Mathematics.

inequalities

Markov's Inequality

Probability25 minDifficulty5 out of 10

You should know: expectation, random variables

Overview

Markov's inequality is one of the simplest and most fundamental tail bounds in probability theory. It states that for any non-negative random variable X with finite expectation E[X], the probability that X exceeds any positive threshold a is at most E[X]/a. Despite its simplicity, Markov's inequality is the foundation from which stronger inequalities — such as Chebyshev's inequality and Chernoff bounds — are derived. It requires almost no assumptions beyond non-negativity and finite mean.

Intuition

If the average household income is $60,000, then at most 1 in 6 households can earn $360,000 or more (since 60000/360000 = 1/6). If more than 1/6 of households earned that much, the average would have to exceed $60,000 — a contradiction. Markov's inequality formalises this simple averaging argument: a distribution cannot concentrate too much mass far above its mean without raising the mean itself.

Formal Definition

Definition

Let X be a non-negative random variable with E[X] < ∞. For any a > 0:

P(Xa)E[X]aP(X \geq a) \leq \frac{E[X]}{a}
Markov's Inequality
P(XkE[X])1kfor k>0P(X \geq k \cdot E[X]) \leq \frac{1}{k} \quad \text{for } k > 0
Equivalent form: probability of exceeding k times the mean

Worked Examples

  1. 1

    Apply Markov's inequality directly with a = 12.

    P(X12)E[X]a=412=13P(X \geq 12) \leq \frac{E[X]}{a} = \frac{4}{12} = \frac{1}{3}

✓ Answer

P(X ≥ 12) ≤ 1/3.

Practice Problems

Easyfree response

A non-negative random variable X satisfies E[X] = 5. Find an upper bound for P(X ≥ 25).

Mediumfree response

For a non-negative random variable with mean μ, what is the maximum probability that X exceeds 10μ?

Mediumfree response

A random variable Y ≥ 0 has E[Y] = 3. Can P(Y ≥ 9) = 0.5? Justify using Markov's inequality.

Quiz

Markov's inequality applies to:
If E[X] = 6 and X ≥ 0, what does Markov's inequality give for P(X ≥ 30)?
Markov's inequality is used as a stepping stone to prove which stronger result?

Summary

  • For any non-negative random variable X with finite mean, P(X ≥ a) ≤ E[X]/a for all a > 0.
  • The inequality requires only non-negativity and a finite mean — no distributional assumptions.
  • Markov's inequality is the building block for Chebyshev's inequality and Chernoff bounds.

References