Markov's Inequality from Threshold Indicators

From a Threshold Event to a Pointwise Bound

A probability such as $P(X\ge t)$ concerns only whether a random variable crosses a threshold. A useful way to connect that event to the distribution of $X$ is to represent it by an indicator. When $X$ is nonnegative, crossing a positive threshold forces $X$ to be at least as large as that threshold. This simple pointwise fact leads to a general upper bound on the event's probability using only the expectation of $X$.

Throughout, $X$ is a measurable, real-valued random variable, $X\ge 0$ almost surely, and $E[X]<\infty$. The threshold $t$ is a fixed real number with $t>0$. Finiteness of the expectation ensures that the bound below is finite; nonnegativity is essential to the pointwise comparison.

Definition: For an event $A$, its indicator is the random variable $1_A$, equal to $1$ when $A$ occurs and $0$ otherwise. In particular, $1_{\{X\ge t\}}$ indicates whether $X$ is at least $t$. Its expectation is $E[1_{\{X\ge t\}}]=P(X\ge t)$.

On the event $\{X\ge t\}$, multiplying the indicator by $t$ gives $t$, which is no greater than $X$. Off that event, the product is zero, which is no greater than $X$ because $X\ge0$. Thus, except possibly on a probability-zero set,

t1_{\{X\ge t\}}\le X.

Taking expectations preserves this inequality. The expectation on the left is $tP(X\ge t)$, so the comparison immediately controls the probability of the threshold event. The next theorem states the resulting bound and proves it explicitly.

Theorem (Markov's inequality): If $X\ge0$ almost surely and $E[X]<\infty$, then for every $t>0$, $$ P(X\ge t)\le \frac{E[X]}{t}. $$

Proof and Interpretation

The event $\{X\ge t\}$ is measurable because $X$ is a random variable and $[t,\infty)$ is a Borel set. Its indicator is therefore measurable. The random variables in the pointwise comparison are integrable: $X$ is integrable by assumption, and $t1_{\{X\ge t\}}$ is bounded. Monotonicity of expectation now gives the proof.

Compare pointwise.
On $\{X\ge t\}$, $t1_{\{X\ge t\}}=t\le X$. On $\{X<t\}$, $t1_{\{X\ge t\}}=0\le X$. Hence $t1_{\{X\ge t\}}\le X$ almost surely.

Take expectations.
By monotonicity and the indicator expectation identity, $tP(X\ge t)=E[t1_{\{X\ge t\}}]\le E[X]$.

Divide by the positive threshold.
Since $t>0$, division preserves the inequality and gives $P(X\ge t)\le E[X]/t$. $\square$

The theorem needs no particular distributional form. It uses only a nonnegative random variable and its expectation. It may therefore be useful when the expectation is known but the exact tail probability is difficult or impossible to calculate. The result is an upper bound, not an assertion that the tail probability is close to the bound.

Worked Example: A Continuous Exponential Distribution

Let $X$ have an exponential distribution with rate $1/2$, so its density is $f(x)=\tfrac12 e^{-x/2}$ for $x\ge0$. Find the Markov bound for $P(X\ge5)$, then compare it with the exact probability.

The mean of an exponential random variable with rate $1/2$ is $1/(1/2)=2$. Markov's inequality gives

P(X\ge5)\le\frac{E[X]}{5}=\frac{2}{5}=0.4.

For comparison, the exact survival probability is

P(X\ge5)=\int_5^\infty \frac12 e^{-x/2}\,dx =\left[-e^{-x/2}\right]_5^\infty =e^{-5/2}\approx0.0821.

The exact probability is below the bound, as required, but it is considerably smaller. Markov's inequality guarantees a valid upper bound from the mean alone; it does not use the exponential distribution's additional structure.

When Is the Bound Informative?

Every probability is at most $1$. Markov's inequality supplies a bound strictly below $1$ precisely when $E[X]/t<1$, or equivalently when $t>E[X]$. If $E[X]/t\ge1$, the theorem remains correct, but the usual probability bound $P(X\ge t)\le1$ is at least as strong. One can combine the two statements as $P(X\ge t)\le\min\{1,E[X]/t\}$.

Thus, saying that Markov's bound is informative will mean here that it is strictly less than $1$. This criterion is about whether the bound improves on the universal bound for a probability; it does not guarantee that the resulting numerical estimate is close to the true probability. When $E[X]=0$, the Markov bound is zero for every $t>0$, so it also implies $P(X\ge t)=0$ at every positive threshold.

Worked Example: A Discrete Distribution and the Bound's Slack

Suppose $X$ takes the values $0$, $3$, and $8$, with probabilities $0.5$, $0.3$, and $0.2$, respectively. Calculate $P(X\ge4)$, its Markov bound, and the difference between the two sides of the expectation comparison.

First calculate the expectation:

$$ E[X]=0(0.5)+3(0.3)+8(0.2)=0+0.9+1.6=2.5. $$

Only the outcome $X=8$ is at least $4$, so $P(X\ge4)=0.2$. Markov's inequality gives

P(X\ge4)\le\frac{2.5}{4}=0.625,

which is valid, though not tight. The pointwise comparison is $4\,1_{\{X\ge4\}}\le X$. Its slack, $X-4\,1_{\{X\ge4\}}$, equals $0$ when $X=0$, $3$ when $X=3$, and $4$ when $X=8$. Its expectation is

$$ 0(0.5)+3(0.3)+4(0.2)=0.9+0.8=1.7. $$

This equals $E[X]-4P(X\ge4)=2.5-4(0.2)=1.7$. The positive slack explains why the expectation comparison, and therefore the probability bound, is not an equality.

Equality and a Sharp Example

The pointwise comparison also identifies exactly when Markov's inequality is an equality. Let $Z=X-t1_{\{X\ge t\}}$, the nonnegative difference between the two sides of that comparison. Since $X$ is integrable and the indicator is bounded, $Z$ is integrable. The difference between the two sides after taking expectations is $E[Z]$. For a nonnegative random variable, expectation zero is equivalent to being zero almost surely: if $P(Z>0)>0$, then at least one event $\{Z\ge1/k\}$, for a positive integer $k$, has positive probability, since $\{Z>0\}=\bigcup_{k=1}^\infty\{Z\ge1/k\}$. That would give $E[Z]\ge P(Z\ge1/k)/k>0$, a contradiction.

Consequently, equality in Markov's inequality holds exactly when $Z=0$ almost surely. On $\{X\ge t\}$, this requires $X=t$; on $\{X<t\}$, it requires $X=0$. The converse follows because those two values make $Z=0$ in both cases.

Equality condition: For $t>0$, $P(X\ge t)=E[X]/t$ if and only if $X$ takes values only in $\{0,t\}$ almost surely. In other words, the distribution places all its probability at zero and at the threshold.

Worked Example: Attaining the Markov Bound

Let $X=10$ with probability $0.2$, and let $X=0$ with probability $0.8$. Apply Markov's inequality at $t=10$, and verify equality.

The expectation is

$$ E[X]=10(0.2)+0(0.8)=2. $$

The event $\{X\ge10\}$ occurs exactly when $X=10$, so $P(X\ge10)=0.2$. The Markov bound is

\frac{E[X]}{10}=\frac{2}{10}=0.2=P(X\ge10).

Equality holds because $X$ takes only the values $0$ and $10$. This example shows that Markov's bound can be attained, so without more information about the distribution, it cannot generally be improved.

Common Mistakes and Practical Use

A common error is to treat the threshold comparison as though it held for every random variable. For example, someone might argue that $t1_{\{X\ge t\}}\le X$ and immediately apply Markov's inequality even when $X$ can be negative. That reasoning fails off the event if $X<0$, because then the indicator term is zero while $X$ is negative. The correction is to check $X\ge0$ almost surely before using this comparison.

Another mistake is to reverse the inequality or to assume it gives an exact tail probability. The pointwise comparison says that the threshold contribution cannot exceed $X$, so it provides an upper bound on $P(X\ge t)$. It does not generally determine that probability. As the exponential example shows, the bound may be quite loose; as the two-point example shows, it can also be exact.

In applications, identify the nonnegative quantity, its finite expectation, and the positive threshold before substituting into the formula. Then check whether $t>E[X]$. If not, the result may still be mathematically true, but it does not improve on the basic probability bound of one. If the threshold does exceed the mean, the ratio $E[X]/t$ quantifies how small the probability must be according to this inequality.

Check Your Understanding

Use the indicator comparison and the theorem to answer these questions.

State the pointwise inequality that leads to Markov's inequality, and explain where nonnegativity is used.
If $E[X]=6$, for which positive thresholds is the Markov bound strictly less than one?
A nonnegative random variable has $E[X]=3$. What upper bound does Markov's inequality give for $P(X\ge12)$?
For a fixed $t>0$, what values can $X$ take almost surely if equality holds in Markov's inequality?
Why is $P(X\ge t)\le E[X]/t$ valid but uninformative when $E[X]/t>1$?