Assume that $S_i(k) =\sum_{t=1}^k X_i(t)$ for $i = 1,2.$ $X_i(t)$ are i.i.d. random variables with positive mean. What is the probability that $\inf_k \{\max_{i} S_i(k)\}< -a$, for some a > 0?

I suppose as $a\to+\infty$, $P(\inf_k \{\max_i S_i(k\} < -a) = P(\inf_k S_1(k)<-a)\cdot P(\inf_k S_2(k)<-a)$. Is that correct? How to prove it?

  • $\begingroup$ It can be proved that with certain constraints of the distribution of X, we have $$\lim_{a\to\infty}P(\inf_k S_i(k)<-a)=Ce^{-a}$$ Do we have $\lim_{a\to\infty} P(inf_k \{max_i S_i(k\} < -a) = Ce^{-2a}$ ? $\endgroup$ – Zishuo Mar 15 at 14:30

Partial Answer showing that your product claim is not correct, but not providing the correct answer $$ \textstyle \text{Saying}\quad \inf_k \max_i S_i(k) \le -a \quad\text{means}\quad \text{there exists a $k$ so that $S_1(k) \le -a$ and $S_2(k) \le -a$}. $$ So we do not simply take the product: we need $S_1$ and $S_2$ to be below $-a$ at the same time. Taking the product would just say "both $S_1$ and $S_2$ individually go below $-a$ at some point".

Note also that you have said "as $a \to \infty$", and then written $a$ on both sides. Do you mean that the two limits are the same? (For a sequence $x_n$, the limit $\lim_n x_n$ must be independent of $n$, if it exists, since $n$ is only a dummy variable.)

  • $\begingroup$ Yes, I mean those two limits are the same as $a\to+\infty$. Or more formally, as $a\to + \infty$ $$\frac{P(\inf_k \{\max_i S_i(k\} < -a)}{ P(\inf_k S_1(k)<-a)\cdot P(\inf_k S_2(k)<-a)}\to 1$$ Those two probability may not equal to each other for every $a$, but the fraction may tends to 1 because difference of them is minor to their value. $\endgroup$ – Zishuo Mar 25 at 9:07
  • $\begingroup$ Ok, good, so as I expected. So yeah, as my answer shows, product form is not correct, but I'm not sure what the actual form is -- it's possibly pretty difficult, and needs quite a refined analysis. I'd suggest trying with a Brownian Motion -- these can be easier to handle, and often give pretty good results $\endgroup$ – Sam T Mar 25 at 23:14
  • $\begingroup$ OK, thank you very much for your friendly reminder. I'll first try to confine it to some specific distributions and find out its properties and then try to generalize it. $\endgroup$ – Zishuo Mar 28 at 8:03

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