Generalized Markov Inequality¶

Theorem¶

Let $f$ be nonnegative and nondecreasing on a domain containing the values of $X$ and the threshold $a$. If $f(a)>0$ and $\mathbb{E}[f(X)]<\infty$$\infty$, then

This follows from $f(X)\ge f(a)\indicator_{\{X\ge a\}}$$f(a){\text{𝟙}}_{\{X\ge a\}}$. For the ordinary Markov inequality $\mathbb{P}(X\ge a)\le\mathbb{E}[X]/a$$\mathbb{E}[X]/a$, take $f(x)=x$ on $[0,\infty)$ and require $X\ge0$$0$, $a>0$$0$.

The Inequality gives an upper bound, not an exact tail probability: the event where a nonnegative variable exceeds $a>0$ contributes at least $a$ times its probability to the expectation.

Interpretation¶

A nonnegative variable with a small average cannot often be large. With mean $10$, at most a fifth of the probability can lie at $50$ or above, because that part alone would contribute at least $50\,\mathbb{P}(X\ge50)$ to the mean. The generalized version applies the same reasoning to $f(X)$ instead of $X$.

The figure compares $f$ with the step function $f(a)\indicator_{\{x\ge a\}}$, which is $0$ below $a$ and $f(a)$ from $a$ onwards. Both $f(x)=x$ and the steeper dashed $f(x)=x^2/a$ have the value $a$ at $x=a$ and lie on or above the step everywhere. Taking expectations of $f(X)\ge f(a)\indicator_{\{X\ge a\}}$ gives $\mathbb{E}[f(X)]\ge f(a)\,\mathbb{P}(X\ge a)$$a)$, which is the theorem.

Each hypothesis is used in that one comparison: nonnegativity of $f$ covers $x<a$$a$, where the step is $0$, and monotonicity covers $x\ge a$$a$, where $f(x)\ge f(a)$$f(a)$. Without nonnegativity the bound fails: if $X$ takes the values $-1$ and $1$ with probability $\tfrac{1}{2}$ each, then $\mathbb{E}X/1=0$ but $\mathbb{P}(X\ge1)=\tfrac{1}{2}$$\frac{1}{2}$.

Different choices of $f$ give the other tail bounds: $f(x)=x^2$ applied to $|X-\mu|$ gives the Chebyshev inequality, and $f(x)=\mathrm{e}^{tx}$ with $t>0$ gives the Chernoff bound. No single choice is always best. For $X\sim\operatorname{Exp}(1)$$\mathrm{Exp}(1)$, with $\mathbb{E}X=1$ and $\mathbb{E}[X^2]=2$$2$, the first-moment bound $1/a$ is smaller for $a<2$ and the second-moment bound $2/a^2$ is smaller for $a>2$$2$.

Key Results¶

• For $X\ge0$ and $a>0$$0$, $\mathbb{P}(X\ge a)\le\mathbb{E}X/a$$\mathbb{E}X/a$.
• For $X\ge0$$0$, $a>0$ and $r>0$$0$, $\mathbb{P}(X\ge a)\le\mathbb{E}[X^r]/a^r$ whenever the moment is finite.

Examples¶

• A nonnegative loss with mean $10$ satisfies $\mathbb{P}(X\ge50)\le10/50=0.2$$0.2$.
• If $X\ge0$ and $\mathbb{E}[X^2]=9$$9$, then $\mathbb{P}(X\ge6)\le9/36=1/4$$1/4$.

Prerequisites¶

• Expectation
• Inequality

Related Concepts¶

• Chebyshev Inequality

Sources¶

• Nicky van Foreest · Probability Distributions · Week 7

Aliases: Markov inequality, generalised Markov inequality, Markov bound Keywords: Markov's inequality Symbols: infinity, mu

Last updated: 27 October 2026 11:52