# If Mean Residual Lifetime is approximately constant, Residual Lifetime is Approximately Exponential in a Strong Sense

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. 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 implies a slightly weaker approximation result.

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$\newcommand{\eps}{\varepsilon} \newcommand{\E}{\mathbb{E}} \renewcommand{\P}{\mathbb{P}}$ 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.

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The idea to use compounding seems right to me. I don't see how the result is implied, however. Taken literally, the displayed conclusion implies that for all t, the absolute value of P(X>t) is bounded from above and below by the corresponding values of the exponential distribution +/- d. The approximation of the exponential distribution is weaker than the approximation from the conclusion of the conjecture. I might not understand the argument fully. Ps: I might really simply overlook how the structure of your argument extends. I very much appreciate your work! Thanks a lot! – Stephan Apr 26 '12 at 15:41

Here is a work which gives general conditions under which your conjecture is true.

"Limiting Properties of the Mean Residual Lifetime Function" by Isaac Meilijson in Ann. Math. Statist. Volume 43, Number 1 (1972), 354-357.

http://projecteuclid.org/euclid.aoms/1177692731

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