Why do the sum and difference of coordinates form independent simple random walk?

Question

Suppose that we are in $\mathbb{Z}^{2}$.A point starts at $(0,0)$ and goes to the left, right, up or down with equal probability $\frac{1}{4}$. If $(X_{n},Y_{n})$ denotes the position of the point at the $n-th$ step, why do $X_{n}+Y_{n}$ , $X_{n}-Y_{n}$ form independent simple random walk?

Answer 1

$(X_n, Y_n) = (X_{n-1}, Y_{n-1}) + (U_{n-1}, V_{n-1})$. That is, $(U_{n-1}, V_{n-1})$ is the increment process.

Clearly, these two $(U_{n-1}, V_{n-1})$ are not independent. Indeed, $$(U_{n-1}, V_{n-1}) = \{ (1,0), (0,1), (0,-1), (-1,0) \}$$ with probability $\frac{1}{4}$ each. (One is not zero implies other is zero).

Note that since the increments are independent, the tuple $(U_{i}, V_{i})$ and $(U_{j}, V_{j})$ are independent for any $ i,j, \quad i \ne j.$

But, consider $Z_{n} = U_{n} + V_{n}$ and $W_{n} = U_{n} - V_{n}$. These two, $Z$ and $W$ are independent. $P(Z_n = 1 | W_n = 1) = P(Z_n = 1) = \frac{1}{2}$. Check other pairs for completeness. They are independent for any $n$.

Now, $X_i+ Y_i$, and $X_i - Y_i$ are incremented by 2 independent random variables $W_i$ and $Z_i$ respectively. Thus, the independence of the sum process follows.