I want an example for a distribution function $F$ with those two conditions for all $\lambda>0$ $$\liminf_{x\to\infty}\frac {1-F(x)} {e^{-\lambda x}}=0 $$ $$\limsup_{x\to\infty}\frac {1-F(x)} {e^{-\lambda x}}=\infty$$ I want one example of a distribution function where those two conditions hold. I may use a $F$ with $F(t)=f_k$ with $t\in [x_k,x_{k+1}]$ and choose fitting $f_k,x_k$, but i have no clue how to go on. $$ $$Thanks Nordmann
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If you pick positive $c_1 < c_2 < \dots$ and $a_1<a_2<\dots$ such that $$ F(x) =\begin{cases} 0, &\qquad \text{if } x < a_1,\\[6pt] 1-e^{-c_k}, &\qquad \text{if } a_k\le x < a_{k+1},\text{ for } k=1, 2, \dots, \end{cases} $$ then to satisfy these conditions it's enough to have $$ c_k/a_k\to\infty, \qquad c_k/a_{k+1}\to 0. $$ So, pick $c_k:=e^{(2k+1)^2}$, $a_k:=e^{4k^2}$. |
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Define $a_n=\exp(3^n)$ and $x_n=a_n^2+1$ and assume that $\mathbb P(X=x_n)=c\mathrm e^{-a_n}$ for every $n\geqslant1$, where $c$ is chosen such that $c\sum\limits_n\mathrm e^{-a_n}=1$. Note that $x_n\ll a_{n+1}$ and that, for every $n$, $\mathrm e^{-a_n}\leqslant\sum\limits_{k\geqslant n}\mathrm e^{-a_k}\leqslant2\mathrm e^{-a_{n}}$. Then $1-F(a_n^2)\gt c\mathrm e^{-a_n}$ hence $\mathrm e^{\lambda a_n^2}(1-F(a_n^2))\gt c\exp(\lambda a_n^2-a_n)\to+\infty$. And $1-F(x_n)\leqslant2c\mathrm e^{-a_{n+1}}$ hence $\mathrm e^{\lambda x_n}(1-F(x_n))\leqslant2c\exp(\lambda x_n-a_{n+1})\to0$. |
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Hint: Take $F(t)$ to be constant on intervals $(x_k, x_{k+1})$ with a jump at each $x_k$ from a value where $\frac{1-F(x)}{e^{-\lambda x}}$ is large (say $k$) to one where it is small (say $ 1/k$). |
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