Do asymmetric random walks also return to the origin infinitely? Do asymmetric random walks also return to the origin infinitely?
 A: This is a consequence of the law of large numbers. The position $S_n$ at time $n$ is the sum of $S_0$ and of $n$ i.i.d. displacements, each with expectation $m\ne0$, hence $S_n/n\to m$ almost surely. In particular, $|S_n|\ge |m|n/2$ for every $n\ge N$ where $N$ is random and almost surely finite, which implies $S_n\ne0$. Since $(S_n)$ does not visit zero after time $N$, the number of visits of zero is almost surely finite. The starting point $S_0$ is irrelevant.
A: No.  Heuristic: If the walk goes right with probability $1/2+\alpha/2>1/2$ then the expected position after $n$ steps is $\alpha n,$ while the expected variation is only $O(\sqrt n).$ Thus the walk crosses the origin only finitely often.
A: It depends. If you consider a Random Walk in a Random Environment, it may be asymmetric and recurrent. See  https://arxiv.org/pdf/0707.3160.pdf
Also, if your walk is homogeneous,
$$ X_i = \begin{cases} +1 \text{ with probability } 2/3\\
 -2 \text{ with probability } 1/3\end{cases}$$ 
let $S_n = \sum_{i=1}^n X_1$ this walk is recurrent. Indeed,  by Donsker Theorem
$$\frac{S_{[tn]}}{\sqrt{n}} \to B_t $$
where $B_t$ is a Brownian motion. This implies that 
$$P(S_n <0 \text{ infinitely often and } S_n >0 \text{ infinitely often} ) = 1$$
Since to cross from the negative to the positive this walk must first reach $0$, we conclude that this asymetric random walk visit $0$ infinitely many times.
A: Proof sketch: let $P(x,y)$ be the generating function of all walks which end up at the origin for the first time, with $x$ meaning left and $y$ meaning right. You can write a recurrence relation for the walks and deduce an expression for $P$ by solving a quadratic. Now substitute $pt$ for $x$ and $1-p$ for $y$.
A: No, it doesn't.
For a random walk, consider point of view $v_k$ as:


*

*let $+1$ and $-k$ be the two "fozen points", $p_k$ is the probability hitting $+1$, $1-p_k$ hitting $-k$, or, equivalently

*let $-1$ and $+k$ be the two "fozen points", $q_k$ is the probability hitting $-1$, $1-q_k$ hitting $+k$.


With such two "forks", we can construct view $v_{k+1}$, such that


*

*let $+1$ and $-(k+1)$ be the two "fozen points", $p_{k+1}$ is the probability hitting $+1$, $1-p_{k+1}$ hitting $-{k+1}$, or, equivalently

*let $-1$ and $+{k+1}$ be the two "fozen points", $q_{k+1}$ is the probability hitting $-1$, $1-q_{k+1}$ hitting $+{k+1}$.


We have: $p_{k+1} = \frac{p_k}{1-(1-p_k)(1-q_k)}$ and $q_{k+1} = \frac{q_k}{1-(1-p_k)(1-q_k)}$.
Obviously, $p_{k+1}/q_{k+1} = p_k/q_k$, so points $(p_k, q_k)$ line up on a line from the origin $(0,0)$.
Also, $1/p_k > p_{k+1}/p_k = q_{k+1}/q_k < 1/q_k$, so points $(p_k,q_k)$ are bounded in the square $(0,0)$, $(1,0)$, $(1,1)$ and $(0,1)$.
So, starting if $p_1 = 1-q_1 < 1/2$, the probability of hitting $+1$ is $p_\infty = 1/q_1 < 1$, vice versa. $p_1 = q_1 = 1/2$ is the only situation that has probability $1$ to hit both $+1$ and $-1$.
Hence, the probability of returning to origin, is less than $1$, unless the random walk is symmetric.
