Question

Let \(\Omega=\left\\{\omega_{1}, \omega_{2}, \ldots\right\\}\) be a countable set and \(\mathscr{F}\) the \(\sigma\)-field of all subsets of \(\Omega\). For a fixed \(N\), let \(X_{0}, X_{1}, \ldots, X_{N}\) be random variables defined on \(\Omega\) and let \(T\) be a Markov time with respect to \(\left\\{X_{n}\right\\}\) satisfying \(0 \leq T \leq N\). Let \(\mathscr{F}_{n}\) be the \(\sigma\)-field generated by \(X_{0}, X_{1}, \ldots, X_{n}\) and define \(\mathscr{F}_{T}\) to be the collection of sets \(A\) in \(\mathscr{F}\) for which \(A \cap\\{T=n\\}\) is in \(\mathscr{F}_{n}\) for \(n=0, \ldots, N\). That is, $$ \mathscr{F}_{T}=\left\\{A: A \in F \quad \text { and } A \cap\\{T=n\\} \in F_{n}, \quad n=0, \ldots, N\right\\} $$ Show: (a) \(\mathscr{F}_{T}\) is a \(\sigma\)-field, (b) \(T\) is measurable with respect to \(\mathscr{F}_{T}\), (c) \(\mathscr{F}_{T}\) is the \(\sigma\)-field generated by \(\left\\{X_{0}, \ldots, X_{T}\right\\}\), where \(\left\\{X_{0}, \ldots, X_{T}\right\\}\) is considered to be a variable-dimensional vector-valued function defined on \(\Omega\).

Short answer

To show the desired properties of \(\mathscr{F}_T\), we proceed as follows: (a) Prove \(\mathscr{F}_T\) is a \(\sigma\)-field by showing that it contains \(\Omega\), is closed under complement, and is closed under countable union. (b) Show that \(T\) is measurable with respect to \(\mathscr{F}_T\) by proving that for any \(A \in \mathscr{F}_T\), \(T^{-1}(A) \in \mathscr{F}_T\). (c) Prove that \(\mathscr{F}_{T}\) is the \(\sigma\)-field generated by \(\left\\{X_{0},\ldots, X_{T}\right\\}\) by demonstrating that \(\mathscr{G}\), the \(\sigma\)-field generated by \(\left\\{X_{0},\ldots, X_{T}\right\\}\), satisfies \(\mathscr{G} \subset \mathscr{F}_T\) and \(\mathscr{F}_T \subset \mathscr{G}\).

Step-by-step solution

(a) Proving \(\mathscr{F}_T\) is a \(\sigma\)-field

To prove that \(\mathscr{F}_T\) is a \(\sigma-\)field, we need to show the following properties: 1. \(\Omega \in \mathscr{F}_T\): Since by definition, \(\mathscr{F}_T\) contains subsets from \(\mathscr{F}\), and \(\Omega\) is in \(\mathscr{F}\), thus \(\Omega \in \mathscr{F}_T\). 2. Closed under complement: Let \(A \in \mathscr{F}_T\). Then, for each \(n = 0, \ldots , N\), \(A \cap \\{T = n\\} \in \mathscr{F}_n\). Since \(\mathscr{F}_n\) is a \(\sigma-\)field, it follows that the complement, \(A^c \cap \\{T = n\\}\), is also in \(\mathscr{F}_n\). Therefore, \(A^c \in \mathscr{F}_T\). 3. Closed under countable union: Let \(\{A_i\}_{i=1}^\infty\) be a countable collection of sets in \(\mathscr{F}_T\). For each \(n = 0, \ldots , N\), the sets \(A_i \cap \\{T = n\\}\) belong to \(\mathscr{F}_n\). Since \(\mathscr{F}_n\) is a \(\sigma-\)field, it follows that the countable union, \(\bigcup_{i = 1}^\infty (A_i \cap \\{T = n\\}) \in \mathscr{F}_n\). We have, \(\bigcup_{i = 1}^\infty A_i \cap \\{T = n\\} \in \mathscr{F}_n\) for each \(n\). Thus, \(\bigcup_{i = 1}^\infty A_i \in \mathscr{F}_T\). Since \(\mathscr{F}_T\) satisfies all three properties, it is a \(\sigma-\)field.

(b) Proving T is measurable with respect to \(\mathscr{F}_T\)

To show that \(T\) is measurable with respect to \(\mathscr{F}_T\), we need to prove that for any \(A \in \mathscr{F}_T\), \(T^{-1}(A) \in \mathscr{F}_T\). By definition, \(A \in \mathscr{F}_T\) if \(A \cap\\{T=n\\} \in \mathscr{F}_n\) for every \(n = 0, \ldots , N\). Let \(A \in \mathscr{F}_T\) and \(n = 0, \ldots , N\), then \(T^{-1}(A) \cap \\{T = n\\} = A \cap \\{T = n\\} \in \mathscr{F}_n\). Therefore \(T\) is measurable with respect to \(\mathscr{F}_T\).

(c) Proving \(\mathscr{F}_{T}\) is the \(\sigma\)-field generated by \(\left\\{X_{0},\ldots, X_{T}\right\\}\)

Let \(\mathscr{G}\) be the \(\sigma\)-field generated by \(\left\\{X_{0},\ldots, X_{T}\right\\}\). We need to prove two things: 1. \(\mathscr{G} \subset \mathscr{F}_T\) For any \(A \in \mathscr{G}\), we have \(A \cap \\{T = n\\}\) is measurable with respect to \(\left\\{X_{0},\ldots, X_{n}\right\\}\). So, \(A \cap \\{T = n\\} \in \mathscr{F}_n\), and thus, \(A \in \mathscr{F}_T\). This implies \(\mathscr{G} \subset \mathscr{F}_T\). 2. \(\mathscr{F}_{T}\subset \mathscr{G}\) For any \(A \in \mathscr{F}_T\), we have \(A \cap \\{T = n\\} \in \mathscr{F}_n\). Since \(\mathscr{F}_n\) is generated by \(\left\\{X_{0},\ldots, X_{n}\right\\}\), we can write \(A \cap \\{T = n\\} = B_n \cap \\{T = n\\}\) for some \(B_n \in \mathscr{G}\). Therefore, \(A = \bigcup_{n = 0}^N (B_n \cap \\{T = n\\}) \in \mathscr{G}\). This implies \(\mathscr{F}_T \subset \mathscr{G}\). From the above arguments, we can conclude that \(\mathscr{F}_T\) is the \(\sigma\)-field generated by \(\left\\{X_{0},\ldots, X_{T}\right\\}\).