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.

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.

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.

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.

Suppose the "expected 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.

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.