The Actuary's Free Study Guide for Exam 3F / Exam MFE - Section 33
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 annualSolution 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
Thus, C(S, K, σ, r, T, ∂) = 1500e-0.03*30.563600238 - 1600e-0.05*30.426102153 =
C(S, K, σ, r, T, ∂) = $185.8385153
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.
Thus, P(S, K, σ, r, T, ∂) = 185.8385153 + 1600e-0.05*3 - 1500e-0.03*3 =
P(S, K, σ, r, T, ∂) = $192.0744997
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.
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Entering "=NormSDist(x)" into a cell in MS Excel for some particular value of x will give you the value of N(x) for that x.



