Higher-Moment Markov Bounds

From Moments to Tail Bounds

A tail probability asks how likely a random variable is to take a value far from zero. Markov’s inequality gives a way to control that probability using the expectation of a nonnegative random variable. Applying it not just to \(|X|\), but to a positive power of \(|X|\), produces a family of bounds. The choice of power can matter: a higher moment may give a sharper bound at one threshold and a weaker one at another.

The method is useful when the distribution is not known in full but one or more absolute moments are available. Its bound is often conservative, so it should be read as an upper limit rather than an estimate of the actual tail probability.

Definition: For a random variable \(X\) and \(p>0\), the \(p\)th absolute moment is \(E[|X|^p]\), when this expectation is finite. For a threshold \(t>0\) and a finite \(p\)th absolute moment, the higher-moment Markov bound is

P(|X|\ge t)\le \frac{E[|X|^p]}{t^p}.

Here \(p\) need not be an integer. The right-hand side is an upper bound, and if it exceeds \(1\), it can be replaced by the trivial bound \(1\).

Derivation and Conditions

The bound is a direct application of Markov’s inequality to \(W=|X|^p\). Because \(p>0\), \(W\) is nonnegative, and the event \(\{|X|\ge t\}\) is exactly the event \(\{W\ge t^p\}\). Markov’s inequality applies when \(E[W]\) is finite and \(t^p>0\).

Theorem: If \(X\) is a random variable, \(p>0\), \(t>0\), and \(E[|X|^p]<\infty\), then

P(|X|\ge t)\le \frac{E[|X|^p]}{t^p}.

Proof. On the event \(\{|X|\ge t\}\), the nonnegative quantity \(|X|^p\) is at least \(t^p\). Pointwise, this gives

|X|^p \ge t^p \mathbf{1}_{\{|X|\ge t\}},

where \(\mathbf{1}_A\) is \(1\) on the event \(A\) and \(0\) otherwise. Taking expectations preserves the inequality. The indicator is measurable because \(X\) is a random variable and \(\{|X|\ge t\}\) is a measurable event. Thus

E[|X|^p]\ge t^p E[\mathbf{1}_{\{|X|\ge t\}}] =t^p P(|X|\ge t).

Dividing by \(t^p>0\) proves the result. \(\square\)

The assumptions are worth keeping distinct. The threshold must be positive for division by \(t^p\), and the moment must be finite for this form of the bound to provide a finite upper bound. If \(E[|X|^p]=0\), then \(|X|^p=0\) almost surely, so \(X=0\) almost surely and the tail probability is zero for every \(t>0\).

Worked Examples With Absolute Moments

Worked Example: Choosing Among Normal Moments

Let \(Z\sim N(0,1)\). Compare the bounds for \(P(|Z|\ge 2)\) using \(p=1,2,4,\) and \(6\).

For \(p> -1\), integrating the standard normal density on the positive half-line and substituting \(u=x^2/2\) gives

E[|Z|^p] =\frac{2}{\sqrt{2\pi}}\int_0^\infty x^p e^{-x^2/2}\,dx =\frac{2^{p/2}\Gamma((p+1)/2)}{\sqrt{\pi}}.

In particular, \(E[|Z|]=\sqrt{2/\pi}\), \(E[|Z|^2]=1\), \(E[|Z|^4]=3\), and \(E[|Z|^6]=15\). Substituting each moment and \(t=2\) into the bound gives

\begin{aligned} p=1 &: \quad \frac{\sqrt{2/\pi}}{2}\approx 0.399,\\ p=2 &: \quad \frac{1}{2^2}=0.25,\\ p=4 &: \quad \frac{3}{2^4}=\frac{3}{16}=0.1875,\\ p=6 &: \quad \frac{15}{2^6}=\frac{15}{64}\approx 0.234. \end{aligned}

Among these choices, \(p=4\) gives the smallest upper bound. The \(p=6\) bound is valid but weaker. This comparison is about the bounds alone; it does not calculate the actual tail probability.

Worked Example: A Uniform Distribution Near Its Support

Let \(X\sim \operatorname{Unif}(-1,1)\) and take \(t=0.8\). For any \(p>0\), symmetry and the uniform density \(1/2\) give

E[|X|^p] =\frac12\int_{-1}^{1}|x|^p\,dx =\int_0^1 x^p\,dx =\frac{1}{p+1}.

For \(p=1,2,\) and \(4\), the resulting bounds are

\begin{aligned} p=1 &: \quad \frac{1/2}{0.8} = 0.625,\\ p=2 &: \quad \frac{1/3}{0.8^2} = \frac{1}{1.92}\approx 0.521,\\ p=4 &: \quad \frac{1/5}{0.8^4} = \frac{1}{2.048}\approx 0.488. \end{aligned}

Here moving from \(p=1\) to \(p=4\) improves the bound. But the support also shows that the actual probability is \(P(|X|\ge 0.8)=0.2\): the interval \([-1,-0.8]\cup[0.8,1]\) has total length \(0.4\) out of a total length of \(2\). The moment bounds are valid but are not exact.

Worked Example: Exponential Moments Can Make a Higher Order Worse

Let \(X\sim\operatorname{Exp}(1)\), with density \(e^{-x}\) for \(x\ge 0\), and set \(t=4\). Since \(X\ge0\), \(|X|=X\). For \(p> -1\),

E[|X|^p]=\int_0^\infty x^p e^{-x}\,dx=\Gamma(p+1).

For integer \(p\ge 1\), this is \(p!\). The bounds at several orders are therefore

\begin{aligned} p=1 &: \quad \frac{1!}{4}=\frac14=0.25,\\ p=2 &: \quad \frac{2!}{4^2}=\frac18=0.125,\\ p=3 &: \quad \frac{3!}{4^3}=\frac{3}{32}=0.09375,\\ p=4 &: \quad \frac{4!}{4^4}=\frac{3}{32}=0.09375,\\ p=6 &: \quad \frac{6!}{4^6}=\frac{45}{256}\approx 0.176. \end{aligned}

The fourth-moment bound is smaller than the sixth-moment bound, even though \(6>4\). The higher order increases the numerator as well as the threshold power in the denominator; which effect dominates depends on the distribution and the chosen threshold.

Comparing Moment Orders

The comparison can be made exactly once the moments are known. Let \(0<p<q\), and suppose \(0<E[|X|^p]<\infty\) and \(E[|X|^q]<\infty\). Write \(B_r(t)=E[|X|^r]/t^r\) for the bound at order \(r\). Then

Comparison rule: Under the stated moment conditions, the order-\(q\) bound is no larger than the order-\(p\) bound exactly when

B_q(t)\le B_p(t) \quad\Longleftrightarrow\quad t^{q-p}\ge \frac{E[|X|^q]}{E[|X|^p]}.

Proof. Both moments are positive and finite, and \(t>0\), so the bounds can be compared by division:

\frac{E[|X|^q]}{t^q}\le \frac{E[|X|^p]}{t^p} \quad\Longleftrightarrow\quad \frac{E[|X|^q]}{E[|X|^p]}\le t^{q-p}.

This is the claimed criterion. \(\square\)

The ratio of moments sets a threshold scale at which the two bounds are equal. Above that scale, the higher-order bound is smaller; below it, the lower-order bound is smaller. This explains why there is no universal rule to choose the largest available \(p\). The moments and the threshold must be considered together.

A useful check is rescaling. If \(Y=cX\), where \(c\ne0\), and the threshold is changed from \(t\) to \(|c|t\), then

\frac{E[|Y|^p]}{(|c|t)^p} =\frac{|c|^p E[|X|^p]}{|c|^p t^p} =\frac{E[|X|^p]}{t^p}.

Thus the bound is unchanged when the variable and threshold are expressed in a different common scale. The units also agree: both \(E[|X|^p]\) and \(t^p\) have units of \(X\) raised to the power \(p\), so their ratio is dimensionless, as a probability must be.

Common Mistakes and Practical Use

Assuming larger \(p\) always gives a smaller bound. The examples show otherwise. Compare the actual moment ratios and threshold using the criterion above.
Forgetting the power on the threshold. The denominator is \(t^p\), not \(t\), unless \(p=1\).
Using a moment that is not finite. The stated bound requires \(E[|X|^p]<\infty\). A formal substitution of an infinite moment gives no useful finite bound.
Treating the upper bound as the tail probability. Markov’s inequality controls the probability from above; it does not claim equality. Equality requires a special distributional arrangement and should not be presumed.
Ignoring that probabilities cannot exceed one. If the calculated right-hand side is larger than \(1\), report the trivial probability bound \(1\) instead, or note that the moment bound is uninformative at that threshold.

Higher-moment bounds are most useful when the available moment is easy to calculate and the target threshold is large enough for its power in the denominator to be helpful. They provide distribution-light control: apart from the moment condition, the proof does not require a particular density or mass function. At the same time, the normal and uniform examples illustrate why knowledge of the full distribution or its support can yield substantially sharper information.

Check Your Understanding

For each question, identify the moment order and threshold before comparing or calculating bounds.

If \(E[|X|^3]=12\), what bound does the higher-moment Markov inequality give for \(P(|X|\ge 2)\)?
For two available orders \(p<q\), what condition determines whether the order-\(q\) bound is no larger than the order-\(p\) bound?
For \(X\sim\operatorname{Unif}(-1,1)\), use \(p=2\) to bound \(P(|X|\ge 0.5)\). Is the bound informative compared with \(1\)?
Why does the inequality require a finite \(p\)th absolute moment and a positive threshold?
Explain why the sixth-moment bound for the rate-one exponential example is weaker than its fourth-moment bound at threshold \(4\).

What You'll Learn