Question

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?

Answer 1

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

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