Recovering Moment Bounds from Tail Probabilities

From Tail Probabilities to Moments

A tail probability describes how often a random variable exceeds a threshold. A moment describes the average size of a power of that variable. These summaries may look different, but for a nonnegative random variable they are linked by an integral: each possible threshold contributes to the moment in proportion to the probability of exceeding it.

This link is useful in both directions. If a tail bound is known, integrating it can establish that a moment exists and give an upper bound for its value. Conversely, the identity makes clear why a tail bound that decays too slowly cannot guarantee a finite moment. The size of the tail for large thresholds is crucial, but the integral near zero must also be well behaved.

Definition: For a random variable \(X\geq 0\) almost surely and a real number \(p>0\), the \(p\)th moment is \(E[X^p]\), with the expectation allowed to be \(+\infty\). The tail function is \(t\mapsto P(X>t)\), for \(t\geq 0\).

The Tail-Integral Identity

For each fixed number \(x\geq 0\), the area under \(p t^{p-1}\) from zero to \(x\) is exactly \(x^p\). The same calculation can be written using an indicator: at a threshold \(t\), the indicator \(\mathbf{1}\{x>t\}\) is one precisely when \(t<x\).

Theorem (tail-integral identity): If \(X\geq 0\) almost surely and \(p>0\), then, with both sides interpreted as extended nonnegative quantities,

E[X^p]=p\int_0^\infty t^{p-1}P(X>t)\,dt.

No finite-moment assumption is needed for this identity. In particular, the integral may diverge, in which case the moment is infinite. The strict inequality in \(P(X>t)\) is consistent with the indicator representation; changing \(>\) to \(\geq\) does not change the integral.

Write a pointwise integral.
For every \(x\geq 0\), \(x^p=\int_0^\infty p t^{p-1}\mathbf{1}\{x>t\}\,dt\). If \(x=0\), both sides are zero; if \(x>0\), the integral is \(\int_0^x p t^{p-1}\,dt=x^p\).

Substitute the random variable.
Pointwise, \(X^p=\int_0^\infty p t^{p-1}\mathbf{1}\{X>t\}\,dt\). The integrand is jointly measurable and nonnegative.

Take expectations and interchange.
Tonelli’s theorem applies to a nonnegative measurable function, even if its integral is infinite. Thus \(E[X^p]=\int_0^\infty p t^{p-1}E[\mathbf{1}\{X>t\}]\,dt\), and \(E[\mathbf{1}\{X>t\}]=P(X>t)\). This gives the identity. \(\square\)

For an arbitrary real-valued \(X\), apply the same identity to the nonnegative random variable \(|X|\):

E[|X|^p]=p\int_0^\infty t^{p-1}P(|X|>t)\,dt,\qquad p>0.

Thus a bound on a two-sided tail can control absolute moments. The identity also applies to truncated variables, such as \(\min(X^p,b^p)\), but the full moment formula is usually the most direct tool when the goal is to establish whether \(E[X^p]\) is finite.

Power-Law Tails and the Critical Exponent

Suppose that, for some constants \(C>0\), \(\alpha>0\), and \(t_0>0\), a nonnegative \(X\) satisfies \(P(X>t)\leq C t^{-\alpha}\) for every \(t\geq t_0\). The tail-integral identity separates the moment integral into a bounded-threshold part and a large-threshold part. On the first part, \(P(X>t)\leq 1\); on the second, use the assumed tail bound.

Theorem (moment bound from a power-law tail): Under the preceding assumptions, if \(0<p<\alpha\), then

E[X^p]\leq t_0^p+\frac{pC}{\alpha-p}t_0^{p-\alpha}<\infty.

To verify the bound, split the integral at \(t_0\). The first part is at most \(p\int_0^{t_0} t^{p-1}\,dt=t_0^p\). For the second part,

p\int_{t_0}^{\infty} t^{p-1}P(X>t)\,dt \leq pC\int_{t_0}^{\infty}t^{p-1-\alpha}\,dt =\frac{pC}{\alpha-p}t_0^{p-\alpha},

where the improper integral converges exactly when \(p-\alpha<0\). The tail bound therefore guarantees every positive moment of order strictly less than \(\alpha\). By itself, it gives no finite bound at \(p=\alpha\) or above: the comparison integral at the endpoint is proportional to \(\int_{t_0}^\infty t^{-1}\,dt\), which diverges.

Worked Example: Moments of a Pareto Distribution

Let \(X\) have survival function \(P(X>t)=1\) for \(0\leq t<1\), and \(P(X>t)=t^{-3}\) for \(t\geq 1\). Find \(E[X^p]\) for \(p>0\), including the orders for which it is finite.

Use the tail-integral identity and split at \(t=1\):

E[X^p] =p\int_0^1t^{p-1}\,dt +p\int_1^\infty t^{p-1}t^{-3}\,dt =1+p\int_1^\infty t^{p-4}\,dt.

The second integral converges if and only if \(p-4<-1\), or \(p<3\). In that case,

E[X^p]=1+\frac{p}{3-p}=\frac{3}{3-p}.

For example, \(E[X]=3/2\) and \(E[X^2]=3\). At \(p=3\), the remaining integral is \(3\int_1^\infty t^{-1}\,dt=\infty\); for \(p>3\), its power is even less integrable. Hence \(E[X^p]=\infty\) for every \(p\geq 3\). This example has the critical tail exponent \(3\): moments below it are finite, while the moment at the critical exponent is already infinite.

Exponential and Gaussian-Type Tails

A rapidly decreasing tail can guarantee moments of every positive order. The substitution used to check this should be explicit: a tail bound must be multiplied by \(p t^{p-1}\), integrated, and then tested for convergence.

Worked Example: An Exponential Variable

Let \(X\) have an exponential distribution with rate \(2\), so \(P(X>t)=e^{-2t}\) for \(t\geq 0\). Find its third moment from the tail.

The identity gives

E[X^3]=3\int_0^\infty t^2e^{-2t}\,dt.

With \(u=2t\), \(t=u/2\), and \(dt=du/2\), this becomes

E[X^3] =\frac{3}{8}\int_0^\infty u^2e^{-u}\,du =\frac{3}{8}\cdot 2 =\frac{3}{4}.

The integral \(\int_0^\infty u^2e^{-u}\,du=2\) follows from two integrations by parts (the boundary terms vanish because \(u^k e^{-u}\to 0\) as \(u\to\infty\)). More generally, every positive power is integrable against this exponential tail, so the exponential tail bound implies finite moments of all positive orders.

A common situation is a two-sided tail bound of Gaussian form. For instance, if \(Z\sim N(0,1)\), its moment generating function is \(E[e^{\lambda Z}]=e^{\lambda^2/2}\). For \(t>0\), exponential Markov applied to \(e^{\lambda Z}\), followed by minimizing over \(\lambda>0\), gives \(P(Z>t)\leq e^{-t^2/2}\); applying the same argument to \(-Z\) gives \(P(|Z|>t)\leq 2e^{-t^2/2}\). This bound is enough to control moments even without integrating the normal density directly.

Worked Example: A Fourth-Moment Bound from a Gaussian Tail

Suppose \(Z\sim N(0,1)\). Use the preceding tail bound to obtain an upper bound for \(E[|Z|^4]\), and check the integral.

For \(t\geq 0\), use \(P(|Z|>t)\leq 2e^{-t^2/2}\). The tail-integral identity yields

E[|Z|^4] =4\int_0^\infty t^3P(|Z|>t)\,dt \leq 8\int_0^\infty t^3e^{-t^2/2}\,dt.

Set \(u=t^2/2\), so \(t\,dt=du\) and \(t^2=2u\). Then

8\int_0^\infty t^3e^{-t^2/2}\,dt =16\int_0^\infty u e^{-u}\,du =16.

The last integral equals \(1\), by integration by parts. Therefore the tail bound proves \(E[|Z|^4]\leq 16\). This is a valid bound, though not sharp: direct evaluation using the normal density gives \(E[Z^4]=3\). The point here is that tail information alone can establish finiteness and produce a usable bound without requiring the exact moment.

Checking Integrability and Avoiding a Common Mistake

A tail bound is useful for moments only after the weighted integral has been checked. For example, if for all sufficiently large \(t\), \(P(X>t)\leq C e^{-ct^r}\), where \(C,c,r>0\), then every \(p>0\) has a finite moment bound: \(t^{p-1}e^{-ct^r}\) is integrable at infinity. Near zero, the probability is at most one, and \(t^{p-1}\) is integrable there for every \(p>0\). Both ends are therefore controlled.

By contrast, a bound \(P(X>t)\leq C t^{-\alpha}\) for large \(t\) cannot establish \(E[X^p]<\infty\) when \(p\geq\alpha\), because the resulting comparison integral diverges. This does not prove that the actual moment must be infinite: the true tail may be much smaller than the bound. It means only that this particular bound does not establish finiteness.

Mistake: “The tail probability tends to zero, so every moment is finite.” Every proper distribution has tail probabilities tending to zero, but that alone says nothing about how quickly the weighted tail integral converges.
Correction: For the desired order \(p\), inspect \(p\int_0^\infty t^{p-1}P(X>t)\,dt\). Check the integral near zero and at infinity, and conclude finiteness only if both parts are finite.

Key takeaway: A moment is a weighted area under the survival function. Tail decay controls the large-threshold contribution, and the tail-integral identity makes the exact integrability condition visible.

Check Your Understanding

Use the tail-integral identity and check convergence explicitly in each problem.

If \(P(X>t)\leq 5t^{-4}\) for \(t\geq 2\), which positive moment orders does this bound guarantee are finite?
For a tail \(P(X>t)=t^{-5}\) when \(t\geq 1\), evaluate \(E[X^2]\) using the identity and the specified tail on \(0\leq t<1\) equal to one.
Why is Tonelli’s theorem applicable in the proof of the tail-integral identity even when \(E[X^p]=\infty\)?
If \(P(|Y|>t)\leq 3e^{-t}\) for all \(t\geq 0\), explain why \(E[|Y|^p]\) is finite for every \(p>0\).
Does a bound \(P(X>t)\leq C t^{-2}\) for large \(t\) prove that \(E[X^2]\) is infinite? Explain what the bound does and does not establish.