I am trying to prove the Exercise 2.15 from the book “Markov Chains”, by Daniel Revuz:

A real random variable ZZ is σ(Xm,m≥n)\sigma (X_m, m \geq n)-measurable if and only if Z=Z′∘θnZ = Z’ \circ \theta^n, where Z′∈FZ’ \in \mathscr{F}.

The definition of σ(Xm,m≥n)\sigma(X_m,m\geq n) is the smallest σ\sigma-algebra such that every random variable Xm,m≥nX_m, m\geq n is measurable. But everywhere I look for, the authors always refer to σ(Xm,m≥n)\sigma(X_m,m\geq n) as the “future” (Definition 2.1).

Definitions:

θn\theta^n is the nn-left-shift operator on a sequence:

θn(x0,x1,…,xn,…)=(xn,xn+1,…)\theta^n(x_0,x_1,\ldots,x_n,\ldots)=(x_n,x_{n+1},\ldots)

F\mathscr{F} is a σ\sigma-algebra on the space of sequences.

Proof (my try):

If ZZ is measurable with respect to σ(Xm,m≥n)\sigma(X_m,m\geq n) then ZZ does not depend on the first nn coordinates, so:

Z(g0,g1,…,gn−1,gn,gn+1,…)=Z(e,e,…,e,gn,gn+1,…).Z(g_0,g_1,\ldots,g_{n-1},g_n,g_{n+1},\ldots)=Z(e,e,\ldots,e,g_n,g_{n+1},\ldots).

Define Z′(gk):=Z(e,e,…,g0,g1,…)Z'(g_k):=Z(e,e,\ldots,g_0,g_1,\ldots) where g0g_0 is on the nn-th coordinate, then:

Z′∘θn(gk)=Z′(gn,gn+1,…)=Z(e,e,…,gn,gn+1,…)=Z(gk).

Z’\circ\theta^n(g_k)=Z'(g_n,g_{n+1},\ldots)=Z(e,e,\ldots,g_n,g_{n+1},\ldots)=Z(g_k).

Conversely, suppose that there exists a random variable Z′Z’ such that Z=Z′∘θnZ=Z’\circ\theta^n. Let us show that ZZ does not depend on the first nn coordinates (Is this a sufficient condition? I couldn’t prove that ZZ is measurable with respect to σ(Xm,m≥n)\sigma(X_m,m\geq n) using inverse images).

Let (gk),(hk)∈Ω(g_k),(h_k)\in\Omega such that gk=hkg_k=h_k for all k≥nk\geq n, then

θn(gk)=θn(g0,g1,…,gn−1,gn,gn+1,…)=(gn,gn+1,…)==(hn,hn+1,…)=θn(h0,h1,…,hn−1,hn,hn+1,…)=θn(hk).\begin{eqnarray*}

\theta^n(g_k)=&\theta^n(g_0,g_1,\ldots,g_{n-1},g_n,g_{n+1},\ldots)=(g_n,g_{n+1},\ldots)&=\\

=&(h_n,h_{n+1},\ldots)=\theta^n(h_0,h_1,\ldots,h_{n-1},h_n,h_{n+1},\ldots)&=\theta^n(h_k).

\end{eqnarray*}

So

Z(gk)=Z′∘θn(gk)=Z′∘θn(hk)=Z(hk).Z(g_k)=Z’\circ\theta^n(g_k)=Z’\circ\theta^n(h_k)=Z(h_k).

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