Markov’s Inequality and Its Proof

Why a Mean Can Bound a Tail Probability

A mean describes an average, not a guarantee about any one outcome. Still, if a random variable cannot be negative, a large value contributes substantially to its mean. This creates a useful relationship: the mean limits how much probability can lie above a positive threshold. Markov’s inequality makes that relationship precise without requiring a particular distribution.

The key is to compare two quantities outcome by outcome. When $X$ reaches a threshold $a$, the indicator of that event is 1, while $X/a$ is at least 1. When $X$ is below $a$, the indicator is 0 and $X/a$ remains nonnegative. Taking expectations of this pointwise comparison proves the result.

Definition: For an event $A$, its indicator $\mathbf{1}_A$ equals 1 when $A$ occurs and 0 otherwise. A random variable $X$ is nonnegative if $P(X\geq 0)=1$. Markov’s inequality applies when $X$ is nonnegative, $a>0$, and $E[X]$ is finite.

Theorem: If $X\geq 0$ with probability 1, $E[X]<\infty$, and $a>0$, then $$ P(X\geq a)\leq \frac{E[X]}{a}. $$

The right side is an upper bound, not generally the probability itself. The inequality does not require $X$ to be discrete or continuous, and it does not require independence from another random variable. It uses only nonnegativity, the positive threshold, and the mean.

Compare the indicator and the scaled variable.
On the event $X\geq a$, $X/a\geq 1$. Outside that event, the indicator is 0 and $X/a\geq 0$. Thus $\mathbf{1}_{\{X\geq a\}}\leq X/a$ for every outcome except possibly a probability-zero set.

Take expectations.
Expectation preserves pointwise inequalities when the expectations exist. The indicator’s expectation is the probability of its event, and scaling a random variable by $1/a$ scales its expectation by $1/a$.

Identify the two sides.
Since $a>0$, the comparison gives $E[\mathbf{1}_{\{X\geq a\}}]\leq E[X/a]=E[X]/a$. Therefore $P(X\geq a)\leq E[X]/a$.

The proof rests on the general facts $E[\mathbf{1}_A]=P(A)$ and $E[cX]=cE[X]$ for a constant $c$, together with the monotonicity of expectation. To see why monotonicity applies, if $U\leq V$ pointwise, then $V-U\geq 0$, so $E[V-U]\geq 0$ and hence $E[U]\leq E[V]$. Applying that fact to the indicator comparison completes the justification.

\mathbf{1}_{\{X\geq a\}}\leq \frac{X}{a} \quad\Longrightarrow\quad P(X\geq a) =E[\mathbf{1}_{\{X\geq a\}}] \leq E\!\left[\frac{X}{a}\right] =\frac{E[X]}{a}.

The nonnegativity condition matters at every outcome below the threshold: it ensures $X/a\geq 0$, so that $X/a$ still bounds the indicator, which is then zero. If $X$ can be negative, the pointwise comparison can fail and the theorem need not hold. The condition $a>0$ is also essential because the proof divides by $a$ and uses its sign.

Worked Applications of the Bound

Worked Example: A Discrete Benefit Amount

Model and assumptions. Consider a hypothetical one-year repair benefit $X$, measured in dollars. Its distribution is $P(X=0)=0.50$, $P(X=200)=0.25$, $P(X=500)=0.20$, and $P(X=1{,}000)=0.05$. These probabilities sum to 1, and $X$ is nonnegative. There is no interest discounting or mortality assumption in this single-payment model. Find the Markov bound for $P(X\geq 500)$, then compare it with the exact probability.

Method. Compute the mean from the probability mass function, use $a=500$, and apply Markov’s inequality. The event includes the value 500 because the inequality uses “greater than or equal to.”

Mean. $$ E[X]=(0)(0.50)+(200)(0.25)+(500)(0.20)+(1{,}000)(0.05) =0+50+100+50=200. $$

The assumptions hold: $X\geq 0$, $a=500>0$, and the finite distribution gives $E[X]=200<\infty$. Therefore $$ P(X\geq 500)\leq \frac{200}{500}=0.40. $$ The exact probability, calculated from the two outcomes at least 500, is $P(X\geq 500)=0.20+0.05=0.25$. The bound is valid because $0.25\leq 0.40$, but it is not exact.

Check and common mistake avoided. The four probabilities sum to 1, and the weighted values sum to 200 dollars. A common mistake is to treat the bound $0.40$ as the requested tail probability. The exact probability is $0.25$; Markov supplies only an upper bound unless additional distribution information is used.

Worked Example: A Continuous Loss Amount

Model and assumptions. A hypothetical covered loss $X$, in dollars, has a uniform distribution on $[0,1{,}200]$, with density $f_X(x)=1/1{,}200$ for $0\leq x\leq 1{,}200$, and zero otherwise. Thus the distribution parameter is the interval $[0,1{,}200]$. There is no interest discounting or mortality assumption. Find a Markov bound and the exact probability for a loss of at least 900 dollars.

Method. First integrate $x f_X(x)$ over the support to obtain the mean; then use $a=900$. The distribution’s support confirms that $X$ is nonnegative.

Mean. $$ E[X]=\int_0^{1{,}200}x\frac{1}{1{,}200}\,dx =\frac{1}{1{,}200}\left[\frac{x^2}{2}\right]_0^{1{,}200} =\frac{1{,}440{,}000}{2{,}400} =600. $$ The mean is 600 dollars. Since $a=900>0$ and the mean is finite, Markov’s inequality gives $$ P(X\geq 900)\leq \frac{600}{900}=\frac{2}{3}\approx 0.6667. $$ For comparison, the exact probability is the length of the interval from 900 to 1,200 divided by the total interval length: $$ P(X\geq 900)=\frac{1{,}200-900}{1{,}200} =\frac{300}{1{,}200}=0.25. $$ Thus the exact probability, 0.2500, is below the Markov bound, approximately 0.6667.

Check and common mistake avoided. The integral gives a mean of 600, not the upper endpoint 1,200. Substituting the endpoint into the inequality would produce a different and unjustified bound; the numerator must be $E[X]$. The comparison $0.25\leq 0.6667$ checks the result.

Worked Example: Keeping Loss Units Consistent

Model and assumptions. A hypothetical reimbursement $X$ is recorded in thousands of dollars. Its distribution is $P(X=0)=0.40$, $P(X=2)=0.30$, $P(X=5)=0.20$, and $P(X=10)=0.10$; these values and probabilities fully specify the model. There is no interest discounting or mortality assumption. Find a Markov bound for a reimbursement of at least 6,000 dollars and compare it with the exact probability.

Method. Because $X$ is measured in thousands, express the dollar threshold in the same units before applying the inequality. Thus 6,000 dollars corresponds to $a=6$, not $a=6{,}000$.

Mean and bound. $$ E[X]=(0)(0.40)+(2)(0.30)+(5)(0.20)+(10)(0.10) =0+0.60+1.00+1.00=2.60 $$ in thousands of dollars. The model is nonnegative, the mean is finite, and $a=6>0$, so $$ P(X\geq 6)\leq \frac{2.60}{6} =0.4333\overline{3}. $$ The exact probability is $P(X\geq 6)=P(X=10)=0.10$, since the only listed value at least 6 is 10. Rounded to four decimals, the Markov bound is 0.4333 and the exact probability is 0.1000.

Check and common mistake avoided. The mean and threshold are both in thousands of dollars, so their ratio is dimensionless as a probability must be. Using $6{,}000$ as the threshold while leaving the mean as $2.60$ thousands would mix units and give a meaningless calculation. The exact probability, 0.1000, is no greater than the bound, 0.4333.

What the Inequality Does—and Does Not—Say

Markov’s inequality is useful when a nonnegative variable’s mean is known but its full distribution is not. It gives a distribution-free upper limit on the probability of reaching a chosen positive level. A larger mean allows a larger upper bound; a larger threshold, with the mean fixed, makes the bound smaller. These statements describe the bound, not necessarily the exact behavior of the tail.

If $E[X]/a$ exceeds 1, the inequality remains mathematically true, but it may provide no useful restriction beyond the fact that a probability is at most 1. Even when the ratio is below 1, the inequality can be conservative, as the examples show. A tight or exact tail probability cannot be inferred from the mean alone.

Equality has a restrictive condition. In the indicator proof, equality of expectations can hold only if the pointwise difference $X/a-\mathbf{1}_{\{X\geq a\}}$ is zero with probability 1. For $X<a$, that requires $X=0$; for $X\geq a$, it requires $X=a$. Thus equality occurs precisely when $X$ takes values only in $\{0,a\}$ with probability 1. Most distributions have outcomes that make the inequality strict.

Common exam trap: Do not apply Markov’s inequality to a variable that can be negative, and do not confuse a tail probability with its upper bound. For example, if $X=-1$ or $X=1$, each with probability $0.5$, then $E[X]=0$. With $a=0.5$, $P(X\geq a)=0.5$, so the proposed bound $P(X\geq a)\leq E[X]/a=0$ is false. The error is not in the arithmetic: this $X$ is not nonnegative, so the theorem does not apply.

Before using the result, check three items: $X$ is nonnegative with probability 1, the threshold $a$ is strictly positive, and $E[X]$ is finite. Then keep the units of $E[X]$ and $a$ consistent, and report the result as an upper bound. These checks protect both the proof’s logic and the calculation.

Check Your Understanding

For each problem, identify the assumptions needed and distinguish the Markov bound from any exact probability requested.

A nonnegative random variable has $E[X]=18$. For $a=30$, what upper bound does Markov’s inequality give for $P(X\geq 30)$? (Answer: $P(X\geq 30)\leq 0.6000$.)
A hypothetical discrete benefit $X$, in dollars, equals 0, 4, or 12 with probabilities 0.50, 0.30, and 0.20, respectively. Find the Markov bound and the exact probability for $X\geq 10$. (Answer: $E[X]=3.6$, Markov bound $0.3600$, exact probability $0.2000$.)
A hypothetical loss is uniform on $[0,40]$, with density $1/40$ on that interval and zero elsewhere. Find the mean, the Markov bound for a loss of at least 30, and the exact probability. (Answer: $E[X]=20$, Markov bound $2/3\approx 0.6667$, exact probability $0.2500$.)
A nonnegative discrete variable takes values 0, 1, 5, and 9 with probabilities 0.20, 0.40, 0.30, and 0.10. Find $E[X]$, then give the Markov bound and exact probability for $X\geq 5$. (Answer: $E[X]=2.8$, Markov bound $0.5600$, exact probability $0.4000$.)
A hypothetical reimbursement $X$ is measured in hundreds of dollars and equals 0, 3, or 8 with probabilities 0.60, 0.25, and 0.15. For a threshold of 500 dollars, give the Markov bound and the exact probability. (Answer: $a=5$ hundreds, $E[X]=1.95$ hundreds, Markov bound $0.3900$, exact probability $0.1500$.)

Markov’s Inequality and Its Proof

What You'll Learn

Why a Mean Can Bound a Tail Probability