5 Qualification

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

Suppose the "expected 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.

4 Formatting

Suppose the "expected residual lifetime," $E[X-x|X≥x]$ \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,∞)$, $$lim_{x→∞}E[X-x|X≥x]= \lim_{x→∞}\mathbb{E}[X-x|X≥x]= \lambda ,$$ then, for all$ε>0$and for all$\Delta>0$, there is some$c$such that$x≥c$implies $$e^{-(1/(λ-ε))t}\leq Pr[X≥x+t|X≥x] e^{-\frac{t}{λ-ε}}\leq \mathbb{P}[X≥x+t|X≥x] \leq e^{-(1/(λ+ε))te^{-\frac{t}{λ+ε}} \qquad ∀t≥\Delta.$$ 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 $$lim_{x→∞}E[f(X-x)|X≥x]=E[f(Y)]$$$\lim_{x→∞}\mathbb{E}[f(X-x)|X≥x]=\mathbb{E}[f(Y)]$$for all nondecreasing functions for which E[f(Y)] \mathbb{E}[f(Y)] exists. In particular, f is allowed to be unbounded. 3 Clarification Suppose the "expected residual lifetime," E[X-x|X≥x] is approximately constant for large x. Then, I believe that the conditional tail distribution is approximately exponential, in the sense of being stochastically dominated by an exponential and dominating a similar exponential. Formally: Conjecture Given any random variable X with support on [0,∞). If, for some \lambda \in(0,∞),$$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}\leq Pr[X≥x+t|X≥x] \leq e^{-(1/(λ+ε))t} \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→∞}E[f(X-x)|X≥x]=E[f(Y)]

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

2 Corrected directions of inequalities in conclusion.
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