I tried to get an intuition for the -1/2*sigma^2*T drift in a geometric brownian motion.Of course, this can be obtained by Ito's Lemma, or otherwise by writing S(T)=S(0)*(1+sigma*sqrt(T/n)*g_1)*(1+sigma*sqrt(T/n)*g_2)*...*(1+sigma*sqrt(T/n)*g_n) and taking 'ln' and doing a Taylor development (N->infinity). I have a naîve interpretation: this drift is due to the fact that negative returns have a stronger effect than positive returns. Suppose you split a time step T in 2, where you have one up move and one down move:S-----up move----->S*(1+sigma*sqrt(T/2))----down move----->S*(1+sigma*sqrt(T/2))*(1-sigma*sqrt(T/2))=S(1-1/2*sigma^2*T)So these ups and downs make the stock lose -1/2*sigma^2*T

I had the same problem. The problem is this sentence:"So these ups and downs make the stock lose -1/2*sigma^2*T "A gbm is a random process, therefore you can't do such a statement . You can say: on average.But on average this doesn't hold truth: While you often have a downward movement, you have onthe other hand greater upward movements, due to the fact, that this process is bounded by zero,but has no upper boundary. So the expected value is V_0, the start value, if you assume zero drift (mu=0).A very simple example:let's say P( up and down) = 0.5, V_0=100, 2 periods.E(V_2)=121+.025+99*0.5+81*0.25 = 100. In 3 of 4 cases you have a lower value as the start value, but theexpected value is V_0.

Regarding this issue of a drift modification, I am beginning to wonder if it is artificial, in the sense that it appears only because of the way Ito defined a stochastic integral. If we consider the SDE, dX = m X dt + v X dB with X(0) = 1then d(ln X) = (1/X) dX + (1/2) (-1 / X^2) dX^2 = (m dt + v dB) - (v^2 / 2) dt = (m - v^2 / 2) dt + v dBTherefore, X(t) = exp( [m - v^2 / 2]*t + v*B(t) )So at this point you say, "Hey what's happening here, the drift has changed!!??" But has it really? What is the expected value of X(t), i.e. E[ X(t) ], then E[ X(t) ] = exp( [m - v^2 / 2]*t ) * E[ exp( v*B(t) ) ]and since E[ v*B(t) ] = exp( t*v^2 / 2 ) we find that E[ X(t) ] = exp( m*t )as one should expect (based on examining the original SDE.) The point is that on one hand it appears as though the drift has changed, but it really hasn't since some is still lurking in the stochastic part of the solution (ie., in the term v*B in the expression for X(t) above.) This apparently seems to happen because of the way Ito defined a stochastic integral. Think about some of the stochastic integrals you've evaluated and found a non-stochastic component being generated. That's exactly what appears to be happening above (the -t*v^2/2 term). So what Ito's integrals seem to be doing is generating an additional component "so as ensure that the expected value of the integral remains unchanged." This line of thought leads me to ask whether this supposed change in drift is physical or not?--------------------------------------------------Just thinking aloud, and I could be wrong ....... thoughts anyone?--------------------------------------------------

I would consider it this way: the -1/2*sigma^2*T drift term is not an aspect of geometric brownian motion at all, really, except as applied to enforce the martingale property on a (usually financial) variable that follows geometrice brownian motion. A variable that follows geometric brownian motion will have whatever drift it has, period. However, in finance it is generally assumed that prices can be modeled as if they are martingales, which means that their expected rate of return (ignoring the risk-less interest rate, as you have done) is zero. The expected value at time T of a variable that follows geometric brownian motion (and that begins with value 1) is exp(mu*T+1/2*sigma^2*T), where mu is the variable's actual drift. In order to turn this into a martingale without fiddling with the sigma term (and the reasons for wanting not to fiddle with the sigma term are a separate discussion), mu must be assumed to be -1/2*sigma^2, so that the expected value at time T is exp(-1/2*sigma^2*T+1/2*sigma^2*T)=exp(0)=1, which gives the expected rate of return of zero that we need to make the process a martingale.

In reply to Waagh Bakri's statement (in reply to my statement "that normal returns are inconsistant with normal percentage changes") that "for small asset moves (changes over a small period of time) returns on assets & percentage changes are nearly equal"; I would like to say the following:His statement is true and it follows that when asset moves are greater than "small" then returns and percentage changes are far from being nearly equal.For example a 50% gain equals a .405 return and a 50% decrease equals a -.693 return. If we assume that these percentage changes are normal ; this is far different from the assumption that returns are normal.Of course, if we restrict ourselves to very small changes over very small time frames, the effect on the theoretical values of the options is very small. So the effect of making both assumptions (i.e. that percentage changes and returns are normal) can be dismissed. But in the real world where we have both volatile stocks and long terms to expiration such as LEAPS, convertible bonds, and employee stock options, the effect of the inconsistent assumptions is large. Subtracting the vol^2/2 from the expected return is based on the assumption that percentage changes are normally distributed.This inconsistency and the fact that stocks are not lognormally distributed is what underlies the so called "volatility skew"." In times of universal deceit, truth is a radical idea." George Orwell

QuoteOriginally posted by: FlexI had the same problem. The problem is this sentence:"So these ups and downs make the stock lose -1/2*sigma^2*T "A gbm is a random process, therefore you can't do such a statement . You can say: on average.You're totally right. Maybe I could remake my statement:These ups and downs make the stock lose 1/2*sigma^2*T in this particular case. In the case of a down followed by up the loss is the same. In the case of up-up, there is a gain > -1/2*sigma^2*T and in case of a down-down, the (negative) gain is < -1/2*sigma^2*T.Conclusion: for 2 periods, -1/2*sigma^2*T seems to be a good median for the return.The error I did came from mistaking median and mean (which are the same for the 2 types of distributions a human brain feels confortable with: multinomial/normal and uniform)

There is only one drift, mu. The expected value at time T isn't exp(mu*T+1/2*sigma^2*T), it is just exp(mu*T). The extra term in the equation S(T) = S(0) exp( [mu - sigma^2 / 2]*T + v*B(T) ) doesn't change that, although it means that the expected overall rate of return, E (log(S(T)/S(0)) , is [mu - sigma^2 / 2]*T and not the instantaneous drift mu.This is basically some AM-GM effect about compounding returns ... as born out by brussels' example. We have outcomes 121, 99, 99, 81, whose average is 100. However the corresponding 'average one-time-step returns' are sqrt(1.21)-1 = 0.1, sqrt(0.99)-1=-0.005, sqrt(0.81)-1=-0.1 which don't average out to zero. As Hull points out (chapter on 'Black-Scholes analysis'), "expected return" is ambiguous, whereas "drift" is always instantaneous. - I don't know how significant Ito's choice of stochastic integral is in this. What's the solution of dS=mu.S.dt+sigma.S.dB under non-Ito calculus?Only if all little-timestep-returns are the same will the overall return be the same as the average of individual returns. That would correspond to zero volatility and that's the only case when E(exp(sigma.B(T))) = exp( E(B(T)) ) (viz, both are 1).

QuoteOriginally posted by: karstyThere is only one drift, mu. The expected value at time T isn't exp(mu*T+1/2*sigma^2*T), it is just exp(mu*T).No. Exp(mu*T) is the median, but not the mean, which is exp(mu*T+1/2*sigma^2*T).Geometric Brownian Motion is something like a variable with equal probabilities of going up by 25% and down by 20% in every time step: after two time steps, it is most likely to be back where it started, but two steps down (64% of original) does not offset two steps up (156.25% of original, if my arithmetic is correct) for the purpose of the arithmetic average.

QuoteOriginally posted by: MarsdenNo. Exp(mu*T) is the median, but not the mean, which is exp(mu*T+1/2*sigma^2*T).Geometric Brownian Motion is something like a variable with equal probabilities of going up by 25% and down by 20% in every time step: after two time steps, it is most likely to be back where it started, but two steps down (64% of original) does not offset two steps up (156.25% of original, if my arithmetic is correct) for the purpose of the arithmetic average.Are you sure? In one time-step, GBM says dS = mu.S.dt + sigma.S.dB so the move is symmetric up or down around the mean because the Brownian motion is. In your example you'd have to have mu=2.5%. In two time-steps, I agree that the returns are positively skewed & two up moves will be bigger than two down moves. However one up-move and one down-move will combine to a small down-move relative to the average. In your example, +25% followed by -20% is +-0: the average final price is (1.5625+1+1+0.64)/4 = 1.050625 which is (1.025)^2 i.e. an average move of +2.5% per timestep: the expected final price is the initial price compounding at the average drift mu. Overall, the solution of the SDE dS = mu*S*dt + sigma*S*dB is S(T) = S(0) * exp( [mu - 1/2*sigma^2]*T + sigma*sqrt(T)*B ) where B comes from normal(0,1)Since the most likely score for B is 0, the most frequent value of S(T) is S(0) * exp([mu-1/2*sigma^2)*T).However with exp being nonlinear, the mean of exp(sigma*sqrt(T)*B) is exp(1/2*sigma^2*T) which gives E(S(T))=S(0) * exp(mu*T).

QuoteOriginally posted by: karstyQuoteOriginally posted by: MarsdenNo. Exp(mu*T) is the median, but not the mean, which is exp(mu*T+1/2*sigma^2*T).Geometric Brownian Motion is something like a variable with equal probabilities of going up by 25% and down by 20% in every time step: after two time steps, it is most likely to be back where it started, but two steps down (64% of original) does not offset two steps up (156.25% of original, if my arithmetic is correct) for the purpose of the arithmetic average.Are you sure? In one time-step, GBM says dS = mu.S.dt + sigma.S.dB so the move is symmetric up or down around the mean because the Brownian motion is. In your example you'd have to have mu=2.5%. In two time-steps, I agree that the returns are positively skewed & two up moves will be bigger than two down moves. However one up-move and one down-move will combine to a small down-move relative to the average. In your example, +25% followed by -20% is +-0: the average final price is (1.5625+1+1+0.64)/4 = 1.050625 which is (1.025)^2 i.e. an average move of +2.5% per timestep: the expected final price is the initial price compounding at the average drift mu. Overall, the solution of the SDE dS = mu*S*dt + sigma*S*dB is S(T) = S(0) * exp( [mu - 1/2*sigma^2]*T + sigma*sqrt(T)*B ) where B comes from normal(0,1)Since the most likely score for B is 0, the most frequent value of S(T) is S(0) * exp([mu-1/2*sigma^2)*T).However with exp being nonlinear, the mean of exp(sigma*sqrt(T)*B) is exp(1/2*sigma^2*T) which gives E(S(T))=S(0) * exp(mu*T).I think we are talking about two different mus. What I mean by "mu" is the drift of the log of the GBM variable (I think I refered to it incorrectly as the drift of the GBM variable at one point); you seem to use "mu" to mean the drift of the GBM variable itself, which is my mu plus 1/2*sigma^2*T. Some texts may try to follow your convention, but I don't think that this is helpful because it leaves you with a normal variable (the log of the GBM variable's increments) whose mean is not mu.

QuoteI don't know how significant Ito's choice of stochastic integral is in this. What's the solution of dS=mu.S.dt+sigma.S.dB under non-Ito calculus?Well, I did think more about it and seem to have concluded that the drift is "physical." More specifically, above I was questioning the "superficial" appearance of the -0.5*sigma^2 term, which of course vanishes when one considers E. Of course, E[ln S], has it's own drift rate. Below, I'm thinking aloud on the question you've posed ....More specifically, the way I see it, we start at point A and our goal is to reach point B, with A being the SDE or local analysis of some phenomena or physical process, and B being the global behaviour/trajectory/closed-form solution. There seem to be several paths for reaching point B from point A. Now SDE's only have meaning through the integral equation counterpart. Which leads us to ask how those integrals are to be defined. Once the integrals are defined, we must be consistent throughout our analysis right upto the end solution. There are several ways to define the integrals encountered in the integral equation (ie SDE), Ito's & Stratonovitch's definition being the most common, and each then specifies the rules of calculus that must be obeyed (ie., each results in its own "stochastic calculus"). However, one might want to be cautious about "superficially" assigning some "physical meaning" to the relations that arise in between the analysis, simply because the Ito's calculus & Stratonovitch's calculus differ. For eg., if one considered the derivation of the SDE for d[ f(S) ] given the SDE for d, noting that Stratonovitch calculus' integration-by-parts rule differs from Ito calculus' integration-by-parts rule by the quadratic covariation term (in addition to the definition of the dB-integrals that arise), one might expect a different form for the SDE. At this point, it would appear the SDE for d[f(S)] under an Ito's & Stratonovitch's transformation are different, but that's only superficial since the SDE is to be strictly interpreted via it's integral equation formulation. The integral equation formulation erases the ambiguity since it forces you to realize that the dB-integrals that are present must be interpreted as either Ito or Stratonovitch dB-integrals. The risk lies in superficially looking at SDE and in particular at the dt-term, and simply claiming, for eg., that the local drift or expected value for d[f(S)] is simply the coefficient*dt term. Such a conclusion would be wrong if you are working with Stratonovitch's calculus, but right if you are working with Ito calculus. Obviously, the subtlety lies in the two definitions of the dB-integrals. In short, it will not matter which approach one uses. One just has to be consistent throughout. Not surprising...

Oh well, so I think.....opinions would be most welcome ....Thanks.