Let $Y_0 = 0$ and $(Y_n)_{n \in \mathbb{N}_0}$ be an iid sequence of random variables with zero mean and attaining exactly values in $\{1, 0, -1\}$ with positive probability. Consider the filtration $F^Y$ generated by $Y = (Y_n)_{n \in \mathbb{N}_0}$ . Define the process $X = (X_n)_{n \in \mathbb{N}_0}$ via $X_n = \sum_{k=0}^n Y_k$ for all $n \in \mathbb{N}_0$. Moreover, define $\tau = \inf \{n \in \mathbb{N}:X_n = 1\}$, which is a stopping time corresponding to $F^Y$. Is $\tau$ finite?
This is an exercise I need to solve. Intuitionally, I think it is not true. I found a similar topic here (Martingale: Show $p\{T<+\infty \}=1$.), but the definition of $\tau$ in that topic is different, and that difference enables us to conclude from the equation $()−(+1)=(−1)−()$ and $(−)=()=1$ that $u(i) = 1$ for all i. In this situation, we only have $u(1) = 1$, so we cannot use the same reasoning to conclude that. Can anyone give me some hint on how to solve this?
Let $p=P[Y_n=1] = P[Y_n=-1]$ (equal because $E[Y_n]=0$), and $r=P[Y_n=0]$ for $n\ge 1$. Clearly $2p+r=1$. Let $b:=P[\tau<\infty]$. By conditioning on the first step of the walk you see that $$ b = p+rb+pb^2. $$ That is, because $1-r = 2p$, $b$ solves the quadratic $$ pb^2-2pb+p=0. $$