Markov's Inequality and Its Sharpness

Why a Mean Can Bound a Tail

A mean describes the overall size of a nonnegative random variable, but it does not specify how its probability is distributed. Even so, if a variable has a small mean, it cannot put too much probability far above zero: enough probability at or above a high threshold would itself contribute substantially to the mean. Markov’s inequality makes that observation precise. Its value is not that it always gives a close estimate, but that it requires very little information about the distribution.

Definition: A random variable \(X\) is nonnegative if \(P(X\geq 0)=1\). It is integrable here if \(E[X]<\infty\). For a nonnegative integrable \(X\) and a number \(t>0\), the event \(\{X\geq t\}\) is the event that \(X\) reaches or exceeds the threshold \(t\).

The threshold must be positive, and nonnegativity is essential to the result. The finite-mean condition ensures the right-hand side is a finite quantity. Under these conditions, the inequality is

Theorem (Markov’s inequality): If \(X\geq 0\) almost surely, \(E[X]<\infty\), and \(t>0\), then

P(X\geq t)\leq \frac{E[X]}{t}.

Proof by a Pointwise Comparison

Write \(\mathbf{1}_{\{X\geq t\}}\) for the indicator of the threshold event: it is 1 when \(X\geq t\), and 0 otherwise. Nonnegativity gives the pointwise inequality

X\geq t\mathbf{1}_{\{X\geq t\}}.

Indeed, when \(X\geq t\), the right side is \(t\) and \(X\) is at least \(t\); when \(X<t\), the right side is 0 and \(X\geq 0\). Taking expectations preserves the inequality. Since the expectation of an indicator is the probability of its event,

E[X]\geq tE[\mathbf{1}_{\{X\geq t\}}] =tP(X\geq t).

Dividing by \(t>0\) proves Markov’s inequality. The two regions in the pointwise comparison also foreshadow exactly when equality is possible: there must be no excess above \(t\) on the threshold event, and no positive values below \(t\).

When Is the Bound an Equality?

The gap in the pointwise comparison is itself nonnegative:

X-t\mathbf{1}_{\{X\geq t\}} = \begin{cases} X-t, & X\geq t,\\ X, & X<t. \end{cases}

Equality in Markov’s inequality holds exactly when the expectation of this gap is zero. A nonnegative random variable \(Y\) has \(E[Y]=0\) only if \(Y=0\) almost surely: for every positive integer \(k\), \(P(Y\geq 1/k)\leq kE[Y]=0\), and the event \(\{Y>0\}\) is the union of the events \(\{Y\geq 1/k\}\). Thus the gap must vanish almost surely. On \(\{X\geq t\}\), that requires \(X=t\); on \(\{X<t\}\), it requires \(X=0\). Conversely, either value makes the gap zero.

Equality characterization: For \(X\geq0\) almost surely, \(E[X]<\infty\), and \(t>0\), equality \(P(X\geq t)=E[X]/t\) holds if and only if \(X\in\{0,t\}\) almost surely.

This is a distributional condition, not merely a condition on the mean. A variable can have the same mean as an equality case and still have strict inequality if it takes other positive values.

Worked Examples

Worked Example: A Discrete Variable With a Loose Bound

Suppose \(X\) has the distribution

P(X=0)=0.5,\qquad P(X=2)=0.3,\qquad P(X=8)=0.2.

Find the actual probability \(P(X\geq 4)\) and the Markov bound at \(t=4\).

First calculate the mean:

$$ E[X]=(0)(0.5)+(2)(0.3)+(8)(0.2)=0+0.6+1.6=2.2. $$

Only the value 8 reaches 4, so \(P(X\geq4)=0.2\). Markov’s inequality gives

P(X\geq4)\leq\frac{2.2}{4}=0.55.

The bound is valid, since \(0.2\leq0.55\), but it is not an equality. In fact \(X\) takes the value 2 with positive probability, so it is not almost surely supported on \(\{0,4\}\). The mean alone does not capture that distributional detail.

Worked Example: A Continuous Variable

Let \(X\) have the uniform distribution on \([0,4]\), with density \(f(x)=1/4\) for \(0\leq x\leq4\). Compare the actual probability of exceeding or reaching 3 with the Markov bound.

The mean, calculated from the density, is

E[X]=\int_0^4 x\frac{1}{4}\,dx =\frac{1}{4}\left[\frac{x^2}{2}\right]_0^4 =2.

The probability of \(X\geq3\) is the probability in the interval from 3 to 4:

P(X\geq3)=\int_3^4\frac{1}{4}\,dx =\frac{1}{4}.

Markov’s inequality instead gives

P(X\geq3)\leq\frac{E[X]}{3}=\frac{2}{3}.

The actual probability \(1/4\) is below the bound \(2/3\). The inequality applies to continuous as well as discrete distributions; it uses only nonnegativity and the mean, not a particular type of probability function.

Worked Example: Constructing an Equality Case

Let \(X=6\) with probability \(1/4\), and \(X=0\) with probability \(3/4\). Take \(t=6\). Calculate the mean and tail probability.

The mean is

E[X]=(6)\left(\frac14\right)+(0)\left(\frac34\right)=\frac{6}{4}=\frac32.

The event \(\{X\geq6\}\) occurs exactly when \(X=6\), so

P(X\geq6)=\frac14 \qquad\text{and}\qquad \frac{E[X]}{6}=\frac{3/2}{6}=\frac14.

The bound is attained. This distribution takes only the values 0 and \(t\), as the equality characterization requires.

Sharpness When Only the Mean Is Known

Sharpness asks whether a smaller bound could be guaranteed using only the stated information. Suppose the known mean is \(m\), where \(m\geq0\), and the threshold is \(t>0\). Since a probability is at most 1, Markov’s inequality and the probability bound together imply

P(X\geq t)\leq \min\left(1,\frac{m}{t}\right).

This is the strongest possible universal bound when all that is known is that \(X\) is nonnegative and \(E[X]=m\). To verify this claim, consider the two cases. If \(0\leq m\leq t\), define \(X=t\) with probability \(m/t\), and \(X=0\) with probability \(1-m/t\). These are valid probabilities because \(0\leq m/t\leq1\). Then

E[X]=t\frac{m}{t}+0\left(1-\frac{m}{t}\right)=m, \qquad P(X\geq t)=\frac{m}{t}.

Thus the Markov bound is attained. This includes \(m=0\), for which \(X=0\) almost surely, and \(m=t\), for which \(X=t\) almost surely. If \(m>t\), take \(X=m\) with probability 1. This variable has mean \(m\) and \(P(X\geq t)=1\), attaining the probability bound 1. No uniformly smaller upper bound can therefore follow from the mean alone.

For \(m>t\), the unmodified expression \(m/t\) exceeds 1, so Markov’s inequality is true but does not provide a useful probability bound. Including the fact that probabilities cannot exceed 1 gives the sharper form \(\min(1,m/t)\). Sharpness of this universal bound does not mean it will be close for every particular distribution. The uniform example, for instance, had a considerably smaller tail probability than the bound.

Common Pitfalls and Practical Meaning

Ignoring nonnegativity. The pointwise comparison fails for negative values below the threshold: \(X\) need not dominate zero there. The theorem requires \(X\geq0\) almost surely.
Forgetting that a bound can exceed 1. If \(E[X]/t>1\), Markov’s inequality remains correct but gives no improvement on the basic probability bound 1.
Assuming equality just because the mean is known. Equality requires \(X\) to be supported on \(\{0,t\}\) almost surely. Having \(E[X]=m\) alone does not establish that condition.
Changing the threshold event without checking the endpoint. The theorem here concerns \(P(X\geq t)\). For \(P(X>t)\), the same numerical upper bound follows because \(\{X>t\}\subseteq\{X\geq t\}\), but the equality condition need not carry over. In particular, the positive value \(t\) is excluded from the strict-threshold event.

Markov’s inequality is useful when only a mean and nonnegativity are available and a guaranteed upper bound is needed. A weak bound can still be informative if it rules out a large tail probability; a bound above 1 cannot do so. Any sharper conclusion requires additional information about the distribution or other properties of \(X\).

Check Your Understanding

Try these questions using the theorem, its equality condition, and the worked calculations above.

State the assumptions on \(X\) and \(t\) needed to apply Markov’s inequality.
If \(E[X]=3\) and \(X\geq0\), what upper bound does Markov’s inequality give for \(P(X\geq 5)\)?
For a nonnegative \(X\), what must be true about its possible values for equality to hold at a threshold \(t>0\)?
If \(E[X]=8\) and \(t=4\), why is the bound \(E[X]/t\) not informative as a probability bound, and what bound follows after using \(P(X\geq t)\leq1\)?
Construct a two-point distribution with mean \(m\), where \(0<m<t\), that attains Markov’s inequality at threshold \(t\).