I am referring to Durrett's " Probability Theory and Examples". I am not including everything that's in the example as given in the book but only the relevant part for which I need explanation
Example 5.2.13:
Consider a simple symmetric random walk $S_n = S_{n-1}+\zeta_n$, where $\zeta_n$ takes values $1$ or $-1$ with probability $\frac{1}{2}$ each. Given $S_0 = 1$
Consider $\tau = inf \{n: S_n = 0\}$. Then, I know that $X_n = S_{min(\tau,n)}$ is a non-negative martingale, (so by martingale convergence theorem) $X_n$ converges to a random variable, call it $X$. Clearly, X is non-negative almost surely, and $X$ is finite almost surely.
The part I don't understand is how the author argues that $X = 0$ almost surely.
The author's argument as given in the book: "$X_n$ converges to a limit $X$, which is finite almost surely, that must be 0, since convergence to $k > 0 $ is impossible (If $X_n = k$, then $X_{n+1} = k+1$ or $k-1$)"
Can anyone clarify how "(If $X_n = k$, then $X_{n+1} = k+1$ or $k-1$)" this implies $X = 0 $ almost surely?
Best Answer
The only way a sequence of integers can converge is if that sequence is eventually constant. If $\lim X_n = X$ a.s. then for almost all $\omega$, we have $X_n(\omega) = X_{n + 1}(\omega) = X_{n + 2}(\omega) = \dots$ for large enough $n$ otherwise the limit would not exist.