Kolmogorov Extension Theorem
- $\mathfrak{S}_{n}$,
- symmetric group with order $n$
With this notation, it is easy to confirm that
\[\begin{eqnarray} \tau \in \mathfrak{S}_{n}, \ A_{i} \in \mathcal{B}(\mathbb{R}), \ \tau(A_{1} \times \cdots \times A_{n}) = A_{\tau(1)} \times \cdots \times A_{\tau(n)} . \nonumber \end{eqnarray}\] \[\begin{eqnarray} \Omega & := & \mathbb{R}^{[0, \infty)} \nonumber \\ \tilde{t} \in \mathcal{T}, \ \omega(\tilde{t}) & := & (\omega(t_{1}), \ldots, \omega(t_{n})) \in \mathbb{R}^{n} \nonumber \end{eqnarray}\] \[\begin{eqnarray} \tilde{t} \in \mathcal{T}^{n}, \ A \in \mathcal{B}(\mathbb{R}^{n}), \ C(A, \tilde{t}) & := & \{ \omega \in \Omega \mid \omega(\tilde{t}) \in A \}, \nonumber \\ \mathcal{C} & := & \{ C(A, \tilde{t}) \mid n \in \mathbb{N}, \ A \in \mathcal{B}(\mathbb{R}^{n}), \ \tilde{t} \in \mathcal{T}^{n} \}, \nonumber \\ \mathcal{B}(\mathbb{R}^{[0, \infty)}) & := & \sigma(\mathcal{C}) \nonumber \end{eqnarray}\]Proposition 1
(1) $\mathcal{C}$ is $\pi$-system.
(2) $\mathcal{C}$ is an algebra on $\Omega$.
(3) If \(C(A_{1}, \tilde{t}_{1}) = C(A_{2}, \tilde{t}_{2})\),
\[\exists n \in \mathbb{N}, \ \exists A \in \mathcal{B}(\mathbb{R}^{n}), \ \exists \tilde{t} \in \mathcal{B}(\mathbb{R}^{n}), \ \text{ s.t. } C(A, \tilde{t}) = C(A_{i}, \tilde{t}_{i}) \quad (i = 1, 2) .\](4) Suppose \(C(A, \tilde{t}) \in \mathcal{C}\). Then
\[\forall \tilde{s} \in \mathcal{T}, \ n := \abs{\tilde{s}}, \ \tilde{u} := (\tilde{t}, \tilde{s}), \ \tilde{u}^{\prime} := (\tilde{s}, \tilde{t}), \ C(A \times \mathbb{R}^{n}, \tilde{u}) = C(A, \tilde{t}) = C(\mathbb{R}^{n} \times A, \tilde{u}^{\prime}) .\](5)
Let \(C(A_{i}, \tilde{t}_{i}) \in \mathcal{C}\). Then
\[C(A_{1} \times A_{2}, \tilde{t}) = C(A_{1} \times \mathbb{R}^{n_{2}}, \tilde{t}) \cap C(\mathbb{R}^{n_{1}} \times A_{2}, \tilde{t}) .\]proof
(1)
(2)
(3)
Let
\[\begin{eqnarray} i = 1, 2, \ n_{i} & := & \abs{\tilde{t}_{i}} \nonumber \\ n & := & n_{1} + n_{2} \nonumber \\ A & := & A_{1} \times A_{2} \nonumber \\ \tilde{t} & := & (\tilde{t}_{1}, \tilde{t}_{2}) . \nonumber \end{eqnarray}\]Let \(\omega \in C(A_{1}, \tilde{t}_{1})\) be fixed. Since $\omega \in C(A_{2}, \tilde{t}_{2})$,
\[\begin{eqnarray} \omega(\tilde{t}_{1}) & \in & A_{1}, \nonumber \\ \omega(\tilde{t}_{2}) & \in & A_{2} . \nonumber \end{eqnarray}\]Hence
\[\omega(\tilde{t}) = (\omega(\tilde{t}_{1}), \omega(\tilde{t}_{1})) \in A_{1} \times A_{2} .\]Therefore, \(C(A_{1}, \tilde{t}_{1}) \subseteq C(A, \tilde{t})\) The opposite inclusion is obvious from the above discussion.
(4)
Let $\tilde{s} \in \mathcal{T}$ be fixed. We will show that the equality holds. Let $\omega \in C(A, \tilde{t})$ be fixed.
\[\omega(\tilde{t}) \in A \Rightarrow \omega(\tilde{u}) = (\omega(\tilde{t}), \omega(\tilde{s})) \in A \times \mathbb{R}^{n}, \ \omega(\tilde{u}^{\prime}) = (\omega(\tilde{s}), \omega(\tilde{t})) \in \mathbb{R}^{n} \times A .\]Hence $C(A, \tilde{t}) \subseteq C(A\times \mathbb{R}^{n}, \tilde{u})$ and $C(A, \tilde{t}) \subseteq C(\mathbb{R}^{n} \times A, \tilde{u}^{\prime})$. Similarly, we can prove the opposite inclusion from the above discussion.
(5)
Note that (3) can be shown from (4) and (5).
Definition consistensy
- \(\{Q_{\tilde{t}}\}_{\tilde{t} \in \mathcal{T}}\),
- $Q_{\tilde{t}}$ is a probability measure on \((\mathbb{R}^{\abs{\tilde{t}}}, \mathcal{B}(\mathbb{R}^{\abs{\tilde{t}}})\),
\(\{Q_{\tilde{t}}\}\) is said to be consistent if
(a)
\[n := \abs{\tilde{t}}, \ \forall A_{i} \in \mathcal{B}(\mathbb{R}), \ Q_{\tilde{t}}(A_{1}\times \cdots \times A_{n}) = Q_{\tilde{t}}(\tau(A_{1}\times \cdots \times A_{n})) .\](b) If $\tilde{t} := (t_{1}, \ldots, t_{n}) \in \mathcal{T}^{n}$ and $\tilde{s} := (t_{1}, \ldots, t_{n-1}) \in \mathcal{T}^{n-1}$,
\[\forall A \in \mathcal{B}(\mathbb{R}^{n-1}), \ Q_{\tilde{t}}(A \times \mathbb{R}) = Q_{\tilde{s}}(A) .\]Proposition 2
- \(\{Q_{\tilde{t}}\}_{\tilde{t} \in \mathcal{T}}\),
- consistent finite-dimentional distributions
(1)
\[\forall n \in \mathbb{N}, \ \forall \tilde{t} \in \mathcal{T}^{n}, \ \forall A \in \mathcal{B}(\mathbb{R}^{n}), \ \forall \tau \in \mathfrak{S}_{n}, \ Q_{\tilde{t}}(A) = Q_{\tau(\tilde{t})}(\tau(A)) .\](2) Suppose that
\[\begin{eqnarray} \tilde{t} & := & (t_{1} ,\ldots, t_{n}) \in \mathcal{T}^{n} \nonumber \\ n_{1}, n_{2} \in \mathbb{N}, \ n_{1} + n_{2} + 1 & = & n \nonumber \\ \tilde{s} & := & (t_{1}, \ldots, t_{n_{1}}, t_{n_{2}}, \ldots, t_{n_{1} + n_{2}}). \nonumber \end{eqnarray}\]Then
\[\forall A \in \mathcal{B}(\mathbb{R}^{n_{1}}), \ \forall B \in \mathcal{B}(\mathbb{R}^{n_{2}}), \ Q_{\tilde{t}}(A \times \mathbb{R} \times B) = Q_{\tilde{s}}(A \times B) .\]proof
Theorem Kolmogorov Extension Theorem
- \(\{Q_{\tilde{t}}\}_{\tilde{t} \in \mathcal{T}}\),
- consistensy finite dimensional distributions
Then there exists a probability measure $P$ on \((\mathbb{R}^{[0, \infty)}, \mathcal{B}(\mathbb{R}^{[0, \infty)}))\) such that
\[\forall \tilde{t} \in \mathcal{T}, \ Q_{\tilde{t}}(A) = P \left( \omega \in \mathbb{R}^{[0, \infty)} \mid \omega(\tilde{t}) \in A \right) .\]proof
\[Q(C(A, \tilde{t})) := Q_{\tilde{t}}(A) .\]Claim: $Q$ is well-defined.
Suppose that \(C(A_{1}, \tilde{t}_{1}) = C(A_{2}, \tilde{t}_{2})\). Let \(n_{i} := \abs{\tilde{t}_{i}}\) and \(\tilde{t} := (\tilde{t}_{1}, \tilde{t}_{2})\). By proposition,
\[\begin{eqnarray} C(A_{1}, \tilde{t}_{1}) & = & C(A_{1} \times \mathbb{R}^{n}, \tilde{t}) \nonumber \\ C(A_{2}, \tilde{t}_{2}) & = & C(\mathbb{R}^{n} \times A_{2}, \tilde{t}) . \nonumber \end{eqnarray}\]Hence,
\[\begin{eqnarray} Q(C(A_{1} \times \mathbb{R}^{n}, \tilde{t})) & = & Q_{\tilde{t}}(A_{1} \times \mathbb{R}^{n}) \nonumber \\ & = & Q_{\tilde{t}_{1}}(A_{1}) \nonumber \\ & = & Q(C(A_{1}, \tilde{t}_{1})) \nonumber \\ Q(C(\mathbb{R}^{n} \times A_{2}, \tilde{t})) & = & Q_{\tilde{t}}(\mathbb{R}^{n} \times A_{2}) \nonumber \\ & = & Q_{\tilde{t}_{2}}(A_{2}) \nonumber \\ & = & Q(C(A_{2}, \tilde{t}_{2})) . \nonumber \end{eqnarray}\]On the other hand,
\[\begin{eqnarray} Q_{\tilde{t}}(A_{1} \times \mathbb{R}^{n_{2}}) - Q_{\tilde{t}}(\mathbb{R}^{n_{1}} \times A_{2}) & = & Q_{\tilde{t}}(A_{1} \times \mathbb{R}^{n_{2}} \setminus \mathbb{R}^{n_{1}} \times A_{2}) \nonumber \\ & = & Q_{\tilde{t}} ( A_{1} \times \mathbb{R}^{n_{2}} \cap \left( \mathbb{R}^{n_{1}} \times A_{2} \right)^{c} ) \nonumber \\ & = & Q_{\tilde{t}}(\emptyset) \nonumber \\ & = & 0 . \nonumber \end{eqnarray}\]Combining the above equations,
\[\begin{eqnarray} Q(C(A_{1}, \tilde{t}_{1})) & = & Q(C(A_{1} \times \mathbb{R}^{n}, \tilde{t})) \nonumber \\ & = & Q(C(\mathbb{R}^{n} \times A_{2}, \tilde{t})) \nonumber \\ & = & Q(C(A_{2}, \tilde{t}_{2})) . \nonumber \end{eqnarray}\]The claim proved.
Claim: $Q(\Omega) = 1$.
\[Q(C(\mathbb{R}, 0)) = Q_{0}(\mathbb{R}) = 1 .\]The claim proved.
Claim: $Q$ is finitely additive on $\mathcal{C}$.
Let \(C(A_{1}, \tilde{t}_{1}), C(A_{2}, \tilde{t}_{2}) \in \mathcal{C}\). Suppose that \(C(A_{1}, \tilde{t}_{1}) \cap C(A_{2}, \tilde{t}_{2}) = \emptyset\). By proposition, letting \(\tilde{t} := (\tilde{t}_{1}, \tilde{t}_{2})\).
\[\begin{eqnarray} C(A_{1} \times A_{2}, \tilde{t}) = C(A_{1} \times \mathbb{R}^{n_{2}}, \tilde{t}) \cap C(\mathbb{R}^{n_{1}} \times A_{2}, \tilde{t}) . \end{eqnarray}\]Thus,
\[\begin{eqnarray} Q_{\tilde{t}}(A_{1} \times A_{2}) & = & Q_{\tilde{t}}(A_{1} \times \mathbb{R}^{n_{2}} \cap \mathbb{R}^{n_{2}} \times A_{1}) \nonumber \\ & = & Q_{\tilde{t}}(A_{1} \times \mathbb{R}^{n_{2}}) + Q_{\tilde{t}}(\mathbb{R}^{n_{2}} \times A_{1}) \nonumber \\ & = & Q_{\tilde{t}_{1}}(A_{1}) + Q_{\tilde{t}_{2}}(A_{2}) \nonumber \end{eqnarray}\]This implies additivity of $Q$. The claim proved.