Why Moment Existence Matters
A severity model can be a valid probability distribution even when its mean is infinite. It can also have a finite mean but an infinite second moment. Before using a moment in an actuarial calculation, first determine whether the defining sum or integral converges. The key question is often what happens far out in the distribution’s tail.
This tutorial develops practical tests for that question. The main tool is comparison with familiar integrals and series; a careful boundary check is essential when the leading power is exactly at its convergence threshold. The examples concern hypothetical loss severities. Interest and mortality assumptions do not apply.
A density must integrate to one for it to define a probability distribution, but that fact does not guarantee that \(\int_0^\infty x^r f_X(x)\,dx\) is finite. On any bounded interval, \(x^r\) is bounded for \(r\geq 0\), so the decisive issue for a proper nonnegative severity is usually the integral over the unbounded tail.
Power-Tail Comparison
Suppose a continuous severity has a density whose tail is asymptotically proportional to a power: \(f_X(x)\sim Cx^{-p}\) as \(x\to\infty\), where \(C>0\). Here “asymptotically proportional” means that the ratio \(f_X(x)/(Cx^{-p})\) tends to one. Multiplying by \(x^r\), the integrand for the \(r\)-th moment behaves like \(Cx^{r-p}\).
For a real exponent \(a\), the improper integral \(\int_M^\infty x^a\,dx\) converges exactly when \(a<-1\). Applying this condition gives the power-tail threshold \(r<p-1\). Equality is not enough: when \(r=p-1\), the comparison integral is proportional to \(\int_M^\infty x^{-1}\,dx\), which diverges.
The result follows from comparison: beyond some sufficiently large \(M\), the ratio of \(f_X(x)\) to \(Cx^{-p}\) is bounded above and below by positive constants. Thus the moment integral over that tail converges or diverges with the comparison integral. The part below \(M\) is finite because \(x^r\leq M^r\) there and the density has total probability at most one. The test is about the density’s tail; it is not a formula for moments of every distribution.
A logarithmic factor can change the answer precisely at the power threshold. If \(f_X(x)\) is asymptotically proportional to \(x^{-p}(\log x)^{-q}\), then at \(r=p-1\) the moment’s tail is proportional to \(\int_M^\infty [x(\log x)^q]^{-1}\,dx\). Substituting \(t=\log x\) reduces this to \(\int_{\log M}^\infty t^{-q}\,dt\): it converges when \(q>1\), and diverges when \(q\leq 1\). Away from the threshold, the power determines convergence.
Worked Example: Finding the Moment Threshold of a Power Tail
A hypothetical claim severity has density \(f_X(x)=3x^{-4}\) for \(x\geq 1\), and zero otherwise. Determine which nonnegative raw moments are finite, and calculate the first two positive integer moments if they exist. The density is already specified; no interest rate or mortality assumption is involved.
Method. First verify the density, then evaluate the defining moment integral. This avoids treating the threshold as inclusive when the boundary integral actually diverges.
The total probability is \( \int_1^\infty 3x^{-4}\,dx =3[-x^{-3}/3]_1^\infty =1\). For \(r\geq 0\),
When \(r<3\), the exponent \(r-4<-1\), and direct integration gives \(E[X^r]=3[x^{r-3}/(r-3)]_1^\infty=3/(3-r)\). When \(r=3\), the integrand is \(3/x\), whose integral diverges. For \(r>3\), the exponent is greater than \(-1\), so the integral also diverges.
Therefore, the finite-moment condition is \(0\leq r<3\). In particular, \(E[X]=3/(3-1)=3/2\), and \(E[X^2]=3/(3-2)=3\). As a check, the variance exists and is \(3-(3/2)^2=3/4\), while the third raw moment does not exist as a finite value. The common mistake is to include \(r=3\) by writing \(r\leq 3\); the direct boundary integral \(3\int_1^\infty x^{-1}\,dx\) shows why that is wrong.
Logarithmic Tails and Boundary Cases
A pure power comparison can settle most cases, but it cannot settle every boundary case. Logarithmic factors decay more slowly than any additional positive power of \(x\), yet their effect can still decide whether the critical integral converges. The following example works directly from the defining integral, including verification that the proposed density is normalized.
Worked Example: A Logarithmic Tail With No Positive Moments
A hypothetical severity has density \(f_X(x)=1/[x(\log x)^2]\) for \(x\geq e\), and zero otherwise. Determine whether its mean and any higher positive raw moment are finite. No interest rate or mortality assumption is relevant.
Method. Use the substitution \(t=\log x\) in both the normalization integral and the moment integral. The common mistake is to assume that the squared logarithm makes every moment finite.
For normalization, \(t=\log x\) gives \(dt=dx/x\), so \( \int_e^\infty f_X(x)\,dx =\int_1^\infty t^{-2}\,dt =[-1/t]_1^\infty=1\). Thus this is a proper density. For any \(r\geq 0\), the same substitution yields
At \(r=0\), this is \(\int_1^\infty t^{-2}\,dt=1\), as required for the zeroth moment. For every \(r>0\), the exponential eventually exceeds \(t^2\): there is a sufficiently large \(T\) such that \(e^{rt}\geq t^2\) for all \(t\geq T\). Consequently \(e^{rt}/t^2\geq 1\) on that tail, and its integral is infinite. The mean, which is the moment at \(r=1\), is therefore infinite, as is every higher positive raw moment. The check is that normalization still holds even though the mean does not.
Rapidly Decreasing Tails
Not every unbounded severity has a power tail. A tail that decreases rapidly enough can make moments of every order finite. For example, the next density has an exponential factor in the square root of the loss. A substitution turns its moment integral into a standard gamma integral; convergence follows directly from that integral.
Worked Example: Checking Every Moment Under a Stretched-Exponential Tail
A hypothetical severity has density \(f_X(x)=\tfrac12 x^{-1/2}e^{-\sqrt{x}}\) for \(x>0\). Determine whether all nonnegative raw moments are finite, and calculate the mean and second raw moment. No interest rate or mortality assumption applies.
Method. Substitute \(t=\sqrt{x}\), so \(x=t^2\) and \(dx=2t\,dt\). This both checks normalization and reduces each moment to a gamma integral. The common mistake is to infer divergence merely because the support is unbounded.
For normalization, the substitution gives \( \int_0^\infty \tfrac12 x^{-1/2}e^{-\sqrt{x}}\,dx =\int_0^\infty e^{-t}\,dt=1\). For a general \(r\geq 0\),
Since \(2r+1>0\), the gamma integral is finite for every \(r\geq 0\). Thus every nonnegative raw moment exists. For the mean, \(E[X]=\Gamma(3)=2!=2\). For the second raw moment, \(E[X^2]=\Gamma(5)=4!=24\). As a check on the calculations, these give \(\operatorname{Var}(X)=24-2^2=20\), a finite variance consistent with the existence of the second moment.
Using Moment Existence Correctly
Moment existence is a prerequisite for some familiar summaries, not a guarantee that every summary is available. A finite first moment gives a finite mean. A finite second raw moment gives a finite variance, since then both \(E[X]\) and \(E[X^2]\) are finite and \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\). A finite mean alone does not guarantee a finite variance. In an exam calculation, check existence before substituting a numerical value into a formula.
Write the requested order \(r\) and the defining integral or sum for \(E[X^r]\).
For a density proportional to \(x^{-p}\), compare \(r\) with \(p-1\). For a discrete model, compare the moment series with a known convergent or divergent series.
At the power threshold, integrate the boundary expression itself. A logarithmic factor may decide convergence, but the power rule alone cannot.
If the defining integral or sum diverges, report that the moment is infinite; do not use a finite-moment formula as though the value existed.
For a discrete model, the same logic applies to the probability-weighted series. If \(P(X=n)\) is proportional to \(n^{-p}\) for positive integers \(n\), then the \(r\)-th moment has terms proportional to \(n^{r-p}\). The series \(\sum_{n=1}^\infty n^a\) converges exactly when \(a<-1\), giving the same threshold \(r<p-1\), with divergence at equality. This is the series counterpart of the integral comparison, not a separate rule for the continuous case.
Check Your Understanding
For each hypothetical severity model, determine the requested moments by checking convergence before calculating. Answers are included so you can verify both the threshold and any finite value.
- A density is \(f_X(x)=2/(1+x)^3\) for \(x\geq 0\). Determine whether its mean and second raw moment are finite, and calculate the mean if it exists. (Answer: The mean is \(1\); the second raw moment is infinite.)
- A density is \(f_X(x)=4x^{-5}\) for \(x\geq 1\). Determine the range of \(r\geq0\) for which \(E[X^r]\) is finite, calculate \(E[X^2]\), and decide whether \(E[X^4]\) is finite. (Answer: Finite for \(0\leq r<4\); \(E[X^2]=2\); \(E[X^4]\) is infinite.)
- A proper density has tail \(f_X(x)\sim Cx^{-3}(\log x)^{-2}\), where \(C>0\). Classify the moments for \(0\leq r<2\), \(r=2\), and \(r>2\). (Answer: Finite for \(0\leq r\leq2\), including the logarithmic boundary case; infinite for \(r>2\).)
- A discrete severity has \(P(X=n)=c/n^4\) for integers \(n\geq1\), where \(c\) is the normalizing constant. Determine which nonnegative integer raw moments are finite. (Answer: The moments of orders \(0,1,2\) are finite; every integer-order moment \(r\geq3\) is infinite.)
- A density is \(f_X(x)=2/[x(\log x)^3]\) for \(x\geq e\). Verify its normalization and determine whether its mean is finite. (Answer: It is normalized because \(2\int_1^\infty t^{-3}\,dt=1\); the mean is infinite.)