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Choquet-Deny: If $left(Y_1,Y_2,...right)$ are iid random variables taking values in a countable abelian group $G$, $X_n=sum_{kle n}Y_k$, and $left(X_1,X_2,...right)$ is irreducible, then every bounded $mathcal{I}$-measurable random variable is constant, where $mathcal{I}={Asubseteq G^mathbb{N}:phi(A)=A}$ is the shift-invariant sigma-algebra and $phi:left(g_0,g_1,g_2,...right)toleft(g_1,g_2,...right)$ for all $(g_0,g_1,g_2,...)in G^mathbb{N}$ is the shift map.

Lemma: For a random walk on a countable abelian group, a random variable is $mathcal{T}$-measurable if and only if it is $mathcal{I}$-measurable, where $mathcal{T}=cap_nmathcal{T}_n$ is the tail sigma-algebra and $mathcal{T}_n=sigmaleft(X_n,X_{n+1},...right)$.

Problem: If $left(X_1,X_2,...right)$ are iid and finitely supported, show that $mathcal{T}$ is trivial using Choquet-Deny and the lemma.

I think I misunderstand what's going on. In the setting of Choquet-Deny, is the following correct: let $Z$ be a bounded $mathcal{I}$-measurable random variable. The only relevance of $left(X_1,X_2,...right)$ to $Z$ is that it affects the probability of the events in $mathcal{I}$. $left(X_1,X_2,...right)$ does not change which events are in $mathcal{I}$.

The only plausible way I see is letting $Y_n=X_{n+1}-X_n$ and applying Choquet-Deny to $left(Y_1,Y_2,...right)$ and $left(X_1,X_2,...right)$. $left(Y_1,Y_2,...right)$ are identically distributed but not independent since knowing $Y_n=X_{n+1}-X_n$ changes the distribution of $Y_{n+1}=X_{n+2}-X_{n+1}$. We also lack irreducibility of $left(X_1,X_2,...right)$.