TITLE: Random walk on $\mathbb{Z}^2$ going forward with probability $p$
QUESTION [5 upvotes]: Consider a random walk on $\mathbb{Z}^2$ which goes forward (i.e. takes a step in the same direction as the last step) with probability $p$ and turns right and left with probability $\frac{1-p}{2}$ respectively. Is it recurrent for all $1 > p \geq 0$?

REPLY [3 votes]: It looks like recurrence follows from Theorem 1 in Bender and Richmond, Correlated Random Walks, Ann. Probab. 12(1) (1984): 274–278 DOI:10.1214/aop/1176993392. (It gets late, though, so I may be getting something wrong.)
If $X_n$ is the "random walk" described in the question and $D_n$ is the "direction" in which the random walk is moving in step $n$, then $(D_n, X_{n+1} - X_n)$ is the "correlated random walk" according to the terminology of Bender and Richmond. Thus, we have to verify three conditions:

The "correlated random walk" is drift-free, that is, $\mathbb E|X_n| = o(n)$. This seems to follow from exponential convergence of the distribution of $D_n$ to the uniform distribution over four admissible directions.

Condition A holds: for some $d$ and $d'$, the support of the conditional distribution $X_{n+1} - X_n$ given $D_n = d$ and $D_{n+1} = d'$ is a linearly dense set in $\mathbb R^2$. This deepends on the exact definition of $X_n$, but even if this condition fails, the "correlated random walk" $(D_{2n}, X_{2n+2} - X_{2n})$ satisfies this condition.

Condition B holds: the Markov chain $D_n$ (or $D_{2n}$) is "strongly irreducible": this is clearly true.