I'm trying to solve this exercise:
$\bullet$ Find mean and variance of the next stochastical process, and prove it is a second order stationary process:
$Y(t)=\displaystyle\int_{t}^{t+1}(W(s)-W(t))\;ds$
To obtain the mean, as $W(s)-W(t)\thicksim N(0,s-t)$, we have that $\mathbb{E}[Y(t)]=\displaystyle\int_{t}^{t+1}\mathbb{E}[W(s)-W(t)]\;ds=\displaystyle\int_{t}^{t+1}0\; ds=0$
To prove it is a second order stationary process, I know I have to show that $\mathbb{E}[Y(t)^2]<\infty$.
As $\mathbb{E}[Y(t)^2]= \Gamma_Y (t,t)$.I think all would be done if I had the covariance, $\Gamma_Y(s,t)$.
But how can I calculate that covariance?
Thanks for any help.
Let $s \leq t$. First of all, we note that by the independence of the increments,
$$\mathbb{E}((W_u-W_s) \cdot (W_v-W_t)) = 0$$
for any $s \leq u \leq t \leq v$. Therefore, Fubini's theorem shows that $\mathbb{E}(Y_t \cdot Y_s)=0$ for any $s \leq t-1$. For $t-1 \leq s \leq t$, we find
$$\begin{align*} \mathbb{E}(Y_t \cdot Y_s) &= \int_t^{t+1} \int_t^{s+1} \mathbb{E}\big(((W_u-W_t)+(W_t - W_s)) \cdot (W_v-W_t)\big) \, du \, dv\\ &=\int_t^{t+1} \int_t^{s+1} \mathbb{E}\big((W_u - W_t) \cdot (W_v-W_t)\big) \, du \, dv \\ &=\int_0^{1} \int_0^{(s+1)-t} \mathbb{E}\big((W_{u+t} - W_t) \cdot (W_{v+t}-W_t)\big) \, du \, dv\end{align*}$$
using that $W_v-W_t$ and $W_t-W_s$ are independent random variables with mean $0$. Note that $B_u := W_{t+u}-W_t$, $u\geq0$, is a Brownian motion and therefore we find
$$\mathbb{E}(Y_t \cdot Y_s) = \int_0^{s+1-t} \int_0^1 \underbrace{\text{cov}(B_u,B_v)}_{\min\{u,v\}} \, dv \, du$$
The latter integral can be easily computated.