# Equivalence of definitions of heavy-tailed distributions

In class we gave two definitions of heavy-tailed distributions:

($\overline{F}(x) = P(X>x)$ denotes the tail of the distribution)

1. $\liminf\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}>0$ , for all $\lambda > 0$

2. $\limsup\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}=\infty$ , for all $\lambda > 0$

However, I am not sure if those two are equivalent. I don't know how to prove it, nor can I find a counterexample. I thought that maybe $X\sim Exp(\mu)$ would not satisfy this as somewhat of an edge case, but it isn't a counterexample, because of the "for all $\lambda>0$" part.

Can you prove the equivalence of those definitions, or disprove it?

• Indeed the "for all $\lambda$" part is crucial. To go from 1. to 2., note that 1. for $\lambda$ implies $$\lim_{x\to\infty}\frac{\overline{F}(x)}{e^{-\mu x}}=\infty$$ for every $\mu<\lambda$ hence 2. holds. To get a feeling of wether the other implication holds, did you try to build some counterexamples?
– Did
Feb 3 '17 at 6:30
• Any complementary CDF $\bar F$ such that $\bar F(x)=e^{-\sqrt{x}}$ for some unbounded values of $x$ and $\bar F(x)=e^{-x^2}$ for some unbounded values of $x$ would disprove the implication $2.\implies 1.$ And it seems possible to build such functions $\bar F$, which leads to the question: Are you sure about the equivalence?
– Did
Feb 3 '17 at 6:38
• @Did Thanks for the comments! In the first one, did you mean $\mu>\lambda$ so we have $e^{\lambda x}\bar{F}(x)e^{(\mu-\lambda) x}$ which goes to $\infty$ as the first part is larger than zero? For the second comment, no, I'm not sure that they are equivalent, that's why I asked here. I don't really understand what you meant with those two complementary cdfs. To combine them somehow into one or what? Also, I should note that we are considering only positive random variables, though I don't think it makes much difference. Feb 3 '17 at 10:20
• Re 1.: Yes, $\mu>\lambda$, sorry about the typo. Re 2.: No, the idea is to have a unique complementary CDF $\bar F$ such that $\bar F(x_n)=e^{-\sqrt{x_n}}$ and $\bar F(y_n)=e^{-y_n^2}$ with $x_n\to\infty$ and $y_n\to\infty$.
– Did
Feb 3 '17 at 10:34
• A specific example is to consider $$x_n=2^{4^n}\qquad a_n=e^{-\sqrt{x_n}}\qquad \frac1c=\sum_{n=0}^\infty a_n$$ and to assume that $$P(X=x_n)=ca_n$$ for every $n\geqslant0$. Then, note that $$x_{n+1}=x_n^4$$ and that $$\bar F(x_n)=\bar F(x_{n+1}-1)=c\sum_{k=n+1}^\infty a_k\sim ca_{n+1}=ce^{-\sqrt{x_{n+1}}}=ce^{-x_n^2}$$ hence $$\bar F(x_n)\sim ce^{-x_n^2}\qquad\bar F(x_n-1)\sim ce^{-\sqrt{x_n}}\sim ce^{-\sqrt{x_n-1}}$$ as desired.
– Did
Feb 3 '17 at 11:22

One direction holds, i.e. the first statement implies the second, as $$\liminf\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}>0\implies(\forall \mu>\lambda)\lim\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\mu x}}=\lim\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}e^{(\mu-\lambda) x}=\infty.$$ The other direction isn't true. A counterexample follows.
We'll define $$x_n=2^{4^n},\quad a_n=e^{-\sqrt{x_n}},\quad \frac1c=\sum\limits_{n=0}^{\infty}a_n,$$
and assume that $$P(X=x_n)=ca_n,\quad \forall n\in\mathbb{N}_0.$$
It follows that $x_{n+1}=x_n^4$ and $$\overline{F}(x_n)=\overline{F}(x_{n+1}-1)=c\sum\limits_{k=n+1}^{\infty}a_k \, \sim\,ca_{n+1}=ce^{-\sqrt{x_{n+1}}}=ce^{-x_n^2}.$$ Hence $$\overline{F}(x_n)\sim ce^{-x_n^2},\quad\overline{F}(x_n-1)\sim ce^{-\sqrt{x_{n}}}\sim ce^{-\sqrt{x_{n}-1}}.$$ Thus we have $$(\forall\lambda>0)\limsup\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}\geq \limsup\limits_{n\to\infty}\frac{\overline{F}(x_n-1)}{e^{-\lambda (x_n-1)}}=\limsup\limits_{n\to\infty}\frac{ce^{-\sqrt{x_{n}-1}}}{e^{-\lambda (x_n-1)}}=\infty,$$ but $$\liminf\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}\leq \liminf\limits_{n\to\infty}\frac{\overline{F}(x_n)}{e^{-\lambda x_n}}=\liminf\limits_{n\to\infty}\frac{ce^{-x_n^2}}{e^{-\lambda x_n}}=0.$$