Solution: Verifying the Martingale Property from First Principles

Exercise: Verifying the Martingale Property from First Principles

Part 1

$S_{n+1} = S_n + X_{n+1}$ . Since $S_n$ is $\mathcal{F}_n$ -measurable and $X_{n+1}$ is independent of $\mathcal{F}_n$ :

\mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = S_n + \mathbb{E}[X_{n+1}] = S_n + (1\cdot p + (-1)\cdot(1-p)) = S_n + (2p - 1).

For $p = 1/2$ : $\mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = S_n$ . Martingale. ✓

Part 2

For general $p$ : $\mathbb{E}[S_{n+1} \mid \mathcal{F}_n] = S_n + (2p - 1)$ .

$p > 1/2$ : RHS $> S_n$ a.s. — submartingale.
$p < 1/2$ : RHS $< S_n$ a.s. — supermartingale.
$p = 1/2$ : martingale.

Part 3

Let $q = (1-p)/p$ . Then $M_n = q^{S_n}$ . Note:

\mathbb{E}[q^{X_{n+1}}] = p\cdot q^{+1} + (1-p)\cdot q^{-1} = p\cdot\frac{1-p}{p} + (1-p)\cdot\frac{p}{1-p} = (1-p) + p = 1.

\mathbb{E}[q^{X_{n+1}}] = 1

for every $p \in (0, 1)$ — the clever choice of the ratio

q = (1-p)/p

is exactly what makes the product have mean 1.

Part 4

$M_{n+1} = M_n\cdot q^{X_{n+1}}$ . Taking conditional expectation:

\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n\cdot \mathbb{E}[q^{X_{n+1}} \mid \mathcal{F}_n] = M_n\cdot \mathbb{E}[q^{X_{n+1}}] = M_n\cdot 1 = M_n.

First equality: $M_n$ is $\mathcal{F}_n$ -measurable, pull it out. Second equality: $X_{n+1}$ independent of $\mathcal{F}_n$ , drop the conditioning. Third: part 3's calculation. Martingale. ✓

The "algebraic identity" that makes the cross-term vanish is simply $\mathbb{E}[q^{X_{n+1}}] = 1$ , which is built into the choice of $q$ .

Takeaways

The martingale property is a drift-cancellation condition. For a random walk with bias, $S_n$ has a drift; we can cancel it either by centering ( $S_n - n(2p-1)$ is a martingale) or by exponentiating with the magic ratio $q = (1-p)/p$ .
Exponential martingales for biased walks are the discrete analogue of the Doléans-Dade exponential. The continuous-time version $\exp(\sigma W_t - \tfrac{1}{2}\sigma^2 t)$ is the same idea — find the exponential transformation that makes the drift vanish.
Independence of $X_{n+1}$ from $\mathcal{F}_n$ is the critical property — it lets us drop the conditioning in the expectation. Without independence, we would need the joint distribution.
The exponential martingale drives gambler's-ruin calculations. Applied with the optional stopping theorem at the exit time, it gives closed-form ruin probabilities for biased walks.