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The example concludes this because the conditional expectation of the discounted time-3 stock price, given information at time 2, equals the discounted time-2 stock price. Specifically, $\mathbb{E}_2\left[\frac{S_3}{(1+r)^3}\right] = 82.64$ and $\frac{S_2}{(1+r)^2} = 82.64$, satisfying the martingale condition $M_n = \mathbb{E}_n[M_{n+1}]$.

Conditions: The expectation is taken under the risk-neutral probability $p = 2/3$.; Discounting uses the interest rate $r = 0.1$.

The example concludes this because the conditional expectation of the discounted time-3 stock price, given information at time 2, equals the discounted time-2 stock price. Specifically, $\mathbb{E}_2\left[\frac{S_3}{(1+r)^3}\right] = 82.64$ and $\frac{S_2}{(1+r)^2} = 82.64$, satisfying the martingale condition $M_n = \mathbb{E}_n[M_{n+1}]$.

Conditions: The expectation is taken under the risk-neutral probability $p = 2/3$.; Discounting uses the interest rate $r = 0.1$.

The risk-neutral probability $p$ of the up move is computed using the formula $p = \frac{(1+r)-d}{u-d}$. Given $r = 0.1$, $u = 1.2$, and $d = 0.9$, the calculation is $p = \frac{1.1 - 0.9}{1.2 - 0.9} = \frac{0.2}{0.3} = \frac{2}{3}$.

Conditions: One-period binomial model with up factor $u$, down factor $d$, and interest rate $r$.; $u > 1+r > d$ to ensure $0 < p < 1$.

The risk-neutral probability $p$ of the up move is computed using the formula $p = \frac{(1+r)-d}{u-d}$. Given $r = 0.1$, $u = 1.2$, and $d = 0.9$, the calculation is $p = \frac{1.1 - 0.9}{1.2 - 0.9} = \frac{0.2}{0.3} = \frac{2}{3}$.

Conditions: One-period binomial model with up factor $u$, down factor $d$, and interest rate $r$.; $u > 1+r > d$ to ensure $0 < p < 1$.