In my course I have been given a proof that simple random walks satisfy the strong Markov property and I'm a bit stuck on a line in the proof. The theorem is stated as follows
Let $\tau$ be a stopping time with respect to the filtration $(\mathcal{G}_n)_{\in\mathbb{N}_0}$. Then on $\{\tau<\infty\}$ the process $(\hat S_n)_{\in\mathbb{N}_0}$, defined by $\hat S_n = S_{n+ \tau} - S_n$ is a SRW starting at the origin and independent of $\mathcal{G}_\tau := \{A \in \mathcal{F} : A \subset \{\tau \leq n\}\in \mathcal{G}_n \forall n \in \mathbb{N}_0 \}$.
In the proof given in my notes we arrive to the following equality which I am unsure why it holds and was wondering if anyone could point out why it holds.
$$\sum_{j=0}^\infty P(S_j-S_j = x_0 , \dots , S_{j+n} - S_j = x_n \cap A \cap \{\tau = j\})\\ =\sum_{j=0}^\infty P(S_0 = x_0 , \dots , S_n = x_n) P(A \cap \{\tau = j\})$$
I understand why the $S_i$'s change the way they do, this simply follows from the way they are defined, but what confuses me is why we can split the two events in the way we do. Since if $A$ was in the "past" of all the $S_i$'s it would follow by the Markov property for SRW but $A$ can't be in the past of $S_0$
Thanks in advance
There are two things happening. First of all, $A\cap \{\tau=j\}\in \mathcal{G}_j$, so it is independent of $(S_{j+1}-S_j,S_{j+2}-S_j,...,S_{n}-S_j)$ and this vector has the same distribution as $(S_1,S_2,...,S_n)$. Now, I did not include $S_0$ and, indeed, the above is not true unless $x_0=0$ and $S_0$ is $0$ almost surely, in which case you simply need to note that degenerate variables are independent of anything.