Return to Question

Qualification

view source

edited Apr 27 2012 at 17:49

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.

Formatting

view source

edited Apr 26 2012 at 18:11

Stephan
99●5

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.

Clarification

view source

edited Apr 26 2012 at 18:02

Stephan
99●5

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