Module 10 · Derivatives

Valuing a Derivative Using a One-Period Binomial Model

EN: Risk-neutral valuation in a single-period two-state model.
VN: Định giá rủi ro trung tính trong mô hình nhị thức 1 kỳ.

1. Setup Core

Stock price moves either up to $S_0 \cdot u$ or down to $S_0 \cdot d$, where $u > 1 + r > d$. Calculate option payoffs at each terminal node, then discount back using risk-neutral probabilities.

2. Risk-Neutral Probability Core

\[ \pi = \frac{(1 + r) - d}{u - d} \]

Components

π Risk-neutral probability of an up move.
u Up multiplier (e.g. 1.20 = +20%).
d Down multiplier (e.g. 0.85 = −15%).
r Risk-free rate per period.

Practice problem

u = 1.25, d = 0.80, r = 4%. Compute π.

Show solution

π = (1.04 − 0.80)/(1.25 − 0.80)

= 0.24/0.45

π ≈ 0.5333

3. Option Value Core

\[ c_0 = \frac{\pi \cdot c^{+} + (1 - \pi) \cdot c^{-}}{1 + r} \]

Components

c⁺ Call payoff if up: $\max(S_0 u - K, 0)$.
c^- Call payoff if down: $\max(S_0 d - K, 0)$.

Practice problem

S = $40, u = 1.25, d = 0.80, K = $42, r = 4%, π from above ≈ 0.5333. Value of call?

Show solution

c+ = max(50 − 42, 0) = 8; c− = 0

c0 = (0.5333×8 + 0.4667×0)/1.04 = 4.2667/1.04

c0 ≈ $4.10

Practice problem

S = $50, u = 1.20, d = 0.80, r = 5%, K = $52, T = 1 period. Value of call?

Show solution

π = (1.05 − 0.80) / (1.20 − 0.80) = 0.25 / 0.40 = 0.625

c⁺ = max(60 − 52, 0) = 8; c⁻ = max(40 − 52, 0) = 0

c = (0.625 × 8 + 0.375 × 0) / 1.05 = 5 / 1.05

≈ $4.76

4. Replication / Hedge Ratio Core

\[ h = \frac{c^{+} - c^{-}}{S_0(u - d)} \]

Number of shares to hold to replicate one call. Combine with borrowing/lending to make a risk-free portfolio.

Practice problem

Same data: S = $40, u = 1.25, d = 0.80, c+ = 8, c− = 0. Compute hedge ratio h.

Show solution

h = (8 − 0) / [40 × (1.25 − 0.80)]

= 8 / 18

h ≈ 0.444 shares per call