Hye,I remember reading an intuitive proof of why dW^2=dt but I just realize there's a point I am missing:The proof says that given that dW=sqrt(dt)*N(0,1), we have dW^2=dt*N(0,1)^2, which impliesE(dW^2)=E(dt*N(0,1)^2)=dtVar(dW^2) ~ dt^2 ~ 0Given that we dismiss terms in dt^2. It means that dW^2 has no variance, hence is deterministic and equal to its mean=dt.I looks like a nice proof, but there's one thing I don't understand: if you use stdev(dW^2) instead of var(dW^2) just see that dW^2 has a stdev that is not negligible w.r.t. to the average so what is meant by 'dW^2' is not random?

hi brussels,You make a good point, but variance is linear while std. dev. is not. In other words, when you consider a finite random walk,you want to what happens to VAR(dW^2_0 + ... dW^2_T)= VAR(dW^2_0) + ... + VAR(dW^2_T) ~ O(1/N). This is because dW^2 ~ O(1/N)^2since dt ~ O(1/N) and T ~ O(N) assuming each step is iid (identical and independent). In this sense, we see that the Chi-Squared distribution formed by sum of N squared normal variables has variance that -> 0 as N->Infinity. Also std. dev. -> sqrt(1/N) -> 0 also. Hence sum ofdW^2_0+...dW^2_T is a sharply peaked (delta function) distribution with mean T. In this sense, we see that dW^2_0+...dW^2_T IS the deterministic value T almost surely with confidence -> 100% as N->Inifinty. In other words, you should look at the entire random walk not just one time step.Peter

Whenever one has a perturbative expansion in some small parameter g, one can neglect terms of order g^2, keeping terms of order g, in the limit g --> 0. Now suppose someone says, "No, no, you cannot neglect the order g^2 term, because its square root is of order g, just like the one we keep".My claim is that that is not a good argument. You have to compare exactly those terms with exactly those powers as they appear side-by-side in the same series expansion. Similarly, in statistics (a subject that I basically know nothing about), the natural objects that appear in expansions (as far as I remember), and that have a direct physical meaning, are the moments. Not the roots of the moments, or powers of the moments, but the very moments. So we wish to understand the properties of a distribution, we need to compare the various moments. That includes comparing the 2nd moment (the variance), rather than its square root, to the 1st moment (the mean). When we do that, we find that it is negligible in the limit dt --> 0.I hope this makes sense.

HyePlee: I like very much your explanation with N->infinity, 1/N->0 rather than dt->0. It is the same as the one that I used to understand why there is a exp(-1/2*sigma^2) drift in a geometric brownian motion and it makes thing more neat.Omar: I agree about your remarks with square roots. Taking dt->0 is like a sort of modulo calculus (e.g. quotient algebra) where dt^2=0 and of course operations like square roots are not well defined in such context. HOWEVER, I am confused by the following example:Suppose Y is a N(0,1), and define the process X(t)=t*(1+Y) . If I apply your reasoning, you can say that E(dX)=dt and Var(dX)=constant.dt^2=0 so you will conclude that dX is non-random (to the first order), so dX=dt (whilst actually it is dt*(1+Y))?Thanks again

Brussels:I think the thing you really need to remember is that the 'dW^2=dt' convention is true only when it appears in an integral over a non-zero interval. The proof is essentially based on the law of large numbers.