Module 9 · Derivatives

Option Replication Using Put-Call Parity

EN: The fundamental relationship between European call, put, underlying, and bond.
VN: Quan hệ căn bản giữa call, put, underlying và bond.

1. Put-Call Parity (European, no dividends) Core

\[ c_0 + \frac{K}{(1 + r)^{T}} = p_0 + S_0 \]

Components

c₀ European call price.
p₀ European put price.
S₀ Spot price of underlying.
K Common strike price.
K/(1+r)^T PV of risk-free zero paying K at expiry.

Both portfolios have payoff $\max(S_T, K)$ at expiry → must have equal price today.

Practice problem

S = $100, K = $100, r = 4%, T = 1 yr, c = $8. Compute p.

Show solution

PV(K) = 100/1.04 ≈ 96.15

p = c + PV(K) − S = 8 + 96.15 − 100

p ≈ $4.15

2. Synthetic Positions Core

\[ \text{Synthetic call} = p_0 + S_0 - K(1+r)^{-T} \] \[ \text{Synthetic put} = c_0 + K(1+r)^{-T} - S_0 \] \[ \text{Synthetic stock} = c_0 - p_0 + K(1+r)^{-T} \] \[ \text{Synthetic bond} = p_0 - c_0 + S_0 \]

Practice problem

c = $5, p = $3, S = $50, K = $50, r = 4%, T = 1y. Construct synthetic stock value.

Show solution

Synthetic stock = c − p + PV(K) = 5 − 3 + 50/1.04

= 2 + 48.08

Synthetic stock ≈ $50.08 (≈ S, parity holds)

3. Put-Call Parity with Dividends Core

\[ c_0 + \frac{K}{(1 + r)^{T}} = p_0 + S_0 - PV(\text{div}) \]

Practice problem

S = $80, K = $80, r = 5%, T = 1y, PV(div) = $2, c = $6. Compute p.

Show solution

p = c + PV(K) − S + PV(div)

= 6 + 80/1.05 − 80 + 2

= 6 + 76.19 − 80 + 2

p ≈ $4.19

Practice problem

S = $50, K = $50, r = 5%, T = 1 year, call premium = $5. By put-call parity, what is the put price?

Show solution

PV(K) = 50 / 1.05 ≈ 47.62

p = c + PV(K) − S = 5 + 47.62 − 50

≈ $2.62