# SDE of Geometric Brownian motion

I was reading The Binomial Asset Pricing Model by Shreve and having some trouble dealing with SDE. On page 169, it surveys the Geometric Brownian motion and tries to computing the SDE of that process. He says: Define $$f(t,x)=S(0)\exp\lbrace \sigma x+(\mu-\frac{1}{2}\sigma^2)t\rbrace$$ so that the GBM is $$S(t)=f(t,B(t)).$$ According to Ito's formula, $$dS(t) =df(t,B(t)) =f_t dt+f_xdB+\frac{1}{2}f_{xx}dt.$$ But the Ito's formula says for twice differentiable function $$F$$, we have $$dF(B(u))=F'(B(u))dB(u)+\frac{1}{2}F''(B(u))du.$$ If it is the case $$F(x)=f(t,x),$$ then where does the $$f_tdt$$ come from?

You added another variable (i.e. explicit time dependence) to the function. For non-stochastic $$x(t),$$ we have $$df(x(t)) = f'(x(t))dx(t).$$ When we add explicit time dependence, $$f(t,x(t)),$$ we need to use the multivariable chain rule, giving $$df(t,x(t)) = f_tdt + f_x dx(t).$$ Similarly, here, now using the fact that the second argument is a drift-diffusion, $$df(t,B(t)) = f_t dt + f_x dx(t) + \frac{1}{2}f_{xx} \sigma^2 dt.$$
• the second argument use the Ito's formula, right? so I guess the function for which Ito's lemma used is $F(x)=f(t,x)$, instead the $t$ is fixed? – Yan Lai Sep 26 '18 at 3:30
• @YanLai Not sure I understand you. Here is an example. If $S(t) = (B(t))^2e^{-at},$ then Ito's lemma gives the SDE $$dS = -a(B(t))^2 e^{-at} dt + 2 B(t) e^{-at}dB(t) + \sigma^2 dt$$ – spaceisdarkgreen Sep 26 '18 at 4:14
• oh ok, btw can you explain the idea of $dB(t)dB(t)=dt?$ it does not make sense to me. Does it depend on a formal definition which is related Ito integral or any facts about absolute volatility? – Yan Lai Sep 26 '18 at 12:41