Updated 12 September 2026

Markov’s inequality

For any non-negative random variable $X$ and any positive threshold $a$:

A finite mean gives a finite upper bound. A nonnegative variable may instead have infinite expectation, in which case the inequality gives no useful finite constraint. No variance or shape assumption is needed.

What to notice

• The bound is usually loose. Pick Exponential(1) and slide a to 5. The true tail is $e^{-5}\approx 0.0067$. The Markov bound is 0.2 — thirty times too large. That looseness is the price of demanding almost no assumptions about X.
• It can be trivial. Whenever $a<E[X]$, Markov says $P(X\ge a)\le E[X]/a > 1$ — a bound bigger than 1, which says nothing. The demo shows those cases clamped at 1.
• A smaller bound need not be relatively tighter. For Exponential(1), the ratio of the Markov bound to the true tail is exp(a)/a, which grows for a > 1. The absolute bound decreases while its relative error worsens.

Why it matters

Markov is a useful starting point for several concentration inequalities. Apply it to $(X-\mu)^{2}$ and you get Chebyshev. Apply it to $e^{tX}$ and you get the Chernoff bound, which leads to Hoeffding and friends. The moment you square, exponentiate, or otherwise transform $X$, Markov’s bound on the transformed variable becomes a bound on the original.

Proof sketch

Since $X\ge a\cdot \mathbb{1}\{X\ge a\}$ (the indicator is 1 when X is at least a, 0 otherwise), taking expectations on both sides gives $E[X] \ge a\, P(X\ge a)$. Divide by a. That’s the whole argument.

Build the worst case

Suppose the mean is 1 and the threshold is 2. An exponential wait has probability $e^{-2}\approx0.135335$ of reaching two time units. But a variable that equals 2 half the time and 0 otherwise has the same mean and reaches 2 with probability 0.5, exactly the Markov bound. The extra exponential assumption provides information a mean alone cannot supply.

For a general finite mean m and threshold a at least m, put probability m/a at a and the remainder at zero. The mean is still m. When a is below m, the constant variable X = m attains the probability cap 1. The demo constructs these alternatives automatically; none is an exponential density. Atoms on the threshold count in the event ≥.

Break an assumption

Let X be −2 or +2, equally likely. Its mean is zero, but P(X ≥ 2) is 1/2. Applying the displayed formula would falsely give zero. The proof fails at the negative outcome: X is then smaller than the zero-valued indicator term. Applying Markov to |X| or X² is valid, but requires the expectation of that transformed variable, not E[X].

References and next steps

Random Services: properties of expectation states Markov and its indicator argument. The two-point alternatives above provide directly checkable sharpness examples. Continue with Chebyshev to use variance, or Hoeffding to add independent bounded observations.