Let a filter with an impulse response $h$ (which is deterministic). Let $X$ be a random signal. The filtered version of $X$ is given by $$ Y = h\star X $$ which admits for expectation $$ \begin{aligned} \mu_Y(t) & = \text{E}(Y(t)) = \text{E}(h\star X (t)) \\& = \text{E}\left(\int_\mathbb{R} h(\tau)x(t-\tau)\mathrm{d} \tau\right) \\& = \int_\mathbb{R} h(\tau)\text{E}(x(t-\tau))\mathrm{d} \tau \\& = \int_\mathbb{R} h(\tau)\mu(t-\tau)\mathrm{d} \tau \\& = h \star \mu_X (t) \end{aligned} $$ and for autocorrelation $$ \begin{aligned} R_Y(t,s) & = \text{E}(Y(t)Y(s)) = \text{E}\left(h\star X(t) h\star X(s)\right) \\ & = \text{E}(\int_\mathbb{R} h(\tau_1)x(t-\tau_1)\mathrm{d} \tau_1 \int_\mathbb{R} h(\tau_2)x(s-\tau_2)\mathrm{d} \tau_1\mathrm{d} \tau_2) \\ & = \int_\mathbb{R}\int_\mathbb{R} h(\tau_1)h(\tau_2)\text{E}(x(t-\tau_1)x(s-\tau_2))\mathrm{d} \tau_1\mathrm{d} \tau_2 \end{aligned} $$
If, in addition, $X$ is a weak sense stationary random signal, then the expectation reads $$ \begin{aligned} \mu_Y(t) & = \int_\mathbb{R} h(\tau)\mu_X(t-\tau)\mathrm{d} \tau \\ & = \mu_X\int_\mathbb{R} h(\tau)\mathrm{d} \tau \\ & = \mu_X \hat h(0) \end{aligned} $$ and the autocorrelation reads $$ \begin{aligned} R_Y(t,s) & = \int_\mathbb{R}\int_\mathbb{R} h(\tau_1)h(\tau_2)\text{E}(x(t-\tau_1)x(s-\tau_2))\mathrm{d} \tau_1\mathrm{d} \tau_2\\ & = \int_\mathbb{R}\int_\mathbb{R} h(\tau_1)h(\tau_2)R_X(t-s-\tau_1+\tau_2)\mathrm{d} \tau_1\mathrm{d} \tau_2\\& = R_Y(t-s) \end{aligned} $$ $Y$ is then also a weak sense stationary random signal.
By applying the Wiener-Khintchine’s theorem, we get $$ S_Y(\nu) = \hat R_Y(t) = |\hat h(\nu)|^2 S_X(\nu) $$
Let $X_1$ and $X_2$ two weak sense stationary random signals. Let two filters with respective impulse responses $h_1$ and $h_2$. We denote by $$ Y_1 = h_1\star X_1\quad\text{ and }\quad Y_2 = h_2\star X_2 $$ We then get the interferences formula $$ R_{Y_1Y_2}(t) = (h_1\star R_{X_1X_2}\star h_2^{-})(t) $$ with $h_2^{-}(t) = h_2(-t)$
In particular, if $X_1=X_2=X$ and $h_1=h$ et $h_2=\delta$, we have $Y=h\star X$, $Y_2=X$ $$ R_{Y_1Y_2}(t) = R_{YX}(t) = (h\star R_{XX})(t) $$ and the Wiener-Khintchine’s theorem leads to $$ S_{Y_1Y_2}(\nu) = h_1(\nu)S_{X_1X_2}(\nu)\overline{h_2(\nu)} $$