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I'm simulating a Brownian motion in MATLAB, however I'm getting a strange outcome where the variance of the increments of the Brownian motion grow over time when it should stay constant. For example I construct a Brownian motion system,

brown_drift = @(t,X) 0;
brown_vol = @(t,X) .2;
brown_sys = bm(brown_drift, brown_vol);

And then I interpolate 1000 trials with timestep 1 and length 10

inter_brown = interpolate(brown_sys, 0:1:10, zeros(1,1,1000),'Times',[0]);
inter_brown = squeeze(inter_brown);

The increments of a Brownian motion should be independent so if I construct a matrix of the increments and take the variance, they should all be the same and equal to the volatility parameter squared.

inc = inter_brown(2:end,:)-inter_brown(1:end-1,:);
var(inc')
ans = 0.0374    0.1184    0.2071    0.2736    0.3516    0.4190    0.5511    0.5891    0.6767    0.7647

However it clearly doesn't satisfy what comes out of the simple theory where the variance should be 0.2^2 for each increment. It looks like each increment in the future adds 2*0.2^2 to the variance of the increment before it. I can't seem to figure out why this is happening when the Brownian motion seems to satisfy the other theory, e.g. the variance of the motion at a given time. Is there something obvious I'm missing here?

Ivan
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1 Answers1

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There are three things I would do differently:

1) Choose your timestep of your Brownian paths smaller. The more steps in one path you simulate the more accurate the calculated variance of your increments in each path will be (check with var(var(inc))). This means for each path you will get more similar results for the variance of the increments.

Shows variance for 100 different paths. One time with tine steps and one time with even tinier steps. You see that the smaller the timestep is, the smaller the variance of the variances gets.

2) Also the variance of the increments is scaling differently if your time is bigger than 1 (see Volatility over time) so you would get different results depending on your initial time scale. i.e 

interpolate(brown_sys, 0:0.1:10, zeros(1,1,1000),'Times',[0]);

would result in a variance of about 0.4 and

interpolate(brown_sys, 0:0.1:100, zeros(1,1,1000),'Times',[0]);

would result in a variance of 4. So just change the time to:

interpolate(brown_sys, 0:dt:1, zeros(1,1,1000),'Times',[0]);

Where you should choose dt very small i.e. dt = 0.001;

3) You should consider the variance of inc and not the transposed one. Because the matrix inter_brown you obtain has on each column a simulated path. So the steps belonging to one path should be in one column. Thus the matrix inc you calculate contains the increments according to one path in a column. The var() of matrices calculates the variance of a sample stored in one column. So if you want to know the variance of the increments of you first path you would have to call something like: var(inc(:,1)). Calling var(inc) gives you the variance for each path. You can plot them with plot(var(inc)). You will then see that the variance of increments is around 0.04as expected. Check it with mean(var(inc)).

v.tralala
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  • Thank you. I don't think I was clear before, I'm not concerned with the variance within a path. I'm concerned about the variance of a given increment from t-1 to t, so I need to take the variance of the transposed matrix. – Ivan Jul 29 '16 at 17:11
  • The variance of the Wiener increment is also dt^2 as long as the time intervalls are disjoint. So you can consider increments from t-1 to t or from t-2 to t-1 and so on. Basically if you calculate 'var(inc(randi(numel(inc(:)),1,100)))' for example then you will obtain correct values. It is however very interesting that by looking at the increments at a specific time over all paths the variance is increasing as the time goes on. I can't explain that but I am quite sure that you don't need to consider `inc` transposed. Are you sure this is what you looking for? – v.tralala Jul 30 '16 at 00:29