Let $(\Omega, \mathcal{F}, P)$ be a probability space and $\{ X_i \}_{i=1}^\infty$ be $\operatorname{i.i.d}$. Assume that $P(X_1 =1) = P(X_1 =-1) = \frac{1}{2} $ and define $S_n$ by
$$\begin{cases} S_0 & = 0\\ S_n :&= \sum_{i=1}^n X_i \,\,\,\, (n \ge 1). \end{cases} $$
Let $A: = \{ S_{100} =0\} $, $B:= \{ S_{200} =0 \}$, are $A$ and $B$ independent? Answer with reason.
Find the value of $\underset{n \to n}{\lim} P(S_n \ge n^{1/3})$. For each $x \in \mathbb{Z}$, $T_x : \inf \{ n \ge 0 | S_n = x \}$, where $\inf \phi = \infty$. Show that $P(T_x < \infty) =1$ for all $x \in \mathbb{Z}$.
For each $n \ge 1$, $M_n : = \frac{\max \{ S_k | 0 \le k \le n\}}{\sqrt{n}}$. Show that $\{P^{M_n} \}_{n=1}^\infty$ convergence weekly as $n \to \infty$ in the topology of probability measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R})).$ ($\mathcal{B}(\mathbb{R})$ is Borel algebra of $\mathbb{R}$).
I have been studying this problem since two weeks and these are my solutions to the three points of the problem. However, the fourth point is not easy, I do not even know where to look fo find the solution of similar solution. I need check for the solution of point 3 and help to solve point four. Thanks in advance.
(1)
Let $ C= \{\sum_{i=101}^{200} X_i =0 \}$. Then $A \cap B = A \cap C$, so $P(A \cap B) = P(A \cap C) = P(A)P(C)$. For this to equal $P(A)P(B)$,we would need $P(B) = P(C)$, but this is false. So $A$ and $B$ are dependent.
(2)-
$\underset{n \to \infty}{\lim} P(S_n \ge n^{1/3}) = \underset{n \to \infty}{\lim} P(\frac{S_n}{\sqrt{n}} - \frac{1}{n^{1/6}} \ge 0)$. By central limit theorem we have $$\frac{S_n}{\sqrt{n}} \xrightarrow{distribution} N(0,1).$$
then
$$\left(\frac{S_n}{\sqrt{n}}, \frac{1}{n^{1/6}} \right) \xrightarrow{d} (N(0,1), 0).$$
By the continuous mapping theorem (applied to $f(x,y) = x-y$).
$$\frac{S_n}{\sqrt{n}} - \frac{1}{n^{1/6}} \xrightarrow{distribution} N(0,1)$$
and then
$$\underset{n \to \infty}{\lim} P(S_n \ge n^{1/3}) = P(N(0,1) \ge 0)=\frac{1}{2}.$$
(3)- Before solving this point we need to recall some facts.
$$T_x = \inf \{n \ge 0: S_n =x \} = \text{Hitting time on x}.$$
Since $S_0=0$ then $T_x$ is also is the first passage time to $x$.
$$V_x = \sum_{n=0}^\infty 1_{\{ S_n=x\}} = \text{ Number of visits to x}.$$
$$f_x = P_x(T_x < \infty) = \text{ return probability to x}.$$
$$m_x = E_x (T_x) = \text{ mean return time to } x.$$
$x$ is \textbf{recurrent} if $P_x (V_x =\infty)=1$. Otherwise $x$ is transient. \begin{theorem}{(1)} \label{1} $$x \text{ is recurrent } \iff f_x =1.$$
$$x \text{ transient } \iff f_x < 1.$$ \end{theorem}
\begin{proof} $$P_x (V_x <\infty) = P_x \left( \bigcup_{k \ge 1} { V_x =k } \right) = \sum_{k=1}^\infty P_x (V_x =k) = \sum_{k=1}^\infty (1- f_x) f_x^{k-1}= \begin{cases} 0, & f_x =1\ 1, & f_x <1 \end{cases}.$$
So $x$ is recurrent iff $f_x=1$, and transient iff $f_x <1$. \end{proof}
\begin{theorem}{(2)} \label{2} $$x \text{ recurrent } \iff \sum_{n=1}^\infty p_{xx}^{(n)} =\infty \iff f_x =1.$$
$$x \text{ transient } \iff \sum_{n=1}^\infty p_{xx}^{(n)} <\infty \iff f_x <1.$$ \begin{proof} If $x$ is recurrent, this means $P_x (V_x =\infty)=1$, and so
$$\sum_{n=1}^\infty p_{xx}^{(n)} = \sum_{n=1}^\infty E_x \left( 1_{\{ S_n=x \}} \right) = E_x \left( \sum_{n=1}^\infty 1_{\{S_n=x \}} \right) = E_x (V_x)=\infty.$$ If $x$ is transient then by Theorem (1) $f_i <1$ and
$$ \sum_{n=1}^\infty p_{xx}^{(n)} = E_x (V_x) = \sum_{r=1}^\infty P_x (V_x > r) = \sum_{r=1}^\infty f_x^r =\frac{1}{1-f_x} < \infty.$$
\end{proof}
\end{theorem}
Now we start solve the problem. By the Markov property we have
$$P(T_x < \infty) = \sum_{y \in \mathbb{Z}} P(X_0=y) P_y (T_x <\infty).$$ so it suffices to show that $P_y(T_x < \infty)=1$ for all $y \in \mathbb{Z}$. Choose $m$ with $p^{(m)}_{xy} > 0$. By Theorem ( 2 ) we have
\begin{align*} 1 &= P_x (S_n = x \text{ for infinitely many } n)\\ &\le P_x (S_n=x \text{ for some } n \ge m+1)\\ & = \sum_{k \in \mathbb{Z}} P_x (S_n = x \text{ for some } n \ge m+1 | S_m = k) P_x (S_m = k)\\ &= \sum_{k \in \mathbb{Z}} P_k ( T_x < \infty) p^{(m)}_{jk} \end{align*}
But $\sum_{k \in \mathbb{Z}} p^{(m)}_{jk}=1$ so we must have $P_y (T_x <\infty)=1$.
(4)-