In Durrett 5.7.2
Let $S_n$ be an asymmetric simple random walk with $1/2 < p < 1$, and let $\sigma^2 = pq$. Use the fact that $X_n = (S_n - (p-q)n)^2 - \sigma^2n$ is a martingale to show $Var(T_b) = b\sigma^2 / (p-q)^3$.
First of all, I suppose there is a sequence $\xi_k$ of i.i.d. increments, s.t. $$\xi_k = \begin{cases}1, & \text{w.p. } ~~p \\ -1, & \text{w.p. } ~~q, \end{cases}$$ for simplicity, $S_0 = 0$. And $$S_n = \sum\limits_{k=1}^{n}\xi_k$$ I'm trying to prove the hint is indeed martingale.
Suppose, $\mathcal{F}_n = \sigma(\xi_1, \xi_2, \ldots, \xi_n)$ - a sequence of $\sigma$-algebras.
$$\mathbb{E}\left(X_{n+1} ~|~ \mathcal{F}_n\right) = \mathbb{E}\left((S_{n+1} - (p-q)(n+1))^2 - \sigma^2(n+1) ~|~ \mathcal{F}_n\right)$$ $$= \mathbb{E}\left(((S_{n} -(p-q)n) + (\xi_{n+1} - (p-q)))^2 - \sigma^2n-\sigma^2 ~|~ \mathcal{F}_n\right)$$
Direct calculations show, $\mathbb{E}(\xi_{k}) = p - q$.
Note, $S_n$ is $\mathcal{F}_n$-measurable, $\xi_{n+1}$ is independent of $\mathcal{F}_n$, then $$= X_n + 2(S_n - (p-q)n)\cdot\require{cancel}\cancelto{0}{\mathbb{E}(\xi_{n+1} - (p-q))} + \mathbb{E}((\xi_{n+1} - (p-q))^2) - \sigma^2$$
but $$\mathbb{E}((\xi_{n+1} - (p-q))^2) = (1 - p + q)^2 \cdot p + (-1 - p + q)^2 \cdot q = 4q^2p + 4p^2q = 4 \sigma^2$$
I don't get it.
Should it be $-4\sigma^2n$ instead of $-\sigma^2n$ ?
Yeah, there is a typo in the problem description; $\sigma^2 = 4pq$ for $\xi_i$
You can compute it as the following: $\mathbb{E}X^2 - (\mathbb{E}X)^2 = 1 - (p - q)^2 = 1 - (2p - 1)^2 = (1 - (2p-1))(1 + 2p - 1) = (2-2p)2p = 4pq$