# Probability of a random walk with positive drift cross a negative threshold

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?

• 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 $\lim_{a\to\infty} P(inf_k \{max_i S_i(k\} < -a) = Ce^{-2a}$ ? – 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.)
• 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. – Zishuo Mar 25 at 9:07