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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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    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 http://marcofrasca.wordpress.com/2012/02/02/numerical-evidence-for-the-square-root-of-a-wiener-process/ and comments there.2012-07-28

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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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    Now I see. Thanks!2012-07-28