Laplace transform of the autocorrelation of a wss random process

Question

Consider a wide-sense-stationary random process $x(t)$. The autocorrelation function is $r(t-\tau):=E[x(t)x(\tau)]$. Let $S(s)$ be the Laplace-transform of $r(t)$. Can I compute $S(s)$ as $S(s)=E[X(s)X(s)^*]$, where $X(s)$ is the Laplace transform of $x(t)$?

Answer 1

Writing the definition of $S$ yields $S(s)=E(X(s)x(0)^*)=s\,E(X(s)X(s)^*)$.

Edit: To show the rightmost identity, start from $$E(X(s)X(s)^*)=\iint E(x(t)x(u)^*)\mathrm e^{-st}\mathrm e^{-su}\mathrm du\mathrm dt=2\int_0^\infty\int_t^\infty r(u-t)\mathrm e^{-s(t+u)}\mathrm du\mathrm dt,$$ and use the change of variable $v=u-t$, $w=t$, to reach the formula $$E(X(s)X(s)^*)=2\int_0^\infty\int_0^\infty r(v)\mathrm e^{-s(2w+v)}\mathrm dv\mathrm dw=2S(s)\int_0^\infty\mathrm e^{-2sw}\mathrm dw,$$ which should be enough to conclude.