Question

Suppose that you are holding a European call option on General Motors stock that expires in 5 years. The option is currently at-the-money. GM's current stock price is $42.50 and the current yield on a 5-year Treasury bond is 2%. The standard deviation of returns on GM stock is 25%. For the purposes of this series of questions, you should assume that GM does not pay dividends.

Please use the field below to provide work associated with your answer to Question 26. You may assume that all of the information above applies.

What is the Black-Scholes value of the European call? use a online calculator

Given your answer to the previous question, what is the value of a European put option on GM stock that expires in 5 years and is also currently at-the-money? (Please express your answer to the nearest cent.)

Suppose that the current price of GM stock were $50 instead of $42.50, but the strike price of the options, the time to expiration, the risk-free rate, and the volatility of returns on GM stock were all the same. How would the values of the put and call option change? Provide a clear explanation for why the values move in the direction that you propose. (No additional calculations are necessary for this problem.)

Answer 1

The formula for European call & Put option using Black Scholes model is given by

Call = S0 x N(d1) - K x e^(-r x t) x N(d2)

Put = K x e^(-r x t) x N(-d2)- (S0 x N(-d1))

Where,

S0 = Price of the underlying, eg stock price.

K = Exercise price.

r = Risk free interest rate

t = Time to expiry

σ = Standard deviation of the underlying asset per annum, eg stock.

N(d1) =  standard normal cumulative distribution function using value of d1

N(d2) =  standard normal cumulative distribution function using the value of d2

The value of d1 & d2 is calculated using the below formula

d1 = (Ln(So/X) + ((r + σ ^2)/2) x t)) / ((σ^2) x t)^1/2)

d2 = (Ln(So/X) + ((r - σ ^2)/2) x t)) / ((σ^2) x t)^1/2)

1.   

In the given problem,

S0 = $42.50

K = $42.50 (Since it is an at the money strike)

r = 2%

t = 5

σ = 25%

We will calculate, d1 & d2

d1 = (Ln(So/X) + (((r + σ ^2)/2) x t) / ((σ^2) x t)^1/2)

= Ln(42.5/42.5) + ((2% +25%^2) /2) x 5 / (((25%^2) x 5)^1/2)

= (0 + 0.2563) / 0.5590

= 0. 4583

d2 = (Ln(So/X) + ((r - σ ^2)/2) x t)) / ((σ^2) x t)^1/2)

= Ln(42.50/42.50) + ((2% -25%^2) /2) x 5 / (((25%^2) x 5)^1/2)

= (0 + (-0.0563)) / 0.5590

= - 0.1006

B. Using the values of d1 & d2 , we find values of N(d1), N(d2), N(-d1) & N(-d2)

N(d1) = 0.6767 & N(d2) = 0.5000  

N(-d1) = 0.3233 & N(-d2) = 0.5000

Call = S0 x N(d1) - K x e^(-rxt) x N(d2)

= 42.50 x 0.6767 - 42.50 X e^(-2% x 5) x0.5000

= 28.76 - 19.23

= 9.53

Put = K x e^(-r xt) x N(-d2)- (S0 x N(-d1))

= 42.50 x e^(-2% x 5) x 0.5000 - 42.50 x 0.3233

= 19.23 - 13.74

= 5.49

2. We can also use, Put Call Parity equation to find the value of Put

Value of Call Option - Value of Put Option = Price of Underlying Asset - Present Value of Strike Price

Value of Put Option = Value of Call Option - Price of Underlying Asset + Present value of Strike

= 9.53 - 42.50 + 42.50 /(1+2%)^5

= 9.53 - 42.50 + 38.49

= 5.52

3. The value of both Call & Put would be higher now.

If the spot price was $50, the 42.50 strike call would have now been an In the Money call, having both intrinsic value as well as the time value. The intrinsic value itself would be 50 - 42.50 = 7.50. There would also be a time value and it would be a long time. Thus value of Call would be higher.

The value of Put will also be higher, although the put would now be an Out of the money Put, with zero intrinsic value. An At the money, also has zero intrinsic value. But the time value would be a multiplier of spot price, thus the time value will increase and so the value of Put will also increase.