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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Hölderlin Jan 4 at 15:10
  • $\begingroup$ 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). $\endgroup$ – tomasz Jan 4 at 17:52
  • $\begingroup$ 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? $\endgroup$ – Hölderlin Jan 4 at 18:44
  • $\begingroup$ 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. $\endgroup$ – tomasz Jan 4 at 18:55
  • $\begingroup$ 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. $\endgroup$ – tomasz Jan 4 at 18:58

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