WSS implies IID?

Question

If I have two WSS (wide-sense stationary) processes, $x(t)$ and $y(t)$, can I state that they are IID (independent and identically distributed) sequences?

Answer 1

A strict-sense stationary process, $x(t)$, has its $n$-order statistics preserved throughout the time, where $n \in \mathbb{N}_{++}$. That is, the $n$-order statistics of the random variables $\left\{x(t_1), x(t_2), \cdots, x(t_n)\right\}$ is identical to the set $\left\{x(t_1+\tau), x(t_2+\tau), \cdots, x(t_n+\tau)\right\}$, for any $\tau \in \mathbb{R}$. Therefore,

$$ p_{x(t_1), x(t_2), \cdots, x(t_k)}(x_1, x_2, \cdots, x_k) = p_{x(t_1+\tau), x(t_2+\tau), \cdots, x(t_k+\tau)}(x_1, x_2, \cdots, x_k), $$ where $p(\cdot)$ is the probability density function (PDF).

However, strict-sense stationary processes are a strong statement, and it is suffice for many applications to guarantee that the first- and second-order statistics are stationary. We refer these type of process as weak-sense stationary process or, for short, WSS process.

For these processes, the mean (its first-order statistic) is simply a constant:

$$ \mu_x(t) = E[x(t)] = \int_{-\infty}^{\infty} x p_{x(t)} (x)\;dx = \mu_x(t+\tau) = \mu_x. $$

And the autocorrelation (a second-order statistic) only depends on $\tau$, that is

$$ R_x(t, t+\tau) = E[x(t)x(t+\tau)] = \iint_{-\infty}^{\infty} x_1x_2 p_{x(t), x(t+\tau)} (x_1, x_2)\;dx_1\;dx_2 = \phi_x(\tau). $$

Similarly, if $x(t)$ and $y(t)$ is jointly WSS, then $$ R_{x,y}(t, t+\tau) = E[x(t)y(t+\tau)] = \iint_{-\infty}^{\infty} x y p_{x(t), y(t+\tau)} (x,y)\;dx\;dy = R_{x,y}(\tau). $$

You just do not realized before that WSS implies IID because the notation $R_{x,y}(t, t+\tau)$ is commonly simplified to $R_{x,y}(\tau)$. You can change $t$ freely, the PDF $p_{x(t), y(t+\tau)} (x,y)$ will be the same :)