The following is stated in a text I am using: Consider the Wiener process (Brownian process) $W(t)$. The Wiener process has no derivative $\xi(t) := \frac{d W}{d t}$, reflecting the fact that it changes randomly and is thus jagged, on even the smallest timescale. The Wiener process is actually an idealization, as no real process changes on infinitely short timescales. It is the fact that we have defined each increment of the Wiener process to be completely independent of the previous increment that makes the Wiener process so jagged. If we demand that it be smooth below some timescale $\Delta t$, then the increments for shorter times could not be independent of each other. Let us discretize time into increments of length $\Delta t$, and denote the Wiener increment in the nth interval by $\Delta W_n$; then the correlation function of $\xi(t)$ is given by $$\langle \xi(n \Delta t) \xi(m \Delta t) \rangle = \bigg\langle \frac{\Delta W_n}{\Delta t} \frac{\Delta W_m}{ \Delta t} \bigg\rangle = \frac{\langle \Delta W_n \Delta W_m \rangle}{(\Delta t)^2} = \frac{\delta_{nm}}{\Delta t}$$ it follows that $\lim\limits_{\Delta t \to 0} \langle \xi(n \Delta t) \xi(m \Delta t) \rangle = \langle \xi(t) \xi(t + \tau) \rangle = \delta(\tau)$.

Question: Can anyone see how we get the final equation involving the delta function and how the final limit follows?

Thanks for any assistance.

  • $\begingroup$ (Note: entry for Wiener process on Wikipedia.) $\endgroup$ – gung Apr 15 '18 at 1:40

$\Delta W_n=W(n\Delta t)-W((n-1)\Delta t)$. By definition, brownian motion has independent differences whenever the two time intervals don't intersect. Otherwise when $n=m$, you're calculating the variance of Brownian motion on an interval $\Delta t$, which is equal to $\Delta t$.

  • $\begingroup$ Thanks for your response, but I'm not following how what you stated implies that $\frac{\langle \Delta W_n \Delta W_m \rangle}{(\Delta t)^2} = \frac{\delta_{nm}}{\Delta t}$? Also, what are you considering as the variance in the equations that I wrote? $\endgroup$ – Tim Davids Aug 4 '17 at 11:23

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