On showing an ''inverting'' non symmetric random walk has $\mathbb{E}(S_n^2)=O(n)$

Question

The random walk position at step $n+1$ is given by $S_{n+1}=S_n+1$ with probability $1-p$ and $S_{n+1}=-S_n$ with probability $p$. For the initial position consider $P(S_0=0)=p, P(S_0=1)=1-p$. We have $\mathbb{E}(S_n^2)=O(n)$ for any $p\in(0,1)$. It would be enough for my purposes to show $\mathbb{E}(|S_n|)=O(n^{1/2})$, but I think that's more difficult.

I was not able to prove this, only verified numerically. Intuitively this is because we have a chance of loosing steps in the right direction after inversion, since for example, $S_{100}=100$ and the event with probability $p$ happens we start loosing $|S_n|$ with -100,-99,...

The motivation to consider this is that the absolute value of $S_n$ gives the number $|n_1|$ of $a$'s in a random product of $a$'s and $b's$ with $a^k b a^k=b, k\in \mathbb{N}$, $b^2=1$. In this product $a$ happens with probability $1-p$ and $b$ with probability $p$. With the relations we can write an n-th product $A^{(n)}= a^{n_1}b^{n_2}$. This appears in the theory of Lyapunov exponents by considering $a$ the matrix diag$(2,1/2)$ and $b$ a rotation by $\frac{\pi}{4}$ ($b^2=-1$ in this case, but we're interested in the norm).

Answer 1

Consider this random walk as a sequence of "runs", i.e. let $\xi_i=0$ with probability $1-p$ and $\xi_i=1$ with probability $p$, and $$ S_{n+1}=\begin{cases} S_n+1, &\text{ if }\xi_i=0;\\ -S_n, &\text{ if }\xi_i=1. \end{cases} $$ Also set $\xi_0=0$. In the sequence $\{\xi_i\}$ consider the terms are equal to $1$, and the numbers of zeros between them, e.g. if $$ \xi_1,\xi_2,\xi_3=0,0,1,0,0,0,0,0,1,1\dots $$ let $\nu_1=2$, $\nu_2=5$, $\nu_3=0$, etc. Formally let $\tau_0=0$ and $$ \tau_k=\inf\{i>\tau_{k-1}:\ \xi_i=1\},\quad k=1,2,\dots $$ and $\nu_k=\tau_k-\tau_{k-1}-1$. Observe that $\nu_k$ are i.i.d. geometric random variables, $\mathbb P(\nu_k\ge \ell)=(1-p)^\ell$, $\ell=0,1,2,\dots$.

Define the embedded process $T_m=S_{\tau_m-1}$. Then¸ assuming $S_0=1$, we have $$ T_m=\nu_1-\nu_2+\nu_3-\dots \pm \nu_m $$ (depending on parity), since when $S$ becomes negative, instead of $S+1$ we get $-S+1=-(S-1)$.

Given that all $\nu$ have the same distribution, we conclude that $$ \mathbb E T_m= \begin{cases} 0, &\text{ if $m$ is even};\\ \frac{1-p}p, &\text{ if $m$ is odd}. \end{cases} $$ and $$ Var(T_m)=\frac{1-p}{p^2}m. $$ Moreover, using large deviation estimates $$ \mathbb P(|T_m|\ge m^{1/2+\delta})\asymp \exp\left(-\frac{p^2}{2^{2\epsilon}(1-p)}m^{2\delta}\right) $$ for $\delta>0$.

Finally, by the strong law of large numbers, $\tau_k=\left(\frac{1-p}p+o(1)\right) k$, hence $S_n$ has the same order in $n$ as $T_k$ in $k$, i.e., $S_n=O(\sqrt{n})$.