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I conjecture a sufficient condition for a distribution to be asymptotically exponential in a strong sense.

Roughly speaking, the idea is this. Suppose the "expected residual lifetime," $E[X-x|X≥x]$ is approximately constant for large $x$. Then, I believe that the conditional tail distribution is approximately exponential, in the sense of being stochastically dominated by an exponential and dominating a similar exponential. Formally:

Conjecture Given any random variable $X$ with support on $[0,∞)$. If $$lim_{x→∞}E[X-x|X≥x]= \lambda ,$$

then for all $ε>0$ and for all $\Delta>0$ there is some $c$ such that $x≥c$ implies $$e^{-(1/(λ-ε))t}≥Pr[X≥x+t|X≥x]≥e^{-(1/(λ+ε))t} \qquad ∀t≥\Delta.$$

Update 2 The first conjecture was wrong. Robert Israel provided a counterexample. The implication is now weaker, restricting $∀t≥\Delta>0$. The weakening takes care of the counterexample. But is it correct?

Update 3 (Removed)

Update 4 I posted the question on MathOverflow.

Update 5 The approximation result is stronger than weak convergence. Let $Y$ be distributed exponentially with parameter $\lambda$. The conclusion of the conjecture implies that

$$\lim_{x→∞}\mathbb{E}[f(X-x)|X≥x]=\mathbb{E}[f(Y)]$$

for all nondecreasing functions for which $\mathbb{E}[f(Y)]$ exists. In particular, $f$ is allowed to be unbounded.

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up vote 1 down vote accepted

What happens as $t \to 0+$? $1 - e^{-t/(\lambda \pm \epsilon))} \approx \frac{t}{\lambda \pm \epsilon}$, so your conclusion would imply that if $x'$ is sufficiently large, the density $f(x')$ exists and $$\frac{1}{\lambda + \epsilon} \le \frac{f(x')}{1-F(x')} \le \frac{1}{\lambda - \epsilon}$$ That is too strong a conclusion, I think. For example, you could try a discrete distribution with possible values, say, $x_n = \sqrt{n}$ for positive integers $n$, and $P(X = x_n) = e^{-x_{n-1}} - e^{-x_n}$ with $P(X=x_1) = 1- e^{-x_1}$. The CDF of this distribution agrees with that of the exponential distribution of parameter $1$ at the points $x_n$, and I think it's easy to see that it will satisfy your hypothesis.

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Ah, I see you now have an update that addresses this objection. –  Robert Israel Apr 25 '12 at 1:09
    
Thank you for the example, the example is neat! I hope that the weaker conclusion is correct. As you write, the weaker statement is now true for your example. -Stephan –  Stephan Apr 25 '12 at 2:31
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