Dear all,Can anyone help me with the following ?I have came across the following from Wilmott's bookLognormal random walkP(1) = P(0)*(1+(Mean*timestep)+(Sqrt(variance)*timestep)*error)However, I have also came across the followingP(1) = P(0)*EXP((Mean*timestep)+(Sqrt(variance)*timestep)*error)error is IID(0,1)So, are they both correct ?Thanks and regards,Nick

P(1) = P(0)*(1 +(Mean*timestep)+(Sqrt(variance)*timestep)*error) P(1) = P(0)*EXP((Mean*timestep)+(Sqrt(variance)*timestep)*error)Take the second, consider infinitismal time steps, expand in a Taylor seies in the infinitismal time steps, keep the first two terms, you get the first.

Hmm. Don't forget Itô! I.e. the error must be Taylorexpanded up to the 2nd derivative (as I am sure Omar knows very well). PW is correct, the other is also correct but only if you take that drift minus variance/2. In symbols we have approximately:P(0)*EXP[(mu-sigma^2/2)*dt + sigma*dX] = P(0)*(1+mu*dt + sigma*dX)The drift in the exp-expression must be lower since high error terms sigma*dX has exponentially higher weight than low (negative) error terms. If you calculate the expectation of both sides, using the normal distribution for dX~N(0,dt), the result will be P(0)*Exp[mu*dt]for the Exp-expression andP(0)*(1+mu*dt)for the other one above, which is just the first order Taylor expansion of the former (correct in this case since there is no error term).

You should do the second one really, since it's exact. However, for most s.d.e. specifications of random walks you have to do something more like the first (unless you want to get very sophisticated) because you can't integrate up the s.d.e.P