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When I look up the definition of 'Wiener process' at Wikipedia, it tells me:

$W(0) = 0$ and $W(t) - W(s) \sim N(0, t-s)$.

When I try to simulate this in matlab, I get different results when I define a vector $W1$ to be like:

$W1 = cumsum(dW)$, where $dW(j) \sim N(0, dt)$,

and a vector $W2$ to be like:

$W2(0) = 0$ and $W2(j) \sim N(0, dt*j)$

$W2$ apparently doesn't look like a Brownian motion, but it is still compliant with the requirements. How comes?

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I don't really understand: in the second one, the mean of an increment is: $E[W2(j) - W2(j-1)]$ with $W2(j) - W2(j-1) \sim N(0, j - (j-1)))$, so that should be zero as well? EDIT: apparently the comment I responded to is removed. – JeffersonA Jul 28 '12 at 10:23
You have to sum all the increments. This is the reason why the first code works ("cumsum") and the the second one does not. – Jon Jul 28 '12 at 10:38
But the idea was that in $W2$, instead of calculating the increments between $j$ and $j-1$, I used the increment between $j$ and 0 (so no additional cumsum step should be necessary) – JeffersonA Jul 28 '12 at 10:45
This is not so. "cumsum" is always needed as you are describing a trajectory that, while has not derivative, is anyway continuous. The other way is just to write dW2(j)=dW2(j)+W2(j-1) and plotting dW2 but is not so efficient. You can find some code at… and comments there. – Jon Jul 28 '12 at 10:52
up vote 1 down vote accepted

In $W_2$ you're plotting a sample from a fresh different instance of the Wiener process for each $j$. The successive values are uncorrelated with each other, so even though things like $W_2(17)-W_2(0)$ have the distribution specified by the definition, $W_2(17)-W_2(16)$ doesn't.

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It looks like I got something fundamentally wrong. I thought that $W_2(17) - W_2(16) \sim N(0, 17*dt) - N(0, 16*dt) \sim N(0, dt)$ (that last step assumes the $W_2(j)$'s are i.i.d.) I still don't see my error. – JeffersonA Jul 28 '12 at 11:04
When the normal distributions are independent, since a normal distribution is symmetric about the mean, $N(0,17)-N(0,16)$ has the same distribution as $N(0,17)+N(0,16)$, which is $N(0,32)$ rather than $N(0,1)$. Your computation produces independent successive values of $W_2$, but the successive values in a Wiener process are not supposed to be independent. – Henning Makholm Jul 28 '12 at 11:09
Now I see. Thanks! – JeffersonA Jul 28 '12 at 11:18

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