Mathematical and Applied Statistics Lesson of the Day – The Motivation and Intuition Behind Chebyshev’s Inequality

In 2 recent Statistics Lessons of the Day, I

Chebyshev’s inequality is just a special version of Markov’s inequality; thus, their motivations and intuitions are similar.

P[|X - \mu| \geq k \sigma] \leq 1 \div k^2

Markov’s inequality roughly says that a random variable X is most frequently observed near its expected value, \mu.  Remarkably, it quantifies just how often X is far away from \mu.  Chebyshev’s inequality goes one step further and quantifies that distance between X and \mu in terms of the number of standard deviations away from \mu.  It roughly says that the probability of X being k standard deviations away from \mu is at most k^{-2}.  Notice that this upper bound decreases as k increases – confirming our intuition that it is highly improbable for X to be far away from \mu.

As with Markov’s inequality, Chebyshev’s inequality applies to any random variable X, as long as E(X) and V(X) are finite.  (Markov’s inequality requires only E(X) to be finite.)  This is quite a marvelous result!

Advertisements

Mathematical Statistics Lesson of the Day – Chebyshev’s Inequality

The variance of a random variable X is just an expected value of a function of X.  Specifically,

V(X) = E[(X - \mu)^2], \ \text{where} \ \mu = E(X).

Let’s substitute (X - \mu)^2 into Markov’s inequality and see what happens.  For convenience and without loss of generality, I will replace the constant c with another constant, b^2.

\text{Let} \ b^2 = c, \ b > 0. \ \ \text{Then,}

P[(X - \mu)^2 \geq b^2] \leq E[(X - \mu)^2] \div b^2

P[ (X - \mu) \leq -b \ \ \text{or} \ \ (X - \mu) \geq b] \leq V(X) \div b^2

P[|X - \mu| \geq b] \leq V(X) \div b^2

Now, let’s substitute b with k \sigma, where \sigma is the standard deviation of X.  (I can make this substitution, because \sigma is just another constant.)

\text{Let} \ k \sigma = b. \ \ \text{Then,}

P[|X - \mu| \geq k \sigma] \leq V(X) \div k^2 \sigma^2

P[|X - \mu| \geq k \sigma] \leq 1 \div k^2

This last inequality is known as Chebyshev’s inequality, and it is just a special version of Markov’s inequality.  In a later Statistics Lesson of the Day, I will discuss the motivation and intuition behind it.  (Hint: Read my earlier lesson on the motivation and intuition behind Markov’s inequality.)

Mathematical and Applied Statistics Lesson of the Day – The Motivation and Intuition Behind Markov’s Inequality

Markov’s inequality may seem like a rather arbitrary pair of mathematical expressions that are coincidentally related to each other by an inequality sign:

P(X \geq c) \leq E(X) \div c, where c > 0.

However, there is a practical motivation behind Markov’s inequality, and it can be posed in the form of a simple question: How often is the random variable X “far” away from its “centre” or “central value”?

Intuitively, the “central value” of X is the value that of X that is most commonly (or most frequently) observed.  Thus, as X deviates farther and farther from its “central value”, we would expect those distant-from-the-centre values to be less frequently observed.

Recall that the expected value, E(X), is a measure of the “centre” of X.  Thus, we would expect that the probability of X being very far away from E(X) is very low.  Indeed, Markov’s inequality rigorously confirms this intuition; here is its rough translation:

As c becomes really far away from E(X), the event X \geq c becomes less probable.

You can confirm this by substituting several key values of c.

 

  • If c = E(X), then P[X \geq E(X)] \leq 1; this is the highest upper bound that P(X \geq c) can get.  This makes intuitive sense; X is going to be frequently observed near its own expected value.

 

  • If c \rightarrow \infty, then P(X \geq \infty) \leq 0.  By Kolmogorov’s axioms of probability, any probability must be inclusively between 0 and 1, so P(X \geq \infty) = 0.  This makes intuitive sense; there is no possible way that X can be bigger than positive infinity.

Mathematical Statistics Lesson of the Day – Markov’s Inequality

Markov’s inequality is an elegant and very useful inequality that relates the probability of an event concerning a non-negative random variable, X, with the expected value of X.  It states that

P(X \geq c) \leq E(X) \div c,

where c > 0.

I find Markov’s inequality to be beautiful for 2 reasons:

  1. It applies to both continuous and discrete random variables.
  2. It applies to any non-negative random variable from any distribution with a finite expected value.

In a later lesson, I will discuss the motivation and intuition behind Markov’s inequality, which has useful implications for understanding a data set.