# Stochastic process theory question

*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 $Y =S^2$ ?

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

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...