If dx(t)=(ax(t)+b)dt+cdW(t) where dW(t) is the brownian motion processI know the E[dW(0)]=0, but does the expectation, E[x(0)]=0??

???can't figure out your question, but if c=0 and you just had the ODEdx = (ax + b).dt,then why would you expect Ex(t) = x(0)? think about it

sorry, should have added that a is not 0 and b and c are constants>0

I cannot understand what do you mean suni: if dx(t) =[ax(t) + b]dt + cdW(t)and W is the Wiener process assuming an initial value condition x=xoexpected value E[x] = xo + [ax(t) + b] tno matter the value of c, supposing E[x(0)]=0 means a and b= 0 too.Does it make sense? Quite strange. Rgds,

E[dw] is not equal to 0 because dw does not exist.The process dx=(ax(t)+b)dt+c dw is just shorthand for,x(t)=x(0)+int_0^t (ax(u)+b) du+int_0^t c dw(u) which does exist because the integral int_0^t c dw(u) can be defined.So E[x(t)]=E[x(0)]+E[int_0^t(ax(u)+b) du]+E[int_0^t c dw(u)].E[x(0)]=x(0)E[int_0^t c dw(u)]=0, (the expected value of an ito integral is zero which is what is usually meant by the statement E[dw]=0),andE[int_0^t(ax(u)+b) du]=int_0^t(aE[x(u)]+b) du, (switching the order of integration). So call E[x(t)]=y(t) and then,y(t)=x(0)+int_0^t(ay(u)+b) du, this is a simple integral equation which you can solve.

Thanks for that spacemonkey