816. The correlation function $K_{x}\left(t_{1}, t_{3}\right)=$ $=\cos \omega t_{1} \cos \omega t_{2}$ of the random function $X(t)$ is given. Find: a) the correlation function; b) the variance of the integral $Y(t)=$ $=\int_{0}^{1} X(s) \mathrm{d} s$.
Solution. a) The correlation function of the integral $Y(t)=$ $=\int_{0}^{t} X(s)$ ds is equal to the double integral of the given correlation function:

$$
\begin{aligned}
& K_{y}\left(t_{1}, t_{2}\right)=\int_{0}^{t_{1}} \int_{0}^{t_{1}} K_{x}\left(s_{1}, s_{2}\right) d s_{1} d s_{2}=\int_{0}^{t_{1}} \int_{0}^{t_{1}} \cos \omega s_{1} \cos \omega s_{2} d s_{1} d s_{2}= \\
& =\int_{0}^{t_{1}} \cos \omega s_{1} d s_{1} \int_{0}^{t_{1}} \cos \omega s_{2} d s_{2}=\left(\sin \omega t_{1} \sin \omega t_{2}\right) / \omega^{2}
\end{aligned}
$$

b) Find the required variance by setting $t_{1}=t_{2}=t$ in the found correlation function:

$$
D_{y}(t)=K_{y}(t, t)=\sin ^{2} \omega t / \omega^{2} .
$$