Somebody posted this problem in another thread.
The time-t price of a stock is S(t). You are given
The risk-neutral process for S(t) is
dS(t) = .15S(t) dt + .32 S(t)d Z_squiggle (t)
Where Z_squiggle (t) is a standard Brownian motion in the risk-neutral measure.
The stock pays dividends of .01S(t) dt between times t and t+dt.
S(0) = 10.
A special put option allows the purchaser to sell S(.25) shares of the stock at time .25 for 100. Determine price of this option.
Solution:
S shares of S are worth S^2. The forward price of S^2, using the risk-neutral process for S, is calculated to be 100e^.1006
We deduce that since .15 = r-dividend, r=.16.
We proceed with the Black-Scholes formula.
N(-d1) = .31762
N(-d2) = .43866
This is what I did and it worked. Why did it work? You could just skip to the very last equation at the bottom.
The time-t price of a stock is S(t). You are given
The risk-neutral process for S(t) is
dS(t) = .15S(t) dt + .32 S(t)d Z_squiggle (t)
Where Z_squiggle (t) is a standard Brownian motion in the risk-neutral measure.
The stock pays dividends of .01S(t) dt between times t and t+dt.
S(0) = 10.
A special put option allows the purchaser to sell S(.25) shares of the stock at time .25 for 100. Determine price of this option.
Solution:
S shares of S are worth S^2. The forward price of S^2, using the risk-neutral process for S, is calculated to be 100e^.1006
We deduce that since .15 = r-dividend, r=.16.
We proceed with the Black-Scholes formula.
N(-d1) = .31762
N(-d2) = .43866
This is what I did and it worked. Why did it work? You could just skip to the very last equation at the bottom.
How does this even work?? S^a, the long way to solve.
