Question

5. You own a call option on the stock with current price So = 4, the "up factor"u 2, the "down factor" d-1/2, the risk-free interest rater 1/4 and the strike price K = 5, You paid the risk-neutral price of $1.20 for this option, and you want to hedge your position (i.e. reduce your risk) so that you end up with $1.50, (as if you had invested S1.2 at the risk free rate) regardless of whether H or T occurs. Assume that you cannot sell the option, and you have no cash. You may only buy (or short sell) shares of stock or bond (borrow or invest at the risk-free interest rate), construct a portfolio to ensure you have S1.50 at t1

Answer 1

Suppose the market evolves over one period, from time 0 to time 1, with no intermediate valuations. At time zero we have a stock with some known price S₀ > 0, while at time 1 the price depends on chance. We model the chance as a binary choice from sample space Ω = {H, T} (here: T stands for ‘tail’, not for ‘maturity’), thus S₁ = S₁(ω) is a random variable depending on ω ∈ Ω. We may think of tossing (perhaps, biased) coin, so that if the coin lands heads the price is S₁(H) = uS₀, and if tails S₁(T) = dS₀. The ‘up factor’ u and ‘down factor’ d are some given positive numbers with d < u. The stock might move up or down with certain ‘true’ or ‘market’ probabilities p and q. However, these are of secondary interest for us, provided both events H, T are possible. For the riskless money market (bonds) we assume (simply compounded) interest rate r, so 1 pound invested at time 0 will yield 1 + r pounds at time 1. To exclude the arbitrage we must assume d < r + 1 < u

Now in the given problem we have,

X₀=$1.2

_X1=$1.5

S₀=4

u=2

d=1/2

r=1/4

K=5

t=1

Then, S₁(H)=uS₀ ; S₁(H)=2*4; S₁(H)=8

S₁(T)=dS₀; S₁(T)=1/2*4; S₁(T)=2

To hedge the option we may start with capital X₀ =$1.2, buying ∆₀ = 1/2 shares at time 0, and investing X₀ − ∆₀S₀ is 1.2-1/2*2

1.2-2=-0.8

At time 1 the cash position will be (1 + r)(X₀ − ∆₀S₀) = −1, and the stock position will be either

1/2 S₀(H) = 4 or 1/2 S₀(T) = 1, leaving us with the wealth either

X₁(H) = 1/2 S₁(H) + (1 + r)(X₀ − ∆₀S₀) = 3 i.e 1/2*8+(1+1/4)(-0.8)

or

X₁(T) = 1/2 S₁(T) + (1 + r)(X₀ − ∆₀S₀) = 0 i.e 1/2*2+(1+1/4)(-0.8)

On the other hand, the payoff of call is (S₁(H) − K)⁺ = 8 − 5 = 3 if ω = H; and (S₁(T) − K)⁺ = (2 − 5)⁺ = 0 if ω = T

We see that a stock-bond portfolio of worth X₀ = 1.2 hedges (replicates) the call. By the ‘no arbitrage’ principle the time-0 value of the call option must be X₀ = 1.2.