Suppose the "mean residual lifetime," $\mathbb{E}[X-x|X≥x]$ 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, for some $\lambda \in(0,∞)$, $$\lim\_{x→∞}\mathbb{E}[X-x|X≥x]= \lambda ,$$


then, for all $ε>0$ and for all $\Delta>0$, there is some $c$ such that $x≥c$ implies $$e^{-\frac{t}{λ-ε}}\leq \mathbb{P}[X≥x+t|X≥x] \leq e^{-\frac{t}{λ+ε}} \qquad ∀t≥\Delta.$$


I posted this question on [StackExchange](https://math.stackexchange.com/questions/136489/sufficient-condition-for-asymptotically-exponential-tail-corrected-but-still-un). [Robert Israel](https://mathoverflow.net/users/13650/robert-israel) provided a counterexample to an earlier conjecture, which was wrong.


**Update** 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*. The first great response by [Ori Gurel-Gurevich](https://mathoverflow.net/users/1061/ori-gurel-gurevich) implies a slightly weaker approximation result.



If Mean Residual Lifetime is approximately constant, Residual Lifetime is Approximately Exponential in a Strong Sense
$\newcommand{\eps}{\varepsilon}
\newcommand{\E}{\mathbb{E}}
\renewcommand{\P}{\mathbb{P}}$
Fix $\eps>0$ and let $a=\E[X]$ and $b=\E[X-\eps \mid X>\eps]$. Let $p=\P(X > \eps)$. Then
$$ (1-p)(b+\eps)\le a \le (1-p)(b+\eps) + p \eps \ .$$


Solving, we get
$$ 1-\frac{a+\eps}{b+\eps}+\frac{\eps}{b+\eps} < p < 1-\frac{a}{b}+\frac{\eps}{b} $$


In other words,
$$ \frac{\eps}{\lambda+\eps} -\delta < p < \frac{\eps}{\lambda} +\delta $$


Where $\delta=\delta(a,b)$ goes to 0 as $a$ and $b$ approach $\lambda$.


Compounding this we get that when $a < \E[X-s \mid X > s] < b$ for all $s>0$ and we take $a$ and $b$ to $\lambda$, we have for any $t>0$


$$(1-\frac{\eps}{\lambda+\eps})^{\frac{t}{\eps}} - \delta\_1 < \P(X > t) < (1-\frac{\eps}{\lambda})^{t/\eps} + \delta\_1$$


where $\delta\_1 \to 0$ when $a,b \to \lambda$.


Taking now $\eps$ small enough yields the desired result.


