Let $X(n)$ be a simple random walk on $\Bbb{Z}^2$. Also we define
- $S_{R} = \inf\{n > 0 : X(n) \notin [-R, R]^2 \} $ : the exit time of the square $[-R, R]^2$,
- $T_{v} = \inf\{n > 0 : X(n) = v\}$ : the hitting time of the lattice point $v \in \Bbb{Z}^2$,
I want to consider two conditional probability
$$ p(w \to v) := \Bbb{P}(X(T_v-1) = w \mid T_{v} < S_{R}). \tag{1} $$
In other words, I want to track the entering direction of my random walk when it hits $v$ before it hits the boundary.
Now fix $x = (a, b)$ in the open square $(-1, 1)^2$. Then $v = v(R) = (\lfloor aR \rfloor, \lfloor bR \rfloor) \in \Bbb{Z}^2$ and we can consider
$$ P(x, e) = p(v+e \to v) \quad \text{for} \quad e \in \{(0, 1), (0, -1), (1, 0), (-1, 0)\} $$
Question. Does $P(x, e)$ converge to $1/4$ as $R \to \infty$ for any $e$? If this is the case, how fast the convergence takes place?
An ideal situation for my case would be that we have
$$ P(x, e) = \tfrac{1}{4} + \mathcal{O}_x (R^{-1}), \tag{2} $$
where the bound for $\mathcal{O}_x$ behaves quite well away from the origin and the boundary of the square. But my Monte-Carlo simulation seems to suggest that convergence would be slower, so I wonder if we have any tool to analyze this probability.
Postscript. I am also interested in the probability of 'last exit direction'
$$ q(v \to w) := \Bbb{P}^{v}(X(1) = w \mid T_{v} > S_{R}), \tag{2} $$
but this is easy to analyze for at least two reasons: it does not depend on the history, and the conditioning event holds with high probability. So I omitted this from my question.
Addendun. From numerical simulations with $R = 500, 1000, 2000, 4000$ and $x = (0, 0.5)$, I obtained the following log-plot for
$$(R, \textstyle \max_e |P(x, e) - 1/4|)$$
as follows:
So it seems not pessimistic to expect that (2) is actually true.