2 dimensional symmetric random walk Let $\left\{\left(X_{n}, Y_{n}\right)\right\}_{n=0}^{\infty}$ be a 2 -dimensional symmetric random walk. Namely, this is a Markov chain where $\left(X_{n+1}, Y_{n+1}\right)$ takes one of the following 4 values with equal probability:
$$\left(X_{n}+1, Y_{n}\right),\left(X_{n}-1,Y_{n}\right),\left(X_{n}, Y_{n}+1\right),\left(X_{n}, Y_{n}-1\right)$$
Suppose that $X_{0}=Y_{0}=0$.
Define $T:=\inf \left\{n \geq 0: \max \left(\left|X_{n}\right|,\left|Y_{n}\right|\right)=3\right\} .$
I want to find the value of $\mathbb{E}[T]$ and $\mathbb{P}\left(X_{T}=3, Y_{T}=0\right)$.
Thanks for your help!
 A: Let $k_{ij} = \mathbb E[T\mid X_0 = i, Y_0=j]$. So we want $k_{00}$. Noting the clear symmetry in both axes, we have the equations
\begin{align*} 
k_{00}&=1+k_{10} \\
k_{10}&=1+\frac{1}{4}k_{00}+\frac{1}{4}k_{20}+\frac{1}{2}k_{11} \\
k_{20}&=1+\frac12k_{12}+\frac14k_{10} \\
k_{11}&=1+\frac12(k_{12}+k_{10}) \\
k_{12}&=1+\frac14(k_{11}+k_{20}+k_{22})\\
k_{22}&=1+\frac12k_{12}
\end{align*}
If you crunch through this system, you will get $\boxed{k_{00}=\tfrac{135}{13}}$.
For the second part, let $p_{ij}$ be the probability, conditional on $X_0=i$, $Y_0=j$, that $(X_T,Y_T)$ is one of $(\pm3,0),(0,\pm3)$. So we want $p_{00}/4$. Similar to above, we can get the system
\begin{align*} 
p_{00}&=p_{10} \\
p_{10}&=\frac{1}{4}p_{00}+\frac{1}{4}p_{20}+\frac{1}{2}p_{11} \\
p_{20}&=\frac14+\frac12p_{12}+\frac14p_{10} \\
p_{11}&=\frac12(p_{12}+p_{10}) \\
p_{12}&=\frac14(p_{11}+p_{20}+p_{22})\\
p_{22}&=\frac12p_{12}
\end{align*}
Solving gives $p_{00}=\frac{4}{13}$, so $\boxed{\mathbb P(X_T=3,Y_T=0)=\tfrac{1}{13}}$.

In case you're interested, similar computations to the above gives $\mathbb P(X_T=3, Y_T=1)=\frac{3}{52}$ and $\mathbb P(X_T=3,Y_T=2)=\frac{3}{104}$.

Another aside: if you have a particular aversion to simultaneous equations, you might like to try some kind of martingale trick. This is especially appealing when you realise that the stopping time can be redefined as $T=\inf\left\{n\geq0\mid X_n^2+Y_n^2\geq9\right\}$. Why is this nice? Well, you can show that $M_n=X_n^2+Y_n^2-n$ is martingale, so by the optional stopping theorem,
$$\mathbb E[M_T]=\mathbb E[M_0]=0\implies\mathbb E[T]=\mathbb E\left[X_T^2+Y_T^2\right].$$
We know that $X_T^2+Y_T^2\in\{9,10,13\}$. So we certainly have the bounds $9<\mathbb E[T]<13$. Unfortunately, I don't think we can glean much more information than this from martingale magic.
A: The answer above is entirely correct. Am just here with some simple code to give you confidence that $\mathbb{E}[T] = \frac{135}{13} $

