Binomial Option Pricing with Puts: Practice Problems and Solutions

This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 3F / Exam MFE, authored by Mr. Stolyarov.

This is Section 19 of the Study Guide. See Section 1 here. See Section 2 here. See Section 3 here. See Section 4 here. See Section 5 here. See Section 6 here. See Section 7 here. See Section 8 here. See Section 9 here. See Section 10 here. See Section 11 here. See Section 12 here. See Section 13 here. See Section 14 here. See Section 15 here. See Section 16 here. See Section 17 here. See Section 18 here.

Binomial option pricing with puts can be done using the exact same formulas and conceptual tools developed in Sections 15-18 except that calculating the put price at expiration uses the formula P = max(0, K – S) instead of C = max(0, S – K).

Here, we will use one and two-period binomial models for all practice problems, because the objective of this section is to establish the conceptual approach to binomial option pricing with puts. Students should be aware that this approach can translate to larger multi-period models as well, using the same essential procedure as the one illustrated here.

Source: McDonald, R.L., Derivatives Markets (Second Edition), Addison Wesley, 2006, Ch. 10, pp. 328-329.

Original Practice Problems and Solutions from the Actuary’s Free Study Guide:

Problem BOPWP1. The stock of Predictable Co. is currently worth $100 per share. In one year, this price can either be $120 or $90. Predictable Co. stock does not pay dividends. The annual continuously compounded risk-free interest rate is 5%. The strike price of a European put option on Predictable Co. stock is $130. Using, the one-period binomial option pricing model, find the price today of one such put option on Predictable Co. stock.

Solution BOPWP1. First, we consider the put option price tree

P – – – P_u

P – – – P_d

In one year, if the stock is worth $120, the put option will be worth P_u = 130 – 120 = 10.

If the stock is worth $90, the put option will be worth P_d = 130 – 90 = 40.

We are given ∂ = 0, r = 0.05, S = 100, h = 1, u = 1.2, and d = 0.9.

We can still use the same formula for the risk-neutral probability of the stock price’s increase next year:

p* = (e^(r-∂)h – d)/(u – d) = (e^0.05 – 0.9)/(1.2 – 0.9) = p* = 0.5042369879

We also note that

P = e^-rh[p*P_u + (1 – p*)P_d] = e^-0.05[0.5042369879*10 + (1 – 0.5042369879)40] =

P = $23.65982519

Problem BOPWP2. Currently, the annual continuously-compounded interest rate is 0.11. Company Co. stock trades for $23 per share, and the annual continuously-compounded dividend yield on Company Co. stock is 0.05. In two months, Company Co. stock will trade for either $18 per share or $29 per share. The strike price of a European put option on Company Co. stock is $30. Using the one-period binomial option pricing model, find the price today of one such put option on Company Co. stock.

Solution BOPWP2.

First, we consider the put option price tree

P – – – P_u

P – – – P_d

In 2 months, if the stock is worth $29, the put option will be worth P_u = 30 – 29 = 1.

If the stock is worth $18, the put option will be worth P_d = 30 – 18 = 12.

We are given ∂ = 0.05, r = 0.11, S = 23, h = 1/6, u = 29/23, and d = 18/23.

p* = (e^(r-∂)h – d)/(u – d) = (e^{(0.11-0.05)/6} – 18/23)/(29/23 – 18/23) = p* = 0.4755594403

P = e^-rh[p*P_u + (1 – p*)P_d] = e^-0.11/6[0.4755594403*1 + (1 – 0.4755594403)12] =

P = $6.645881266

Problem BOPWP3. The stock of Reputable LLC will sell for either $130 or $124 one year from now. The annual continuously compounded interest rate is 0.11. The risk-neutral probability of an increase in the stock price (to $130) is 0.77. Using the one-period binomial option pricing model, find the current price of a one-year European put option on Reputable LLC stock with a strike price of $160.

Solution BOPWP3.

We note that P_u = 160 – 130 = P_u = 30 and

P_d = 160 – 124 = P_d = 36.

Here, p* = 0.77, h = 1, and r = 0.11.

So P = e^-rh[p*P_u + (1 – p*)P_d] = e^-0.11[0.77*30 + (1 – 0.77)36] = P = $28.11127517

Problem BOPWP4. Gregarious, Inc., stock is currently worth $56. Every year, it can change by a factor of 0.9 or 1.3. The stock pays no dividends, and the annual continuously-compounded risk-free interest rate is 0.04. Using a two-period binomial option pricing model, find the price today of one two-year European put option on Gregarious, Inc., stock with a strike price of $120.

Solution BOPWP4. In one year, the stock will either be worth S_u = 1.3*56 = 72.8, or it will be worth S_d = 0.9*56 = 50.4. In two years, the stock will either be worth

S_uu = 1.3²*56 = 94.64 or S_ud = S_du = 1.3*0.9*56 = 65.52 or S_dd = 0.9*0.9*56 = 45.36.

At S_uu = 94.64, the put is worth P_uu = 120 – 94.64 = P_uu = 25.36

At S_du = 65.52, the put is worth P_du = 120 – 65.52 = P_du = 54.48

At S_dd = 45.36, the call is worth P_dd = 120 – 45.36 = P_dd = 74.64

Now we calculate p* = (e^(r-∂)h – d)/(u – d) = (e^0.04 – 0.9)/(1.3 – 0.9) = 0.3520269355.

We note that there can be a direct calculation of the put price today using the three possible put prices two periods from now using the binomial model. The formula might make intuitive sense to you if you consider the way a binomial probability distribution works:

P = e^-2rh[(p*)²P_uu + (p*)(1-p*)P_ud + (1 – p*)²P_dd]

P = e^-2*0.04[(0.3520269355)²25.36 + (0.3520269355)(1- 0.3520269355)54.48 + (1 – 0.3520269355)²74.64] = P = $43.30229835.

This approach is much faster than finding the intermediate put prices. The identical kind of formula can be applied to call pricing using the two-period binomial model as well.

Problem BOPWP5. Complicated, Inc., pays dividends on its stock at an annual continuously compounded yield of 0.06. The annual effective interest rate is 0.09. Complicated, Inc., stock is currently worth $100. Every two years, it can change by a factor of 0.7 or 1.5. Using a two-period binomial option pricing model, find the price today of one four-year European put option on Gregarious, Inc., stock with a strike price of $130.

Solution BOPWP5. We are given that r = 0.09, ∂ = 0.06, and h = 2. Thus,

(r-∂)h = (0.09 – 0.06)*2 = 0.06.

p* = (e^(r-∂)h – d)/(u – d). Here, for every time period, p* = (e^0.06 – 0.7)/(1.5 – 0.7) = p* = 0.4522956832.

We find

S_uu = 1.5²*100 = 225, which implies that P_uu = 0

S_ud = S_du = 1.5*0.7*100 = 105, which implies that P_ud = 130 – 105 = P_ud = 25

S_dd = 0.7²*100 = 49, which implies that P_dd = 130 – 49 = P_dd = 81.

Now we use the great time-saving formula

P = e^-2rh[(p*)²P_uu + (p*)(1-p*)P_ud + (1 – p*)²P_dd] =

e^-2*0.09*2[0 + (0.4522956832)(1-0.4522956832)25 + (1 – 0.4522956832)²81] =

P = $21.27319

See other sections of The Actuary’s Free Study Guide for Exam 3F / Exam MFE.