Question: Let $W$ a Brownian motion. If $X(t) = \int_0^t W(s)\cos (t-s) ds$ compute $$P(X(t) \le x)=??$$
Attempt: Since that Brownian motion has continous paths, the Riemann integral $X(t)$ is well defined for all random path as we integrate path by path. Then $$ X(t)=\lim_{n\to \infty} \sum_{i=1}^{n} W(s_i)\cos(t-s_i)(s_i-s_{i-1})$$ where $s_0,\dots,s_n$ partition of $[0,t]$.
Denoted the Riemann sum by $S_n$. Fixed $n$ and partition $(s_i)$, $S_n$ is linear combination of the Gaussian variables $(W(s_i))$ and $(W(s))$ is Gaussian process thus $S_n$ is Gaussian variable. We want to see that effectively $S_n$ converge in distribution.
For simplicity, take $s_i =\frac{t}{n^2}i$, $i=0,1,\dots,n^2$. Then, $$ S_n = \frac{t}{n^2} \sum_{i=1}^{n^2} W\left(\frac{t}{n^2} i\right)\cos\left(t-\frac{t}{n^2}i\right). $$ Since $$ \left( W\left(\frac{t}{n^2}\right), \dots, W\left(\frac{t}{n^2} n^2\right) \right) \sim \frac{1}{n} \left( W(t), \dots, W(n^2t) \right)= \frac{1}{n} Z$$ we have to $$ S_n = \frac{t}{n^3} \pmb{a}\cdot Z$$ where $$ \pmb{a}=\left(\cos\left(t-\frac{t}{n^2}\right), \cdots, \cos\left(t-\frac{t}{n^2}n^2\right) \right).$$ Thus $$ S_n \sim \mathcal{N}\left(0, \frac{t^2}{n^6} \pmb{a}\mathrm{cov}(Z) \cdot \pmb{a}\right).$$
Indeed, the limit of Gaussian variables is a Gaussian, then $$ X(t)\sim \mathcal{N}(0,\sigma^2).$$
To calculate the variance, note that $$ \sigma^2 = \mathbb{E}\left[ \int_0^t W(u)\cos(t-u)\, du \int_0^t W(v)\cos(t-v)\, dv \right] = \mathbb{E}\left[ \int_0^t \int_0^t W(u)W(v)\cos(t-u)\cos(t-v)\, dudv \right]. $$ Then, Fubini's theorem follows $$ \sigma^2 = \int_0^t \int_0^t \mathbb{E}\left[ W(u)W(v) \right] \cos(t-u)\cos(t-v)\, dudv = \int_0^t \int_0^t \min(u,v) \cos(t-u)\cos(t-v)\, dudv. $$
The justification that the limit of Gaussian variables is Gaussian is deducted from Levy's Theorem of Continuity? Thanks for attention.
