A maximal inequality

Question

Let $\{X_i\}_{i\in\mathbb{N}}$ be i.i.d. symmetric random variables, with $-1\leq X_i\leq 1$, $\mathbb{E}(X_i) =0$, $\mathbb{E}(X_i^2) = 1$. We have that: $$ P\left(\bigcap_{k = 1}^{n}\frac{|\sum_{i = 1}^{k}X_i|}{\sqrt{k}} \leq 1\right) = P\left(\max_{k = 1,\dots,n}\frac{|\sum_{i = 1}^{k}X_i|}{\sqrt{k}} \leq 1\right) \geq g(n)\to 0 \mbox{ as $n\to\infty$} $$ I would like to find a function $n\mapsto g(n)$ such that $g(n) > 0$ which satisfy the condition above $\forall n\geq n_0$, with $n_0$ sufficiently large. Is there a "known" inequality that could help to solve this question?

Update

Using the notations given in the first answer to this post, is it possible to bound $p_{1,2^{r+1}}$ in terms of $p_{1,2^r}$? For example something like: $$ p_{1,2^{r+1}} \geq p_{1,2^r}h(r) $$ where $h(r)$ is non trivial (non identically zero, at most it can approaches $0$ as $r\to\infty$, although it would be better if $h(r) \geq c>0$). Note that in general $p_{1,2^{r+1}} \leq p_{1,2^{r}}$, so I'm asking for a function $h$ that "flips" the inequality.

An example for $h(r)$ is the following: $$ h(r) = \min_{x\in[-\sqrt{2^{r}},\sqrt{2^r}]}P\left(|S_k -S_{2^r}+ x| \leq\sqrt{2^r} ,\forall m\in\{2^{r}+1,\dots,2^{r+1}\}\right) $$ Can we do better? Because in general this function goes to $0$ really quickly.

Important

Probably I solved it, I'll post an answer soon.

Answer 1

$\newcommand\ol\overline$In accordance with the comment by the OP, consider \begin{equation*} p_{N,n}:=P\Big(\max_{k\in\ol{N,n}}\frac{|S_k|}{\sqrt{k/2}}\le1\Big), \end{equation*} where $\ol{A,B}:=\{k\in\Bbb Z\colon A\le k\le B\}$ and $S_k:=\sum_1^k X_i$.

Let us show that for all large enough $N$ and all $n\ge N$ \begin{equation*} p_{N,n}=n^{-c}; \tag{10}\label{10} \end{equation*} here and in what follows $c$ denotes various expressions that are $\asymp1$.

Indeed, without loss of generality $N=2^{r_0}$ for some natural $r_0\ge2$ and $n=2^r$ for some natural $r>r_0$. Then \begin{align*} p_{N,n}&= P(|S_k|\le\sqrt{k/2}\ \ \forall j\in\ol{r_0+1,r}\ \forall k\in\ol{2^{j-1},2^j}) \\ &\ge P(|S_{2^{j-1}}|\le b_{j-1}, |S_k|\le b_j\ \ \forall j\in\ol{r_0+1,r}\ \forall k\in\ol{2^{j-1}+1,2^j}), \end{align*} where \begin{equation*} b_j:=\sqrt{2^{j-2}+1/2}. \end{equation*} So, \begin{equation*} p_{N,n}\ge\prod_{j=r_0+1}^r q_j, \end{equation*} where \begin{align*} q_j&:=\min_{x\in[-b_{j-1},b_{j-1}]}P(|S_k-S_{2^{j-1}}+x|\le b_j\ \ \forall k\in\ol{2^{j-1}+1,2^j}) \\ &:=\min_{x\in[-b_{j-1},b_{j-1}]}P(|S_m+x|\le b_j\ \ \forall m\in\ol{1,2^{j-1}}) \\ &\to\min_{u\in[-1/2,1/2]}P(|W_t+u|\le1/\sqrt2\ \ \forall t\in[0,1])=:q>0 \end{align*} as $j\to\infty$ by the invariance principle. So, for all large enough $N$ and all $n\ge N$ \begin{equation*} p_{N,n}\ge c_1(q/2)^{r-r_0}= c_1n^{-c}, \tag{20}\label{20} \end{equation*} where $c_1:=P(|S_{2^{r_0}}|\le b_{r_0})(q/2)^{-r_0}>0$ (by the central limit theorem) if $N=2^{r_0}$ is large enough.

On the other hand, again for $N=2^{r_0}$ and $n=2^r$ with $r>r_0$, \begin{equation*} p_{N,n}\le\prod_{j=r_0+1}^r Q_j, \end{equation*} where \begin{align*} Q_j&:=P(|S_{2^j}-S_{2^{j-1}}|\le\sqrt{2^{j-1}/2}+\sqrt{2^j}/2)\to Q\in(0,1) \end{align*} as $j\to\infty$ by the central limit theorem. So, for $Q_1:=(1+Q)/2\in(0,1)$ and all large enough $N$ and all $n\ge N$ \begin{equation*} p_{N,n}\le Q_1^{r-r_0}=n^{-c}. \tag{30}\label{30} \end{equation*}

Now \eqref{10} follows from \eqref{20} and \eqref{30}. $\quad\Box$.