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 ?


2 Answers 2


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".


Yes, conditional on ARMA(p,q) being the true model, what you said is correct.


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