Ok, a reallly simple question here but one I can't figure out.I understand that a weiner process is conceptually just a random walk where the step sizes are random. However I have trouble relating it's variance to that of a random walk.For a random walk with steps +r and -r with probs p and q, the var after one step is r-squared.Accordingly after n steps, the var is n*r-squared and the std dev is sqrt(n)*r.But then for the wiener process the var becomes dt (std dev = sqrt(dt)). How does one pass from step size (squared), a distance, to time? The number of steps n also drops out of the equation.

Try doing something similar, but have two parameters - the number of steps per second, and the number of seconds. Then you will have your variance as a function of these two parameters, and see what happens as the first goes to infinity.

what the person said about 2 parameters and letting one go to infinity is fine but in regard to what you said,the expression for a weiner process isds = u*dt + sigma*dw = u*dt + sigma*epsilon*radical t ( discretizing the brownian motion where epsilon is normal(0,1) ).so, the variance is sigma^2* t which is time ( as you say ) if you get rid of the sigma squared which is just a scaling factor.

markpark: the variance is indeed time, but I'm trying to find it starting from a random walk rather than the definition?gjlipman: ok, say the size of a step is +- 1, there are N seconds (fixed quantity) and n steps per second.If n = 1 the total var is N after the N seconds.If we let n vary, then the var after 1 sec is n, and the total var is n*NIf n -> infinity then the var blows up.Lets say N=1 for simplicity. The argument is that we then scale the size of the step taken during the fraction of the second 1/n by 1/sqrt(n).1/sqrt(n) = sqrt(dt) so the var after 1/n sec is dt and after the 1 sec the var is n*dt = TIs this the correct thinking? To quibble abit more, the scaling of 1/sqrt(n) seems abit arbitrary (added to make it work)?

Suppose you have n steps per second, and N seconds. Now, we are increasing n to make things more accurate, not to fundamentally change the situation, so we don't want the variance to be a function of n.So, as we increase n, we want to decrease the up and down magnitudes, in order to leave the variance per 1 second equal to 1.So, the variance per step must equal 1/n. Then, the variance per second is 1, and the variance for the whole period is N.The way to get a variance per step of 1/n is to have each up and down movement equal to sqrt(1/n).So, having got n steps per second with variance per step of 1/n and variance per second of 1, well, this converges by definition on dS=dz (where z is a brownian motion).

Ok got it, thanks