Revised Exam-Style Questions on Option Elasticity, Option Volatility, and the Black-Scholes Formula

The Actuary's Free Study Guide for Exam 3F / Exam MFE - Section 46 (Version 2.0)

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 the revised Section 46 of the Study Guide. See an index of all sections by following the link in this paragraph.

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 ESQOEOVBSF1.

Similar to Question 22 from the Casualty Actuarial Society's Fall 2007 Exam 3:

Evasive Co. stock currently trades for $77 per share. The stock pays no dividends, and the annual continuously-compounded risk-free interest rate is 0.05. The stock price volatility is 0.44. A call option on Evasive Co. stock has a strike price of $74 and time to expiration of 2 years. Calculate Ω, the elasticity of this call option within the Black-Scholes framework.

Solution ESQOEOVBSF1.

In order to use the formula Ω = S∆/C, we first need to find C and ∆.

To find C, we use the Black-Scholes formula, where

d₁ = [ln(S/K) + (r - ∂ + 0.5σ²)T]/[σ√(T)] = [ln(77/74) + (0.05 - 0 + 0.5*0.44²)2]/[0.44√(2)] =

d₁ = 0.5356981973

d₂ = d₁ - σ√(T) = 0.5356981973 - 0.44√(2) = d₂ = -0.0865557701

In MS Excel, using the input "=NormSDist(0.5356981973)", we find that N(d₁) = 0.70391645

In MS Excel, using the input "=NormSDist(-0.0865557701)", we find that N(d₂) = 0.465512247

Now we use the Black-Scholes formula:

C(S, K, σ, r, T, ∂) = Se^-∂TN(d₁) - Ke^-rTN(d₂) = 77*0.70391645 - 74*e^-2*0.050.465512247 =

C = 23.03181208

Also, ∆_call = e^-∂TN(d₁), which in this case is just N(d₁) = 0.70391645. Thus,

Ω = S∆/C = 77*0.70391645/23.03181208 = Ω = 2.353334877

Problem ESQOEOVBSF2.