Proof of Brownian Increments After a Time are Independent of the Model Past

Throughout, $\lVert\cdot\rVert_{2}$ is the mean-square norm of Square-Integrable Random Variables and the Mean-Square Inner Product, and the covariance is bilinear in each argument (by linearity of the integral, Linearity and Monotonicity of the Lebesgue Integral, applied to its defining formula) and satisfies, for square-integrable $X$ and $Y_n\to Y$ in mean square, $\operatorname{Cov}(X,Y_n)\to\operatorname{Cov}(X,Y)$: indeed $|\operatorname{Cov}(X,Y_n-Y)|=|\mathbb{E}[X(Y_n-Y)]-\mathbb{E}[X]\mathbb{E}[Y_n-Y]|\le2\lVert X\rVert_{2}\lVert Y_n-Y\rVert_{2}$ by Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm (the constant $\mathbb{E}[X]$ having $\lVert\mathbb{E}[X]\rVert_2\le\lVert X\rVert_2$).

Claim 1. Step 1: vanishing cross-covariances. Fix an increment $D=W^{j}_{v}-W^{j}_{u}$ with $s\le u<v\le T$. We check that $D$ is uncorrelated with every member of the generating family of $\mathcal{H}_s$.

(a) For $\xi^{i}$: $\sigma(\xi^{1},\dots,\xi^{l})$ is independent of $\sigma(W^{j'}_r:1\le j'\le m,\ r\ge0)$ by the model, so $\xi^{i}$ is independent of $W^{j}_{v}$ and of $W^{j}_{u}$; by Expectation of a Product of Independent Random Variables and bilinearity, $\operatorname{Cov}(D,\xi^{i})=\operatorname{Cov}(W^{j}_v,\xi^{i})-\operatorname{Cov}(W^{j}_u,\xi^{i})=0$.

(b) For $W^{j'}_r$ with $0\le r\le s$: by the covariances of claim 3 of Wiener Integrals Against a Vector Brownian Motion are Jointly Gaussian, if $j'=j$ then $\operatorname{Cov}(D,W^{j}_r)=\min(v,r)-\min(u,r)=r-r=0$ since $r\le s\le u<v$, and if $j'\ne j$ both covariances vanish.

(c) For $u^{j''}_q$ with $0\le q\le s$: by claim 1 of Gaussian and Span Structure of the Linear-Gaussian State-Observation Model, $u^{j''}_q$ is a mean-square limit of finite linear combinations of the $\xi^{i}$ and the $W^{j'}_r$ with $r\le q\le s$; by (a), (b), bilinearity, and the mean-square continuity of the covariance stated above, $\operatorname{Cov}(D,u^{j''}_q)=0$.

Step 2: independence from finite generator blocks. Fix increments $D_1,\dots,D_n$ as in the claim and finitely many random variables $g_1,\dots,g_d$, each of which is one of the $\xi^{i}$, one of the $W^{j'}_r$ ($r\le s$), or one of the $u^{j''}_q$ ($q\le s$). By claim 2 of Gaussian and Span Structure of the Linear-Gaussian State-Observation Model, the combined family of all $\xi^{i}$, $W^{j'}_r$, $X^{i}_r$, and $u^{j''}_r$ is jointly Gaussian; hence the finite tuple $(W^{j_1}_{v_1},W^{j_1}_{u_1},\dots,W^{j_n}_{v_n},W^{j_n}_{u_n},g_1,\dots,g_d)$ is a Gaussian random vector, and so is its linear image $(D_1,\dots,D_n,g_1,\dots,g_d)$ by Affine Transformations of Gaussian Random Vectors are Gaussian. Every $\operatorname{Cov}(D_p,g_{d'})$ vanishes by Step 1, so by Uncorrelated Jointly Gaussian Blocks are Independent the $\sigma$-algebras $\sigma(D_1,\dots,D_n)$ and $\sigma(g_1,\dots,g_d)$ are independent.

Step 3: independence from $\mathcal{H}_s$. Let $\mathcal{P}$ be the collection of all finite intersections of events of the form $g^{-1}(B)$, where $g$ ranges over the generating family of $\mathcal{H}_s$ and $B$ over Borel sets. Then $\mathcal{P}$ is a $\pi$-system in the sense of Dynkin's Pi-Lambda Theorem: it is nonempty ($\Omega$ is the preimage of $\mathbb{R}$ under $\xi^{1}$) and closed under finite intersections by construction; and the generated $\sigma$-algebra $\sigma(\mathcal{P})$ equals $\mathcal{H}_s$, since each contains the other's generators. Let

Then $\mathcal{L}$ is a $\lambda$-system in the sense of Dynkin's Pi-Lambda Theorem: $\Omega\in\mathcal{L}$; if $B\subseteq B'$ both lie in $\mathcal{L}$ then $P(A\cap(B'\setminus B))=P(A\cap B')-P(A\cap B)=P(A)\bigl(P(B')-P(B)\bigr)=P(A)P(B'\setminus B)$; and for an increasing sequence $B_n\uparrow B$ in $\mathcal{L}$, continuity of the measure from below gives $P(A\cap B)=\lim P(A\cap B_n)=P(A)\lim P(B_n)=P(A)P(B)$. Every member of $\mathcal{P}$ lies in $\sigma(g_1,\dots,g_d)$ for the finite family of generators appearing in it, so $\mathcal{P}\subseteq\mathcal{L}$ by Step 2. By Dynkin's Pi-Lambda Theorem, $\mathcal{H}_s=\sigma(\mathcal{P})\subseteq\mathcal{L}$, which is exactly the asserted independence.

Claim 2. $\mathbb{E}[I]=0$ since Wiener integrals have expectation zero (claim 2 of Wiener Integrals Against a Vector Brownian Motion are Jointly Gaussian) and expectation is linear. For the span assertion, recall from Wiener Integrals Against a Vector Brownian Motion are Jointly Gaussian that $\int_0^{t}f(r)\,dW^{j}_r$ is the It^{o} integral of $(f(r))_{r\in(0,t]}$ with respect to the It^{o} integrator $(W^{j},1)$ of Brownian Motion is an Ito Integrator with Unit Intensity, for the natural filtration of $W^{j}$, and that $\int_0^{s}f(r)\,dW^{j}_r$ is the It^{o} integral of the restriction of $f$ to $(0,s]$, by the subinterval paragraph of Ito Integrable Process and the Ito Integral (for $s=0$ it is $0$ and the assertion reduces to the case treated below with no pieces to the left of $s$).

For $n\ge1$, let $0=y_0<y_1<\dots<y_{N(n)}=t$ be the partition of $[0,t]$ whose point set is $\{pt/n:0\le p\le n\}\cup\{s\}$ (listed strictly increasingly), and define the deterministic simple adapted process $f_n$ on $(0,t]$ by $f_n(r):=f(y_{p-1})$ for $r\in(y_{p-1},y_p]$. Since $f$ is uniformly continuous on the compact interval $[0,t]$ by Continuity on a Closed Interval Implies Uniform Continuity, and the mesh of these partitions tends to $0$, we have $\epsilon_n:=\sup_{r\in(0,t]}|f_n(r)-f(r)|\to0$. Hence $(f_n)$ is an approximating sequence for $(f(r))_{r\in(0,t]}$ in the sense of Existence and Uniqueness of the Mean-Square Extension of the Elementary Stochastic Integral: for its Cauchy-type condition (a), the intensity of the integrator $(W^{j},1)$ is the constant $1$, and for all $n,n'$ the deterministic simple processes satisfy $\int_{\mathbb{R}}\mathbf{1}_{(0,t]}(r)\,\mathbb{E}\bigl[(f_n(r)-f_{n'}(r))^{2}\bigr]\,dr\le t\,(\epsilon_n+\epsilon_{n'})^{2}\to0$; and for its pointwise condition (b), at each $r\in(0,t]$ we have $\lVert f_n(r)-f(r)\rVert_{2}=|f_n(r)-f(r)|\le\epsilon_n\to0$. The restrictions of the $f_n$ to $(0,s]$ form an approximating sequence for the restriction of $f$, by the subinterval paragraph of Ito Integrable Process and the Ito Integral. By claims 1-2 of Existence and Uniqueness of the Mean-Square Extension of the Elementary Stochastic Integral (claim 1 producing a limit of the elementary stochastic integrals, claim 2 identifying it, up to almost sure equality, with any version of the It^{o} integral), the elementary stochastic integrals of $f_n$ over $(0,t]$ and over $(0,s]$ converge in mean square to $\int_0^t f\,dW^{j}$ and $\int_0^s f\,dW^{j}$ respectively. By Elementary Stochastic Integral of a Simple Adapted Process, the difference of the two elementary integrals is

a finite linear combination of increments $W^{j}_v-W^{j}_u$ with $s\le u<v\le t$ (the partition contains $s$, so every subinterval lies entirely in $[0,s]$ or in $[s,t]$). Then $\lVert I_n-I\rVert_2\to0$, so $I$ lies in the closed mean-square span of these increments.

Claim 3. Let $\tilde Q$ be an $\mathcal{H}_s$-measurable random variable almost surely equal to $Q$. With $I_n$ as in claim 2: each $I_n$ is measurable with respect to the $\sigma$-algebra generated by the finitely many increments appearing in it (preimages under a continuous, hence Borel, linear combination), which is independent of $\mathcal{H}_s$ by claim 1; hence $I_n$ and $\tilde Q$ are independent, and by Expectation of a Product of Independent Random Variables, $\mathbb{E}[I_n\tilde Q]=\mathbb{E}[I_n]\,\mathbb{E}[\tilde Q]=0$, each increment having expectation $0$ (claim 2 of Wiener Integrals Against a Vector Brownian Motion are Jointly Gaussian and linearity of the expectation). By Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, $|\mathbb{E}[I\tilde Q]-\mathbb{E}[I_n\tilde Q]|\le\lVert I-I_n\rVert_2\lVert\tilde Q\rVert_2\to0$, so $\mathbb{E}[I\tilde Q]=0$; the product $I\tilde Q$ is integrable by Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm. Since $IQ=I\tilde Q$ almost surely, $IQ$ is integrable with $\mathbb{E}[IQ]=0$, and $\operatorname{Cov}(I,Q)=\mathbb{E}[IQ]-\mathbb{E}[I]\mathbb{E}[Q]=0$ by claim 2. $\square$