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I found in some lecture notes that Brownian motion is defined by its Fourier series:

$$ B(t) = \sum_{m \in \mathbb{Z},\; m \neq 0} \frac{a_m}{m} e^{2\pi i \,mt} $$

Then I would get that its derivative is white noise:

$$ W(t) = \sum_{m \in \mathbb{Z},\; m \neq 0} a_m e^{2\pi i \,mt} $$

What is the random Fourier series if we truncate it to the positive half? Is there a name for that random process?

$$ f(t) = \sum_{ m \geq 0} a_m e^{2\pi i \,mt} $$

Here $a_m$ are standard identical, independently distributed Gaussian random variables with mean 0 and variance 1.

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  • $\begingroup$ The series you wrote after "$B(t)=$" is a periodic function, and it is for such functions that Fourier series are defined. But the usual Brownian motion (or Wiener process) is not periodic. $\endgroup$
    – Michael Hardy
    Aug 6, 2013 at 4:01
  • $\begingroup$ @MichaelHardy The notes say $B(t) - b_0(t)$ is the Brownian motion. The variance of Brownian motion is linear in the time $t$. Then $B(0) = 0$, $B(t)$ is a.s. continuous, independent increments, and $B(t_2) - B(t_1)$ is normal. $\endgroup$
    – john mangual
    Aug 6, 2013 at 10:27
  • $\begingroup$ @john : according to the notes (which have some confusing typos), brownian motion (starting at $0$) is given by $a_0t+B(t)-B(0)$ but only for $0\leq t\leq 1$. Here $a_0$ is another standard gaussian. $\endgroup$
    – BS.
    Aug 6, 2013 at 13:55
  • $\begingroup$ @BS With independent increments, you could pick new Gaussian variables at t=1 and keep going. $\endgroup$
    – john mangual
    Aug 6, 2013 at 20:17

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