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Could someone show me the steps to obtain E = s(0) exp[muT]? if ln S is distributed as Normal[ ln s(0) + (mu - (sig^2)/2), sigT^(1/2)) Thanks

do we obtain s(0) exp (mu*T) imposing mu = (mu - sig^2)T?

Start w/ dp = p (mu dt + sig dz).Use Ito to get d ln(p) = (mu - sig^2/2) dt + sig dzIntegrate and exponentiate to get P(t) = P(0) exp( mu *t) exp (-sig^2/2 t + sig Z(t) )So your question really is: Does 1 = E( exp(-sig^2/2 t + sig Z(t) )?To see that this is so, just integrate against pdf for Z(t) which is Gaussian w/ mean zero, variance t (easiest to complete the square in the exponential, giving you 1 = exp(-sig^2/2 *t ) exp (+sig^2/2 *t) * (integral over gaussian pdf = 1).

thanks a lot

Apologies to raffapi if I am usurping your thread. I have another simple question.One of the best know properties of Brownian motion is that almost all its sample paths have infinite variation over any time interval of positive length. Thus Brownian sample paths are emphatically not variation finite functions. Very good statement of this is the so-called quadratic variation of Brownian motion paths. The most difficult part of this result is that does the quadratic variation exists for all Brownian paths? Is there a proposition which explains above claims? Or easy calculations which may help?When I say simple question I meant for the maestros!!Many thanksMabadu

I would be grateful to your reply guys. Mabadu