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Karamata's Theorem

Statement

Let LL be eventually positive, measurable, locally bounded, and slowly varying at infinity. If ρ>−1\rho>-1, then

∫axtρL(t) dt∼xρ+1L(x)ρ+1,x→∞.\int_a^x t^\rho L(t)\,dt \sim \frac{x^{\rho+1}L(x)}{\rho+1}, \qquad x\to\infty.

If ρ<−1\rho<-1, then

∫x∞tρL(t) dt∼xρ+1L(x)−ρ−1,x→∞.\int_x^\infty t^\rho L(t)\,dt \sim \frac{x^{\rho+1}L(x)}{-\rho-1}, \qquad x\to\infty.

Here a>0a>0 is fixed in a region where LL is locally bounded, and A(x)∼B(x)A(x)\sim B(x) means A(x)/B(x)→1A(x)/B(x)\to1. Both integral statements are cited from Bingham et al. (1987), Theorem 1.5.11, p. 28 (direct half); the special case for LL and powers also appears as Propositions 1.5.8 and 1.5.10, pp. 26–27.

For a nonnegative random variable XX with regularly varying survival function Fˉ(x)=x−αL(x)\bar F(x)=x^{-\alpha}L(x) and α>0\alpha>0, integration by parts gives the moment-test consequences

E[Xp1{X>x}]∼αα−pxpFˉ(x),0<p<α,\mathbb E[X^p\mathbf 1_{\{X>x\}}] \sim \frac{\alpha}{\alpha-p}x^p\bar F(x), \qquad 0<p<\alpha,

and

E[Xp1{X≤x}]∼αp−αxpFˉ(x),p>α.\mathbb E[X^p\mathbf 1_{\{X\le x\}}] \sim \frac{\alpha}{p-\alpha}x^p\bar F(x), \qquad p>\alpha.

Thus a Pareto-type tail has finite ppth moment for 0<p<α0<p<\alpha and infinite ppth moment for p>αp>\alpha. The boundary case p=αp=\alpha is not decided by this theorem alone; it depends on the slowly varying factor LL.

Here Fˉ(x)=P(X>x)\bar F(x)=\mathbb P(X>x), α\alpha is the positive tail exponent, and pp is the moment order. The integration parameters aa and ρ\rho belong to the integral theorem.

Intuition

Karamata’s theorem says that, for regularly varying functions, integrals are asymptotically governed by the endpoint where the mass accumulates. When ρ>−1\rho>-1, the integral up to xx is controlled by the upper endpoint. When ρ<−1\rho<-1, the remaining tail integral beyond xx is controlled by the lower endpoint.

This is the bridge between tail shape and moment existence. Once the survival tail behaves like x−αL(x)x^{-\alpha}L(x), multiplying by xpx^p tests whether the ppth moment is still dominated by ordinary observations or by the far tail.

Examples

Deriving the moment consequences

We use Karamata’s integral asymptotics as a cited theorem, then derive the two moment consequences by integration by parts. The boundary case p=αp=\alpha is not decided by Karamata alone; it depends on the slowly varying factor and is handled through examples and caveats.

The integral asymptotics are the classical form of Karamata’s theorem for regularly varying functions. The full proof uses uniform convergence of slowly varying functions on compact multiplier intervals and a split of the integral into a near-endpoint part and a negligible remainder; this page cites the standard theorem rather than reproducing that argument here.

The moment consequences follow from integration by parts. For 0<p<α0<p<\alpha,

E[Xp1{X>x}]=xpFˉ(x)+p∫x∞tp−1Fˉ(t) dt.\mathbb E[X^p\mathbf 1_{\{X>x\}}] =x^p\bar F(x)+p\int_x^\infty t^{p-1}\bar F(t)\,dt.

Since tp−1Fˉ(t)=tp−α−1L(t)t^{p-1}\bar F(t)=t^{p-\alpha-1}L(t) and p−α−1<−1p-\alpha-1<-1, Karamata’s tail-integral form gives

∫x∞tp−1Fˉ(t) dt∼xp−αL(x)α−p=xpFˉ(x)α−p.\int_x^\infty t^{p-1}\bar F(t)\,dt \sim \frac{x^{p-\alpha}L(x)}{\alpha-p} = \frac{x^p\bar F(x)}{\alpha-p}.

Therefore

E[Xp1{X>x}]∼(1+pα−p)xpFˉ(x)=αα−pxpFˉ(x).\mathbb E[X^p\mathbf 1_{\{X>x\}}] \sim \left(1+\frac{p}{\alpha-p}\right)x^p\bar F(x) = \frac{\alpha}{\alpha-p}x^p\bar F(x).

For p>αp>\alpha,

E[Xp1{X≤x}]=p∫0xtp−1Fˉ(t) dt−xpFˉ(x).\mathbb E[X^p\mathbf 1_{\{X\le x\}}] =p\int_0^x t^{p-1}\bar F(t)\,dt-x^p\bar F(x).

The part of the integral over any fixed bounded interval is negligible relative to xpFˉ(x)x^p\bar F(x). Applying Karamata to tp−α−1L(t)t^{p-\alpha-1}L(t) gives

p∫0xtp−1Fˉ(t) dt∼pp−αxpFˉ(x),p\int_0^x t^{p-1}\bar F(t)\,dt \sim \frac{p}{p-\alpha}x^p\bar F(x),

so subtracting xpFˉ(x)x^p\bar F(x) leaves

E[Xp1{X≤x}]∼αp−αxpFˉ(x).\mathbb E[X^p\mathbf 1_{\{X\le x\}}] \sim \frac{\alpha}{p-\alpha}x^p\bar F(x).

For the exact Pareto special case, LL is constant and the same formulas are obtained by direct power integration.

Moment-test calculation

For a slowly varying example take L(t)=log⁡tL(t)=\log t on t>1t>1 and ρ>−1\rho>-1. Integration by parts gives the exact identity

∫1xtρlog⁡t dt=xρ+1log⁡xρ+1−xρ+1−1(ρ+1)2.\int_1^x t^\rho\log t\,dt =\frac{x^{\rho+1}\log x}{\rho+1} -\frac{x^{\rho+1}-1}{(\rho+1)^2}.

Dividing by Karamata’s leading term yields

∫1xtρlog⁡t dtxρ+1log⁡x/(ρ+1)=1−1−x−(ρ+1)(ρ+1)log⁡x⟶1.\frac{\int_1^x t^\rho\log t\,dt} {x^{\rho+1}\log x/(\rho+1)} =1-\frac{1-x^{-(\rho+1)}}{(\rho+1)\log x}\longrightarrow1.

For exact Pareto with xm=1x_m=1 and p>αp>\alpha, the ratio of the truncated moment to its leading term is 1−x−(p−α)→11-x^{-(p-\alpha)}\to1. At p=αp=\alpha, the truncated moment is αlog⁡x\alpha\log x instead. These are exact calculations illustrating the cited theorem.

Conditional size and the boundary moment

For α>1\alpha>1, write e(u)=E[X−u∣X>u]e(u)=\mathbb E[X-u\mid X>u] for the mean excess. Set p=1p=1 in the proved tail-moment consequence and divide by Fˉ(u)>0\bar F(u)>0. It gives

E[X∣X>u]u⟶αα−1,e(u)u⟶1α−1.\frac{\mathbb E[X\mid X>u]}u\longrightarrow\frac{\alpha}{\alpha-1}, \qquad \frac{e(u)}u\longrightarrow\frac1{\alpha-1}.

This is a consequence under a regular-variation assumption, not a finite threshold criterion for fitting that class. It gives the precise forward implication relevant to SCoFT v4 §5.0.2, pp. 93–94 Taleb, 2025. The Pareto note owns the exact version. The chapter guide gives a subexponential Weibull example whose conditional-mean ratio tends to one.

For a complete example with a finite boundary moment, take the survival equal to one below ee and, for x≥ex\ge e, set Fˉ(x)=eαx−α/(log⁡x)2\bar F(x)=e^\alpha x^{-\alpha}/(\log x)^2. It is continuous at ee and strictly decreasing afterwards. Tonelli gives

EXα=α∫0∞tα−1Fˉ(t) dt=eα+αeα∫e∞dtt(log⁡t)2=(1+α)eα<∞.\mathbb EX^\alpha =\alpha\int_0^\infty t^{\alpha-1}\bar F(t)\,dt =e^\alpha+\alpha e^\alpha\int_e^\infty\frac{dt}{t(\log t)^2} =(1+\alpha)e^\alpha<\infty.

This proves why the exact-Pareto boundary cannot be transferred to every regularly varying tail. For α=2\alpha=2 the answer is 3e23e^2, whereas an exact Pareto of exponent 2 has infinite second moment.

The chapter 5 integral figure compares the logarithmic integral in the earlier moment-test calculation with its Karamata leading term. The demo also evaluates the boundary example in logarithmic coordinates, integrating to infinity.

Caveats

References

References
  1. Bingham, N. H., Goldie, C. M., & Teugels, J. L. (1987). Regular Variation. Cambridge University Press. 10.1017/CBO9780511721434
  2. Taleb, N. N. (2025). Statistical Consequences of Fat Tails (3rd ed.). https://arxiv.org/abs/2001.10488v4
  3. Embrechts, P., Klüppelberg, C., & Mikosch, T. (1997). Modelling Extremal Events. Springer. 10.1007/978-3-642-33483-2
  4. Resnick, S. I. (2007). Heavy-Tail Phenomena: Probabilistic and Statistical Modeling. Springer. 10.1007/978-0-387-45024-7