LTI output of a WSS input is WSS

Let a WSS process $X(t)$ (mean $m_X$, autocorrelation $R_X(\tau)$, PSD $S_X(f)$) be applied to a stable LTI system with impulse response $h(t)$ and $H(f)$:

$$Y(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha$$

  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y

1. Mean of output. Expectation and integration can be interchanged:

$$\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\ &= m_X\,H(0) \quad(\text{a constant, independent of } t) \end{aligned}$$

2. Autocorrelation of output.

$$\begin{aligned} R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}$$

because $X$ is WSS, its autocorrelation depends only on the time difference $(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta$. The result depends only on $\tau$, not on $t$:

$$R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)$$

3. Mean-square value $E[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta$ is finite and constant.

Since the mean is constant and the autocorrelation depends only on $\tau$, $Y(t)$ is also wide-sense stationary.

4. Output PSD. Taking the Fourier transform of $R_Y(\tau)$ ($h(\tau)\to H(f)$, $h(-\tau)\to H^*(f)$):

$$S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)$$

and the output power is $P_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df$. The phase of $H(f)$ has no effect on the output PSD.

Example. White noise ($S_X=N_0/2$, WSS) through an RC low-pass filter, $\lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2}$:

$$S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2},\qquad R_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}$$

$R_Y$ depends only on $\tau$, confirming that the output is WSS; its power $R_Y(0)=N_0/4RC$ is now finite.