Dtyjlui

Let us consider a biased random walk in $mathbf{Z}$ whose step-lengths satisfy $mathbf{P}(xi = +1) = p > mathbf{P}(xi = -1) = q$ (with $p+q = 1$).

A value $k in mathbf{Z}$ is a "separator" of the random walk if the random walk started before $k$ passes through $k$ only once.

By the strong Markov property $mathbf{P}_k(0 text{ is separator}) = mathbf{P}_0(text{The random walk never returns to zero}).$ Denote by $A_k$ the probability of the random walk, started at $k neq 0,$ to never visit zero and $A_0$ be the probability of never return to zero. By the strong law of large numbers, $A_k = 0$ for all $k$ negative. Now, using first step analysis is easy to deduce $A_{k + 1} = q A_k + pA_{k + 2}$ for $k geq 1,$ $A_0 = p A_1$ and $A_1 = p A_2.$ Then, we have
$$A_{k + 2} - A_{k + 1} = dfrac{q}{p}(A_{k + 1} - A_k) quad (k geq 1).$$
It is easy to show $A_{k+2}-A_{k+1}=(frac{q}{p})^{k+1} A_1$ which then leads to, by means of a telescopic sum,
$$A_k=dfrac{1}{p} left[ sum_{j=0}^k left( dfrac{q}{p} right)^j right] A_0.$$
This shows that $A_k to dfrac{1}{p} dfrac{1}{1 - frac{q}{p}} A_0 = dfrac{1}{p-q}A_0.$

My intuition suggests strongly that $A_k to 1.$ How to prove this formally?

Having this, I can conclude $A_0 = p - q,$ which is very reasonable. So, the probability of zero being separator, starting at $k < 0$ is simply $p - q.$

Consider now the set $mathrm{X}_a = {atext{ is a separator}}.$ The previous can be restated as $$mathbf{P}_0(mathrm{X}_a) = p - q$$ for every $a geq 0.$

Can I use the previous formula to conclude $mathbf{P}_0left(bigcuplimits_{a = 0}^N mathrm{X}_aright) to 1$ as $N to infty$?

You can show that
$$
(1-A_k)=(1-A_1)^kqquadtext{for all $kge 1$}tag{$*$}
$$
Why? You know $1-A_k$ is the probability of starting from $k$ and eventually hitting $0$. This is equal to the probability of starting from $k$ and moving eventually to $k-1$, then starting from $k-1$ and moving eventually to $k-2$, then $dots$ then starting from $1$ and moving to $0$. But moving from $i$ to $i-1$ is the same as moving from $1$ to $0$, so each event in that list of $k$ events has probability $(1-A_1)$.

Plugging $(*)$ with $k=2$ and $k=3$ into
$$
A_2=pA_3+qA_1,
$$
you can solve for $A_1$. You will find that $A_1<1$, so that $(*)$ implies $A_kto 1$ as $ktoinfty$ .

I found a nice solution after a while.

My first question is yes. The reason is much simpler than not. It is well known that for an irreducible Markov chain that is is recurrent we have that: for whatever the initial state $k$ may be and whatever another stat $o$ may be, the probability of starting at $k$ and eventually reaching $o$ is one.

To my second question. We consider $alpha_n = mathbf{P}_0left( bigcuplimits_{a=0}^n mathrm{X}_a^complementright)$ and notice that $(alpha_n)$ is increasing and therefore convergent. We show now that $(alpha_{n^2})$ converges to 1. To see this simply notice that

$$0 leq 1 - alpha_{n^2} leq mathbf{P}_0left( bigcuplimits_{a=0}^n mathrm{X}_{an}^complementright).$$

We can define now stopping times $tau_p$ to be the first entrance to state $p$ and measurable times $theta_p$ which are the first returns to $p.$ Then we divide
$$begin{align*}
mathbf{P}_0left( bigcuplimits_{a=0}^n mathrm{X}_{an}^complementright) &= mathbf{P}_0left( bigcuplimits_{a=0}^n mathrm{X}_{an}^complement cap{theta_0 < tau_n leq theta_n < tau_{2n}leq theta_{2n}<ldots<tau_{n^2}leqtheta_{n^2}<infty}right)\
&+mathbf{P}_0left( bigcuplimits_{a=0}^n mathrm{X}_{an}^complement capbigcup_{j=1}^n{tau_j < theta_{(j-1)n}}right)\
&leqmathbf{P}_0left(theta_0 < tau_n leq theta_n < tau_{2n}leq theta_{2n}<ldots<tau_{n^2}leqtheta_{n^2}<inftyright) + sum_{j=1}^nmathbf{P}_0(tau_j<theta_{(j-1)n})\
&mathop{=}^triangle gamma_n +sum_{j=1}^n omega_j.
end{align*}$$

To deal with $gamma_n$ simply use conditional expectation conditioning on the sigma field generated by $S_0,xi_1, ldots, xi_{tau_n}$ to reach the inequality $gamma_nleq(1-(p-q))gamma_{n-1}$ and so $gamma_n to 0$ (exponentially fast!)

Observe that $omega_j leq mathbf{P}_0(text{The random walk will ever hit} -n)$ and this is well known to be $(frac{q}{p})^n$ and clearly $n(frac{q}{p})^n to 0$ as well. Q.E.D.