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hershey79
November 19th 2009, 04:44 PM
Hi all, I'm new to the forum, but have this question that has been torturing me all week from class. Your input would be greatly appreciated...

Assume that the stock price, S, evolves according to the following process
dS = uSdt + oSe(dt)^1/2

where u is mu, o is sigma and e is the error term

(1) Consider the function G= S (exp(r(T-t)) -1). Compute the process for G.

(2) An option on the stock S, can be written as a function of S and time:
f(S,t). Compute the process for f.

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I know G= S exp(r(T-t)) is the function for the forward price of a stock but where does the negative one come in??

NoMoreExams
November 19th 2009, 05:01 PM
You sure you mean dt^(1/2)?

The Ito processes that are covered on MFE (i.e. not on FM) are of the form

df = alpha dt + sigma dZ

Where f(.) is some function for example.

hershey79
November 19th 2009, 07:24 PM
Yes dz= e(dt)^1/2

Does that make a difference with the ultimate question?

NoMoreExams
November 19th 2009, 07:41 PM
Yes dz= e(dt)^1/2

Does that make a difference with the ultimate question?

I suppose I didn't recognize that e must be there, from what I understand (dZ)^2 = dt but anyways. Is your question on how to calculate a process for G or why there's a "-1"?

hershey79
November 19th 2009, 08:12 PM
The question is how to compute that given equation for G in the question according to the process given.

I'm in an Actuarial class at present and we were asked to do that derivation.

NoMoreExams
November 19th 2009, 10:11 PM
The question is how to compute that given equation for G in the question according to the process given.

I'm in an Actuarial class at present and we were asked to do that derivation.

Fine so you are told

dS = mu*S dt + sigma*S dZ

you are also told that

G = S[e^{r(T-t)} - 1]

Ito's Lemma tells you that

dG = G_S dS + 1/2 * G_{SS} (dS)^2 + G_{t} dt