Choosing an Inequality for Moment and Tail Bounds

Start With the Quantity You Need to Bound

Expectation inequalities are most useful when chosen to match the target. A tail probability is an expectation of an indicator; a product expectation calls for a product inequality; and an expectation of a function of a random variable may be controlled by convexity. The available information matters just as much: knowing a mean, a variance, a pair of moments, or a support restriction leads to different bounds.

There is no inequality that is automatically strongest in every problem. A sound choice begins by identifying the target, rewriting it in a form an inequality can control, and verifying that the required moments exist. The inequalities below are stated for real-valued random variables; all almost-sure statements are with respect to their underlying probability measure.

A Selection Guide

Target	Useful information	First candidate
$P(Y\ge a)$, with $Y\ge0$	$E[Y]$	Markov or pointwise positivity
$P(\|X-E[X]\|\ge t)$	$\operatorname{Var}(X)$	Chebyshev
$E[\|UV\|]$	Second moments of $U,V$	Cauchy–Schwarz
$E[\|UV\|]$	Conjugate moments of $U,V$	Hölder
$E[\phi(X)]$	$E[X]$ and convexity or concavity	Jensen

The table is a starting point, not a ranking. For example, Markov applied to a square gives Chebyshev’s inequality, and Cauchy–Schwarz is the $p=q=2$ case of Hölder’s inequality. The useful question is which assumptions match the information in the problem.

Definition: If $Y\ge0$ almost surely and $E[Y]<\infty$, then for every $a>0$, $$ P(Y\ge a)\le \frac{E[Y]}{a}. $$ This is Markov’s inequality. If $X$ has finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$, then for every $t>0$, $$ P(|X-\mu|\ge t)\le \frac{\sigma^2}{t^2}. $$ This is Chebyshev’s inequality.

Why Markov and Chebyshev Fit Tail Events

A tail event can be represented by an indicator. If $Y\ge0$, then pointwise, $a\,\mathbf{1}_{\{Y\ge a\}}\le Y$. Taking expectations turns this pointwise comparison into a probability bound, because $E[\mathbf{1}_{\{Y\ge a\}}]=P(Y\ge a)$.

Theorem: For an integrable random variable $Y\ge0$ almost surely and $a>0$, $$ aP(Y\ge a) =aE[\mathbf{1}_{\{Y\ge a\}}] \le E[Y]. $$ Consequently, Markov’s inequality holds.

The inequality follows because $Y$ is at least $a$ on the event $\{Y\ge a\}$, and is nonnegative elsewhere. Thus no negative contribution outside the event can invalidate the comparison.

For Chebyshev’s inequality, apply Markov to the nonnegative variable $(X-\mu)^2$, with threshold $t^2$. Since $E[(X-\mu)^2]=\sigma^2$, this gives

P(|X-\mu|\ge t) =P((X-\mu)^2\ge t^2) \le \frac{E[(X-\mu)^2]}{t^2} =\frac{\sigma^2}{t^2}.

This derivation also explains the choice: use Chebyshev when the available information is a variance and the target is a deviation from the mean. If the relevant mean is not finite, or the variance is infinite, the stated bound does not apply.

Worked Example: A Tail Bound From the Mean

Let $Y$ take the values $0,2,5$ with probabilities $0.5,0.3,0.2$, respectively. Bound $P(Y\ge4)$ using the mean.

First verify the required nonnegativity: every possible value of $Y$ is at least zero. Its expectation is $$ E[Y]=0(0.5)+2(0.3)+5(0.2)=0+0.6+1=1.6. $$ Markov’s inequality with $a=4$ gives $$ P(Y\ge4)\le \frac{1.6}{4}=0.4. $$ In this distribution the event occurs exactly when $Y=5$, so its actual probability is $0.2$; the bound is valid but not exact. Directly, $4\mathbf{1}_{\{Y\ge4\}}\le Y$ for each possible value, which is the positivity argument behind the same bound.

Worked Example: A Deviation Bound From the Variance

Suppose $X\sim N(10,9)$, where the second parameter is the variance. Bound the probability that $X$ is at least $6$ units from its mean.

Here $\mu=10$, $\sigma^2=9$, and $t=6$. The event is $\{|X-10|\ge6\}$. Chebyshev’s inequality gives $$ P(|X-10|\ge6)\le\frac{9}{6^2} =\frac{9}{36} =\frac14. $$ Equivalently, Markov applied to $(X-10)^2$ uses the threshold $36$ and gives $E[(X-10)^2]/36=9/36$. The calculation needs only the mean and variance, not the detailed normal shape; Chebyshev’s bound is therefore also valid for any distribution with those finite moments.

Product Expectations: Cauchy–Schwarz or Hölder

A product expectation has a different structure from a tail probability. If $U$ and $V$ are square-integrable, Cauchy–Schwarz gives $$ E[|UV|]\le \sqrt{E[U^2]E[V^2]}. $$ If the available moments have other exponents, use Hölder’s inequality: for $p,q>1$ with $1/p+1/q=1$, if $E[|U|^p]<\infty$ and $E[|V|^q]<\infty$, then $$ E[|UV|]\le (E[|U|^p])^{1/p}(E[|V|^q])^{1/q}. $$ Cauchy–Schwarz is the special case $p=q=2$. These bounds do not require independence.

Worked Example: Matching Unequal Moments With Hölder

On a probability space, suppose $A\subseteq B$, $P(A)=1/8$, and $P(B)=1/2$. Let $U=\mathbf{1}_A$ and $V=\mathbf{1}_B$. Use Hölder with $p=3$ and $q=3/2$ to bound $E[|UV|]$.

The exponents are conjugate because $1/3+1/(3/2)=1/3+2/3=1$. Since indicators take only the values zero and one, $U^3=U$ and $V^{3/2}=V$. Therefore $$ E[|U|^3]=P(A)=\frac18, \qquad E[|V|^{3/2}]=P(B)=\frac12. $$ Hölder gives $$ E[|UV|] \le \left(\frac18\right)^{1/3}\left(\frac12\right)^{2/3} =\frac12\left(\frac12\right)^{2/3} =\frac{1}{2^{5/3}} \approx0.315. $$ Because $A\subseteq B$, $UV=\mathbf{1}_A$, so the actual expectation is $P(A)=1/8=0.125$, below the bound. The bound is not sharp here; the point is that Hölder uses the specified third and $3/2$-order moments. When second moments are the information supplied, Cauchy–Schwarz is the direct choice.

Expectations of Functions: Jensen’s Inequality

When the target is $E[\phi(X)]$, inspect the curvature of $\phi$. Jensen’s inequality says that if $X$ is integrable, takes values almost surely in an interval $I$, and $\phi$ is convex on $I$, then $$ \phi(E[X])\le E[\phi(X)]. $$ For a concave $\phi$, the inequality is reversed. To use the inequality as a finite numerical bound, the relevant expectation $E[\phi(X)]$ must be well-defined; in the examples below it is finite. Jensen is useful when the function of the mean is easier to compute than the expectation of the transformed variable.

Worked Example: A Lower Bound for a Reciprocal Expectation

Let $X$ equal $1$ or $4$, each with probability $1/2$. Find a Jensen lower bound for $E[1/X]$.

On the positive interval, $\phi(x)=1/x$ is convex because $\phi''(x)=2/x^3>0$. The mean is $$ E[X]=\frac12(1)+\frac12(4)=\frac52. $$ Jensen’s inequality therefore gives $$ E[1/X]\ge\frac{1}{E[X]}=\frac{2}{5}=0.4. $$ Checking the expectation directly, $$ E[1/X]=\frac12(1)+\frac12\left(\frac14\right)=\frac58=0.625, $$ which is indeed greater than $0.4$. Convexity determines the direction: using a concave-function version here would reverse the bound incorrectly.

A Practical Decision Process

Name the target.
Decide whether the unknown quantity is a tail probability, a product expectation, or an expectation of a transformed variable.

Rewrite it in a compatible form.
For a tail, use an indicator or a nonnegative power. For a product, identify its factors. For a transformed expectation, identify the function and its curvature.

Match the assumptions to the information.
Check nonnegativity and integrability for Markov, finite variance for Chebyshev, the required moments for Cauchy–Schwarz or Hölder, and the domain and curvature conditions for Jensen.

Calculate, then assess the result.
Substitute the actual moment values, check the direction and arithmetic, and note whether the bound is informative. A probability bound above one is valid but can be replaced by the trivial bound of one.

Common Mistake and Why the Choice Matters

Wrong reasoning: “The target is a tail probability, so I can always apply Markov directly to $X$.” This fails if $X$ can be negative, because Markov requires a nonnegative random variable. For example, knowing $E[X]$ does not by itself make $P(X\ge a)\le E[X]/a$ a valid Markov step when $X$ can take negative values.

Correction: Check nonnegativity before applying Markov. For a deviation event, the variable $(X-\mu)^2$ is nonnegative, so Markov applies to that square and yields Chebyshev’s bound when the variance is finite. For a product expectation, use Cauchy–Schwarz or Hölder only after checking the corresponding moments. For Jensen, establish the direction from convexity or concavity before substituting the mean.

A bound is a guarantee based on stated information, not necessarily a close approximation to the true quantity. If a problem supplies a variance, Chebyshev uses that information without distributional assumptions. If it supplies a particular distribution, a direct calculation might be sharper. The right inequality is the one whose hypotheses are met and whose inputs answer the question being asked.

Check Your Understanding

For each question, identify the relevant inequality and state the assumptions you need before calculating.

If $Y\ge0$ and $E[Y]=6$, what Markov bound follows for $P(Y\ge12)$?
If $E[X]=3$ and $\operatorname{Var}(X)=16$, what Chebyshev bound controls $P(|X-3|\ge8)$?
Which inequality directly bounds $E[|UV|]$ when $E[U^2]$ and $E[V^2]$ are finite?
For conjugate exponents $p=4$ and $q=4/3$, what moment conditions are needed to apply Hölder to $E[|UV|]$?
If $X>0$, $E[X]$ is finite, and $1/x$ is considered on the support of $X$, which direction does Jensen give for $E[1/X]$ relative to $1/E[X]$?

Target	Useful information	First candidate
\(P(Y\ge a)\), with \(Y\ge0\)	\(E[Y]\)	Markov or pointwise positivity
\(P(\|X-E[X]\|\ge t)\)	\(\operatorname{Var}(X)\)	Chebyshev
\(E[\|UV\|]\)	Second moments of \(U,V\)	Cauchy–Schwarz
\(E[\|UV\|]\)	Conjugate moments of \(U,V\)	Hölder
\(E[\phi(X)]\)	\(E[X]\) and convexity or concavity	Jensen