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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 $$\mathbb{P}(\max_{1\leq n\leq m}|Z_n|\geq b)\leq \frac{\mathbb{E}Z_m^2}{b^2}$$

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  • $\begingroup$ As formulated, $M=\infty$ almost surely (unless the distribution is degenerate, i.e. $X_k\equiv 0$). $\endgroup$ – zhoraster Apr 30 at 15:23
  • $\begingroup$ 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. $\endgroup$ – Zishuo May 1 at 9:00
  • $\begingroup$ 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. $\endgroup$ – Zishuo May 1 at 9:16

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