# Maximum of a zero-mean random walk

Assume $$S_n=\sum_{k=1}^n X_k$$ where $$X_k$$ is i.i.d distributed and $$\mathbb{E}X_1=0$$ and $$\mathbb EX_1^2<\infty$$. Let $$M_n=\max_{1\leq k \leq n}\{S_k\}$$. What is the exponential asymptotic probability of $$M_n>a$$ as $$a\to+\infty$$? Or what is the limit of the following fraction: $$\lim_{a\to+\infty}\frac{-\log\mathbb{P}(M_n>a)}{a}$$

Does it exist? What is the value if it does(may be represented by $$n$$)? Or maybe easier, is there any positive lower bound of this limitation? Does it requires more restrictions on $$X_k$$ to get useful results?

Some ideas:

1.I have some results based on Cramer's Theorem of Large Deviation Theory. However it is the probability characteristic of $$S_n$$ as $$n\to\infty$$:

$$\limsup_{n\to\infty}\log \mathbb{P}(\frac{S_n}{n}\in F)\leq -\inf_{x\in F} I(x)$$ where $$I(x)=\sup_\theta\{\theta x-\log M(\theta)\}$$ and $$M(\theta)$$ is moment generating function of $$X_1$$.

2.Maybe we can treat $$M_n$$ as a sub-martingale and get some results based on characteristic of martingales.

3.Kolmgorov's submartingale inequality with L2 restrictions

$$\{Z_n\}_1^\infty$$is a martingale with $$\mathbb{E}Z_n^2 < \infty$$ for all $$n\geq 1$$. Then $$\mathbb{P}(\max_{1\leq n\leq m}|Z_n|\geq b)\leq \frac{\mathbb{E}Z_m^2}{b^2}$$

• As formulated, $M=\infty$ almost surely (unless the distribution is degenerate, i.e. $X_k\equiv 0$). – zhoraster Apr 30 at 15:23
• Thanks for pointing it out! I wonder whether it is well-defined if $M$ is change to $M_n$, the maximum of some finite sequence? As changed in the question. – Zishuo May 1 at 9:00
• Idea 3 is useful but we can only derive some asymptotic decrease rate with square (power) level rather than exponential level. Is this the best we can do? Because our $Z_n$ is more than just a martingale, it's a zero-mean random walk. – Zishuo May 1 at 9:16