1
$\begingroup$

*S follows a process $dS= mSdt + oSdz$ where m and o are constant.

What is the probability followed by $ Y=(Se)^{(r-t)} $.

If S follows a process $ dS= k (b-S) dt + oSdz $ where k, b, o are constant.

What’s the process followed by $Y =S^2$ ?

$\endgroup$
  • $\begingroup$ This looks like an application for Ito's Lemma. $\endgroup$ – SmallChess Dec 8 '15 at 5:22
  • $\begingroup$ what is $〖Se〗^{(r-t)}$? Is it $(Se)^{(r-t)}$? $\endgroup$ – Gordon Dec 8 '15 at 14:11
  • $\begingroup$ In addition, what does probability mean? Do you mean probability distribution? or the dynamics? $\endgroup$ – Gordon Dec 8 '15 at 14:58
  • $\begingroup$ yes it is written as you posted it. I mean the probability distribution $\endgroup$ – Sandro Dec 9 '15 at 15:21
1
$\begingroup$

Not sure I fully understand your question. However, I'd suggest using the Ito's lemma (second equation on wikipedia page https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma) to solve for dY. In both cases, dY will have both a drift term and a stochastic term. The coefficient of the stochastic term will indicate what sort of probability process Y follows.

e.g., in the first case, you'll get something like dY = [ (m-1)Y ]dt + [ rY ]dW, implying log-normal distribution for Y.

$\endgroup$
1
$\begingroup$

The first part has already been answer by @Uditg_ucla, so I am only providing answer of your 2nd part.

Rewriting your SDE in more sophisticated way: $$dS=k(b-S)dt+\sigma S dz$$ You want SDE for $S^2$. Using Taylor series, it can be written as: $$df(S)=f'(S)dS + \frac{1}{2!}f''(S)(dS)^2+\cdots$$ $$df(S)=2SdS+(dS)^2$$ $$df(S)=2S[k(b-S)dt+\sigma S dz]+\sigma^2 S^2 dt$$ $$df(S)=\bigg(2Sk(b-S)+\sigma^2S^2\bigg)dt+2\sigma S^2dz$$ since $Y=S^2$, so replacing $S^2$ from $Y$, $$dY=\bigg(2k(b\sqrt{Y}-Y)+\sigma^2Y\bigg)dt+2\sigma Y dz$$ desired SDE...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.