More Exam-Style Questions on Binomial Option Pricing for Actuaries

The Actuary's Free Study Guide for Exam 3F / Exam MFE - Section 24

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 24 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. See Section 19 here. See Section 20 here. See Section 21 here. See Section 22 here. See Section 23 here.

The problems in this section were designed to be similar to problems from past versions of Exam 3F / Exam MFE. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Problem MESQBOP1.

Similar to Question 18 from the Casualty Actuarial Society's Fall 2007 Exam 3:

The stock of Devious Co. currently trades for $65 per share. The annual continuously compounded risk-free interest rate is 0.10. Every 2 years, the stock price either increases by 30% or decreases by 20%. The stock pays no dividends. Using a two-period binomial model, calculate the price of a 4-year American put option on Devious Co. stock with a strike price of $74.

Solution MESQBOP1.

We cannot bypass the intermediate values of put prices here, because for an American put, at each node in the binomial tree, we must choose the higher of two values, K - S or the otherwise equivalent European option price.

We are given that u = 1.3 and d = 0.8. Also, S = 65, r = 0.10, ∂ = 0, h = 2, and K = 74.

We first find p* = (e^(r-∂)h - d)/(u - d) = (e^(0.1)2 - 0.8)/(1.3 - 0.8) = p* = 0.8428055163

S_uu = 1.3²*65 = 109.85, implying that P_uu = 0

S_du = 1.3*0.8*65 = 67.6, implying that P_du = 74 - 67.6 = P_du = 6.4