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 $X_n = \sum_{i=1}^{n} Y_i$.
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.
Let $T$ be a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time. Prove that $\forall n \ge 1$
$$E[X_{T \wedge n}] = (p-q)E[T \wedge n]$$
What I tried:
$$E[X_{T \wedge n}] = [\sum_{i=0}^{n-1} E[X_{T \wedge n} 1_{T=i}]] + E[X_{T \wedge n} 1_{T \ge n}]$$
$$ = [\sum_{i=0}^{n-1} E[X_{T \wedge n} |T=i]P(T=i)] + E[X_{T \wedge n} | T \ge n]P(T \ge n)$$
$$ = [\sum_{i=0}^{n-1} E[X_i]P(T=i)] + E[X_n]P(T \ge n)$$
$$ = [\sum_{i=0}^{n-1} (i)(p-q)P(T=i)] + (n)(p-q)P(T \ge n)$$
$$ = [\sum_{i=0}^{n-1} (i)(p-q)P(T=i)] + (n)(p-q)\sum_{i=n}^{\infty} P(T=i)$$
$$ = [\sum_{i=0}^{n-1} (i)(p-q)P(T=i)] + \sum_{i=n}^{\infty} (n)(p-q)P(T=i)$$
$$ = \sum_{i=1}^{\infty} (n \wedge i)(p-q)P(T=i)$$
$$ = (p-q) \sum_{i=1}^{\infty} (n \wedge i)P(T=i)$$
$$ = (p-q) E[n \wedge T]$$
Is that right?
