My apologies in advance if this question has come up before, but I couldn't find it answered directly in any of the threads that I have looked at. From the sde for dSI understand that in order to derive the equation for S(t), one uses F = ln S(t) and then uses Ito's Lemma, and that from the resulting sde we get the following:However, in my naivety, I would have first tried the following:and would have tried to intregrate both sides to get: This is obviously wrong, so could someone please explain to me why I assume that it has something to do with S(t) being a stochastic random variable, but I would like to understand this better so that I can be put out of my misery!Thanks.

Because you are applying classical calculus to random variable, which does not really make sense. For a deterministic function d(f(S)) = df/dS . dS, so int ( df/ds . ds) = f(s), int being your favorite integration theory. But you cannot apply rieman/lebesgue to random variables. You have to use another integration theory, which main result can can be written :df(S) = df/ds dS + 1/2 d²F/dS² d<S,S>One way to get the intuition, is to think of dW as having the same order of magnitude as sqrt(dt), which is really what previous equation tells you, so if you do a taylor exapnsion of df, you have to go to the second order in df/dW to get all the terms in dt, hence the integration result.W moves super quickly, you can't control it in dt, which is one of the reason why the stop loss hedging strategy fails.

QuoteOriginally posted by: tjuMy apologies in advance if this question has come up before, ... The standard example of your "conundrum" is:The extra term -T/2 as compared to ordinary calculus is sometimes called the "Ito correction". islington gave some intuition for this.

QuoteOriginally posted by: tjuThis is obviously wrong, so could someone please explain to me why I assume that it has something to do with S(t) being a stochastic random variable ...yes!the differentiald ln(S)is not onlydS /Syou have to add the Ito 'correction'- 1/2 sigma^2 dtso thatdS/S = mu dt + sigma dXbecomesd ln(S) + 1/2 sigma^2 dt = mu dt + sigma dXthe solution obviously satisfies this equation ...

TJU,the result of naive integration, namely ln(St) is your first guess in obtaining the strong solution to Stso the next step is to sayby ito (you know Ito's lemma, right?):d(ln(St)) = ...= (mu - 0.5(sigma^2))dt +sigma(dX) where X is BMand then only you can integrate both sides to getln(St) - ln(S0) = (mu-0.5sigma^2)t +sigma(dX - 0)in the last step, you treat LHS as Stieltjes, the first integrand on RHS as Riemann and the second integrand as.,.,., Stieltjes again?? -- someone correct me if im wrong here..

It was discussed in several threads in the forum.I tried to make a summary on how stochastic integration differs and counter intuitive to the Riemann.It happens simply because the functions are not smooth when they jump and randomly oscillate.http://www.wilmott.com/messageview.cfm? ... SGDBTABLE=

Quoteand would have tried to intregrate both sides to get: This is obviously wrong, so could someone please explain to me why I assume that it has something to do with S(t) being a stochastic random variable, but I would like to understand this better so that I can be put out of my misery!Thanks.You are right!If S(t) would be a deterministic function, you could use the Stieltjes-integral to get the equation. Nevertheless S(t) as solution of a SDE is a random variable and so you can not use deterministic Stieltjes-integration, but you have to use Ito-integration (or Stratonovich without martingale property).Additionally you obtain the so called Ito 'correction' "- 1/2 sigma^2 dt" by using the stochastic chain rule of the Ito-lemma.