779. Find the mutual correlation function of two random functions: $X(t)=t^{2} U$ and $Y(t)=t^{3} U$, where $U$ is a random variable, and $D(U)=5$.
Solution. Let's find the mathematical expectations:

$$
m_{x}(t)=M\left(t^{2} U\right)=t^{2} m_{z}, \quad m_{y}(t)=M\left(t^{3} U\right)=t^{3} m_{x}
$$

Let's find the centered functions:

$$
\begin{aligned}
& \dot{X}(t)=X(t)-m_{x}(t)=t^{2} U-t^{2} m_{a}=t^{2}\left(U-m_{a}\right) \\
& \dot{Y}(t)=Y(t)-m_{y}(t)=t^{2} U-t^{3} m_{a}=t^{2}\left(U-m_{a}\right)
\end{aligned}
$$

Let's find the cross-correlation function:

$$
\begin{gathered}
R_{x y}=M\left[\dot{X}\left(t_{1}\right) \dot{Y}\left(t_{2}\right)\right]=M\left\{\left[t_{1}^{2}\left(U-m_{u}\right)\right]\left[t_{2}^{3}\left(U-m_{u}\right)\right]\right\}= \\
=t_{1}^{2} t_{2}^{3} M\left[\left(U-m_{u}\right)^{2}\right]=t_{1}^{2} t_{2}^{3} D(U)=5 t_{1}^{2} t_{2}^{3} .
\end{gathered}
$$

Thus, \( R_{x y}=5 t_{1}^{2} t_{2}^{3} \).