What is the cross-correlation of a WSS signal and the LTI filtered signal?
If the linear system (does not have to be time-invariant) is bounded-input bounded-output (BIBO) stable (meaning that if $|x(t)| \leq M$ for all $t$ for some finite positive 
number $M$, then there is another finite positive number $M^\prime$ such that 
$|y(t)| \leq M^\prime$ for all $t$), then when the input is a finite-variance
random process (a.k.a. second-order proecess), one can interchange the order of
the expectation operation and the integration/sum for the convolution.  That is, if
$$Y(t) = \int_{-\infty}^{\infty} h(s)X(t-s)\,\mathrm ds$$
where $h(t)$ is the impulse response of the LTI system, then 
$$\begin{align*}
E[Y(t)] &amp;= E\left[\int_{-\infty}^{\infty} h(s)X(t-s)\,\mathrm ds\right ]\\
&amp;= \int_{-\infty}^{\infty} E\left[h(s)X(t-s)\,\mathrm ds\right ]\\
&amp;= \int_{-\infty}^{\infty} h(s)E\left[X(t-s)\right ]\,\mathrm ds\\
\end{align*}
$$
and so if the input is a WSS process with $E[X(t)] = \mu_X$ for all $t$,
$$E[Y(t)] = \mu_X\int_{-\infty}^{\infty} h(s)\,\mathrm ds = \mu_Y$$
is also a constant.  In fact, $\{Y(t)\}$ is also a WSS process.  If we define the crosscorrelation function $R_{X,Y}(\tau)$ as 
$R_{X,Y}(\tau) = E[X(t-\tau)Y(t)]$, then
$$\begin{align*}
R_{X,Y}(\tau) &amp;= E\left[X(t-\tau)\int_{-\infty}^{\infty} h(s)X(t-s)\,\mathrm ds\right]\\
&amp;= \int_{-\infty}^{\infty} h(s)E[X(t-\tau)X(t-s)]\,\mathrm ds\\
&amp;= \int_{-\infty}^{\infty} h(s)R_X(\tau-s)\,\mathrm ds.
\end{align*}$$
In short,
$$R_{X,Y} = h*R_X.$$
For completeness, 
$$R_Y = \tilde{h}*h*R_X = (\tilde{h}*h)*R_X = R_h*R_X$$
where $\tilde{h}(t) = h(-t)$ for all $t$ is the time-reversed impulse response
and $R_h = \tilde{h}*h$ is the autocorrelation function of the 
deterministic signal $h(t)$.  Translated to the frequency domain, this gives the
power spectral density relationship
$$S_Y(f) = |H(f)|^2 S_X(f).$$