Suppose there is a random variable, $X$, with finite variance, and c.d.f. $F(x)$. Does this imply that the upper Matuszewska index of $\bar F(x)$ exists and is strictly smaller than $-2$?

The upper Matuszewska index of $f(x)$ is defined as the infimum of $\alpha$ such that there exists a $C$ such that for each $\Lambda > 1$,

$$ f(\lambda x)/f(x) \le C(1 + o(1))\lambda^\alpha \,\,\,(x \rightarrow \infty)\text{ uniformly in }\lambda \in [1, \Lambda]$$

I ask this because I was reading a paper where a result was proven for distributions with upper Matuszewska index less than $-2$, but in the conclusion the authors stated that the result held for finite variance distributions, seemingly without justification.


1 Answer 1


The answer is no: $\bar F$ does not have to have any negative Matuszewska index (any tail function $\bar F$ trivially has any nonnegative Matuszewska index).

Indeed, take any negative real $\alpha$. Suppose that $P(X=n^n)=c/n^{3n}$ for natural $n$, where $c:=1/\sum_1^\infty1/n^{3n}$. Then $EX^2<\infty$. However, $$\bar F(n^n/2)\sim c/n^{3n}\sim \bar F(n^n/(2\lambda))$$ for any real $\lambda>1$ as $n\to\infty$. Taking here any large enough $\lambda>1$ such that $C\lambda^\alpha<1$, we see that $\bar F$ does not have the upper Matuszewska index $\alpha$, for any negative real $\alpha$.


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