2
$\begingroup$

Let $T = \inf\{ n : S_n = a \text{ or } S_n = -b\}$ be a stopping time, where $S_n = X_1 + \dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < \infty$ almost surely. They state:

$$P(T = \infty) \leq P(T > n) \leq P(|S_n| \leq \text{max}\{a,b\})$$

Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| \leq \text{max}\{a,b\}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ T\not > n$?

$\endgroup$
4
  • $\begingroup$ But if $S_n=a$ or $S_n=-b$, then $T\not>n$ $\endgroup$
    – Jimmy R.
    Commented Mar 30, 2016 at 19:30
  • 1
    $\begingroup$ @JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter. $\endgroup$
    – Ian
    Commented Mar 30, 2016 at 19:31
  • $\begingroup$ everything else is copied up correctly $\endgroup$
    – lampj20la
    Commented Mar 30, 2016 at 19:32
  • $\begingroup$ Do you mean that $S_n$ is a martingale? Otherwise it doesn't make much sense to say that each $X_n$ is a martingale, unless I don't know something. $\endgroup$
    – baibo
    Commented Oct 9, 2019 at 10:40

1 Answer 1

Reset to default
5
$\begingroup$

To have $T=\infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k \leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $\max_{k \leq n} |S_n|$ has to be less than $\max \{ a,b \}$.

So $\{ T=\infty \} \subseteq \{ T>n \} \subseteq \{ \max_{k \leq n} |S_n| \leq \max \{ a,b \} \}$. Then use monotonicity of $P$ to get the result.

By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.

$\endgroup$
6
  • $\begingroup$ Thanks for your response - I don't follow why $|S_n| \leq \text{max}\{a,b\}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$? $\endgroup$
    – lampj20la
    Commented Mar 30, 2016 at 19:35
  • 1
    $\begingroup$ @lampj20la That's true, but we're saying $T>n \Rightarrow |S_n| \leq \max \{ a,b \}$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that. $\endgroup$
    – Ian
    Commented Mar 30, 2016 at 19:36
  • 1
    $\begingroup$ I see - thanks for the quick answer! $\endgroup$
    – lampj20la
    Commented Mar 30, 2016 at 19:37
  • 1
    $\begingroup$ @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b. $\endgroup$
    – Dole
    Commented Dec 2, 2018 at 4:04
  • 2
    $\begingroup$ I'm also curious as to how we know that $P(\tau = \infty)=0$ from this inequality. I believe that's what we need to show in order to prove that this stopping time is a.s finite. $\endgroup$
    – dp1221
    Commented Apr 28, 2020 at 20:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .