Markov's Inequality

Motivation

Can we bound how often a non-negative random variable takes large values? Markov’s inequality gives a simple but powerful bound using only the mean.

Statement

If $X$ is a non-negative random variable (i.e., $X \geq 0$ with probability 1) with finite expectation $E[X]$, then for any $a > 0$:

$P(X \geq a) \leq \frac{E[X]}{a}$

Proof

Since $X \geq 0$:

$X \geq a \cdot \mathbb{1}_{X \geq a}$

where $\mathbb{1}_{X \geq a}$ is the indicator of the event $\{X \geq a\}$.

Taking expectations:

$E[X] \geq a \cdot E[\mathbb{1}_{X \geq a}] = a \cdot P(X \geq a)$

Therefore:

$P(X \geq a) \leq \frac{E[X]}{a}$

Visual Intuition

The area under the tail beyond $a$ cannot exceed $E[X]/a$, since otherwise the total area would exceed the mean.

Alternative Form

For non-negative $X$:

$P(X \geq k E[X]) \leq \frac{1}{k}$

Interpretation: At most $1/k$ of the distribution lies at $k$ or more times the mean.

Example

If $E[X] = 10$, then:

$P(X \geq 30) = P(X \geq 3 \cdot E[X]) \leq \frac{1}{3}$

At most one-third of values are at least 30.

Examples

Example 1: Mean 5

A non-negative random variable has mean 5. What can we say about $P(X \geq 10)$?

$P(X \geq 10) \leq \frac{E[X]}{10} = \frac{5}{10} = 0.5$

At most 50% of probability lies at or above 10.

Example 2: Exam Scores

Scores are non-negative with mean 70. What proportion score at least 90?

$P(X \geq 90) \leq \frac{70}{90} \approx 0.78$

At most 78% can score 90 or higher. (The true proportion is likely much smaller, but this is the bound.)

Example 3: Something True for Exponential

If $X \sim \text{Exp}(\lambda)$, then $E[X] = 1/\lambda$.

Markov: $P(X \geq 1) = P(X \geq \lambda E[X]) \leq 1/\lambda$.

Actual: $P(X \geq 1) = e^{-\lambda} \leq \lambda$.

For $\lambda = 1$: Markov gives $P(X \geq 1) \leq 1$. True but useless!

Markov’s bound can be very loose.

Corollary: Bounded Variables

If $X$ is bounded in $[0, M]$, then:

$P(X \leq M - a) \leq \frac{M - E[X]}{a}$

Or equivalently:

$P(X \geq M - a) \geq \frac{E[X] + a - M}{a}$

Limitations

1. Only uses the mean: Much stronger bounds are possible with more information (like variance)
2. Often loose: The bound can be far from the actual probability
3. Non-negative only: Only applies to $X \geq 0$

Why Useful?

Despite being loose, Markov’s inequality is:

• Simple to apply
• Requires only the mean
• Foundation for stronger bounds (Chebyshev)

Summary

Property	Statement
Condition	$X \geq 0$ with finite mean
Inequality	$P(X \geq a) \leq \frac{E[X]}{a}$
Scaled form	$P(X \geq k E[X]) \leq \frac{1}{k}$
Limitation	Can be very loose