Exam-Style Questions on Binomial Option Pricing for Actuaries

The Actuary's Free Study Guide for Exam 3F / Exam MFE - Section 23

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 23 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.

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 ESQBOP1.

Similar to Question 14 from the Casualty Actuarial Society's Spring 2007 Exam 3:

A European call option on Opaque LLC stock has the following specifications:
Strike price = $578, current stock price = $581, time to expiration = 4 years, annual continuously compounded interest rate = 0.12, dividend yield = 0. You can use a two-period binomial model to calculate the option's price.

This is the binomial tree for the stock price movements:

----------------------981.89

----------755.3 -----------

581 -------------- 604.24

---------464.8 ------------

--------------------371.84

Find the price of one European call option on Opaque LLC stock.

Solution ESQBOP1.

First, it is useful to find u and d. Here, u = 755.3/581 = u = 1.3 and d = 464.8/581 = d = 0.8.

Since this is a two-period model, we know that h = 2. Furthermore, r = 0.12, and ∂ = 0.

We calculate p* = (e^(r-∂)h - d)/(u - d) = (e^(0.12)2 - 0.8)/(1.3 - 0.8) = p* = 0.9424983006

We note that C_uu = 981.89 - 578 = C_uu = 403.89 and C_du = 604.24 - 578 = C_du = 26.24