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For a square-integrable time series, let ${\mathcal{H}}_{t-1}$ be the closed linear span of its past. The innovation ${\epsilon }_t=X_t-{\mathrm{proj}}_{{\mathcal{H}}_{t-1}}{X}_t$ is orthogonal to every past linear observation. A causal invertible autoregressive moving-average model is driven by these innovations, which are weak white noise for a weakly stationary process.

Here the innovation process consists of linear innovations, the linear one-step prediction errors: ${\epsilon }_t=X_t-{\mathrm{proj}}_{{\mathcal{H}}_{t-1}}{X}_t$, where ${\mathcal{H}}_{t-1}$ is the closed linear span of the past $X$'s in $L^2$. For Gaussian processes this is also the conditional expectation prediction error. For general non-Gaussian strong white noise these two notions can differ.

The given ${\eta }_t$ is not the linear innovation process. The moving-average factor has its zero at $1/3$, inside the unit disk, and is noninvertible as a causal moving-average filter. The identityis the moving-average root reflection that places this zero outside the unit disk. Thus the causal invertible representation has innovation variance $9$, rather than the given variance $1$.

For an explicit verification, defineThe filter has constant squared modulus $9$, so $\epsilon$ is weak white noise with variance $9$. The new moving-average factor has root $3$ and is invertible, while its autoregressive factor is causal. Hence the past spans of $X$ and $\epsilon$ agree, and $X_t-{\epsilon }_t$ belongs to that past span. Orthogonality of ${\epsilon }_t$ to past $\epsilon$ therefore identifies it as the linear innovation process. The variance difference proves that it cannot be ${\eta }_t$. If the original noise is Gaussian, the new linear innovations are independent Gaussian variables; without Gaussianity they need only be uncorrelated random variables.

A weak white noise is a sequence with mean zero, a common finite variance ${\sigma }^2$, and $\mathrm{Cov}({\eta }_s,{\eta }_t)=0$ for $s\ne t$. A strong white noise is a sequence of independent and identically distributed random variables with mean zero and finite variance. Strong white noise is therefore weak white noise, but the converse need not hold. Neither definition, by itself, requires a Gaussian distribution.

A sequence of independent and identically distributed random variables with zero mean and finite variance is strong white noise. It is weak white noise, but need not have a normal distribution.