# Expansion of Brownian Motion

Let $$\{\eta_k : k=0,\cdots\}$$ be a collection of mutually independent standard Gaussian random variable with mean zero and variance one. Define
\begin{align*}
X(t) =\frac{t}{\sqrt\pi} \eta_0 + \sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty \frac{\sin(k t)}{k} \eta_k \;.
\end{align*}

1. Show that on the interval $$[0,\pi]$$, $$X(t)$$ has the same mean, variance and covariance as Brownian motion. (In fact, it is Brownian motion. )
2. ** Prove it is Brownian motion.

There are a number of ways to prove it is Brownian motion.. One is to see $$X$$ as the limit of the finite sums which are each continuous functions. Then prove that $$X$$ is the uniform limit of these continuous functions and hence is itself continuous.

Then observe that “formally” the time derivative of $$X(t)$$ is the sum of all frequencies with a random amplitudes which are independent and identical $$N(0,1)$$ Gaussian random variables. This is the origin of the term “white noise” since all frequencies are equally represented as in white light.

In the above calculations you may need the fact that
\begin{align*}
\min(t,s)= \frac{ts}\pi +\frac{2}\pi \sum_{k=1}^\infty \frac{\sin(k t)\sin(k s)}{k^2}\;.
\end{align*}

If you are interested, this can be shown by periodically extending $$\min(t,s)$$ to the interval $$[-\pi,\pi]$$ and then showing that it has the same Fourier transform as the right-hand side of the above expression. Then use the fact that two continuous functions with the same Fourier transform are equal on $$[-\pi,\pi]$$.)