my question is the following:
Consider the symmetric random walk on $\mathbb{Z}$, denoted by $(X_n)$, starting at point $0$. Taking $a<0<b$, $a,b \in \mathbb{Z}$, define $T := \min(T_a,T_b)$, where $T_a, T_b$ are hitting times.
Now,the question I try to solve is the following: Show $\lim_{n \rightarrow \infty} E(X_{min(T,n)}) = E(X_T)$.
What is known: $\lim_{n \rightarrow \infty} X_{\min(T_k,n)} = X_{T_k} =k$ for some hitting time on $k$. However, I cannot seem to use this to solve the problem.
What I tried: At first, I constructed a counter-example for a stopped martingale which is not a martingale anymore. This follows simply from above, as $E(X_{T_k})=k$, but $E(X_0)=0$. Therefor, it is generally not true that one can use monotone or dominated convergence (which was my first intuition). Then I tried to split the expectation of $E(X_{\min(T,n)})= E(X_n)P(T>n)+E(X_T)P(T\leq n)$. However, I am quite sure this is very wrong, as this would lead to a contradiction in that $E(X_{min(T,n)})=0$ $\forall n \in \mathbb{N}$ (via OST). But taking the limit on both sides one would obtain $0$ for the L.H.S. and because $E(X_n)=0$ (martingale property) the R.H.S. is given by $\lim E(X_T)P(T<n) \rightarrow E(X_T)$, as $P(T<n) \rightarrow 1$, which is clearly wrong as $0 \neq 4$. Unfortunately, I cannot seem to find the mistake in that "proof" either, so if someone could point out my error I'd love to hear it.
I am also aware of Wald's Equality, but I don't see how one would apply it here.
Another approach would maybe be to use some sort of "symmetry" approach, i.e. by moving the problem onto the positive axis. However, I am not sure if (and especially how) this would help me. Any help is greatly appreciated!