Expectation of the derivative of a random process

Question

Let's have a Random Process $Y(t) = X(t) + 0.3 X'(t)$

Mean of $X(t) = 5t$

Question : Find the mean function of $Y(t)$

What I did : $E(Y) = E(X) + 0.3\cdot E(X')$ ?

I don't know if I have to take the derivative out of the expectation and if yes, then why (what are the cases where I can do it)?

Answer 1

Suppose $X$ is m.s. differentiable. Then for any $t$ fixed, $$ \frac{X(s)-X(t)}{s-t}\to_{m.s.}X'(t)\quad \text{as }s\to t $$ It thus follows that $$ \frac{\mu_X(s)-\mu_X(t)}{s-t}\to_{m.s.}\mu_{X'}(t)\quad \text{as }s\to t \tag 1 $$ because the limit of the means is the mean of the limit, for a m.s. convergent sequence (in fact if $X_n\to_{m.s.}X$ then $\mathbb{E}(X_n)\to_{m.s.}\mathbb{E}(X)$)

But (1) is just the definition of the statement that the derivative of $\mu_X$ at $t$ is equal to $\mu_{X'}(t)$.That is,$\frac{d X}{dt}(t) =\mu_{X'}(t)$ for all $t$, or more concisely, $\mu'_X(t)=\mu_{X'}(t)$

So you have $$\mathbb{E}(Y)=\mu_X+\mu_X'=5t+\frac{3}{2}$$