Continue with the previous question. A stock is worth $60 today. In a year the stock price can rise or fall by 15 percent. The interest rate is 6%. A put option expires in two years and has an exercise price of $60. What is the number of shares needed to construct a risk-free hedge at each point in the binomial tree? Use 10,000 puts.

Correct Answer: See below for explanation.

The risk-neutral probability is π = (1.06 - 0.85) / (1.15 - 0.85) = 0.7, and 1 - π = 0.3.

Stock prices in the binomial tree one and two years from now are:

• S⁺ = 60 (1.15) = $69
• S^- = 60 (0.85) = $51
• S⁺⁺ = 60 (1.15) (1.15) = $79.35
• S^+- = S-+ = 60 (1.15) (0.85) = $58.65
• S^-- = 60 (0.85) (0.85) = $43.35

Put option values at expiration two years from now are:

• p⁺⁺ = Max (0, 60 - 79.35) = $0
• p^+- = p-+ = Max (0, 60 - 58.65) = $1.35
• p^-- = Max (0, 60 - 43.35) = $16.65

The option prices at the end of year 1:
p⁺ = (0.7 x 0 + 0.3 x 1.35)/(1.06) = $0.3821
p^- = (0.7 x 1.35 + 0.3 x 16.65)/(1.06) = $5.60

The put price today is p = (0.7 x 0.3821 + 0.3 x 5.6)/1.06 = $1.83.

At the current price of $60, n = (p^- - p⁺) / (S⁺ - S^-) = (5.6 - 0.3821) / (69 - 51) = 0.2899.

At the end of year 1:

• If the stock price is $69, n⁺ = (p^+- - p⁺⁺) / (S⁺⁺ - S^-+) = (1.35 - 0) / (79.35 - 58.65) = 0.065.
• If the stock price $51, n^- = (p^-- - p^-+) / (S^+- - S^--) = (16.65 - 1.35) / (58.65 - 43.35) = 1. This means that the risk-free hedge would consist of a long position in 10,000 puts and a long position in 10,000 shares.