Comparing Exact Tails With Markov, Chebyshev, and Chernoff Bounds

Why Compare a Bound With the Exact Tail?

A tail inequality replaces a probability that may be difficult to calculate with a quantity that is guaranteed to be at least as large. When a distribution is known, however, its exact tail can often be computed too. Putting the two side by side answers a practical question: how much probability does the bound leave unused? A valid bound need not be close to the exact answer, and a more detailed method does not automatically produce a smaller bound at every threshold.

This tutorial compares upper tails of three standard distributions. Markov’s inequality uses a nonnegative variable and its mean; Chebyshev’s inequality uses a mean and variance; a Chernoff bound uses a moment-generating function (MGF) at a valid positive argument. Each method gives a universal guarantee under its stated assumptions. The examples show what those guarantees look like when the exact distribution is also available.

The Bounds Being Compared

Theorem: If $Z\geq 0$ almost surely and $E[Z]<\infty$, then for $a>0$, $$ P(Z\geq a)\leq \frac{E[Z]}{a}. $$ If $X$ has finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$, then for $r>0$, $$ P(|X-\mu|\geq r)\leq \frac{\sigma^2}{r^2}. $$ In particular, if $a>\mu$, $$ P(X\geq a)\leq \frac{\sigma^2}{(a-\mu)^2}. $$ If $M_X(t)=E[e^{tX}]$ is finite for some $t>0$, then $$ P(X\geq a)\leq e^{-ta}M_X(t). $$ Consequently, for any $a$ and any nonempty set of valid positive MGF arguments, $$ P(X\geq a)\leq \inf_{\substack{t>0\\M_X(t)<\infty}} e^{-ta}M_X(t). $$

The one-sided Chebyshev display follows because $X\geq a>\mu$ implies $|X-\mu|\geq a-\mu$. It is a convenient application of the two-sided inequality, not a claim that Chebyshev is always the best bound obtainable from a variance. The Chernoff expression is obtained by optimizing the exponential Markov bound over its valid parameter values.

Why the Inequalities Hold

Each bound begins with a pointwise comparison. For Markov’s inequality, $Z\geq a$ on the event being bounded. For Chebyshev’s inequality, the squared deviation is large on a large-deviation event. For the Chernoff bound, exponentiation turns an upper-tail event into one to which Markov’s inequality applies.

Proof: Since $Z$ is nonnegative, $$ Z\geq a\,\mathbf{1}_{\{Z\geq a\}}. $$ Taking expectations gives $E[Z]\geq aP(Z\geq a)$, and division by $a>0$ proves Markov’s inequality. Apply Markov’s inequality to $Z=(X-\mu)^2$, which is nonnegative and has expectation $\sigma^2$, to obtain $$ P(|X-\mu|\geq r) =P((X-\mu)^2\geq r^2) \leq \frac{\sigma^2}{r^2}. $$ For the MGF bound, fix a $t>0$ such that $M_X(t)<\infty$. On $\{X\geq a\}$, $e^{tX}\geq e^{ta}$. Thus Markov’s inequality applied to $e^{tX}$ yields $$ P(X\geq a)\leq \frac{E[e^{tX}]}{e^{ta}} =e^{-ta}M_X(t). $$ This holds for every valid $t>0$; taking the infimum over those $t$ proves the stated Chernoff bound. $\square$

For a fixed distribution and threshold, the exact tail is at most every valid bound above. There is no general ordering among the numerical bounds: a mean-based Markov bound, a variance-based Chebyshev bound, and an optimized MGF bound use different information. The minimum of valid bounds is also a valid bound, but no one method must produce that minimum for every problem.

Worked Comparisons

Worked Example: A Binomial Upper Tail

Let $X\sim\operatorname{Binomial}(10,0.5)$. Compare the exact probability $P(X\geq 8)$ with Markov, Chebyshev, and Chernoff bounds.

Exact tail. The binomial probabilities give $$ P(X\geq 8) =\frac{\binom{10}{8}+\binom{10}{9}+\binom{10}{10}}{2^{10}} =\frac{45+10+1}{1024} =\frac{56}{1024} =0.0546875. $$

Markov. Since $X\geq0$ and $E[X]=10(0.5)=5$, $$ P(X\geq8)\leq \frac{5}{8}=0.625. $$

Chebyshev. The variance is $10(0.5)(0.5)=2.5$, and $X\geq8$ implies $|X-5|\geq3$. Therefore $$ P(X\geq8)\leq P(|X-5|\geq3)\leq\frac{2.5}{3^2} =\frac{2.5}{9}\approx0.2778. $$

Chernoff. The MGF is $M_X(t)=((1+e^t)/2)^{10}$, finite for all real $t$. For $t>0$, the bound is $e^{-8t}((1+e^t)/2)^{10}$. Its logarithm has derivative $$ -8+\frac{10e^t}{1+e^t}. $$ Setting this to zero gives $e^t=4$, or $t=\ln4>0$. The second derivative of the log bound is $10e^t/(1+e^t)^2>0$, so this critical point is the minimum. Hence $$ P(X\geq8)\leq4^{-8}\left(\frac{5}{2}\right)^{10} \approx0.1455. $$

All three bounds exceed the exact probability; among these calculated bounds, Chernoff is smallest. Even that bound is about $2.66$ times the exact tail. Markov’s mean alone is particularly uninformative here because $E[X]$ is not far below the threshold relative to the threshold itself.

Worked Example: A Poisson Upper Tail

Let $Y\sim\operatorname{Poisson}(4)$, and compare the probability $P(Y\geq8)$ with the same three bounds.

Exact tail. From the Poisson mass function, $$ P(Y\geq8)=e^{-4}\sum_{k=8}^{\infty}\frac{4^k}{k!}. $$ The terms in the sum begin $1.62540,\ 0.72240,\ 0.28896,\ 0.10508,\ 0.03503,\ 0.01078,\ldots$; summing the tail gives approximately $2.7918$. Thus $$ P(Y\geq8)\approx e^{-4}(2.7918)\approx0.0511. $$

Markov. Here $E[Y]=4$, so $$ P(Y\geq8)\leq\frac{4}{8}=0.5. $$

Chebyshev. A Poisson variable has variance $4$. Since $Y\geq8$ implies $|Y-4|\geq4$, $$ P(Y\geq8)\leq P(|Y-4|\geq4)\leq\frac{4}{16}=0.25. $$

Chernoff. The MGF is $M_Y(t)=\exp(4(e^t-1))$, finite for every real $t$. The log bound is $4(e^t-1)-8t$, whose derivative is $4e^t-8$. Its unique minimum occurs at $t=\ln2>0$; the second derivative $4e^t$ is positive. Therefore $$ P(Y\geq8)\leq \exp(4-8\ln2)\approx0.2131. $$

The exact probability is close to $0.05$, while the optimized Chernoff bound is about $0.21$. It is sharper than the displayed Markov and Chebyshev bounds, but still more than four times the exact tail. Comparing against the exact answer reveals the size of this gap; the bound alone does not.

Worked Example: When Chebyshev Beats the Chernoff Bound

Let $W\sim\operatorname{Exp}(1)$, with density $f(w)=e^{-w}$ for $w\geq0$. Find the exact upper tail at $3$ and compare the three bounds.

Exact tail. Integrating the density gives $$ P(W\geq3)=\int_3^\infty e^{-w}\,dw=e^{-3}\approx0.0498. $$

Markov. Since $E[W]=1$, $$ P(W\geq3)\leq\frac{1}{3}\approx0.3333. $$

Chebyshev. The variance is $1$, so the event $W\geq3$ implies $|W-1|\geq2$. Thus $$ P(W\geq3)\leq\frac{1}{2^2}=0.25. $$

Chernoff. For $t<1$, $M_W(t)=1/(1-t)$; in particular the positive MGF domain is $0<t<1$. The log of the Chernoff bound is $-3t-\ln(1-t)$. Its derivative is $-3+1/(1-t)$, which vanishes at $t=2/3$, inside the valid domain. The second derivative $1/(1-t)^2$ is positive, so this is the minimum: $$ P(W\geq3)\leq e^{-2}\frac{1}{1-2/3}=3e^{-2}\approx0.4060. $$

Here Chebyshev’s bound is smaller than the optimized Chernoff bound, even though Chernoff uses the MGF. Both remain well above the exact tail. This example demonstrates why the computed values—not a presumed ranking—should guide a comparison.

Interpreting Sharpness and Avoiding a Common Mistake

A useful comparison reports four things together: the event and threshold, the exact probability when available, each bound with its assumptions, and the numerical gap between each bound and the exact answer. A bound of $0.2$ for an exact probability of $0.05$ is valid, but it gives a much less precise numerical description than a bound of $0.06$. Bound sharpness can also change with the threshold: a method that is tighter at one threshold may be looser at another.

The available information matters. Markov’s inequality only needs a finite mean and nonnegativity. Chebyshev’s inequality additionally uses a finite variance. Chernoff’s bound uses a finite MGF at positive arguments and may benefit from information about its shape over that domain. The exact calculations above use still more: the specified full distribution. These are different information sets, not a universal ladder of guaranteed numerical performance.

Common mistake: “Chernoff uses the MGF, so its optimized bound must always be smaller than Chebyshev’s bound.” This reasoning is incorrect. The bounds arise from different inequalities, and their numerical values have no universal ordering. In the exponential example, the optimized Chernoff bound is about $0.4060$, whereas Chebyshev gives $0.25$. Calculate both bounds under their respective assumptions, check the MGF domain, and compare their values rather than inferring a ranking from the method’s name.

The exact tail is a benchmark for understanding a bound, not a prerequisite for using one. In applications, the full distribution may be unknown or exact calculation may be impractical; an inequality can still provide a rigorous guarantee from limited information. When the distribution is known, as in these examples, comparison shows precisely what that guarantee costs.

Check Your Understanding

For each question, distinguish the exact probability from the guarantee supplied by a bound.

For a nonnegative random variable with finite mean, what pointwise inequality leads to Markov’s upper-tail bound?
If $E[X]=2$, $\operatorname{Var}(X)=9$, and $a=8$, calculate the one-sided Chebyshev bound for $P(X\geq8)$.
Why must a candidate Chernoff parameter satisfy both $t>0$ and $M_X(t)<\infty$?
In the exponential example, which of the three calculated bounds was smallest, and what does that show about a presumed universal ranking?
For $X\sim\operatorname{Binomial}(10,0.5)$, the exact probability $P(X\geq8)$ is $0.0546875$. Is the Chernoff value $0.1455$ an estimate of that probability or an upper bound? Explain.