## Formalized Name
Uniqueness of the Wold Components
## Formalized Statement
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $(X_t)_{t\in\mathbb{Z}}$ be a real-valued second-order process with $X_t\in L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$. For each $t\in\mathbb{Z}$, let $\mathcal{R}_t$ be a closed linear subspace of $L^2(\Omega,\mathcal{F},\mathbb{P})$, interpreted as the remote linear past at time $t$. Suppose that a Wold-type decomposition is defined by
\begin{align*}
D_t=P_{\mathcal{R}_t}X_t,\qquad Y_t=X_t-D_t,
\end{align*}
where $P_{\mathcal{R}_t}$ is the orthogonal projection onto $\mathcal{R}_t$. If another decomposition
\begin{align*}
X_t=D'_t+Y'_t
\end{align*}
satisfies $D'_t\in\mathcal{R}_t$ and $Y'_t\perp\mathcal{R}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$, then
\begin{align*}
D'_t=D_t,\qquad Y'_t=Y_t
\end{align*}
in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$. Hence the deterministic component and the purely nondeterministic component are uniquely determined in $L^2$ by $(X_t)$ and the remote-past subspaces $(\mathcal{R}_t)$.
## Proof
[proofplan]
Fix a time $t$. Orthogonal projection onto a closed subspace in a Hilbert space is characterized uniquely by two conditions: the projected point lies in the subspace, and the residual is orthogonal to the subspace. The pair $(D_t,Y_t)$ satisfies these conditions by definition. Any other pair $(D'_t,Y'_t)$ satisfying the same conditions gives another orthogonal projection of $X_t$ onto $\mathcal{R}_t$, so uniqueness of projection gives $D'_t=D_t$ in $L^2$. Subtracting from $X_t$ then gives $Y'_t=Y_t$ in $L^2$.
[/proofplan]
[step:Use the projection characterization at one fixed time]
Fix $t\in\mathbb{Z}$. The space $L^2(\Omega,\mathcal{F},\mathbb{P})$ is a Hilbert space with inner product
\begin{align*}
\langle U,V\rangle=\mathbb{E}[UV].
\end{align*}
Because $\mathcal{R}_t$ is a closed linear subspace, the orthogonal projection $P_{\mathcal{R}_t}X_t$ is the unique element $D\in\mathcal{R}_t$ such that
\begin{align*}
X_t-D\perp\mathcal{R}_t.
\end{align*}
By definition,
\begin{align*}
D_t=P_{\mathcal{R}_t}X_t,\qquad Y_t=X_t-D_t,
\end{align*}
so $D_t\in\mathcal{R}_t$ and $Y_t\perp\mathcal{R}_t$.
[guided]
At the fixed time $t$, the problem is a Hilbert-space projection problem. The closed subspace is $\mathcal{R}_t$, and the vector being decomposed is $X_t\in L^2(\Omega,\mathcal{F},\mathbb{P})$. The projection theorem says there is exactly one vector $D\in\mathcal{R}_t$ with residual $X_t-D$ orthogonal to $\mathcal{R}_t$. Since
\begin{align*}
D_t=P_{\mathcal{R}_t}X_t
\end{align*}
and
\begin{align*}
Y_t=X_t-D_t,
\end{align*}
the pair $(D_t,Y_t)$ satisfies
\begin{align*}
D_t\in\mathcal{R}_t,\qquad Y_t\perp\mathcal{R}_t.
\end{align*}
[/guided]
[/step]
[step:Compare with any other admissible decomposition]
Now suppose
\begin{align*}
X_t=D'_t+Y'_t,
\end{align*}
where $D'_t\in\mathcal{R}_t$ and $Y'_t\perp\mathcal{R}_t$. Since $Y'_t=X_t-D'_t$, the element $D'_t\in\mathcal{R}_t$ also has residual orthogonal to $\mathcal{R}_t$:
\begin{align*}
X_t-D'_t=Y'_t\perp\mathcal{R}_t.
\end{align*}
Therefore $D'_t$ satisfies the same projection characterization as $D_t=P_{\mathcal{R}_t}X_t$. By uniqueness of orthogonal projection,
\begin{align*}
D'_t=D_t
\end{align*}
in $L^2(\Omega,\mathcal{F},\mathbb{P})$.
[/step]
[step:Recover the purely nondeterministic component]
Since both decompositions sum to $X_t$,
\begin{align*}
Y'_t=X_t-D'_t.
\end{align*}
Using $D'_t=D_t$ in $L^2$, we get
\begin{align*}
Y'_t=X_t-D_t=Y_t
\end{align*}
in $L^2(\Omega,\mathcal{F},\mathbb{P})$. The argument holds for every $t\in\mathbb{Z}$, so the two component processes are uniquely determined in $L^2$ at each time index.
[/step]
