Show that, for a random walk $X$ on $\mathbb Z_2$, $X_\infty$ is independent of the past $\left\{ X_t \right\}_{0\le t<\infty}$

Question

Define the *-finite time set $T=\{0,1,\ldots,n\}$, with $n \in {^*\mathbb N} \setminus \mathbb N$.
Let $X=\left( X_t \right)_{t \in T}$ be a random walk on the cyclic group $\mathbb Z _ 2 = \mathbb Z / 2 \mathbb Z$, where $X_0$ is uniformly drawn from $\mathbb Z_2$. That is $$ P\left(X_{t} - X_{t-1} = 0 \right) =\frac12 =P\left(X_{t} - X_{t-1} = 1 \right) , \text{ for } t = T \setminus \{0\}, $$
with $P$ the *-finite measure.
Show $X_n$ is independent of the finite time part of the process, that is $$ \mathbf{P}\left(\left. X_n = x \right| X_t, t \in {^\circ T}\right) = \mathbf{P}\left( X_n = x \right), $$ where ${^\circ(\cdot)}=\text{st}(\cdot)$ is the standard part map, and $\mathbf{P}$ is the Loeb measure.

Define $A = \left\{ X_t, t \in {^\circ T} \right\}$. My approach would be to repetitively conditioned on the penultimate time \begin{align} P\left(X_n=0 \mid A \right) &= E\left( P\left(X_n=0 \mid X_{n-1}, A \right) \right) \\ &= \frac12 P\left(X_{n-1} =0\mid A \right) + \frac12 P\left(X_{n-1} =1\mid A \right) \\ &= \frac12\left( \frac12 P\left(X_{n-2} =0\mid A \right) + \frac12 P\left(X_{n-2} =1\mid A \right) \right) + \frac12\left( \frac12 P\left(X_{n-2} =0\mid A \right) + \frac12 P\left(X_{n-2} =1\mid A \right) \right) \\ &= \ldots \end{align} So, I end up with $$ P\left( X_n=0 \mid A \right) = \frac12, $$ and $X_n$ is independent of $A$.
I don't see how to 'jump' from infinite numbers of the form $n-t, t \in {^\circ T}$, to the finite numbers $t, t\in{^\circ T}$.
I guess I should use saturation or something like this.

Answer 1

Not sure what the *-things and "standard part map" refer to but anyway, you might want to note (and prove) the following:

Every random walk on $\mathbb Z_2$ with uniform initial distribution and uniform transition probabilities coincides with an i.i.d. sequence with uniform distribution on $\mathbb Z_2$.