I wrote the following proof and wanted to know if there is anything I can improve in terms of maths but also writing :
Let $S_n$ be a simple random walk in $\Bbb Z^d$. For all $n, m, k \in \Bbb N$, $x,y \in \Bbb Z^d$ $$\mathbb P (S_{n+k} = y \lvert S_n = x) = \mathbb P (S_{m+k} = y \lvert S_m = x).$$
Proof : We have by definition of the SRW and that of conditional probability that \begin{align*} \mathbb P (S_{n+k} = y \lvert S_n = x) &= \mathbb P \Bigg( \sum_{i=0}^{n+k} X_i = y \Bigg| \sum_{i=0}^{n} X_i = x \Bigg) \\ \mathbb P \Bigg( \sum_{i=0}^{n+k} X_i = y \Bigg| \sum_{i=0}^{n} X_i = x \Bigg) &= \dfrac{\mathbb P \big((\sum_{i=0}^{n+k} X_i = y )\cap (\sum_{i=0}^{n} X_i = x) \big)}{\mathbb P (\sum_{i=0}^{n} X_i = x )} \\ \mathbb P \Bigg( \sum_{i=0}^{n+k} X_i = y \Bigg| \sum_{i=0}^{n} X_i = x \Bigg) &= \dfrac{\mathbb P \big((\sum_{i=n+1}^{n+k} X_i = y-x )\cap (\sum_{i=0}^{n} X_i = x) \big)}{\mathbb P (\sum_{i=0}^{n} X_i = x )} \end{align*} By independence of the steps $X_j,X_k$ for $j\neq k$, we get \begin{align*} \dfrac{\mathbb P \big((\sum_{i=n+1}^{n+k} X_i = y-x )\cap (\sum_{i=0}^{n} X_i = x) \big)}{\mathbb P (\sum_{i=0}^{n} X_i = x )} &= \dfrac{\mathbb P \big(\sum_{i=n+1}^{n+k} X_i = y-x \big) \mathbb P \big(\sum_{i=0}^{n} X_i = x\big) }{\mathbb P (\sum_{i=0}^{n} X_i = x )}\\ \implies \mathbb P (S_{n+k} = y \lvert S_n = x) &= \mathbb P \Bigg( \sum_{i=n+1}^{n+k} X_i = y -x\Bigg) \end{align*} Since each step $X_j$ is identically distributed, $$\mathbb P \Bigg( \sum_{i=n+1}^{n+k} X_i = y -x\Bigg)=\mathbb P \Bigg( \sum_{i=1}^{k} X_i = y -x\Bigg) = \mathbb P (S_{k} = y -x \lvert S_0 = 0).$$
We get a value that is independent of $n$, meaning that $\forall n,m \in \Bbb N$ $$\mathbb P (S_{n+k} = y \lvert S_n = x) = \mathbb P (S_{m+k} = y \lvert S_m = x)= \Bbb P (S_{k} = y -x \lvert S_0 = 0).$$
I also have two questions :
I supposed that my random walk is starting at $X_0=0$, but it is not necessary right ?
In the 3rd equality, I divide by $\mathbb P (\sum_{i=0}^{n} X_i = x )$ but that quantity might be $0$ if it is impossible to reach $x$ from my starting point in $n$ steps, right ?