Given random variables $Y_1, Y_2, ... \stackrel{iid}{\sim} P(Y_i = 1) = p = 1 - q = 1 - P(Y_i = -1)$ where $p > q$ in a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^Y$,
define $X = (X_n)_{n \ge 0}$ where $X_0 = 0$ and $X_n = \sum_{i=1}^{n} Y_i$
Let $b$ be a positive integer and $T:= \inf\{n: X_n = b\}$.
It can be shown that the stochastic process $M = (M_n)_{n \ge 0}$ where $M_n = X_n - n(p-q)$ is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$-martingale.
Prove that $T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time.
What I tried based on my previous question:
Case 1: b is odd
$$\emptyset = \{T = 0\} = \{T = 1\} = ... = \{T = b-1\} = \{T = b+1\} = ... = \{T = 2n\} = ... \in \mathscr F_0 \subseteq \mathscr F_i \ (i = 0, 1, ..., b-1, b+1, ..., 2n, ...)$$
$$\{T = b\} = \{Y_1 = ... = Y_b = 1 \} \in \mathscr F_b$$
$$\{T = b+2\} = \{Y_1 + ... = Y_{b+2} = b \} \setminus \{T = b\} \in \mathscr F_{b+2}$$
$$\vdots$$
$$\{T = 2n+1\} = \{Y_1 + ... = Y_{2n+1} = b \} \setminus (\{T = b\} \cup \{T = b+1\} \cup \{T = 2n - 1\})\in \mathscr F_{2n+1}$$
Case 2: b is even
$$\emptyset = \{T = 0\} = \{T = 1\} = ... = \{T = b-1\} = \{T = b+1\} = ... = \{T = 2n+1\} = ... \in \mathscr F_0 \subseteq \mathscr F_i \ (i = 0, 1, ..., b-1, b+1, ..., 2n+1, ...)$$
$$\{T = b\} = \{Y_1 = ... = Y_b = 1 \} \in \mathscr F_b$$
$$\{T = b+2\} = \{Y_1 + ... = Y_{b+2} = b \} \setminus \{T = b\} \in \mathscr F_{b+2}$$
$$\vdots$$
$$\{T = 2n\} = \{Y_1 + ... = Y_{2n} = b \} \setminus (\{T = b\} \cup \{T = b+1\} \cup \{T = 2n - 2\})\in \mathscr F_{2n}$$
QED
Is that right?
