I need to prove that for independent standard normal distributed $X_{n}$ and $t\in [0,1]$ the series $$ tX_{0} + \sum_{n=1}^{\infty}\sqrt{2}\frac{\sin{(n\pi t)}}{n\pi}X_{n}$$ converges and that it's limit $B_{t}$ fulfills $EB_{t} = 0$ and $E(B_{t}B_{s}) = \operatorname{min}(t,s)$
I really don't even know how to do the first part, even though it apparently shouldn't be that hard...
Thanks for your help
On
Let's start with showing that the sum converges by considering the mgf. Denote this sum as $B_t$. We know each $X_n$ has expected value $\mu = 0$ and variance $\sigma^2 = 1$. So the movement generating function $M_{X_n}(z) = \exp(\mu z + \frac{1}{2}\sigma^2z^2) = \exp(z^2/2).$ Then \begin{align*}M_{B_t}(z) = E[e^{zB_t}] &= E[e^{ztX_0 + z\sum_{n=1}^\infty \sqrt{2} \frac{\sin(n\pi t)}{n\pi}X_n}]\\&= E[e^{ztX_0} \cdot \prod_{n=1}^\infty e^{z\sqrt{2} \frac{\sin(n\pi t)}{n\pi}X_n}]\\&= E[e^{ztX_0}]\cdot \prod_{n=1}^\infty E[e^{z\sqrt{2} \frac{\sin(n\pi t)}{n\pi}X_n}], \quad\text{see note 1}\\ &= M_{X_0}(zt) \cdot \prod_{n=1}^\infty M_{X_n}(z\sqrt{2} \frac{\sin(n\pi t)}{n\pi})\\ &= \exp(z^2t^2/2) \cdot \prod_{n=1}^\infty \exp(z^2 \frac{\sin^2(n\pi t)}{n^2\pi^2})\\ &= \exp\left(z^2t^2/2 + z^2 \sum_{n=1}^\infty \frac{\sin^2(n\pi t)}{n^2 \pi^2}\right).\end{align*}
Now notice that $$\sum_{n=1}^\infty \frac{\sin^2(n\pi t)}{n^2 \pi^2} \leq \sum_{n=1}^\infty \frac{1}{n^2 \pi^2} = \frac{\pi^2}{\pi^2 6} = \frac{1}{6}$$ so \begin{align*}\exp\left(z^2t^2/2 + z^2 \sum_{n=1}^\infty \frac{\sin^2(n\pi t)}{n^2 \pi^2}\right) = \exp(z^2 t^2/2 + z^2(2ct + c^2)/2) = \exp(z^2(t+c)^2/2)\end{align*} which means $B_t$ is a normal random variable with mean $0$ and variance $(t+c)^2$. Here $c$ is chosen such that $$\sum_{n=1}^\infty \frac{\sin^2(n\pi t)}{\pi^2\pi^2} = \frac{2ct + c^2}{2}$$ which must exist because the sum converges and we can find the $c$ by quadratic formula. Note that $c$ does depend on $t$.
Thus $E[B_t] = 0$. We could also verify this after knowing the sum converges by applying linearity of expectation as follows: $$E[B_t]= tE[X_0] + \sum_{n=1}^\infty \sqrt{2}\frac{\sin^2(n \pi t)}{n\pi}E[X_n] = 0.$$
Lastly, \begin{align*} B_tB_s = stX_0^2 + \sum_{k=1}^\infty s\sqrt{2} \frac{\sin(n\pi t)}{n\pi}X_0X_k + \sum_{k=1}^\infty t\sqrt{2} \frac{\sin(n\pi s)}{n\pi}X_0X_k + \sum_{n,m \geq 1} 2\frac{\sin(n\pi t)\sin(m\pi s)}{nm\pi^2}X_nX_m\end{align*} and $E[X_i X_j] = E[X_i]E[X_j] = 0$ for $i \neq j$ so $$E[B_tB_s] = st E[X_0^2] + \sum_{n=1}^\infty 2 \frac{\sin(n\pi t)\sin(n\pi s)}{n^2\pi^2}E[X_n^2].$$ Since $1 = \mathrm{Var}[X_i] = E[X_i^2] - E[X_i^2] = E[X_i^2]$ we can simplify to \begin{align*}E[B_sB_t] &= st + \sum_{n=1}^\infty 2 \frac{\sin(n\pi t)\sin(n\pi s)}{n^2\pi^2}\\ &= st + \sum_{n=1}^\infty\frac{\cos(\pi n (s-t))\cos(\pi n (s+t))}{\pi^2 n^2},\quad \text{WLOG assume }s\geq t\\ &= st + \frac{\mathrm{Li}_2(e^{i(s-t)\pi})+ \mathrm{Li}_2(e^{-i(s-t)\pi}) - \mathrm{Li}_2(e^{i(s+t)\pi})- \mathrm{Li}_2(e^{-i(s+t)\pi})}{\pi^2n^2},\quad \mathrm{Li}_2 \text{ is the polylogarithm}\\ &= st + \frac{\frac{-(2\pi i)^2}{2!}B_2\left(\frac{s-t}{2}\right)-\frac{-(2\pi i)^2}{2!}B_2\left(\frac{s+t}{2}\right)}{\pi^2 n^2},\quad B_2 \text{ is the Bernoulli polynomial}\\ &= st +B_2\left(\frac{s-t}{2}\right) - B_2\left(\frac{s+t}{2}\right)\\ &= st - st + t\\ &= \min(s,t).\end{align*}
This is the Fourier construction of Brownian motion. Define $$ B_t^n:=tX_0+\sum_{i=1}^n \frac{\sqrt{2}\sin(i\pi t)}{i\pi}X_i. $$ The sequence $\{B_t^n\}$ converges in $L^2$ because $$ \operatorname{Var}(B_t^{n}-B_t^{m})=\sum_{i=m\wedge n+1}^{m\vee n}\frac{2\sin^2(i\pi t)}{i^2\pi^2}\to 0 \quad\text{as }n,m\to\infty, $$ i.e., $\{B_t^n\}$ is $L^2$-Cauchy, and, thus, has an $L^2$-limit $B_t$.
For $s\ne t$, $(B_s^{n},B_t^{n})\xrightarrow{p}(B_s,B_t)$, which implies that the limit is Gaussian with mean $0$, and $\operatorname{Cov}(B_s,B_t)=s\wedge t$ because $$ \operatorname{Cov}(B_s^{n},B_t^{n})=st+\sum_{i=1}^n \frac{2\sin(i\pi s)\sin(i\pi t)}{i^2\pi^2}\to s\wedge t $$ as $n\to\infty$.