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 33 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. See Section 24 here. See Section 25 here. See Section 26 here. See Section 27 here. See Section 28 here. See Section 29 here. See Section 30 here. See Section 31 here. See Section 32 here.
The Black-Scholes formula for option pricing is valid under the following six assumptions, as stated by R. L. McDonald:
“1. Continuously compounded returns on the stock are normally distributed and independent over time.
“2. The volatility of continuously compounded returns is known and constant.
“3. Future dividends are known, either as a dollar amount or as a fixed dividend yield.
“4. The risk-free interest rate is known and constant.
“5. There are no transaction costs or taxes.
“6. It is possible short-sell costlessly and to borrow at the risk-free rate.”
The Black-Scholes formula for the call price is
C(S, K, σ, r, T, ∂) = Se-∂TN(d1) – Ke-rTN(d2)
where d1 = [ln(S/K) + (r – ∂ + 0.5σ2)T]/[σ√(T)] and d2 = d1 – σ√(T)
The Black-Scholes formula for the put price is
P(S, K, σ, r, T, ∂) = Ke-rTN(-d2) – Se-∂TN(-d1)
We can also get the put formula via put-call parity:
P(S, K, σ, r, T, ∂) = C(S, K, σ, r, T, ∂) + Ke-rT – Se-∂T
A note on the function N(x): N(x) is called the cumulative normal distribution function. It is the probability that a randomly chosen number in the standard normal distribution (where mean = 0 and variance = 1) is less than x. It is impossible to directly integrate the normal probability distribution function to find N(x). On the exam, you will be given a table of values for N(x), so such integration will not be necessary. Here, however, we will use the Microsoft Excel function NormSDist. Try entering “=NormSDist(1)” into a cell in MS Excel. The result should be 0.84134474. Using this method will give us greater accuracy than using a table would.
We also note that N(-x) = 1 – N(x), since the probability of being below -x within the standard normal distribution is the same as the probability of being above x within the standard normal distribution.
We will proceed gently with applying the formula, its components, and its variants for different underlying assets. There will be plenty of uses for this formula throughout future sections of this study guide.
Original Practice Problems and Solutions from the Actuary’s Free Study Guide:
Problem BSF1. The stock of Blackscholesian Co. currently sells for $1500 per share. The annual stock price volatility is 0.2, and the annual continuously compounded risk-free interest rate is 0.05. The stock’s annual continuously compounded dividend yield is 0.03. Find the value of d1 in the Black-Scholes formula for the price of a call option on Blackscholesian Co. stock with strike price $1600 and time to expiration of 3 years.
Solution BSF1. We use the formula d1 = [ln(S/K) + (r – ∂ + 0.5σ2)T]/[σ√(T)], where we are given that S = 1500, K = 1600, r = 0.05, ∂ = 0.03, σ = 0.2, and T = 3.
Problem BSF2. The stock of Blackscholesian Co. currently sells for $1500 per share. The annual stock price volatility is 0.2, and the annual continuously compounded risk-free interest rate is 0.05. The stock’s annual continuously compounded dividend yield is 0.03. Find the value of d2 in the Black-Scholes formula for the price of a call option on Blackscholesian Co. stock with strike price $1600 and time to expiration of 3 years.
Solution BSF2. We use the formula d2 = d1 – σ√(T), where, from Solution BSF1, d1 = 0.1601034988 and we are given that T = 3 and σ = 0.2. Thus, d2 = 0.1601034988 – 0.2√(3) =
d2 = -0.1863066628
Problem BSF3. The stock of Blackscholesian Co. currently sells for $1500 per share. The annual stock price volatility is 0.2, and the annual continuously compounded risk-free interest rate is 0.05. The stock’s annual continuously compounded dividend yield is 0.03. Use the Black-Scholes formula to find the price of a call option on Blackscholesian Co. stock with strike price $1600 and time to expiration of 3 years.
Solution BSF3. We use the formula C(S, K, σ, r, T, ∂) = Se-∂TN(d1) – Ke-rTN(d2), where we know that S = 1500, K = 1600, r = 0.05, ∂ = 0.03, σ = 0.2, and T = 3. From Solution BSF1, d1 = 0.1601034988. From Solution BSF2, d2 = -0.1863066628.
We find N(d1) via MS Excel using the input “=NormSDist(0.1601034988)” = N(d1) = 0.563600238.
We find N(d2) via MS Excel using the input “=NormSDist(-0.1863066628)” = N(d2) = 0.426102153
Problem BSF4. The stock of Blackscholesian Co. currently sells for $1500 per share. The annual stock price volatility is 0.2, and the annual continuously compounded risk-free interest rate is 0.05. The stock’s annual continuously compounded dividend yield is 0.03. Find the price of a put option on Blackscholesian Co. stock with strike price $1600 and time to expiration of 3 years.
Solution BSF4. We recall that, since the call price is known from Solution BSF3, we can use put-call parity to get the put price: P(S, K, σ, r, T, ∂) = C(S, K, σ, r, T, ∂) + Ke-rT – Se-∂T
where C(S, K, σ, r, T, ∂) = 185.8385153, S = 1500, K = 1600, r = 0.05, ∂ = 0.03, and T = 3.
Problem BSF5. The stock of Blackscholesian Co. currently sells for $1500 per share. The annual stock price volatility is 0.2, and the annual continuously compounded risk-free interest rate is 0.05. The stock’s annual continuously compounded dividend yield is 0.03. Within the Black-Scholes formula for the price of a put option on Blackscholesian Co. stock with strike price $1600 and time to expiration of 3 years, find the value of N(-d2).
Solution BSF5. We use the formula N(-x) = 1 – N(x). We know from Solution BSF3 that N(d2) = 0.426102153. Thus, N(-d2) = 1 – N(d2) = 1 – 0.426102153 = N(-d2) = 0.573897847
See other sections of The Actuary’s Free Study Guide for Exam 3F / Exam MFE.
