Exercise: Hedging-error integral on a binomial path

Consider a simplified two-step binomial model: $S_0 = 100$ , at each step $S$ goes up by factor $u = 1.10$ or down by $d = 0.909$ with equal probability. $r = 0$ , so no discounting.

A trader sells a European call with $K = 100$ , $T = 2$ steps. She hedges using Black-Scholes deltas calibrated with $\sigma_h$ such that $u = e^{\sigma_h\sqrt{\Delta t}}$ for $\Delta t = 1$ . So $\sigma_h = \ln(1.10) \approx 0.0953$ .

Under the risk-neutral measure for this model,

q = (1 - d)/(u - d) = 0.091/0.191 \approx 0.4764

. The call price at

t = 0

C_0 = q^2 \cdot (121 - 100)^+ + 2q(1-q)\cdot (100 - 100)^+ + (1-q)^2 \cdot (82.6 - 100)^+ = q^2 \cdot 21 \approx 4.77.

Tasks

Compute the Black-Scholes call price using $\sigma_h = 0.0953$ , $r = 0$ , $T = 2$ , $S_0 = K = 100$ . Does it agree approximately with $4.77$ ?
Compute the Black-Scholes delta $\Delta_0$ at $t = 0$ . This is the number of shares the trader holds initially.
Trace one path: $S_0 = 100 \to S_1 = 110 \to S_2 = 121$ . Write out the hedger's cash position at each time step (receive premium, buy $\Delta_0$ shares, rebalance to $\Delta_1$ at $t = 1$ , deliver or collect at $t = 2$ ).
Compute the net P&L on this path.
Repeat for the path $S_0 = 100 \to S_1 = 90.9 \to S_2 = 100$ . Compute the net P&L.
Compare the two P&Ls. Is the sum of their probabilities times P&L approximately zero (as Black-Scholes predicts)?

Hint

Black-Scholes ATM call formula: $C = S_0(\Phi(d_1) - \Phi(d_2))$ when $r = 0$ , $K = S_0$ ; with $d_1 = \sigma_h\sqrt T / 2$ , $d_2 = -d_1$ .