X)}$ 

Dividing by ${\displaystyle a}$  yields the desired result.

Chebyshev's inequality edit

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

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

for any a > 0.^[3] Here Var(X) is the variance of X, defined as:

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

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

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

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

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

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

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

Other corollaries edit

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

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

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

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

   the proof using

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

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

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

   can be shown in a similar manner.

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.^[4]

Another simple example is as follows: Andrew makes 4 mistakes on average on a random test his Statistics course tests. The best upper bound on the probability that Andrew will do at least 10 mistakes is 0.4 as ${\displaystyle \operatorname {P} (X\geq 10)\leq {\frac {\operatorname {E} (X)}{\alpha }}={\frac {4}{10}}.}$  Note that Andrew might do exactly 10 mistakes with probability 0.4 and make no mistakes with probability 0.6; the expectation is exactly 4 mistakes.

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

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