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In a number of different contexts, I have wanted to estimate hitting times for a monotonic process $(T_n)$ taking values in the reals (or sometimes a process $(T_n,X_n)$ taking values in $\mathbb R^2$ where the first component is monotonic). I'm assuming the step size is small, and am interested in the first time, $\tau_1$, that $T_n$ exceeds some threshold $a$ (and in the 2d case, possibly the distribution of $X_n$ at this hitting time).

For some concreteness, assume $a=1$, $T_0=0$ and $(T_n)$ is Markov, where the jump size, $T_{n+1}-T_n$ is much smaller than 1. If necessary, we can assume that the jump sizes are i.i.d., although I'd ultimately prefer to have something more flexible where the distribution of $T_{n+1}-T_n$ is in some sense continuously dependent on $T_n$. I would like to obtain reasonably precise information on the distribution of $\tau_1$.

Is there an established machinery that can address questions of this type?

Here is a specific simple (made up) instance that I would be very interested to see a clean answer to (especially, as indicated above, an answer that might be generalizable away from the i.i.d. jump case): Let $T_{n+1}-T_n\sim \exp(\text{Unif}[-2k,-k])$ and $T_0=0$. What can be said about the distribution of $\tau_1$? How does the distribution of $\tau_1$ depend on $k$?

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    $\begingroup$ I am not sure if I understood the question correctly. The distribution of $\tau_1$ is given by the formula $P(\tau_1>n)=P(T_n\le a)$ for all natural $n$. So, the problem reduces to that of the distribution of $T_n$. Already in the iid jump case, there are a million papers on that. $\endgroup$ – Iosif Pinelis Nov 18 at 22:34
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    $\begingroup$ In the i.i.d. case, this is precisely what renewal theory studies. Its extension to the "Markovian" setting (although perhaps slightly different from what you wrote; I would have to check that) is called Markov renewal theory; Çinlar wrote a survey of this subject long ago. $\endgroup$ – Mateusz Kwaśnicki Nov 18 at 22:58
  • $\begingroup$ Thanks @IosifPinelis. I think you have understood my question. I agree with your reduction, and this is a helpful observation. Do you know any papers where this is actually used to get explicit about the distribution on $\tau$? $\endgroup$ – Anthony Quas Nov 19 at 0:22
  • $\begingroup$ @AnthonyQuas : I think I saw this reduction somewhere long ago, when I was a student. I don't remember where, we almost used no textbooks. Or maybe my memory is mistaken. I think the answer will depend on what properties of the distribution of $\tau_1$ you want to study. $\endgroup$ – Iosif Pinelis Nov 19 at 0:41
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    $\begingroup$ I think this paper by Reinert and Yang begins with an up-to-date discussion of known quantitative bounds on the distribution of $\tau$. $\endgroup$ – Mateusz Kwaśnicki Nov 20 at 9:13
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$\newcommand{\de}{\delta} \newcommand{\be}{\beta} \newcommand{\si}{\sigma}$ Consider the iid case, when $T_n=X_1+\dots+X_n$, where $X_1,X_2,\dots$ are positive iid random variables with, say, $EX_1=\de\to0$, $Var\,X_1=\de^2\si^2$, $E|X_1|^3=\de^3\be<\infty$ (scaling with $\de$) such that \begin{equation} \be\sqrt\de/\si^3\to0,\quad \de\si^2\to0. \tag{0} \end{equation}

The simple key observation is that \begin{equation} P(\tau_1\le n)=P(T_n>1). \tag{1} \end{equation} Note that $ET_n=n\de$ and $Var\,T_n=n\de^2\si^2$.

So, by Chebyshev's inequality, for any fixed $t\in(0,1)$ and any $n\le(1-t)/\de$, \begin{equation} P(\tau_1\le n)=P(T_n>1)\le\frac{n\de^2\si^2}{(1-n\de)^2} \le\frac{(1-t)\de\si^2}{t^2}\to0; \end{equation} similarly, $P(\tau_1\ge n)\to0$ if $n\sim(1+t)/\de$ and hence if $n\gtrsim(1+t)/\de$. So, \begin{equation} \de\tau_1\to1 \end{equation} in probability and hence without loss of generality we need consider only \begin{equation} n\sim1/\de \end{equation} in (1).

Next, by the Berry--Esseen inequality, \begin{equation} P(\tau_1\le n)=P(T_n>1)=P\Big(Z>\frac{1-n\de}{\de\si\sqrt n}\Big)+R =P\Big(Z\le\frac{n-1/\de}{\si\sqrt n}\Big)+R, \end{equation} where $Z\sim N(0,1)$, \begin{equation} |R|\le C\frac\be{\si^3\sqrt n}\sim C\frac{\be\sqrt\de}{\si^3}\to0 \end{equation} by (0), and $C$ is a universal constant. Also, assuming $|\frac{n-1/\de}{\si\sqrt n}|=O(1)$, we have \begin{equation} P\Big(Z\le\frac{n-1/\de}{\si\sqrt n}\Big) =P\Big(Z\le\frac{n-1/\de}{\si/\sqrt\de}\Big)+o(1). \end{equation}

So, $\tau_1$ is asymptotically normal with asymptotic mean $1/\de$ and asymptotic variance $\si^2/\de$.

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  • $\begingroup$ Thanks. I think this is very much in the spirit of what I was hoping for. $\endgroup$ – Anthony Quas Nov 19 at 23:10

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