Using Markov’s Inequality for Probability Bounds

From an Expected Amount to a Tail Bound

An expectation describes a long-run average, not the probability of any particular large outcome. Still, if a random quantity cannot be negative, a large outcome must contribute substantially to its average. Markov’s inequality turns that observation into a general upper bound on the probability of exceeding a chosen threshold. It requires only nonnegativity and a finite expectation; it does not require a particular distributional model.

For an actuarial application, imagine a hypothetical insurer modeling a nonnegative annual payment total. If its expected value is known but its full distribution is not, Markov’s inequality can give a conservative bound on the chance that payments exceed a specified amount. The bound may be informative, or it may be too large to guide a decision. Interpreting its size is part of using it correctly.

Definition: Let $X$ be a nonnegative random variable with finite expectation $E[X]$, and let $a>0$. Markov’s inequality states that $$ P(X\geq a)\leq \frac{E[X]}{a}. $$ Since a probability cannot exceed 1, a convenient capped version is $$ P(X\geq a)\leq \min\left(1,\frac{E[X]}{a}\right). $$ The same upper bound applies to $P(X>a)$, because $\{X>a\}$ is contained in $\{X\geq a\}$. The ratio and bound are meaningful as a tail bound only for a nonnegative $X$, a finite $E[X]$, and a positive threshold $a$.

The capped form is often the most useful to report. If $E[X]/a$ is greater than 1, that ratio alone is not a probability-sized bound; the statement $P(X\geq a)\leq 1$ is true but conveys no information. If the ratio is below 1, it gives a nontrivial bound. A nontrivial bound is not necessarily a close estimate of the actual probability.

Why the Inequality Holds

The key is a pointwise comparison. Whenever $X\geq a$, the value of $X$ is at least $a$; outside that event, the indicator of the event is zero. Thus $X$ is at least $a$ times the indicator of the event. Taking expectations preserves the inequality.

X\geq a\,\mathbf{1}_{\{X\geq a\}} \quad\Longrightarrow\quad E[X]\geq aE[\mathbf{1}_{\{X\geq a\}}] =aP(X\geq a).

Dividing by $a>0$ gives $P(X\geq a)\leq E[X]/a$. This argument explains why nonnegativity matters: if $X$ can be negative, values outside the event may reduce the expectation, so the pointwise comparison need not hold. For a strict-tail event, the same reasoning applies because $X\geq a$ whenever $X>a$; alternatively, event inclusion gives $P(X>a)\leq P(X\geq a)$.

Worked Applications

Worked Example: Bounding a Hypothetical Annual Payment Total

A hypothetical benefit program models its total payments in one year by a nonnegative random variable $X$, measured in dollars. The model specifies $E[X]=\$1{,}800$, but gives no further distributional information. Find an upper bound on the probability that payments are at least $\$7{,}500$, and assess what happens if the threshold is raised to $\$9{,}000$.

Method and conditions. $X$ is nonnegative, its expectation is finite, and each threshold is positive. Markov’s inequality therefore applies. For $a=\$7{,}500$,

P(X\geq \$7{,}500) \leq \frac{\$1{,}800}{\$7{,}500} =0.24 =0.2400.

Thus the model guarantees only that the probability is at most $0.2400$, or 24.00%. For the higher threshold,

P(X\geq \$9{,}000) \leq \frac{\$1{,}800}{\$9{,}000} =0.20 =0.2000.

The units cancel in each ratio, as they must for a probability. The larger threshold produces a smaller uncapped ratio, and here both ratios are already below 1, so the capped bounds also decrease. Common mistake avoided: treating the bound as an estimate of the actual probability. The information given establishes an upper limit, not the actual tail probability.

Check. Both reported bounds lie between 0 and 1, and the bound for the higher threshold is smaller, consistent with the nested events $\{X\geq \$9{,}000\}\subseteq\{X\geq \$7{,}500\}$.

Worked Example: Computing the Expectation Before Bounding a Tail

For a hypothetical claim payment $X$, measured in thousands of dollars, a model assigns the following probabilities: $P(X=0)=0.50$, $P(X=2)=0.30$, $P(X=8)=0.15$, and $P(X=20)=0.05$. Find a Markov upper bound on $P(X>\$10{,}000)$. Then use the given distribution to compare the bound with the actual probability.

Step 1: Calculate the expectation. The values are in thousands of dollars, so the weighted sum is also in thousands of dollars.

$$ E[X]=(0)(0.50)+(2)(0.30)+(8)(0.15)+(20)(0.05) =0+0.60+1.20+1.00 =2.80. $$

Therefore $E[X]=\$2{,}800$. The event $X>\$10{,}000$ is $X>10$ in the stated units. Since it is contained in $X\geq 10$, Markov’s inequality gives

P(X>\$10{,}000) \leq \frac{2.80}{10} =0.28 =0.2800.

Step 2: Compare with the actual probability. Of the four possible values, only $X=20$ is greater than 10. Hence the actual probability from the supplied distribution is $0.05=0.0500$. The Markov bound is valid but substantially conservative: $0.2800$ is 5.6 times $0.0500$.

Common mistake avoided: including the outcome $X=8$ in the event $X>10$. It is below the threshold, and the strict inequality excludes it. Check. The expectation calculation includes every possible value with its probability, and the probabilities sum to $0.50+0.30+0.15+0.05=1.00$.

Worked Example: Interpreting the Capped Bound and Choosing a Threshold

A hypothetical reserve requirement $X$, measured in thousands of dollars, is nonnegative and has $E[X]=18$. Calculate the capped Markov bounds for $P(X\geq 12)$, $P(X\geq 15)$, and $P(X\geq 24)$. Then find a threshold that makes the Markov upper bound no greater than 0.10.

Method. For each positive threshold $a$, calculate $18/a$ and cap the result at 1.

\begin{aligned} a=12:&\quad \min(1,18/12)=\min(1,1.5)=1,\\ a=15:&\quad \min(1,18/15)=\min(1,1.2)=1,\\ a=24:&\quad \min(1,18/24)=\min(1,0.75)=0.75. \end{aligned}

The corresponding bounds are $1.0000$, $1.0000$, and $0.7500$. The first two are valid but uninformative: each says only that a probability is at most 1. To make the bound no greater than 0.10, solve

\frac{18}{a}\leq 0.10 \quad\Longleftrightarrow\quad 18\leq 0.10a \quad\Longleftrightarrow\quad a\geq 180.

Thus a threshold of at least 180 thousand dollars is required for this method to give an upper bound of 0.1000 or less. At $a=180$, the bound is exactly $18/180=0.1000$.

Common mistake avoided: claiming that the capped bound must strictly decrease whenever the threshold increases. The uncapped ratio $E[X]/a$ decreases as $a$ increases, but the capped bound $\min(1,E[X]/a)$ is only nonincreasing: it can remain at 1 until $a$ exceeds $E[X]$. Here the thresholds 12 and 15 both give a capped bound of 1. Check. At 24, the threshold exceeds the expectation, so the ratio is below 1 and the cap no longer changes it.

Using the Bound Carefully

The value of Markov’s inequality is that it needs little information. If a nonnegative quantity has a known finite mean, the inequality gives a valid tail bound even when its distribution is unknown. That same generality explains why it can be loose: many distributions with the same expectation can have very different tail probabilities. When a full distribution is supplied and exact calculation is appropriate, the exact probability can be more informative than a general bound.

The uncapped ratio $E[X]/a$ decreases as the positive threshold $a$ increases while the expectation stays fixed. The capped bound $\min(1,E[X]/a)$, however, is nonincreasing rather than necessarily strictly decreasing. It stays equal to 1 whenever $a\leq E[X]$, then decreases as $a$ rises above $E[X]$. Also, increasing the threshold cannot increase the actual tail probability: the event at the higher threshold is contained in the event at the lower threshold.

A Markov bound is most immediately informative when the threshold is appreciably larger than the expectation. If $a\leq E[X]$, the capped bound is 1 and gives no nontrivial restriction. If $a>E[X]$, the bound is below 1; the larger the threshold relative to the mean, the smaller the bound. Even then, a bound such as 0.60 may not be useful for a decision that requires a much smaller probability.

The expectation can also come from a sum of nonnegative costs. If $X=X_1+X_2+\cdots+X_n$, each component is nonnegative, and each expectation is finite, linearity of expectation gives $E[X]=\sum_{j=1}^{n}E[X_j]$. Markov’s inequality can then be applied to $X$. No independence assumption is required for this expectation calculation or for the resulting bound.

Common Exam Traps

Ignoring nonnegativity. A finite expectation alone is not enough. Markov’s inequality in this form requires $X\geq 0$.
Using a zero or negative threshold. The ratio requires $a>0$. Check the threshold before substituting.
Reporting a value above 1 as a useful probability bound. Use $\min(1,E[X]/a)$; if the result is 1, recognize that it is uninformative.
Mixing units. Convert the expectation and threshold to the same units before dividing. For example, divide thousands of dollars by thousands of dollars, not by dollars.
Calling the upper bound the actual probability. Markov’s inequality establishes a ceiling, not an equality in general.
Claiming strict decrease for the capped bound. The capped bound is nonincreasing and can remain at 1 over a range of thresholds.

Key takeaway: For a nonnegative random variable $X$ with finite expectation and a positive threshold $a$, $P(X\geq a)\leq \min(1,E[X]/a)$. Calculate the ratio in consistent units, apply the cap, and judge whether the resulting upper bound is informative for the question at hand.

Check Your Understanding

For each problem, identify the expectation and threshold carefully before applying Markov’s inequality. Answers are given after each question for checking.

A hypothetical nonnegative annual expense $C$ has $E[C]=\$240$. Find a Markov upper bound for $P(C>\$600)$. (Answer: $0.4000$.)
A nonnegative payment $Y$, measured in thousands of dollars, has $E[Y]=3.6$. Give a Markov upper bound for the probability that the payment is at least $\$9{,}000$. (Answer: $0.4000$.)
A nonnegative random variable $W$ has $E[W]=14$. Find the capped Markov bounds for $P(W\geq 10)$ and $P(W\geq 20)$. (Answer: $1.0000$ and $0.7000$, respectively.)
A hypothetical total cost is $T=A+B+C$, where $A,B,C$ are nonnegative and have expectations 2.5, 4.0, and 1.5 thousand dollars. Find a Markov upper bound for $P(T\geq \$20{,}000)$, and state whether independence is needed. (Answer: $0.4000$; independence is not needed.)
A nonnegative payment $Z$, measured in thousands of dollars, equals 0, 5, or 15 with respective probabilities 0.60, 0.30, and 0.10. Calculate $E[Z]$, then find the Markov bound and the exact probability for $P(Z\geq 10)$. (Answer: $E[Z]=3$ thousand dollars; Markov bound $0.3000$; exact probability $0.1000$.)

What You'll Learn