# Hitting time estimates

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 $$k$$?

• 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. – Iosif Pinelis Nov 18 at 22:34
• 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. – Mateusz Kwaśnicki Nov 18 at 22:58
• 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 $\tau$? – Anthony Quas Nov 19 at 0:22
• @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. – Iosif Pinelis Nov 19 at 0:41
• I think this paper by Reinert and Yang begins with an up-to-date discussion of known quantitative bounds on the distribution of $\tau$. – Mateusz Kwaśnicki Nov 20 at 9:13

The simple key observation is that $$$$P(\tau_1\le n)=P(T_n>1). \tag{1}$$$$ Note that $$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$$, $$$$P(\tau_1\le n)=P(T_n>1)\le\frac{n\de^2\si^2}{(1-n\de)^2} \le\frac{(1-t)\de\si^2}{t^2}\to0;$$$$ similarly, $$P(\tau_1\ge n)\to0$$ if $$n\sim(1+t)/\de$$ and hence if $$n\gtrsim(1+t)/\de$$. So, $$$$\de\tau_1\to1$$$$ in probability and hence without loss of generality we need consider only $$$$n\sim1/\de$$$$ in (1).
Next, by the Berry--Esseen inequality, $$$$P(\tau_1\le n)=P(T_n>1)=P\Big(Z>\frac{1-n\de}{\de\si\sqrt n}\Big)+R =P\Big(Z\le\frac{n-1/\de}{\si\sqrt n}\Big)+R,$$$$ where $$Z\sim N(0,1)$$, $$$$|R|\le C\frac\be{\si^3\sqrt n}\sim C\frac{\be\sqrt\de}{\si^3}\to0$$$$ by (0), and $$C$$ is a universal constant. Also, assuming $$|\frac{n-1/\de}{\si\sqrt n}|=O(1)$$, we have $$$$P\Big(Z\le\frac{n-1/\de}{\si\sqrt n}\Big) =P\Big(Z\le\frac{n-1/\de}{\si/\sqrt\de}\Big)+o(1).$$$$
So, $$\tau_1$$ is asymptotically normal with asymptotic mean $$1/\de$$ and asymptotic variance $$\si^2/\de$$.