In: Finance
Consider a 1-year option with exercise price $100 on a stock with annual standard deviation 15%. The T-bill rate is 2% per year. Find N(d1) for stock prices (a) $95, (b) $100, and (c) $105. (Do not round intermediate calculations. Round your answers to 4 decimal places.)
S | N(d1) |
$95 | |
$100 | |
$105 |
As per Black Scholes Model | ||||||
Value of call option = (S)*N(d1)-N(d2)*K*e^(-r*t) | ||||||
Where | ||||||
S = Current price = | 95 | |||||
t = time to expiry = | 1 | |||||
K = Strike price = | 100 | |||||
r = Risk free rate = | 2.0% | |||||
q = Dividend Yield = | 0% | |||||
σ = Std dev = | 15% | |||||
d1 = (ln(S/K)+(r-q+σ^2/2)*t)/(σ*t^(1/2) | ||||||
d1 = (ln(95/100)+(0.02-0+0.15^2/2)*1)/(0.15*1^(1/2)) | ||||||
d1 = -0.133622 | ||||||
d2 = d1-σ*t^(1/2) | ||||||
d2 =-0.133622-0.15*1^(1/2) | ||||||
d2 = -0.283622 | ||||||
N(d1) = Cumulative standard normal dist. of d1 | ||||||
N(d1) =0.4469 |
As per Black Scholes Model | ||||||
Value of call option = (S)*N(d1)-N(d2)*K*e^(-r*t) | ||||||
Where | ||||||
S = Current price = | 100 | |||||
t = time to expiry = | 1 | |||||
K = Strike price = | 100 | |||||
r = Risk free rate = | 2.0% | |||||
q = Dividend Yield = | 0% | |||||
σ = Std dev = | 15% | |||||
d1 = (ln(S/K)+(r-q+σ^2/2)*t)/(σ*t^(1/2) | ||||||
d1 = (ln(100/100)+(0.02-0+0.15^2/2)*1)/(0.15*1^(1/2)) | ||||||
d1 = 0.208333 | ||||||
d2 = d1-σ*t^(1/2) | ||||||
d2 =0.208333-0.15*1^(1/2) | ||||||
d2 = 0.058333 | ||||||
N(d1) = Cumulative standard normal dist. of d1 | ||||||
N(d1) =0.5825 |
As per Black Scholes Model | ||||||
Value of call option = (S)*N(d1)-N(d2)*K*e^(-r*t) | ||||||
Where | ||||||
S = Current price = | 105 | |||||
t = time to expiry = | 1 | |||||
K = Strike price = | 100 | |||||
r = Risk free rate = | 2.0% | |||||
q = Dividend Yield = | 0% | |||||
σ = Std dev = | 15% | |||||
d1 = (ln(S/K)+(r-q+σ^2/2)*t)/(σ*t^(1/2) | ||||||
d1 = (ln(105/100)+(0.02-0+0.15^2/2)*1)/(0.15*1^(1/2)) | ||||||
d1 = 0.533601 | ||||||
d2 = d1-σ*t^(1/2) | ||||||
d2 =0.533601-0.15*1^(1/2) | ||||||
d2 = 0.383601 | ||||||
N(d1) = Cumulative standard normal dist. of d1 | ||||||
N(d1) =0.7032 |