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Precompact Stochastic Process 📂Probability Theory

Precompact Stochastic Process

Theorem

Let’s define a function space consisting of continuous functions going from measurable space (S,S)(S, \mathcal{S}) to (S′,S′)(S ', \mathcal{S} ') as H:=C(S,S’)\mathscr{H}:= C \left( S,S’ \right), and say that {h−1(A’):h∈H,A′∈S′}\left\{ h^{-1}(A’): h \in \mathscr{H} , A ' \in \mathcal{S} ' \right\} is a separating class of (S,S)(S , \mathcal{S}). Here, XX is a probability element defined in SS, and {Xn}n∈N\left\{ X_n \right\}_{n \in \mathbb{N}} is a stochastic process defined in SS.

If

  • (i) {Xn}\left\{ X_{n} \right\} is pre-compact.
  • (ii) For all h∈Hh \in \mathscr{H}, h(Xn)→Dh(X)h \left( X_{n} \right) \overset{D}{\to} h(X)

then, Xn→DXX_{n} \overset{D}{\to} X holds.

Explanation

The continuous mapping theorem required the condition P(X∈Ch)=1P(X \in C_{h})=1 to demonstrate that Xn→DX  ⟹  h(Xn)→Dh(X)X_{n} \overset{D}{\to} X \implies h(X_{n}) \overset{D}{\to} h(X) is true, likewise, to prove its converse, the condition of being pre-compact is necessary.

Proof

Under assumption (i), that the stochastic process {Xn}\left\{ X_{n} \right\} is pre-compact means that for every subsequence {Xn′}⊂{Xn}\left\{ X_{n '} \right\} \subset \left\{ X_{n} \right\}, there exists a further subsequence {Xn′′}⊂{Xn′}⊂{Xn}\left\{ X_{n ''} \right\} \subset\left\{ X_{n '} \right\} \subset \left\{ X_{n} \right\} that converges to some Y∈SY \in S. In other words, Y∈SY \in S and {Xn′′}\left\{ X_{n ''} \right\} satisfying Xn′′→DYX_{n ''} \overset{D}{\to} Y exist, and then for all h∈Hh \in \mathscr{H}, h(Xn′′)→Dh(Y)h \left( X_{n ''} \right) \overset{D}{\to} h \left( Y \right) and as per assumption (ii), h(Xn)→Dh(X)h \left( X_{n} \right) \overset{D}{\to} h(X), so h(Xn′′)→Dh(X)h \left( X_{n ''} \right) \overset{D}{\to} h \left( X \right) Meanwhile, since {h−1(A’):h∈H,A′∈S′}\left\{ h^{-1}(A’): h \in \mathscr{H} , A ' \in \mathcal{S} ' \right\} is placed as a separating class, for all A∈S′A \in \mathcal{S} ' and h∈Hh \in \mathscr{H}, h(X)=Dh(Y)  ⟺  P(X∈h−1(A))=P(Y∈h−1(A))  ⟺  P∘X−1=P∘Y−1, on {h−1(A’):h∈H,A′∈S′}  ⟺  P∘X−1=P∘Y−1, on (S,S)  ⟺  X=DY \begin{align*} & h(X) \overset{D}{=} h(Y) \\ \iff & P \left( X \in h^{-1}(A) \right) = P \left( Y \in h^{-1}(A) \right) \\ \iff & P \circ X^{-1} = P \circ Y^{-1} \qquad \text{, on }\left\{ h^{-1}(A’): h \in \mathscr{H} , A ' \in \mathcal{S} ' \right\} \\ \color{red}{\iff}& P \circ X^{-1} = P \circ Y^{-1} \qquad \text{, on } (S,\mathcal{S}) \\ \iff &X \overset{D}{=} Y \end{align*}

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