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### Subject 4. The two-period binomial model

The principles we developed for the one-period binomial model also apply to a two-period framework. Over two periods, the underlying price must follow one of four patterns: up-up, up-down, down-up, and down-down. Assuming fixed down and up percentages, the up-down and down-up sequences result in the same terminal underlying price. For each terminal stock price, a call option has a specific value.

To clarify the notation, S++ indicates the terminal underlying price if the price goes up in both periods, and c++ is the resulting call price at expiration when the underlying price goes up in both periods. Other patterns are defined accordingly.

As stated in los a, the valuation process is iterative, starting at each final node, and then working backwards through the tree to the first node (valuation date), where the calculated result is the value of the option.

Suppose the price of a stock is \$100. It can either rise or fall by 10% in any period. Assume the exercise price of a call option on the stock is \$100. The call is two periods from expiration. The risk-free rate is 6%.

The formula for π (pi) is still the same:
π = (1 + r - d) / (μ - d) = (1.06 - 0.9) / (1.1 - 0.9) = 0.8.

The stock price at expiration will be

• S++ = 100 x 1.1 x 1.1 = \$121
• S+- = S-+ = 100 x 1.1 x 0.9 = \$99.

• S-- = 100 x 0.9 x 0.9 = \$81.

When the option expires, it will be worth

• c++ = Max (0, S++ - 100) = \$21
• c+- = c-+ = Max (0, S+- - 100) = \$0.
• c-- = Max (0, S-- - 100) = \$0.

Now we move forward to the end of the first period. Suppose we are at the point where the stock price is S+. There is one period to go and two outcomes. The call price is c+ and can go up to c++ or down to c+-. Using what we know from the one-period model, the call price must be:

c+ = [π c++ + (1 - π) c+-] / (1 + r) = (0.8 x 21 + 0.2 x 0) / 1.06 = \$15.85.
c- = ... = \$0.

Now we step back to the starting point and find that the option price is still given as c = [π c+ + (1 - π) c-] / (1 + r) = (0.8 x 15.85 + 0.2 x 0) / 1.06 = \$11.96.

Recall that the hedge ratio, n, was given as the differences in the next two call prices divided by the differences in the next two underlying prices. This will be true in all cases throughout the binomial tree. Hence, we have different hedge ratios at each time point:

• n = (c+ - c-) / (S+ - S-)
• n+ = (c++ - c+-) / (S++ - S+-)
• n- = (c-+ - c--) / (S-+ - S--)

#### Practice Question 1

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.

Use the two-period binomial model to calculate the put option price.

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

#### Practice Question 2

Continue with question 1. 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: 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.

#### Practice Question 3

A stock is worth \$60 today. In a year the stock price can rise or fall by 15 percent. If the interest rate is 6%, what is the price of a call option that expires in two years and has an exercise price of \$55?

A. \$17.11.
B. \$11.98.
C. \$2.41.

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.
Call option values at expiration two years from now are:
• c++ = Max (0, 79.35 - 55) = \$24.35.
• c+- = c-+ = Max (0, 58.65 - 55) = \$3.65.
• c-- = Max (0, \$43.35 - 55) = \$0.
The option prices at the end of year 1:
c+ = (0.7 x 24.35 + 0.3 x 3.65)/(1.06) = \$17.11.
c- = (0.7 x 3.65 + 0.3 x 0)/(1.06) = \$2.41.

The call price today is c = (0.7 x 17.11 + 0.3 x 2.41)/1.06 = \$11.98.

#### Practice Question 4

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 call option expires in two years and has an exercise price of \$55. Today at time 0, a risk-free hedge consists of a short position in 10,000 calls and a long position in ______ shares of the stock.

A. 5865.
B. 7935.
C. 8167.

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.
Call option values at expiration two years from now are:
• c++ = Max (0, 79.35 - 55) = \$24.35.
• c+- = c-+ = Max (0, 58.65 - 55) = \$3.65.
• c-- = Max (0, \$43.35 - 55) = \$0.
The option prices at the end of year 1:
c+ = (0.7 x 24.35 + 0.3 x 3.65)/(1.06) = \$17.11.
c- = (0.7 x 3.65 + 0.3 x 0)/(1.06) = \$2.41.

At the current price of \$60, n = (c+ - c-) / (S+ - S-) = (17.11 - 2.41) / (69 - 51) = 0.8167.

#### Practice Question 5

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 call option expires in two years and has an exercise price of \$55. What is the number of calls that would be sold to construct a risk-free hedge at the end of year 1 if the stock price becomes \$69 (115% of the current price)? Use 1,000 shares.

A. 1224.
B. 1478.
C. 1000.

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.
Call option values at expiration two years from now are:
• c++ = Max (0, 79.35 - 55) = \$24.35.
• c+- = c-+ = Max (0, 58.65 - 55) = \$3.65.
• c-- = Max (0, \$43.35 - 55) = \$0.
The option prices at the end of year 1:
c+ = (0.7 x 24.35 + 0.3 x 3.65)/(1.06) = \$17.11.
c- = (0.7 x 3.65 + 0.3 x 0)/(1.06) = \$2.41.

n+ = (c++ - c+-) / (S++ - S+-) = (24.35 - 3.65) / (79.35 - 58.65) = 1.