Looking for a heavy-tailed distribution

I want to find a probability distribution (given by its CDF $$F$$) such that $$\liminf_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = 0 \quad \text{for all } \lambda > 0$$ and $$\limsup_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = \infty \quad \text{for all } \lambda > 0.$$

The only thing that I could guess almost immediately is that this distribituion has to be heavy-tailed. Otherwise I have no clue.

It suffices to find a non-negative integrable function $$f$$ such that for all $$c>1$$, there are sequences $$(x_n'),(x_n'')$$ tending to infinity such that you have $$c^{x_n'}\int_{x_n'}^\infty f(x)\,\mathrm{d}x$$ grows to infinity and $$c^{x_n''}\int_{x_n''}^\infty f(x)\,\mathrm{d}x$$ decays to zero.

The trick to deal with all $$c$$ at the same time is to make $$\int_{x_n''}^\infty f(x)\,\mathrm{d}x$$ decay superexponentially, while making $$\int_{x_n'}^\infty f(x)\,\mathrm{d}x$$ decay only subexponentially, for instance by ensuring that $$\int_{x_n'}^\infty f(x)\,\mathrm{d}x\geq 1/(2x'_n)$$, while $$\int_{x_n''}^\infty f(x)\,\mathrm{d}x\leq 2^{-2^{x''_n}}$$.

I'll leave the details to you, but with a clever choice of $$(x_n')$$ and $$(x_n'')$$, you can ensure that for $$A=\bigcup_n (x_n',x_n'')$$ (the union of intervals), this is true for the function $$f(x)=1/x^2\cdot \mathbf 1_A(x)$$ (where $$\mathbf 1_A$$ is the indicator function of $$A$$).

Then you only need to rescale $$f$$ to ensure that it is a pdf. $$F$$ can be recovered by integrating the resulting function, if needed.

• Sorry, my calculus here is sketchy. With this choice of $f$ and ignoring $A$ this yields $c^x/x$ in the LHS. The lim sup is obvious. But how is the lim inf different from that and why is your sequence criterion equivalent to taking the lim inf? – Hölderlin Jan 4 at 15:10
• The whole point is that taking the $A$ will allow you to make large "gaps" where $f$ is zero, which allows you to make the integral arbitrarily small. The criterion is equivalent to taking lim sup and lim inf pretty much by definition (modulo a mild application of axiom of choice, maybe - this is pretty much the same as equivalence of Cauchy/$\varepsilon-\delta$ and Heine/sequential definition of the limit). – tomasz Jan 4 at 17:52
• I guess I'll look into the definition again. Your idea is that I should come up with the sequences and then get $A$ from that, right? – Hölderlin Jan 4 at 18:44
• Yes, I think that is the easiest solution. You could probably cook up something without specifically using integration, but I think it's easier to think of it in these terms. – tomasz Jan 4 at 18:55
• Oh, I neglected to include that: the sequences $(x_n'), (x_n'')$ should tend to infinity for the equivalence to be true. I modified the answer to reflect that. – tomasz Jan 4 at 18:58