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Ito question

B

barbados

Member
Hi,

Suppose S follows a geometric Brownian motion, i.e. dS = mu S dt + sigma S dz

What is the process followed by y = exp(S).

I come this far:

dy = (Dy/DS) dS + 0.5 (D^2 y/ DS^2) (dS)^2

Since both (Dy/Ds) = (D^2 y/ DS^2) = exp(S) = y, it follows that:

dy = y dS + 0.5 y (dS)^2 = y [mu S dt + sigma S dz] + 0.5 y sigma^2 S^2 dt.

My question, how do I get rid of the S in the final dy formula - simply replace it by lny?

Thanks.
 
Ito for y=exp(S)

Yes, I'd substitute in S=ln(y) and then group the dt and dz terms together.

It seems an unusual process to look at though. S(t) is already an exponential process, so y would be doubly exponential. Is there a particular reason you're looking at this function or is it just for Ito practice (which is fine - nothing wrong with that)?
 
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