# Difficulty in understanding the proof of the Wold decomposition theorem

The proof of the Wold Decomposition [1] of $$x_t$$ involves the definition of the process

$$w_t = x_t - P_{\mathcal{M}_{t-1}^x} x_t,$$

where $$x_t$$ is a stationary zero-mean process, $$\mathcal{M}_n^x = \overline{{\mathrm{span}}}\{x_t, -\infty \lt t \le n\}$$, and $$P$$ denotes the projection.

If $$x_t$$ is a real-valued process (as opposed to a vector-valued one) and $$x_i \ne 0$$ for some $$-\infty \lt i \le (t-1)$$, isn't $$\mathcal{M}_{t-1}^x = \mathbb{R}$$ and, consequently, $$w_t=0$$?

I guess it doesn't, since the rest of the proof would not make sense if it did. Could you please point out where my interpretation is wrong?

[1] Shumway, RH, Stoffer DS. Time Series Analysis and Its Applications with R Examples. 4th ed.

You project on an infinite dimensional space spanned by $$x_{t-1}, x_{t-2},\ldots$$ The fact that a single $$x_{t-k}$$ for some $$k$$ is zero does not make such infinite dimensional space become the real line, it continues to be an infinite dimensional space.
• Your answer allows me to get to the heart of the question. Why is the space spanned by $x_{t−1},x_{t−2},...$ infinite dimensional? Since $x_{i}$ are real numbers, aren't all $x_{i}$ colinear, in the same way that two vectors (a,0) and (b,0) would be in $\mathbb{R}^2$? If not, could you please illustrate by showing what the projection of 5 is on the space spanned by 1 and 2? – toliveira Jun 25 at 16:00
• No, $x_i$ are not real numbers, are random variables and if no linear combination of the past can predict perfectly the future, as is the case in regular processes, the "past" is infinite dimensional. You might want to read chapter 7 of Anderson, T.W.(1970) The Statistical Analysis of Time Series. It requires a bit lengthy explanation that I cannot attempt in a comment. – F. Tusell Jun 26 at 7:34
Reading the proof in more detail, it looks like the condition $$\sigma_x^2 >0$$ is required for the Wold decomposition theorem to apply. Above the theorem in the book this quantity is defined such that the situation you're describing falls under the case $$\sigma_x^2 = 0.$$
• I don't see how the situation I am describing falls under the case $\sigma^2=0$. Maybe the grammar in the original question was ambiguous. I have improved it after your answer. – toliveira Jun 25 at 15:54