If $X(t)$ is wide sense stationary (WSS), and f(.) is a real-valued monotonic function, then is $f(X(t))$ and X (t) jointly WSS?

Question

If $X(t)$ is a continuous wide sense stationary (WSS) process, we know that E(X(t)) is independent of time R_xx (t1,t2) depends on t1-t2 only.

Now if we have a function f(z) where z is any real number, inverse of f(z) exists and f(z) is monotonic and bounded (particularly, f(z)=tanh(z) ) can we say that X(t) and f(X(t)) are jointly WSS?

Particularly, if X(t) is a gaussian process or of the form X(t)=A(sin(wt+\theta)) with \thata ~ uniform(0,2\pi) Update: found that for gaussian process, you can use Hermite polynomials to show that f(X(t)) is 2nd order stationary

Answer 1

No, for two reasons.

One is pathological. Suppose $f()$ is chosen so $f(X(t))$ has no finite moments, then $f(X(t))$ cannot be WSS. For example, if $X$ is Normal, I think $f(X) = \exp(\exp(X))$ would do. Even if $X(t)$ is strong-sense stationary, $f(X(t))$ may not be weak-sense stationary

The second reason is that WSS is a restriction only on two moments. You don't say whether you're interested in discrete or continuous time here. I'll start with discrete, because it's easier Suppose $X(t1)$ is independent of $X(t2)$, but for some $t$, $X(t)\sim N(0,1)$ and for other $t$ $X(t)=\pm 1$ with equal probability. The mean and variance are constant in time and the autocovariance is zero at all lags except 0. So $X(t)$ is WSS. But $\exp(X(t))$ is different for the Normal and discrete distributions, so $f(X(t))$ is not WSS.

If you want an example for continuous time that has nice behaviour -- like a.s. continuous sample paths -- it's harder to come up with, but the basic idea is the same: knowing the mean and autocovariance doesn't pin down the marginal distribution, and different marginal distributions will give different variances under monotone transformation.