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I would like to ask a question related to Geometric Brownian Motion again. Thanks in advance.

Question:

Suppose that $S(t)$ follows a geometric Brownian Process:

              dS(t) = μS(t)dt + σS(t)dW(t)

What is the process ( that is, $dY(t)$) followed by $$Y(t) = \frac{e^{r(T-t)}}{S(t)}$$

Arnaldo
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skatip
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1 Answers1

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Hint

You will need Ito's Lemma:

$$dY=\left(Y_{t}+\mu S Y_{S}+\frac{1}{2} \sigma^2 S^2Y_{SS} \right)dt+(\sigma SY_{S})dW$$

You just have to calculate the derivatives: $Y_{t}, Y_{S}, Y_{SS}$.

Can you finish?

Arnaldo
  • 21,342
  • $$dY = df(t,S(t)) so, /frac{-re^(r(T-t))}{S(t)}dt + /frac{2e^(r(T-t))}{S^2(t)}dS(t) + /frac{e^(r(T-t))}{S^3(t)}dS(t)dS(t)$$ Do you think, I am right? I tknow dS(t) as given so I think I can put it directly on my formula and remove some of the variables from there. When I do this I end up with $$/frac{dY(t)}{Y(t)} = (σ^2-r+2μ)dt + 2σdw(t)$$ – skatip May 05 '17 at 07:51
  • @skatip: please use https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. Without a better formating will be hard to understand what you are trying to say. – Arnaldo May 05 '17 at 12:40