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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

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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.

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  • $\begingroup$ Thank you very much for your answer. It certainly clarified things. Yes I should have been more specific. Particularly, I am looking for continuous case. Seems like in general they are not jointly WSS. However, if f(z) is bounded (i.e, f(z)=tanh(z)), will they be jointly WSS? I have updated the question description for more information. Meanwhile, I am not very proficient in statistics so let me think about your answer. $\endgroup$ May 29, 2020 at 14:26
  • $\begingroup$ Actually, I found that if X(t) is gaussian process, then by using Hermite polynomials, I can show that f(X(t)) is 2nd order stationary. Let me update my question $\endgroup$ May 29, 2020 at 16:36
  • $\begingroup$ I can certainly believe it's true for Gaussian processes if you make the additional assumption that f(X(t)) has two finite moments. It's not true without that assumption. $\endgroup$ May 30, 2020 at 6:31
  • $\begingroup$ You are right. That is correct. Thank you very much. $\endgroup$ May 30, 2020 at 7:37

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