Sweep-segment bot: Will this random walk sweep the plane?

Let us call $[A_n,B_n]$ the segment at time $n$. Then, for all $n \ge 1$:

1) $A_n$ is chosen uniformly on the circle of radius $1$ centred at $A_{n-1}$ (independently of the past).

2) $B_n=A_{n-1}$.

Therefore $A_n$ performs a simple random walk.

Once Didier touched it, let me comment that we do not need the full $[-\pi,\pi]$ swing here. Any fixed non-degenerate sectors for $A$ and $B$ rotations would do the job perfectly well.

As Didier proposed, we consider any "reinforced random walk with translation and circular symmetry" that operates like that: at each moment, we have a point $x_k$ (with $x_0=0$, for definiteness) and a rotation $R_k$ ($R_0$ can be anything). Each step consists of adding to $x_k$ the vector $R_k w_k$ and multiplying $R_k$ by $S_k$ to get $x_{k+1}$ and $R_{k+1}$ respectively where $(w_k,S_k)$ are i.i.d. pairs with some joint distribution satisfying some natural assumptions (we do not need much here: to know that $w_k$ are bounded and that the joint density is bounded is more than enough and our process falls under those assumptions if we agree to view 10 steps as one)

The proof starts the same way as Polya's recurrence argument: if the probability fof $x$ to return to some neighborhood of the origin is less than $1$, then, for any bounded region $U$, the probability for $x$ to visit $U$ more than $k$ times decays exponentially with $k$, so all moments of the number of visits are finite, etc.

Now, cover the plane by unit disks in some nice way. Two cases are possible: A) The probability to ever visit the disk $Q$ decays to $0$ as the distance from $Q$ to the origin tends to $0$. B) There are arbitrarily far disks $Q$ (depending on $R_0$, of course) that are hit with probability at least $q$ for some fixed $q>0$.

We'll show that both (A) and (B) imply recurrence.

In case (A), notice that the expectation $E|x_k|^2\le Cn$ for $k\le n$ (just estimate scalar products of increments $(R_kw_k,R_mw_m)$. they decay exponentially with $|k-m|$) , so if we denote by $v(Q)$ the number of visits to $Q$ after $n$ steps, we'll have $E\sum_Q v(Q)d(Q)^2\le Cn^2$ where $d(Q)$ is the distance from $Q$ to the origin. Thus, the sum of $v(Q)$ over the disks with $d(Q)<C\sqrt n$ is at least $n/2$.

On the other hand, under the non-recurrence assumption, $E\sum_{Q:d(Q)\le C\sqrt n}v(Q)^2\le C_1 n$ (each term is bounded). It follows that, after $n$ steps, the number of visited unit disks with $d(Q)\le C\sqrt n$ is at least $cn$ with noticeable probability. But the assumption (A) immediately implies that the expected number of disks with that property that we ever visit is $o(n)$ - Contradiction.

In case (B), we note first that if $Q_0$ is hit with probability $\ge q$ by the process starting with some $R_0$, then $CQ_0$ is hit with probability $\ge q/2$ starting with any $R_0$ if $C$ is big enough. Indeed, if the initial rotations are close enough, we have some small difference in the first increments but then we can use the representations of these 2 processes in which all other increments are the same outside a very small probability event. To pass from the small differences to everything, it is enough to note that it is highly unlikely not to have $R_k$ in any given open set at least once during the first $M$ steps if $M$ is big enough no matter what rotation we start with. But the first $M$ steps can shift us by $CM$ at most and we can compensate for that by expanding the disk.

The moral now is that every disk $CQ$ at the same distance as $CQ_0$ from the origin is hit with probability $\ge q/2$ no matter what rotation we start with. Then every disk $2CQ$ whose distance from the origin is not bigger than twice that of $CQ_0$ is hit with probability $\ge q^2/4$. Since the distance $d(Q_0)$ could be taken arbitrarily large, we eventually hit every disk of radius $2C$ with fixed positive probability no matter how we start. This is the same as recurrence for the processes under consideration.

Of course, this all is well-known in much higher generality but it was easier to present this (sketch of a) proof than to dig through the literature. Now it just remains to prove that I'm a human being (Grrrr... That captcha really acts on my nerves).

Call $[A_n,A_n+V_n]$ the segment at time $n$ assuming that $A_n$ is the end of the segment around which it rotates between times $n$ and $n+1$. Thus $A_n$ and $V_n$ are complex numbers, $A_0=1=V_0$, and $V_n$ has modulus $1$ for every $n$. Then (I believe that) for every $n\ge0$, $$ A_{n+1}=A_n+U_{n+1}V_n,\qquad V_{n+1}=-U_{n+1}V_n,$$ where $(U_n)_{n\ge0}$ is an i.i.d. sequence with a given distribution on the unit circle.

This shows that $(A_n)_n$ is not a Markov chain in general except for the case you said you were interested in, where the random variables $U_n$ are uniform on the unit circle, that $(A_n,V_n)_n$ is always a Markov chain, and that $$ V_n=(-U_1)\cdots(-U_n),\qquad A_n=1-\sum_{k=1}^n(-U_1)\cdots(-U_k). $$ We seem to be back to some known territory here...