Given random variables $Y_1, Y_2, \ldots \stackrel{\mathrm{iid}}{\sim} P(Y_i = 1) = p = 1 - q = 1 - P(Y_i = -1)$ 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:
$T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time s.t. $E[T] < \infty (\to T < \infty \ \text{a.s.})$.
$M = (M_n)_{n \ge 0}$ where $M_n = X_n - n(p-q)$ is a ($\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)-$martingale.
Prove $E[T] = \frac{b}{p-q}$. Hint: Consider $|M_{n+1} - M_n|$
I'm not sure how to use the hint, but I think I was able to prove the proposition anyway:
Attempted Proof 1:
$$\because E[X_{T \wedge n}] = (p-q)E[T \wedge n]$$
we have by DCT that
$$E[X_{T}] = (p-q)E[T]$$
where
$$E[X_{T}] = E[b] = b$$
$$\therefore, E[T] = \frac{b}{p-q} \ QED$$
Is that right?
Attempted Proof 2:
$$M_T = X_T - T(p-q) = b - T(p-q)$$
$\because M$ is a martingale and $E[M_i] = 0$, we have $E[M_T] = 0$.
$$\to E[T] = \frac{b}{p-q} \ QED$$
Is that right?
Attempted Proof 3:
$$X_{T \wedge n} = \sum_{i=1}^{T \wedge n} Y_{i+1}$$
$$M_{T \wedge n} = \sum_{i=1}^{T \wedge n} (Y_{i+1} - (p-q))$$
$$\to M_{T \wedge n} = X_{T \wedge n} - (T \wedge n)(p-q)$$
$$\to \lim M_{T \wedge n} = \lim X_{T \wedge n} - \lim (T \wedge n)(p-q)$$
$$\to M_{T} = X_{T} - (T)(p-q)$$
Thus, I reinvented the wheel. Is my reinventing right?
I'm not sure what the hint is supposed to do:
$$|M_{n+1} - M_n| = |Y_{n+1} - (p-q)| \le 1 + p - q$$
$$\to M_{T \wedge n} \le (1 + p - q)(T \wedge n) = T \wedge n + E[X_{T \wedge n}]$$
$$(?) \ \to M_{T} \le (1 + p - q)(T)$$
How can I use the hint?
