Determine the distribution of $\int_0^t (W_s-\frac{s}{t}W_t) ds$, where $(W_s)_{s\geq 0}$ is a brownian motion

Question

I have to find the distribution of $X_t:=\int_0^t (W_s-\frac{s}{t}W_t) ds$ where $(W_s)_{s\geq 0}$ is a brownian motion.

I already showed the first integral $\int_0^t W_s ds$ is $\mathcal{N}(0,t^3/3)$. How should I proceed with $\int_0^t sW_tds$?

Answer 1

The main goal of the exercise is to make you realize (and use the fact) that the random variable $X$ defined as $$ X=\int_0^tY_s\mathrm ds, \qquad Y_s=W_s-\frac{s}{t}W_t, $$ is a linear combination of the gaussian family $(W_s)_{0\leqslant s\leqslant t}$ and that, as such, $X$ is itself gaussian. Hence, to fully determine the distribution of $X$, all there is to do is to compute $E(X)$ and $\mathrm{var}(X)$.

Let me assume that you can show that $E(X)=0$ and let me turn to the computation of the variance. Note that $$ E(X^2) = E\left(\int_0^t\int_0^tY_sY_u\mathrm ds\mathrm du\right)=2\int_0^t\int_0^uE(Y_sY_u)\mathrm ds\mathrm du. $$ The computation of $E(Y_sY_u)$ for every $0\leqslant s\leqslant u\leqslant t$, then of $\mathrm{var}(X)$, should not be a problem now.