# Random walk on $\mathbb{Z}^2$ going forward with probability $p$

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 $$1 > p \geq 0$$?

• this can be diractlly calculate by solve a equation come from markov property. Nov 24, 2020 at 19:23
• Well, on $\mathbb{Z}^2$ there is right, left, forward and back. Nov 24, 2020 at 19:33
• uh, what is the mean of turns right and left. it go through two lattice? Nov 24, 2020 at 19:35
• Yes, it continues the same direction as the last step or it starts to walk in a new direction. Nov 24, 2020 at 19:44
• This is an interesting example of a discrete time process, which is not a Markov chain. Of course it can be formulated as a Markov chain with state space f.i. $S = \mathbb{Z}^2 \times\{u,d,l,r\}$, with $u$ representing the upper direction, $\ldots$. The formulation of the transition law is a little bit cumbersome. I think that this MC is recurrent but I don't know. Nov 24, 2020 at 21:59

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.