d(lnS) = r*dt + sigma*dX?

It's getting from the PDE: dV/dt + rS * dV/dS + 0.5 sigma^2 d2V/dS2 - rV = 0

是dlnS=(r-q-sigma^2/2)dt+sigmadz
这个是怎么推导的

1) The below PDE is obtained using non-arbitrary and dynamic hedging.
=> dV/dt + (r-q)*S*dV/dS + 0.5*sigma^2*d2V/dS2 - rV = 0

2) The below SDE is derived from the above PDE using Feynman-Kay Theorem.
=> dS = (r-q)*S*dt + sigma*S*dX

3) The above SDE can be transformed to below form using Ito lemma, where F = lnS
=> dF = (r-q-0.5*sigma^2)dt + sigma*dX

找本随机过程的书看看吧~
3# crazygod

是dlnS=(r-q-sigma^2/2)dt+sigmadz
这个是怎么推导的