LTI and Deterministic Channel
If your input "random signal" is modeled as a zero-mean wide-sense-stationary random process $\{X(t)\}$ with autocorrelation function $R_X(t)$, then the output of the channel (which is basically an LTI system with impulse response $h(t)$) is also a zero-mean wide-sense-stationary process $\{Y(t)\}$ with autocorrelation function $R_Y(t)$ given by 
$$R_Y = R_h \star R_X$$
where $R_h$ is the autocorrelation function of the impulse response $h(t)$ $\big($that is, $R_h(t) = \int_{-\infty}^\infty h(t)h(t+\tau) \, \mathrm d\tau$ or $\int_{-\infty}^\infty h(t)h(t-\tau) \, \mathrm d\tau$ for left-handed folks$\big)$ and $\star$ notes convolution. Indeed, since both $R_h$ and $R_X$ are even functions, that convolution can be written as a cross-correlation if you like, which makes for a nice mantra to murmur to impress your boss during your presentation: "The output autocorrelation function is found by cross-correlating the input autocorrelation with the channel autocorrelation".

If you prefer power spectral densities, then 
$$S_Y(f) = |H(f)|^2S_X(f)$$ 
where $H(f)$ is the channel transfer function.