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)

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

If X(t) is wide sense stationary (WSS), we know that E(X(t)) is independent of time R_xx (t1,t2) is also independent of time

Now if we have a function f(z) where z is any real number, inverse of f(z) exists and f(z) is monotonic can we say that X(t) and f(X(t)) are 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)