Distribution of integral with respect to Brownian motion

Question

Let $B$ be a Brownian motion and define the complex sequence $(X(n))_{n\in \mathbb Z}$ as

$$ X(n) := \int^\pi_{-\pi} e^{inx} dB(\pi + x)$$

What is the distribution of $X(n), n\in \mathbb Z$?

Answer 1

First I'll recast the problem as $$ X(n) = e^{-in\pi}\int_{0}^{2\pi} e^{inx} dB(x) = e^{-in\pi}\left(\int_{0}^{2\pi} \cos(nx) dB(x) + i \int_{0}^{2\pi} \sin(nx) dB(x) \right)$$ Each of the two stochastic integrals of determinstic continuous functions with respect to the BM. Hence they are Gaussian random variables, so we only need to know their mean and variance. Since they are stochastic integrals their mean it is zero. By Ito's isometry their variance can be expressed in terms of the quadratic variation (remember $\langle B \rangle(x) = x$) \begin{align*} E((\int_{0}^{2\pi} \sin(nx) dB(x))^2) = \int_{0}^{2\pi} \sin(nx)^2 dx &= \pi - \frac{\sin(4\pi n)}{4n} \\ E((\int_{0}^{2\pi} \cos(nx) dB(x))^2) = \int_{0}^{2\pi} \cos(nx)^2 dx &= \pi + \frac{\sin(4\pi n)}{4n} \end{align*} We can also calculate their covariance $$ \int_{0}^{2\pi} \sin(nx)\cos(nx) dx = \pi + \frac{\sin^2(2\pi n)}{2n} $$ And since they are normal we can conclude that they are not independent. I think you can continue from here ;)