In: Finance
You are given the following information concerning options on a particular stock:
Stock price = $62
Exercise price = $60
Risk-free rate = 5% per year, compounded continuously
Maturity = 3 months
Standard deviation = 45% per year
What is the time value of each option?
Call option:
Option price= | SN(d1) - Xe-r t N(d2) | |||
d1 = | [ ln(S/X) + ( r+ v2 /2) t ]/ v t0.5 | |||
d2 = | d1 - v t0.5 | |||
Where | ||||
S= | Current stock price= | 62 | ||
X= | Exercise price= | 60 | ||
r= | Risk free interest rate= | 5% | ||
v= | Standard devriation= | 45% | ||
t= | time to expiration (in years) = | 0.2500 | ||
d1 = | [ ln(62/60) + ( 0.05 + (0.45^2)/2 ) *0.25] / [0.45*0.25^ 0.5 ] | |||
d1 = | [ 0.03279 + 0.037813 ] /0.225 | |||
d1 = | 0.313788 | |||
d2 = | 0.313788 - 0.45 * 0.25^0.5 | |||
0.088788 | ||||
N(d1) = | N( 0.313788 ) = | 0.62316 | ||
N(d2) = | N( 0.088788 ) = | 0.53537 | ||
Option price= | 62*0.623159009009996-60*(e^-0.05*0.25) *0.535374842989488 | |||
6.91 |
Call option price is $6.91
Put option:
Option price= | = Xe –rt × N(-d2) – S × N(-d1) | |||
d1 = | [ ln(S/X) + ( r+ v2 /2) t ]/ v t0.5 | |||
d2 = | d1 - v t0.5 | |||
Where | ||||
S= | Current stock price= | 62 | ||
X= | Exercise price= | 60 | ||
r= | Risk free interest rate= | 5% | ||
v= | Standard devriation= | 45% | ||
t= | time to expiration (in years)= | 3/12 = | 0.25 | |
d1 = | [ ln(62/60) + ( 0.05 + (0.45^2)/2 ) *0.25] / [0.45*0.25^ 0.5 ] | |||
d1 = | [ 0.03279 + 0.0378125 ] /0.225 | |||
d1 = | 0.3137881 | |||
d2 = | 0.31379 - 0.45 * 0.25^0.5 | |||
0.088788101 | ||||
N(-d1) = | N( - 0.31379 ) = | 0.37684 | ||
N(-d2) = | N( - 0.08879 ) = | 0.46463 | ||
60 × e^(-0.05 × 0.25) ×(1- N( 0.08879)) -62× (1-N(0.31379)) | ||||
Option price= | 4.17 |
Put option price is $4.17
Please rate.