# Wold Decomposition Unique?

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?

• 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? – whuber Apr 25 at 13:22
• @whuber I hope my edits help to clarify the question – Joogs Apr 25 at 13:27