Difference between the Wold Decomposition and MA representation The Wold Theorem states that any (weakly) stationary process $(x_t)_{t=-\infty}^{+\infty}$ with zero mean can be decomposed into 
$$
   x_t=\sum_{j=0}^{\infty} b_j\epsilon_{t-j}\ + \eta_t ,
$$
where the first summand is the stochastic and the second term a determinstic part.
Let $x_t$ be any (weakly stationary) ARMA(p,q)-process and consider I would like to find the Wold representation of that process. Is it true that the Wold decomposition always coincides with the Moving Avergage MA($\infty$)-representation of that ARMA(p,q)-model, such that $\eta_t=0$ and  $b_j$ being the MA weights ?
 A: The Wold decomposition does not say what you state. It says that any weakly stationary $(x_t)_{t=-\infty}^{\infty}$, there exists a white noise process $\{\epsilon_t\}_{t=-\infty}^{+\infty}$ such that $(x_t)_{t=-\infty}^{\infty}$ has  two-sided MA representation
$$
x_t=\sum_{-\infty < j < \infty} b_j\epsilon_{t-j}.
$$
What you're asking is whether the two-sided MA representation is unique. 
No. Existence does not imply uniqueness, i.e. there is no "the Wold decomposition." Given a  two-sided MA representation
$$
x_t=\sum_{-\infty < j < \infty} b_j\epsilon_{t-j},
$$
it is easy to find another MA representation---a different white noise process $\{\epsilon'_t\}_{t=-\infty}^{+\infty}$ and a different sequence $\{b'_t\}_{t=-\infty}^{+\infty}$ such that 
$$
x_t=\sum_{-\infty < j < \infty} b'_j\epsilon'_{t-j}.
$$
So it does not make sense to speak of "the Wold decomposition." 
Given a stationary ARMA $(x_t)_{t=-\infty}^{\infty}$, you can write down an MA(∞) representation 
$$
x_t=\sum_{-\infty < j < \infty} \psi_j\epsilon_{t-j}.
$$
But it does not make sense to ask whether "...the Wold decomposition...coincides with the MA(∞)-representation". 
A: Yes, conditional on ARMA(p,q) being the true model, what you said is correct.
