Question

An economist writes a 1-period expectation model for valuing options. The model assumes that the stock starts at S and moves to 2S or 0.5S

in 1 year’s time with equal probability. Assume rates are zero.

(a) Using this expectation model what is the value of a call option struck at K?

(b) Now use the 1-period binomial model to calculate the risk-neutral probabilities and thus calculate the risk-neutral value of this call op- tion?

(c) Is there an arbitrage between these models? How could you capture it?

Answer 1

Part(a)

Probability of up state, P_u = 0.5, S_u = 2S, Hence value of Call in up state, C_u = max(S_u - K, 0) = max (2S - K, 0) = 2S - K

Probability of down state, P_d = 0.5, S_d = 0.5S, Hence value of Call in up state, C_d = max(S_d - K, 0) = max (0.5 - K, 0) = 0

Hence, expected value of call at the end of period 1 = P_uC_u + P_dC_d = 0.5 x (2S - K) + 0.5 x 0 = S - 0.5K

Assume continuous compounding of interest rate.

Hence value of call today = PV of expected value of call at the end of 1 year = e^-rT(S - 0.5K)

T = 1 and r = risk free rate = 0 (question says assume rates are zero), Hence, e^-rT = e^-0x1 =1.

Hence, value of call today = e^-rT(S - 0.5K) = S - 0.5K

Part (b)

Let's say the risk neutral probability of stock price going up is p. Hence the risk neutral probability of stock going down is (1-p).

Expected value of the stock at t=1 will be

= p x S_u + (1-p) x S_d = p x 2S + (1 - p) x 0.5S = 0.5S + 1.5pS

In the risk neutral world, the expected price of stock one year down the line will always be S₀e^rT where r = risk free rate. As r = 0 and T = 1, expected value of stock = S₀e^rT = S x e^0x1 = S

Hence, 0.5S + 1.5pS = S

Hence, p = 0.50 / 1.50 = 1/3

Hence expected value of the call at the end of t=1 in the risk neutral world = pC_u + (1-p)C_d = 1/3 x (2S - K) + (1 - 1/3) x 0 = 1/3 x (2S - K)

Hence value of call today = PV of expected value of call at the end of 1 year = e^-rT x 1/3 x (2S - K)

T = 1 and r = risk free rate = 0 (question says assume rates are zero), Hence, e^-rT = e^-0x1 =1.

Hence, value of call today = e^-rT x 1 / 3 x (2S - K) = 1 / 3 x (2S - K) = 2 / 3 x (S - 0.5K)

Part (c)

As we can see, the value of the call option today in part (a) = S - 0.5K

Value of the call option today in part (b) = 2 / 3 x (S - 0.5K)

Since the values are not equal, there is an arbitrage opportunity.

In order to capture the arbitrage:

1. First create the replicating portfolio - a portfolio of borrowing and buying stocks that matches the payoff from the call option.
2. We will have to buy that portfolio today and short the call at the price derived in part (a)
3. This will result into a surplus cash in hand at the start and result into a positive gain or no gain in future.
4. Thus we have created a state where there is a gain without any investment. This is the arbitrage.