The wold decomposition theorem says that any weakly stationary process $X_t$ can be written as $$X_t=\sum_{i=1}^\infty \psi_j\epsilon_{t-j}+W_t$$ where $\epsilon_t$ is a white noise process and $W_t$ is a stationary deterministic process uncorrelated with $\epsilon_t.$

Now if I have a weakly stationary process $X_t$ with a certain mean and autocovariance function and I have another stationary process $Y_t$ in the form of the wold decomposition theorem$$Y_t=\sum_{i=1}^\infty \psi_j^{*}\epsilon_{t-j}^{*}+W_t^{*}$$

,where the process $\epsilon_{t-j}^{*}$ is white noise uncorrelated with the stationary process $W_t^{*}$, and we know that $Y_t$ has the same mean and autocovariance function as $X_t$, does this imply that we have found the wold decomposition for $X_t$ as $X_t=\sum_{i=1}^\infty \psi_j^{*}\epsilon_{t-j}^{*}+W_t^{*}$?

Edit I have a hard time pinning down the question exactly. Maybe these related questions will help:

Is the Wold decomposition unique? Can we identify a weakly stationary process by its mean and autocovariance function? If not, what if its a linear weakly stationary process?

  • $\begingroup$ Because the right hand sides of your two equations are literally identical, then $X_t$ and $Y_t$ must be the same thing (at least if mathematical notation is intended to make any sense). I suspect you mean something else, such as that the $\epsilon_{t-j}$ and $W_t$ might not actually refer to the same things in both equations, but really refer to distributions rather than random variables. Could you clarify what you mean? $\endgroup$
    – whuber
    Apr 25, 2019 at 13:22
  • $\begingroup$ @whuber I hope my edits help to clarify the question $\endgroup$
    – Joogs
    Apr 25, 2019 at 13:27
  • $\begingroup$ Just as an answer to your last questions. The mean and autocovariance functions do not describe a process fully. In fact, an MA(1) process with coefficient $\theta$ and one with coefficient $\frac{1}{\theta}$ will have the same ACF. However, these are two different processes and their Wold representation is different. $\endgroup$
    – Lorenzo
    Jan 25, 2021 at 17:10


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