Expected value and variance of random process

Question

Let $U,V$ be random variables with distributions $\mathcal{U}(-1,1)$ ,$\mathcal{E}(2)$ (uniform and exponential). If $U$ and $V$ are independent what is the variance and expectation of the random process $X_t = f(t)U + g(t)V$. I don't understand how is this calculated, how do we treat the functions of $t$ in front of the random variables? Can $X_t$ be treated as a random variable with constant parameter $t$?

Answer 1

Hint: In full generality (if $U$ and $V$ are integrable), $$E(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}\cdot E(U)+\color{blue}{3}\cdot E(V).$$ If $U$ and $V$ are independent (and square integrable), $$\mathrm{var}(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}^2\cdot \mathrm{var}(U)+\color{blue}{3}^2\cdot \mathrm{var}(V).$$