I have trouble understanding the proof of the Wold decomposition theorem

Question

I'm trying to understand the proof of the Wold decomposition theorem in [1, p.187]. I find a few things about it very irritating. The theorem states:

Theorem 5.7.1 (The Wold Decomposition). Let $X_t$ be a stationary process with $$\sigma^2 = E|X_{n+1}-P_{M_n}X_{n+1}|^2 >0,$$ where $$M_n=\overline{span} \lbrace X_t: -\infty<t\leq n \rbrace.$$

Then $X_t$ can be decomposed as $$X_t=\sum_{j=0}^\infty \psi_j Z_{t-j} + V_t$$ where

(1) $\psi_0=1, \sum_{j=0}^\infty \psi^2< \infty$

(2) $E[Z_t]=0$, $var(Z_t)=\sigma^2$, $t \in \mathbb{Z}$ and $Z_t$ is uncorrelated

(3) $Z_t \in M_t$, $t \in \mathbb{Z}$

(4) $E(Z_tV_s)=0$, $t,s \in \mathbb{Z}$

(5) $V_t$ is deterministic

Here $P$ denotes the projection.

The proof starts by setting $$Z_t:=X_t-P_{M_{t-1}}X_t,$$ $$\psi_t:=\langle X_t, Z_{t-j}\rangle/\sigma^2,$$ $$V_t:=X_t-\sum_{j=0}^\infty \psi_j Z_{t-j}.$$ We have to show that these sequences satisfy (1)-(5).

Since $Z_t \in M_t$ and $Z_t \in M_{t-1}^\bot$ by definition, we have that $$Z_t \in M_{t-1}^\bot \subset M_{t-2}^\bot \subset...$$ which shows that $E(Z_sZ_t)=0$ for $s<t$. This establishes the last part of (2). Furthermore, an exercise in the book establishes that $var(Z_t)=\sigma^2$. What I am having trouble with is another part:

My question: Why is $E[Z_t]=0$? And why does that not yield that $V=0$?

My idea was to use the zero-mean-property of $X_t$. We know that $Z_t \in M_t$, i.e., in the smallest closed subspace that contains all $X_t$, $t<n$. Therefore, $Z_t$ is the limit of a subsequence $\lbrace X_{t_j} \rbrace$ of zero-mean and so $$E[Z_t]=E[\lim_{j \to \infty} X_{t_j}] = \int\lim_{j \to \infty} X_{t_j}dP = \lim_{j \to \infty} E[X_{t_j}]=0.$$ But why can I interchange the limit and the integral in the third inequality?

And even worse: If $X_t$ as well as $Z_t$ indeed have zero-mean, why does that not result in $E[V_t]=0$ and therefore -by determinancy- $V_t=0$?

Any help is much appreciated!

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[1] Brockwell, Peter J.: Time series: theory and methods. Second Edition. 2006, Springer.

Answer 1

For any closed subspace $F \subseteq L^2(\mathbb{P})$ the projection of a (square-integrable) random variable $X$ equals

$$P_F(X) = \mathbb{E}(X \mid F)$$

where $\mathbb{E}(X \mid F)$ denotes the conditional expectation with respect to $F$. In particular, one can show that

$$\mathbb{E}(P_F(X)) = \mathbb{E}\bigg[ \mathbb{E}(X \mid F) \bigg] = \mathbb{E}X;$$

this is a special case of the so-called tower property. Applying this to $X=X_t$ and $F=M_{t-1}$ yields $\mathbb{E}Z_t=0$.

Concerning your second question: This is a matter of interchanging limit and summation. Note that $\mathbb{E}(Z_t)=0$ does not necessarily imply

$$\mathbb{E} \left( \sum_{j=0}^{\infty} \psi_j \cdot Z_{t-j} \right)=0.$$

If $\sum_{j \geq 0} |\psi_j|<\infty$, then this equality holds by the dominated convergence theorem.