Question

Suppose you are given the following data: 2-month option on XYZ stock:

Underlying S = 118.49

Strike X = 120

Put price = $2.1

A. What should be the price of call to prevent arbitrage if 2-month interest rate is 6% p.a.?

B. If the actual call price was $1.0, how would you implement an arbitrage opportunity?

C. Compute your payoff at maturity.

Answer 1

All financials below are in $.

(A) S = 118.49; X = 120, P = 2.10, Risk free rate, R = 6%, Time, t = 2 months = 2/12 = 1/6 year

Call Put Parity Equation:

C + X / (1 + r)^t = S + P

Hence, C = S + P - X / (1 + r)^t = 118.49 + 2.10 - 120 / (1 + 6%)^1/6 = 1.75

The price of call to prevent arbitrage = C = 1.75

(B) Since actual price of the call is 1.00 < No arbitrage call price, we will buy the actual call and short the synthetic call. So, we will implement the following strategy:

• Short 1 stock; Cash flow today = +S = +118.49
• Short 1 put option; Cash flows today = +P = +2.10
• Lend an amount = X / (1 + r)^t = 120 / (1 + 6%)^1/6 = 118.84; Cash flow today = - 118.84
• Buy 1 Call option; Cash flows today = -C = -1.00

Net cash flows at t= 0 i.e. today = 118.49 + 2.10 - 118.84 - 1 = 0.75

(C) Payoff on maturity = Payoff from short position in stock + payoff from short position in Put + Receipt of amount lent along with interest + Payoff from long position in Call = - S_t - max (X - S_t, 0) + 118.84 x (1 + r)^t + max (S_t - X, 0)

Case 1: If S_t ≥ X then, max (X - S_t, 0) = 0; max (S_t - X, 0) = S_t - X

Hence, payoff on maturity = - S_t + 0 + 120 + S_t - 120 = 0

Case 2: If S_t < X then, max (X - S_t​​​​​​​, 0) = X - S_t; max (S_t - X, 0) = 0

Hence, payoff on maturity = - S_t - (X - S_t) + 120 + 0 = 0

Hence, payoff on maturity = 0

So, the arbitrage is we make $ 0.75 at t = 0 without any liability in future.