Let $w(t)$ be a random process (for example a noisy signal) with the probability density function $f_w(w,t)$. Is it true that the expected value of the Laplace transform (with respect to time) is equal to the Laplace transform of the expected value? That is,
$$ E\left [\mathscr{L}\{w(t)\}\right ] = \mathscr{L} \{ E\left [w(t) \right ] \} \;\;? $$
Here is my attempt to prove it but I am not sure if it is correct. Especially when rearranging terms inside the integrals.
\begin{eqnarray} E\left[ \mathscr{L}\{w(t)\}\right ] &=& E\left [ \int_{0}^{\infty } w(t) e^{-st}dt \right ] \\ &=& \int_{-\infty}^{\infty} \int_{0}^{\infty } w(t) e^{-st}dt \, f_w(w,t)dw \\ &=& \int_{0}^{\infty } \int_{-\infty}^{\infty} w(t)f_w(w,t) e^{-st} dw\, dt \\ &=& \int_{0}^{\infty } e^{-st} \underbrace{ \int_{-\infty}^{\infty} w(t)f_w(w,t) dw}_{E[w(t)]} \, dt \\ &=& \mathscr{L} \{ E\left [w(t) \right ] \} \end{eqnarray}
Am I allowed to move $f_w(w,t)$ inside the time integral and then switch the order of integration since the limits do not depend on either $t$ or $w$?
Intuitively I think should be able to interchange the operators since they are both linear and the Laplace transform operates on time while the expectation uses the other variable on which the probability density function depends. So they don't mess with each other.
I am aware of the case:
$$\mathscr{L} \{ f_X(\overset{\downarrow}{x} ,t) \} = \int_{0}^{\infty } e^{-sx} f_X(x,t) dx = E \left[ e^{-sX(t) } \right] $$
for a non-negative random variable $X$. This is when the Laplace transform operates on the sample space as opposed to my situation.
Any insight to this issue will be greatly appreciated!