Ito's Lemma

[hershey79]

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.

-------------------
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]

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]

Yes dz= e(dt)^1/2

Does that make a difference with the ultimate question?

[NoMoreExams]

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]

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]

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

So answer the following:

G_S = ?
G_{SS} = ?
(dS)^2 = ?
G_{t} = ?

Actually if you can answer what G_{SS} = , you should see that it doesn't matter what (dS)^2 =

[hershey79]

Thanks No more Exams, I'll work on that this weekend.