Markov's inequality

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 gives an upper bound for the measure of the set (indicated in red) where exceeds a given level . The bound combines the level with the average value of .

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.

StatementEdit

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]

 

Let   (where  ); then we can rewrite the previous inequality as

 

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

 

This measure-theoretic definition is sometimes referred to as Chebyshev's inequality.[2]

Extended version for monotonically increasing functionsEdit

If φ is a monotonically increasing nonnegative function for the nonnegative reals, X is a random variable, a ≥ 0, and φ(a) > 0, then

 

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

 

ProofsEdit

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.

IntuitiveEdit

  where   is larger than 0 as r.v   is non-negative and   is larger than   because the conditional expectation only takes into account of values larger than   which r.v   can take.

Hence intuitively  , which directly leads to  .

Proof in the language of probability theoryEdit

Method 1: From the definition of expectation:

 

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

 

From this we can derive,

 

From here, dividing through by   allows us to see that

 

Method 2: For any event  , let   be the indicator random variable of  , that is,   if   occurs and   otherwise.

Using this notation, we have   if the event   occurs, and   if  . Then, given  ,

 

which is clear if we consider the two possible values of  . If  , then  , and so  . Otherwise, we have  , for which   and so  .

Since   is a monotonically increasing function, taking expectation of both sides of an inequality cannot reverse it. Therefore,

 

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

 

Thus we have

 

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

In the language of measure theoryEdit

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

 

Then  . By the definition of the Lebesgue integral

 

and since  , both sides can be divided by  , obtaining

 

CorollariesEdit

Chebyshev's inequalityEdit

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

 

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

 

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

 

and the constant   for which Markov's inequality reads

 

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

 

Other corollariesEdit

  1. The "monotonic" result can be demonstrated by:
     
  2. The result that, for a nonnegative random variable X, the quantile function of X satisfies:
     
    the proof using
     
  3. Let   be a self-adjoint matrix-valued random variable and a > 0. Then
     
    can be shown in a similar manner.

ExamplesEdit

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 alsoEdit

ReferencesEdit

  1. ^ "Markov and Chebyshev Inequalities". Retrieved 4 February 2016.
  2. ^ Stein, E. M.; Shakarchi, R. (2005), Real Analysis, Princeton Lectures in Analysis, 3 (1st ed.), p. 91.

External linksEdit