# 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?

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

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.