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Can someone prove to me how $dW_t=W_t-W_s$, where $t=s+1$, the difference of the Wiener process eventually equates to $dW_t=z*(dt)^{(1/2)}$ where $z$ is standard normal, $N(0,1)$ in the following SDE: $dS_t=S_t μ dt+S_t σdW_t$?

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The question could use a little rewording. You should define what "eventually equates" means. Anyway, I think the point is that the drift term decays at a faster rate than the volatility. –  quasi Mar 7 at 4:43
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2 Answers 2

The question is not 100% clear.

If you set $X = W_t-W_s$ where $t-s = 1$ then this is equal in distribution to $W_1-W_0$ and the defining property of Brownian motion is that increments are normally distributed.

In the general case $W_t-W_s$ is $N(0,t-s)$, where the second parameter is variance.

If you set $dW_t = W_{t+dt}-W_t = Z \sqrt{dt}$, where $dt>0$ is not infinitesimal but just some positive real number and $Z$ is standard normal (mean $0$ and variance $1$), then the first "$=$" is a definiton and the second "$=$" is "equal in distribution".

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ach okey it was supposed to be $(dt)^{1/2}$ ... –  Probilitator Mar 7 at 10:38
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I think this question might be asking for the central limit theorem.

If we consider a process W which varies as a series of independent random steps, then the Central Limit Theorem tells us that after many steps, the value of W will be normally distributed.

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