Let's say a standard Brownian motion is written as $B(t)$. Then, define $W(t) = \sigma B(t)$. I know that $\int_0^1 W(t)dt $ has the distribution Normal(0,$\sigma^2/3$). But, what about $\int_0^1 tW(t)dt$?
Also, it seems that the two distributions $\int_0^1 W(t)dt $ and $\int_0^1 tW(t)dt$ are correlated. Is there a way to calculate their covariance?
Thanks!
$\newcommand{\E}{\mathbb{E}}$ Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $\sigma^2 = 1$, as we can simply multiply by $\sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $\int_0^1 t W(t) \,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.
We compute \begin{align*} \E\left[\int_0^1W(t)\, dt \cdot \int_{0}^1 sW(s)\,ds \right] &= \E\left[ \iint s W(s) W(t)\, ds\, dt\right] \\ &= \int_{0}^1 \int_0^s s\E[W(s)W(t)]\,dt\,ds + \int_{0}^1 \int_0^t s\E[W(s)W(t)]\,ds\,dt \\ &=\int_0^1\int_0^s ts \,dt\,ds + \int_0^1 \int_0^t s^2 \,ds\,dt\\ &= \frac{1}{8} + \frac{1}{12} \\ &=\frac{5}{24}\,. \end{align*}
