Solution:

Stationary random process A random process s(t) is said to be stationary if it is statistically indistinguishable from a delayed version of itself. That is, s(t) and $s(t-d)$ have the same statistics for any delay $d \in(-\infty, \infty)$.

For a stationary random process s, the mean function satisfies

$$ m_s(t)=m_s(t-d) $$

for any t, regardless of the value of d. Choosing d=t, we infer that

$$ m_s(t)=m_s(0) . $$

That is, the mean function is a constant. Similarly, the autocorrelation function satisfies

$$ R_s\left(t_1, t_2\right)=R_s\left(t_1-d, t_2-d\right) $$

for any $t_1, t_2$, regardless of the value of d. Setting $\tau=t_2$, we have

$$ R_s\left(t_1, t_2\right)=R_s\left(t_1-t_2, 0\right) . $$

That is, the autocorrelation function depends only on the difference in its arguments.

Stationarity is a stringent requirement that is not always easy to verify. However, the preceding properties of the mean and autocorrelation functions can be used as the defining characteristics for a weaker property termed wide sense stationarity.

Wide sense stationary (WSS) random process A random process s is said to be WSS if

$$ m_s(t) \equiv m_s(0) \text { for all } t $$

and

$$ R_s\left(t_1, t_2\right)=R_s\left(t_1-t_2, 0\right) \text { for all } t_1, t_2 . $$

In this case, we change the notation and express the autocorrelation function as a function of $\tau=t_1-t_2$ alone. Thus, for a WSS process, we can define the autocorrelation function as

$$ R_s(\tau)=E\left[s(t) s^*(t-\tau)\right] \text { for } s \text { WSS } $$

with the understanding that the expectation is independent of t.