We have a process $\{X\_{t}\}\_{t\geq 0}$ ,with fixed parameter $\epsilon>0$, starting from zero that satisfies


* The process is strictly monotone $X\_{t+r}-X\_{t}>0$ with moments existing $p\in(-\infty, \beta)$ for some $\beta>0$. (In the interval [0,1] we also have the lower bound $X\_{t+r}-X\_{t}>cr^{b}$ where $b>1$ and c is a random constant).
* $X\_{t+r}-X\_{r}\stackrel{d}{=}X\_{t}$ for $r\geq 0$ (stationary increments).
* For $a<b<c<d$ the increments $X\_{b}-X\_{a}$ and $X\_{d}-X\_{c}$ are independent when $b+\epsilon<c$ (independence when epsilon away).


So we have Markov up to the recent past:
$$P(X\_{t+r}-X\_{t}|\sigma(X\_{s}),s\leq t)=P(X\_{t+r}-X\_{t}|\sigma(X\_{t}-X\_{s}),s\in [t-\epsilon,t]).$$


Given the hitting times $T\_{a\_{k}}:=\inf\{s: X\_{s}\geq a\_{k}\}$ for points $a\_{k}\geq 0$, we want to test if they satisfy some weak stationarity. The stationarity
$$T\_{c+b}-T\_{c}\stackrel{d}{=}T\_{b}$$


is not true because contrary to Brownian motion we can have $c>b$ but the increments are dependent when $b+\epsilon>c >b$.



> 
> Q1: Have you seen an analogous process to $X\_{t}$ anywhere else that has been studied?
> 
> 
> Q2:Can we make some deterministic choice of $a\_{k}\geq 0$ to give 
>  $$T\_{a\_{k}+t}-T\_{a\_{k}}\stackrel{d}{=}T\_{t}$$
>  or having comparability for the moments
>  $$c\_{1} E[(T\_{t})^{p}] \leq E[(T\_{a\_{k}+t}-T\_{a\_{k}})^{p}]\leq c\_{2} E[(T\_{t})^{p}]$$
>  Q3: Is there a quantitative Strong Markov property 
>  $$ |P[x>X\_{T\_{a}+t}-X\_{T\_{a}}]-P[x>X\_{t}]|=g(\epsilon,a,t,x),$$ 
>  where $g(\epsilon,a,t,x)\to 0$ as $\epsilon\to 0$. 
> 
> 
> 


**Attempts**


1) Can we estimate the difference
$$ |P[T\_{a\_{k}+x}- T\_{a\_{k}}>t]-P[T\_{x}>t]|=g(\epsilon,a\_{k},t),$$ 
where $g(\epsilon,a)\to 0$ as $\epsilon\to 0$. By increasing monotonicity, we have that $X\_{y}$ can be inverted as the inverse to $T\_{x}$ i.e. $X\_{T\_{x}}=x$ and so we study the difference:
$$ |P[x>X\_{T\_{a}+t}-X\_{T\_{a}}]-P[x>X\_{t}]|=g(\epsilon,a),$$ 
where $g(\epsilon,a)\to 0$ as $\epsilon\to 0$. 


2)Using a proof similar to the one of the strong Markov property for Brownian motion we obtain:
$$ X\_{T\_{a}+\epsilon+t}-X\_{T\_{a}+\epsilon}\stackrel{d}{=}X\_{t}$$
or equivalently
$$T\_{a\_{\epsilon}+x}-T\_{a\_{\epsilon}}\stackrel{d}{=}T\_{x},$$
where $a\_{\epsilon}:=X\_{T\_{a}+\epsilon}$. So then the question becomes whether we have 
$$c\_{1} E[(T\_{a\_{\epsilon}+x}-T\_{a\_{\epsilon}})^{p}] \leq E[(T\_{a+x}-T\_{a})^{p}]\leq c\_{2} E[(T\_{a\_{\epsilon}+x}-T\_{a\_{\epsilon}})^{p}]$$
since 
$$ E[(T\_{a\_{\epsilon}+x}-T\_{a\_{\epsilon}})^{p}]= E[(T\_{x})^{p}].$$



Markov with epsilon memory and Quantitative Strong Markov property
Q1: I have not seen such a process, but I can easily imagine one: Let $Z\_t$ be any non-negative Lévy process with positive drift, and let $X\_t = \int\_{t-\epsilon}^t Z\_s ds$.


Q2: No, unless we know something more about $X\_t$. Imagine the process $X\_t$ as above, with $Z\_t$ having frequent extremely large jumps. As $a \to \infty$, the random variable $T\_{a+b} - T\_a$ will converge to zero (for typically $\tfrac{d}{dt} X\_t$ at $t = T\_a$ will grow indefinitely as $a \to \infty)$, so there is no hope that $T\_{a+b} - T\_a$ will ever has equal distribution as $T\_b$.


(This is related to the overshoot of $Z\_t$ when it crosses the level $a$: as $a$ goes to infinite, $Z\_t$ will typically pass through $a$ with a very large jump, so that $T^Z\_{a+b} - T^Z\_a$ will be equal to zero with an increasing probability. In this case $T^X\_{a+b} - T^X\_a$ will be very small, too. Here of course $T^Z\_a$ denotes the first passage time for $Z\_t$.)


