Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Got something from an answer below? – Did Feb 8 '13 at 7:32

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 $c_k:=e^{(2k+1)^2}$, $a_k:=e^{4k^2}$.

share|cite|improve this answer
You don't seem to be naive in the use of TeX like so many people here, but you use the antiquated \ {\rm blah\ blah\ blah\ } instead of \text{ blah blah blah } and the complicated {array} instead of the far simpler {cases} when it's about cases (i.e. piecewise definition). Do you prefer those for a reason? – Michael Hardy Jan 31 '13 at 1:26
I guess it just shows you how long ago I learned TeX. – David Moews Jan 31 '13 at 1:27
Im new to Tex, but i appreciate every advice – nordmann Jan 31 '13 at 1:28

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 $\mathrm e^{\lambda x_n}(1-F(x_n))\leqslant2c\exp(\lambda x_n-a_{n+1})\to0$.

share|cite|improve this answer

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$).

share|cite|improve this answer
Sure. But why does this yield a nondecreasing F? – Did Feb 8 '13 at 7:32
If $\dfrac{1-F(x_k-)}{e^{-\lambda x_k}} = k$ and $\dfrac{1-F(x_k+)}{e^{-\lambda x_k}} = 1/k$ with $k > 1$, you must have $F(x_k+) > F(x_k-)$. – Robert Israel Feb 8 '13 at 8:16
This does not answer my question. One would also need that $\mathrm e^{\lambda (x_{k+1}-x_k)}\geqslant k^2$ for every $k$, a condition which seems difficult to achieve for every positive $\lambda$. – Did Feb 8 '13 at 10:24
Just take $x_{k+1} \ge x_k + \log(k^2)/\lambda$. – Robert Israel Feb 8 '13 at 18:51
Quote: difficult to achieve for every positive λ. Unquote. – Did Feb 8 '13 at 20:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.