jerrylicious/diffusion
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Added the abstract wiener space section.

Browse Source
This commit is contained in:
Bokuan Li
2026-05-31 14:12:39 -04:00
parent 5f68dff976
commit 89c1804ceb
11 changed files with 842 additions and 4 deletions
Download Patch File Download Diff File Expand all files Collapse all files

1
.gitignore vendored Normal file
View File

@@ -0,0 +1 @@
spec.db

31
preamble.sty
View File

@@ -110,7 +110,6 @@
\newcommand{\proj}{\text{proj}}
\newcommand{\tr}{\text{tr}}
\newcommand{\spec}[1]{\text{Spec}\paren{#1}}
\newcommand{\ba}{\mathbf{A}}
% Dual Pairings & Inner Products
@@ -151,7 +150,6 @@
\newcommand{\cm}{\mathcal{M}}
\newcommand{\cf}{\mathcal{F}}
\newcommand{\ce}{\mathcal{E}}
\newcommand{\ci}{\mathcal{I}}
\newcommand{\pow}[1]{\mathcal{P}\paren{#1}}
\newcommand{\cb}{\mathcal{B}}
\newcommand{\cn}{\mathcal{N}}
@@ -167,7 +165,6 @@
\newcommand{\cd}{\mathcal{D}}
\newcommand{\calr}{\mathcal{R}}
\newcommand{\scp}{\mathscr{P}}
\newcommand{\wien}{\mathcal{W}}
% Jokes
\newcommand{\lol}{\boxed{\text{LOL.}}}
@@ -224,3 +221,31 @@
% Real or Complex Numbers
\newcommand{\RC}{\bracs{\real, \complex}}
% AWS
\newcommand{\ci}{\mathcal{I}}
\newcommand{\ba}{\mathbf{A}}
% Words
\newcommand{\hldr}{Hölder\ }
% Wiener Space
\newcommand{\wien}{\mathcal{W}}
\newcommand{\sps}{{\Theta(\real^d)}} % Sample Path Space
\newcommand{\ccm}{{\mathbf{H}^1(\real^d)}} % Classical Cameron-Martin Space
% Dual Pairings & Inner Products
\newcommand{\floor}[1]{\lfloor #1 \rfloor}
\newcommand{\ceil}[1]{\lceil #1 \rceil}
\newcommand{\seqfn}[1]{\bracsn{#1}_1^{n}}
\newcommand{\seqfm}[1]{\bracs{#1}_1^{m}}
\newcommand{\seqfmz}[1]{\bracs{#1}_0^{m}}
\newcommand{\rad}[1]{\text{rad}\!\left(#1\right)}
\newcommand{\fp}{\mathfrak{p}}
\newcommand{\fq}{\mathfrak{q}}

24
refs.bib
View File

@@ -47,3 +47,27 @@
year={1985},
publisher={Springer New York}
}
@book{stroock, place={Cambridge}, edition={3}, title={Probability Theory, An Analytic View}, publisher={Cambridge University Press}, author={Stroock, Daniel W.}, year={2024}}
@book{folland,
title={Real Analysis: Modern Techniques and Their Applications},
author={Folland, G.B.},
isbn={9781118626399},
series={Pure and Applied Mathematics: A Wiley Series of Texts, Monographs and Tracts},
url={https://books.google.ca/books?id=wI4fAwAAQBAJ},
year={2013},
publisher={Wiley}
}
@book{measure,
title={Measure Theory},
author={Bogachev, V.I.},
number={v. 1},
isbn={9783540345145},
lccn={2006933997},
series={Measure Theory},
url={https://books.google.com/books?id=CoSIe7h5mTsC},
year={2007},
publisher={Springer Berlin Heidelberg}
}

88
src/aws/aws.tex Normal file
View File

@@ -0,0 +1,88 @@
\section{Abstract Wiener Spaces}
\label{section:aws}
\begin{lemma}[{{\cite[Lemma 8.2.2]{stroock}}}]
\label{lem:banach-hilbert}
Let $E$ be a separable Banach space over $\real$ and $H$ be a Hilbert space over $\real$ that is continuously embedded into $E$, then
\begin{enumerate}
\item The adjoint of the inclusion $\iota^*: E^* \to H$ is injective and continuous with respect to the weak* topology on $E$ and the weak topology on $H$. In addition, if $L \subset E^*$ is weak*-dense, then $\iota^*(L)$ is dense in $H$.
\item For each $x \in E$, $x \in H$ if and only if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E < \infty$. In which case, $\norm{x}_H = \sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E$.
\item For each weak*-dense subspace $L \subset E^*$, there exists $\bracs{x^*_n}_1^\infty \subset L$ such that $\bracsn{\iota^*x^*_n}_1^\infty \subset H$ is an orthonormal basis.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)$: Since $\iota: H \to E$ is continuous, $\iota^*: E^* \to H$ is continuous with respect to the strong and weak* topologies. As $H \subset E$ is dense, $\iota^*: E^* \to H^*$ is injective.
Let $h \in H$ such that $h \perp \iota^*(L)$, then by the weak*-density of $L \subset E^*$, $\angles{h, x^*}_E = 0$ for all $x^* \in E$. Therefore $h = 0$, and $\iota^*(L)$ is dense in $H$.
$(2)$: By $(1)$, the image $\iota^*(E^*) \subset H$ is dense. Hence if $x \in H$, then
\[
\norm{x}_H = \sup_{\substack{h \in H \\ \norm{h}_H = 1}}\angles{x, h}_H = \sup_{\substack{h \in \iota^*(E^*) \\ \norm{h}_H = 1}}\angles{x, h}_H = \sup_{\substack{x^* \in E^* \\ \norm{\iota^*x^*}_H = 1}}\angles{x, x^*}_E
\]
On the other hand, if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E < \infty$ then the map $\iota^*(E^*) \to \real$ with $\iota^*x^* \mapsto \angles{x, x^*}_E$ is well-defined as $\iota^*$ is injective, and continuous by assumption. By density of $\iota^*(E^*) \subset H$, there exists $h \in H$ such that $\anglesn{h, \iota^*x^*}_H = \angles{h, x^*}_{E} = \angles{x, x^*}_E$ for all $x \in E^*$. In which case, $x = h \in H$.
$(3)$: Since $E^*$ is separable with respect to the weak* topology, there exists a linearly independent set $\seq{x_n^*} \subset L$ whose span is dense in $E$. By $(1)$, $\seq{\iota^*x_n^*} \subset H$ has dense span in $H$. Using the Gram-Schmidt process, there exists an orthonormal basis $\seq{h_n} \subset \text{span}\bracs{\iota^*x_n^*: n \in \nat}$ for $H$. Since $\iota^*$ is injective and $L$ is a subspace, $\bracsn{(\iota^*)^{-1}(h_n)}_1^\infty \subset L$ is the desired sequence.
\end{proof}
\begin{definition}[Abstract Wiener Space]
\label{define:aws}
Let $E$ be a separable Banach space over $\real$, $H \subset E$ be a Hilbert space over $\real$ that is continuously embedded into $E$, and $\wien: \cb(E) \to [0, 1]$ be a Gaussian measure, then the triple $(H, E, \wien)$ is an \textbf{abstract Wiener space} if
\[
\wh \wien(x^*) = e^{-\norm{\iota^* x^*}_H^2/2} \quad \forall x^* \in E^*
\]
\end{definition}
\begin{theorem}[{{\cite[Theorem 8.3.2]{stroock}}}]
\label{thm:gaussian-aws}
Let $E$ be a separable Banach space over $\real$, and $\wien: \cb(E) \to [0, 1]$ be a Borel probability measure, then the following are equivalent:
\begin{enumerate}
\item $\wien$ is a non-degenerate centred Gaussian measure.
\item There exists a Hilbert space $(H, \inp_H)$ contained in $E$ such that $(H, E, \wien)$ is an abstract Wiener space.
\end{enumerate}
If $(G, \inp_G)$ is another Hilbert space in $E$ such that $(G, E, \wien)$ is an abstract Wiener space, then $G = H$ and $\inp_G = \inp_H$.
\end{theorem}
\begin{proof}[Proof of equivalence.]
$(1) \Rightarrow (2)$: The requirement on the characteristic function of $\wien$ defines the desired inner product for us, thus the work lies in finding an appropriate subspace and completing it as a Hilbert space. To this end, observe that since each $x^* \in E^*$ viewed as a random variable is Gaussian, the inner product can be taken in $L^2(\wien; \real)$, where we can take the completion, and then recover an element of $E$ by integrating against the identity.
We now begin the construction. By Fernique's theorem, $\int_E \norm{x}_E^2 d\wien(x) < \infty$. Thus there exists a bounded linear map $R: E^* \to L^2(\wien; \real)$ defined by identifying each $x^* \in E^*$ as a $L^2$-random variable on $E$. In addition, since $\text{Id}_E$ itself is an $E$-valued random variable, there exists a bounded linear map
\[
I: L^2(\wien; \real) \to E \quad Y \mapsto \int_E Y(x)xd\wien(x)
\]
then for any $x^* \in E^*$ and $Y \in L^2(\wien; \real)$,
\[
\angles{IY, x^*}_E = \int_E Y(x)\angles{x, x^*}d\wien(x) = \angles{Y, Rx^*}_{L^2(\wien; \real)}
\]
In particular, $IY = 0$ if and only if $Y \perp R(E^*)$, and $I|_{\overline{R(E^*)}}$ is injective.
Let $H = I(\overline{R(E^*)})$ and define $\angles{x, y}_H = \anglesn{I^{-1}x, I^{-1}y}_{L^2(\wien; \real)}$. Since $\wien$ is non-degenerate, for any $x^* \in E^*$, $\angles{IRx^*, x^*}_E = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)} > 0$. Thus $H$ is dense in $E$ with $IR = \iota^*: E^* \to H$ corresponding to the adjoint of the inclusion.
Lastly, since $\wien$ is a centred Gaussian measure, for any $x^* \in E^*$, $\wh \wien (x^*) = e^{-\norm{Rx^*}_{L^2(\wien; \real)}^2/2} = e^{-\normn{\iota^*x^*}_{H}^2/2}$.
$(2) \Rightarrow (1)$: Since $\wh \wien(x^*) = e^{-\normn{\iota^*x^*}_H^2/2}$ and $\iota^*$ is injective, $x^*$ is a non-degenerate centred Gaussian. Thus $\wien$ is non-degenerate.
\end{proof}
\begin{proof}[Proof of uniqueness.]
Let $R: E^* \to L^2(\wien; \real)$ and $(H, \inp_H)$ be as in the proof of $(1) \Rightarrow (2)$, and $\iota^*_G: E^* \to G$ be the adjoint of the inclusion $\iota_G: G \to E$. Since $\wh \wien(x^*) = e^{-\norm{\iota_G^*x^*}_G^2/2}$ for all $x^* \in E^*$,
\[
\angles{\iota_G^*x^*, \iota_G^*x^*}_G = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)}
\]
and by polarisation, for any $y^* \in E^*$,
\[
\angles{\iota_G^*x^*, y^*}_E = \angles{\iota_G^*x^*, \iota_G^*y^*}_G = \angles{Rx^*, Ry^*}_{L^2(\wien; \real)} = \angles{IRx^*, y^*}_E
\]
Therefore $\iota_G^*x^* = IRx^*$ for all $x^* \in E^*$, and the following diagram commutes
\[
\begin{CD}
E^* @>R>> \overline{R(E^*)} \\
@A{(\iota_G^*)^{-1}}AA @VV{I}V \\
\iota_G^*(E^*) @>{\iota}>> H
\end{CD}
\]
By density of $\iota^*(E^*)$, the composition $IR(\iota_G^*)^{-1}: \iota_G^*(E^*) \to H$ admits a unique extension to $G$ as an isometry. Therefore $G = H$ with the same inner product.
\end{proof}

307
src/aws/classical.tex Normal file
View File

@@ -0,0 +1,307 @@
\section{Classical Wiener Space}
\label{sec:classical}
Let $C([0, \infty); \real^d)$ be the space of $\real^d$-valued continuous functions on $[0, \infty)$, equipped with the topology of uniform convergence on compact sets. The space $C_0((0, \infty); \real^d)$ of $\real^d$-valued continuous functions on $(0, \infty)$ that vanish at $0$ and $\infty$ is equipped with the uniform norm.
\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
\label{lem:path-space}
For each $\theta \in C([0, \infty); \real^d)$, let
\[
\norm{\theta}_{\sps} = \sup_{t \ge 0}\frac{\abs{\theta(t)}}{1 + t}
\]
and
\[
\sps = \bracs{\theta \in C([0, \infty); \real^d): \theta(0) = 0, \lim_{t \to \infty}\frac{\theta(t)}{t} = 0}
\]
then
\begin{enumerate}
\item $\norm{\cdot}_{\sps}: C([0, \infty): \real^d) \to [0, \infty]$ is lower semicontinuous.
\item $\iota: \sps \to C([0, \infty); \real^d)$ is continuous.
\item The mapping $\Lambda: \sps \to C_0((0, \infty); \real^d)$ defined by $(\Lambda \theta)(t) = \theta(t)/(1 + t)$ is an isometric isomorphism, and $\sps$ is a separable Banach space.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)$: Let $R \ge 0$, then
\[
\bracs{\theta \in C([0, \infty); \real^d): \norm{\theta}_{\sps} \le R} = \bigcap_{t \ge 0}\bracs{\theta \in C([0, \infty); \real^d): \frac{\abs{\theta(t)}}{1 + t} \le R}
\]
where each $\bracs{\theta \in C([0, \infty); \real^d): \abs{\theta(t)}/(1+t) \le R}$ is a closed subset of $C([0, \infty); \real^d)$, and their intersection is also closed. Thus $\norm{\cdot}_{\sps}$ is lower semicontinuous.
$(2)$: Let $K \subset [0, \infty)$ be compact, then $M = \sup_{t \in K}(1 + t) < \infty$. Thus $\norm{\theta|_K}_u \le M\norm{\theta}_{\sps}$.
$(3)$: Since $\theta(0) = 0$ and $\limv{t}\theta(t)/t = 0$ for all $\theta \in \sps$,
\[
\lim_{t \to 0}(\Lambda \theta)(t) = \limv{t}(\Lambda \theta)(t) = 0
\]
and $\norm{\Lambda \theta}_{u, (0, \infty)} = \norm{\theta}_{\sps}$. On the other hand, for any $f \in C_0((0, \infty); \real^d)$, let $\Lambda^{-1}f$ be defined by $(\Lambda^{-1}f)(0) = 0$ and $(\Lambda^{-1}f)(t) = f(t)(1 + t)$ for $t > 0$. Since $\lim_{t \to 0}f(t) = \limv{t}f(t) = 0$,
\[
\lim_{t \to 0^+}(\Lambda^{-1}f)(t) = \limv{t}\frac{(\Lambda^{-1}f)(t)}{t} = 0
\]
so $\Lambda^{-1}f \in \sps$ with $\normn{\Lambda^{-1}f}_{\sps} = \norm{f}_{u, (0, \infty)}$.
As $C_0((0, \infty); \real^d)$ is a separable Banach space, so is $\sps$.
\end{proof}
\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
\label{lem:thetadual}
For each $\phi \in \sps^*$, there exists a unique $\real^d$-valued Borel measure $\lambda$ on $[0, \infty)$ such that
\begin{enumerate}
\item $\lambda(0) = 0$.
\item $\angles{\theta, \phi}_{\sps} = \int_{[0, \infty)} \theta d\lambda$ for all $\theta \in \sps$.
\item $\norm{\phi}_{\sps^*} = \int_{[0, \infty
)}(1 + t) d\abs{\lambda}(t) < \infty$.
\end{enumerate}
If a finite $\real^d$-valued Borel measure $\lambda$ on $[0, \infty)$ satisfies the above, then $\lambda$ defines a linear functional on $\sps^*$.
\end{lemma}
\begin{proof}
Let $\Lambda: \sps \to C_0((0, \infty); \real^d)$ be the isomorphism defined above, then $\phi \circ \Lambda^{-1} \in C_0((0, \infty); \real^d)^*$. By the Riesz representation theorem, there exists a vector measure $\lambda_0 \in M((0, \infty); \real^d)$ such that
\[
\anglesn{\Lambda^{-1}f, \phi}_{\sps} = \anglesn{f, \phi \circ \Lambda^{-1}}_{C_0((0, \infty); \real^d)} = \int_{(0, \infty)}f d\lambda_0
\]
and $\norm{\lambda_0}_{\text{var}} = \normn{\phi \circ \lambda^{-1}}_{\sps}$. Thus for any $\theta \in \sps$,
\[
\angles{\theta, \phi}_{\sps} = \anglesn{\Lambda \theta, \phi \circ \Lambda^{-1}}_{C_0((0, \infty); \real^d))} = \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\lambda_0(t)
\]
and
\[
\lambda: \cb([0, \infty)) \to \real^d \quad A \mapsto \int_{A \setminus \bracs{0}}\frac{1}{1 + t}d\lambda_0(t)
\]
is the desired measure.
\end{proof}
\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
\label{lem:sigma-algebra}
The space $\sps$ is a Borel set in $C([0, \infty); \real^d)$, and $\cb(\sps)$ is the restriction of the Borel $\sigma$-algebra on $C([0, \infty); \real^d)$ to $\sps$.
\end{lemma}
\begin{proof}
Firstly, $\lim_{t \to \infty}\theta(t)/t = 0$ if and only if for every $n \in \nat$, there exists $k \in \nat$ such that $\abs{\theta(s)}/s \le 1/n$ for all $s \ge k$. Thus
\begin{align*}
\Theta_\infty &= \bracs{\theta \in C([0, \infty); \real^d): \limv{t}\frac{\theta(t)}{t} =0} \\
&= \bigcap_{n \in \nat}\bigcup_{k \in \nat}\underbrace{\bigcap_{s \ge k}\bracs{\theta \in C([0, \infty); \real^d): \frac{\abs{\theta(s)}}{s} \le \frac{1}{n}}}_{\text{closed}}
\end{align*}
and
\[
\Theta_0 = \bracs{\theta \in C([0, \infty); \real^d): \theta(0) = 0}
\]
are Borel. Thus $\sps = \Theta_0 \cap \Theta_\infty \cap \Theta_b$ is Borel in $C([0, \infty); \real^d)$.
Now, by lower semicontinuity of $\norm{\cdot}_{\sps}$, every closed set in $\sps$ is the intersection of $\sps$ and a Borel set in $C([0, \infty), \real^d)$. Therefore $\cb(\sps)$ contains the induced $\sigma$-algebra from $\cb(C([0, \infty); \real^d))$. On the other hand, the inclusion $\iota: \sps \to C([0, \infty); \real^d)$ is continuous. Thus $\cb(\sps)$ contains the induced $\sigma$-algebra.
\end{proof}
\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
\label{lem:path-justify}
Let $(\Omega, \cf, \bp)$ be a probability space and $(B_t, \bracs{\cf_t: t \ge 0})$ be a standard Brownian motion, then
\begin{enumerate}
\item Almost every sample path of $B$ is in $\sps$.
\item $\ev[\norm{B}_{\sps}^2] \le 32d$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)$: By the Brownian Strong Law of Large Numbers, $B(\omega) \in \sps$ for almost every $\omega$.
$(2)$: By Doob's maximal inequality, for any $r \ge 0$,
\begin{align*}
\ev\braks{\norm{B}_{u, [0, 2]}^2}^{1/2} &\le 2\ev\braks{\norm{B_r}_{\real^d}^2}^{1/2} \\
\ev\braks{\norm{B}_{u, [0, 2]}^2} &\le 4\ev\braks{\norm{B_r}_{\real^d}^2}
\end{align*}
and for any $\theta \in \sps$,
\begin{align*}
\norm{\theta}_{\sps}^2 &\le \sum_{n = 0}^\infty \sup_{t \in [0, 2^n]}\frac{\abs{\theta(t)}^2}{(1 + t)^2} \le \sum_{n = 0}^\infty \frac{1}{2^{2(n-1)}}\norm{\theta}_{u, [0, 2^n]}^2\\
&\le \sum_{n = 0}^\infty 4^{1 - n}\norm{\theta}_{u, [0, 2^n]}^2
\end{align*}
Thus
\begin{align*}
\ev[\norm{B}_{\sps}^2] &\le \sum_{n = 0}^\infty 4^{1 - n}\ev\braks{\norm{\theta}_{u, [0, 2^n]}^2}\le \sum_{n = 0}^\infty 4^{1 - n}\ev\braks{\norm{B_{2^n}}_{\real^d}^2} \le 32d
\end{align*}
\end{proof}
\begin{definition}[Characteristic Function {{\cite[Lemma 8.1.2]{stroock}}}]
\label{def:charfunc}
Let $(E, \norm{\cdot})$ be a separable Banach space over $\real$, and $\mu$ be a Borel probability measure on $E$. Define
\[
\wh \mu: E^* \to \complex \quad \wh x^* \mapsto \int_E e^{i\angles{x, x^*}_E}d\mu(x)
\]
as the \textbf{characteristic function} of $\mu$, then
\begin{enumerate}
\item $\wh \mu$ is \textit{sequentially}\footnote{Most sources I found only claimed and proved the sequential case. Since Stroock used Dominated Convergence Theorem, I assume it should be sequential too.} continuous with respect to the weak* topology on $E^*$.
\item If $\nu$ is a Borel probability measure on $E$ such that $\wh \mu = \wh \nu$, then $\nu = \mu$.
\end{enumerate}
\end{definition}
\begin{proof}
$(1)$: Dominated Convergence Theorem.
$(2)$: Let $\bracsn{x_j^*}_1^n \subset E^*$ and $\mu_{(x_1^*, \cdots, x_n^*)}$ be the distribution of $(x_1^*, \cdots, x_n^*): E \to \real^n$ under $\mu$, then the characteristic function $\mu_{(x_1^*, \cdots, x_n^*)}$ is
\begin{align*}
\wh \mu_{(x_1^*, \cdots, x_n^*)}(\xi) &= \int_{\real^n}e^{i\angles{\xi, x}_{\real^n}}d\mu_{(x_1^*, \cdots, x_n^*)}(x) \\
&= \int_{E}e^{i\angles{x, \sum_{j = 1}^n \xi_j x_j^*}}d\mu(x) = \wh \mu\paren{\sum_{j = 1}^n \xi_jx_j^*}
\end{align*}
Thus $\wh \mu$ uniquely determines the joint characteristic function of $(x_1^*, \cdots, x_n^*)$. So for any $\bracsn{x_j^*}_1^n \subset E^*$ and Borel sets $\seqf{B_j} \subset \cb(\real)$,
\[
\mu\paren{\bigcap_{j = 1}^n (x_j^*)^{-1}(B_j)} = \nu\paren{\bigcap_{j = 1}^n (x_j^*)^{-1}(B_j)}
\]
As these cylinder sets form a $\pi$ system that generates $\cb(E)$, $\mu = \nu$ by Dynkin's Uniqueness Theorem.
\end{proof}
\begin{lemma}[{{\cite[Equation 8.1]{stroock}}}]
\label{lem:classical-ft}
Let $\wien: \cb(\sps) \to [0, 1]$ be the classical Wiener measure on $(\sps, \cb(\real^d))$, then for every $\lambda \in \sps^*$,
\begin{enumerate}
\item $\lambda$ viewed as a random variable on $\sps$ is a centred Gaussian.
\item $\ev^\wien[\abs{\lambda}^2] = \iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.
\item $\wh \wien(\lambda) = \exp\braks{-\frac{1}{2}\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)}$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)$, Finite: First suppose that there exists $0 = t_0 \le t_1 < \cdots < t_n < \infty$ and $\seqf{\xi_j} \subset \real^d$ such that
\[
\angles{\theta, \lambda}_{\sps} = \sum_{j = 1}^n \angles{\theta(t_j), \xi_j}_{\real^d}
\]
for all $\theta \in \sps$. This allows decomposing
\begin{align*}
\angles{\theta, \lambda}_{\sps} &= \sum_{j = 1}^n \sum_{k = 1}^{j}\angles{\theta(t_k) - \theta(t_{k - 1}), \xi_j}_{\real^d} \\
&= \sum_{k = 1}^n \angles{\theta(t_k) - \theta(t_{k - 1}), \sum_{j = k}^n\xi_j}_{\real^d}
\end{align*}
as a sum of independent centred Gaussian random variables. Thus $\theta \mapsto \angles{\theta, \lambda}_{\real^d}$ is a centred Gaussian random variable.
$(1)$, with Riemann-Stieltjes: Let $\mu: \cb((0, \infty)) \to \real$ be a finite signed measure, and $F(t) = \mu((0, t])$ be the distribution function of $\mu$, then $\norm{F}_{\text{var}} = \norm{\mu}_{\text{var}} < \infty$.
For any indicator function of the form $\one_{(a, b]}$ with $0 \le a < b < \infty$,
\[
\int_0^\infty \one_{(a, b]}(t)dF(t) = F(b) - F(a) = \mu((0, t]) = \int_{(0, \infty)}\one_{(a, b]}(t)d\mu(t)
\]
where the integral on the left is a Riemann-Stieltjes integral. As simple functions that vanish at $0$ are dense in $C_0((0, \infty); \real)$, and
\[
\abs{\int_0^\infty \phi(t)dF(t)} \le \norm{\phi}_u\norm{F}_{\text{var}} \quad \abs{\int_{(0, \infty)}\phi(t)d\mu(t)} \le \norm{\phi}_u\norm{\mu}_{\text{var}}
\]
for any simple function $\phi: (0, \infty) \to \real$,
\[
\int_0^\infty \phi(t)dF(t) = \int_{(0, \infty)}\phi(t)d\mu(t)
\]
for all $\phi \in C_0((0, \infty); \real)$.
As seen in the proof of \autoref{lem:thetadual}, there exists a $\real^d$-valued finite Borel measure $\mu: \cb((0, \infty)) \to \real^d$ such that for any $\theta \in \sps$,
\[
\angles{\theta, \lambda}_{\sps} = \int_{(0, \infty)} \Lambda \theta d\mu = \int_{(0, \infty)} \frac{\theta(t)}{1 + t}d\mu(t)
\]
Let $F(t) = \mu((0, t])$, then since $\Lambda \theta \in C_0((0, \infty); \real)$,
\[
\angles{\theta, \lambda}_{\sps} = \int_0^\infty \frac{\theta(t)}{1 + t}dF(t)
\]
can be expressed as a Riemann-Stieltjes integral. Thus
\[
\angles{\theta, \lambda}_{\sps} = \lim_{t \to \infty}\limv{n}\sum_{j = 1}^{2^n}\theta(2^{-n}jt)[F(2^{-n}jt) - F((2^{-n} - 1)jt)]
\]
As the above holds for all $\theta \in \sps$, $\lambda$ viewed as a random variable is a\footnote{I'm pretty sure I can just use one single limit here, but just to be safe I nested it.} sequential limit of Gaussian random variables, so it is also Gaussian.
$(1)$, with Approximations: As seen in \autoref{lem:thetadual}, there exists a $\real^d$-valued Borel measure $\mu: \cb((0, \infty)) \to \real^d$ such that
\[
\angles{\theta, \lambda}_{\sps} = \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\mu(t) \quad \forall \theta \in \sps
\]
Since there exists $\real^d$-valued Borel measures $\seq{\mu_k}$ with finite support such that $\mu_k \Rightarrow \mu$, each $\theta \mapsto \int \frac{\theta(t)}{1 + t}d\mu_k(t)$ is Gaussian random variable and $\theta \mapsto \angles{\theta, \lambda}_{\sps}$ is a pointwise limit of Gaussian random variables, thus it is also Gaussian.
$(2)$: Firstly, for any $0 \le s, t < \infty$ and $\xi, \eta \in \real^d$,
\[
\ev^\wien(\angles{\theta(s), \xi} \cdot \angles{\theta(t), \eta}) = (s \wedge t)\angles{\xi, \eta}
\]
Let $F: [0, \infty) \to \real^d$ be measurable such that $d\lambda = Fd\abs{\lambda}$, then
\begin{align*}
\angles{\theta, \lambda}_{\sps}^2 &= \braks{\int_{[0, \infty)} \theta(t) d\lambda(t)}^2 = \int_{[0, \infty)}\theta(t)\int_{[0, \infty)} \theta(s)d\lambda(s)d\lambda(t) \\
&= \int_{[0, \infty)^2} \angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t)
\end{align*}
By Fubini's theorem, the expectation can be pushed inside the integral:
\begin{align*}
\ev^\wien[\abs{\lambda}^2] &= \iint_{[0, \infty)^2} \ev^\wien\braks{\angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}}d\abs{\lambda}(s)d\abs{\lambda}(t) \\
&= \iint_{[0, \infty)^2} (s \wedge t)\angles{F(s), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t) \\
&= \iint_{[0, \infty)^2} s \wedge t d\lambda(s)d\lambda(t)
\end{align*}
$(3)$: $\lambda$ is a Gaussian random variable with variance $\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.
\end{proof}
\begin{definition}[Classical Cameron-Martin Space]
\label{define:cameron-martin-classical}
Let $\ccm$ be the set of absolutely continuous functions $\theta: [0, \infty) \to \real^d$ such that $\theta(0) = 0$ and $D\theta \in L^2([0, \infty); \real^d)$, equipped with the inner product
\[
\angles{\theta, \eta}_{\ccm} = \angles{D\theta, D\eta}_{L^2([0, \infty); \real^d)} \quad \forall \theta, \eta \in \ccm
\]
then
\begin{enumerate}
\item For each $\theta \in \ccm$, $\norm{\theta}_{\sps} \le \frac{1}{2}\norm{\theta}_{\ccm}$.
\item $\ccm$ is a dense subspace of $\sps$.
\end{enumerate}
Thus $\ccm$ is continuously embedded in $\sps$ as a dense subspace, known as the \textbf{classical Cameron-Martin space} for the classical Wiener measure.
\end{definition}
\begin{proof}
$(1)$: Let $\theta \in \ccm$, then for each $t \ge 0$,
\[
\abs{\theta(t)} = \abs{\int_{[0, \infty)} \one_{[0, t)} D\theta ds} \le \normn{\one_{[0, t)}}_{L^2([0, \infty))}\normn{D\theta}_{L^2([0, \infty); \real^d)} = \sqrt{t} \norm{\theta}_\ccm
\]
Thus
\[
\frac{\abs{\theta(t)}}{1 + t} \le \norm{\theta}_\ccm \frac{\sqrt{t}}{1 + t} \le \frac{\norm{\theta}_\ccm}{2}
\]
$(2)$: Follows from the density of $C_c^\infty(\real^d)$ in $C_0((0, \infty); \real^d)$ and thus in $\sps$.
\end{proof}
\begin{lemma}
\label{lem:classical-cameron-martin}
Let $\lambda \in \sps^*$, then there exists $h_\lambda \in \ccm$ such that $\angles{\theta, \lambda}_\sps = \angles{\theta, h_\lambda}_\ccm$ for all $\theta \in \ccm$, which is given by
\[
h_\lambda(t) = \int_{(0,t]}\lambda((s, \infty))ds
\]
with
\[
\norm{h_\lambda}_\ccm^2 = \iint_{[0, \infty)}(s \wedge t)d\lambda(s)d\lambda(t) = \ev^\wien(\abs{\lambda}^2)
\]
thus
\[
\wh \wien(\lambda) = \exp\braks{-\frac{\norm{h_\lambda}_\ccm}{2}^2}
\]
Moreover, for each $\theta \in \sps$,
\[
\angles{\theta, \lambda}_\sps = \int_0^\infty Dh_\lambda d\theta
\]
is an improper Riemann-Stieltjes integral.
\end{lemma}
\begin{proof}
Firstly, $\lambda$ is bounded, so $h_\lambda$ is absolutely continuous and $Dh_\lambda(s) = \lambda((s, \infty))$ by the Lebesgue differentiation theorem. From here,
\begin{align*}
\norm{h_\lambda}_\ccm^2 &= \int_{[0, \infty)}\abs{\lambda((t, \infty))}^2dt = \int_{[0, \infty)}\iint_{[t, \infty)^2}d\lambda(r)d\lambda(s)dt \\
&= \iiint_{[0, \infty)^3}\one_{\bracs{r, s \ge t}}(r, s, t)d\lambda(r)d\lambda(s)dt \\
&= \iint_{[0, \infty)^2} \int_{[0, \infty)}\one_{\bracs{t \le r \wedge s}}(r, s, t)dtd\lambda(r)d\lambda(s) \\
&= \iint_{[0, \infty)^2}s \wedge t d\lambda(r)d\lambda(s) = \ev^\wien[\abs{\lambda}^2]
\end{align*}
so $h_\lambda \in \ccm$ and
\begin{align*}
\angles{h, \lambda}_\sps &= \int_{[0, \infty)} h d\lambda = \int_{[0, \infty)} \int_{[0, t]}Dh(s)dsd\lambda(t) \\
&= \int_{[0, \infty)} \int_{[0, \infty)}\one_{\bracs{s \le t}}(s, t)Dh(s)d\lambda(t) ds \\
&= \int_{[0, \infty)}Dh(s)\lambda((s, \infty))ds = \angles{\theta, h_\lambda}_\ccm
\end{align*}
Now let $F(t) = \lambda((0, t])$, then $Dh_\lambda(t) = \lambda((0, \infty)) - F(t)$ and
\[
\int_0^\infty \theta(t) d\lambda(t) = \int_0^\infty \theta(t)dF(t) = -\int_0^\infty \theta(t) dDh_\lambda(t)
\]
can be expressed as improper Riemann-Stieltjes integrals. Using the by parts formula for Riemann-Stieltjes integrals,
\[
\int_0^T \theta(t) dDh_\lambda(t) = \theta(T)Dh_\lambda(T) - \int_0^TDh_\lambda(t)d\theta(t)
\]
where since $\theta(T)/T \to 0$ as $T \to \infty$ and $(1 + t) \in L^1(\abs{\lambda})$,
\[
\limv{T}\abs{\theta(T)}\abs{\lambda}((T, \infty)) \le \limv{T}\frac{\abs{\theta(T)}}{1 + T}\int_{(0, \infty)}(1 + t)d\abs{\lambda}(t) = 0
\]
Therefore
\[
\angles{h, \lambda}_\sps = -\int_0^\infty \theta(t)dDh_\lambda(t) = \int_0^\infty Dh_\lambda(t)d\theta(t)
\]
\end{proof}

93
src/aws/cm.tex Normal file
View File

@@ -0,0 +1,93 @@
\section{Cameron-Martin Space}
\label{section:cm-space}
\begin{proposition}[{{\cite[Theorem 8.2.6]{stroock}}}]
\label{prop:aws-compact}
Let $(H, E, \wien)$ be an abstract Wiener space, then the operators
\[
\begin{CD}
E^* @>{\iota^*}>> H @>{\iota}>> E
\end{CD}
\]
are compact. In particular, $H \in \cb(E)$.
\end{proposition}
\begin{proof}
It is sufficient to show that both operators map weakly convergent sequences to strongly convergent sequences.
Let $\seq{x_n^*} \subset E^*$ such that $x_n^* \to x \in E^*$ weakly, then since $\wh \wien: E^* \to \complex$ is continuous with respect to sequential weak* convergence,
\[
e^{-\norm{\iota^*x_n^* - \iota^*x^*}_H^2/2} = \wh \wien (x^*_n - x^*) \to 1
\]
as $n \to \infty$. Thus $\iota^* x_n^* \to \iota^*x^*$ strongly in $H$.
To see the compactness of $\iota$, let $\seq{h_n} \subset H$ such that $h_n \to 0$ weakly. By the Uniform Boundedness Principle, there exists $R \ge 0$ such that $\sup_{n \in \nat}\norm{h_n}_H \le R$. Let $\seq{x_n^*} \subset E^*$ be a weak*-dense subset of $E^*$ with $\norm{x_n^*}_{E^*} = 1$ for all $n \in \nat$. Let $\eps > 0$, then by compactness of $\seq{\iota^*x^*_n} \subset H$, there exists $N \in \nat$ such that $\seq{\iota^*x^*_n} \subset \bigcup_{n = 1}^N B_H(\iota^*x^*_n, \eps)$. In which case,
\[
\norm{h_n}_E = \sup_{n \in \nat}\angles{h_n, x_n^*}_{E} \le 2\max_{1 \le k \le N}\angles{h_n, x_k^*}_E + R\eps
\]
Given that $h_n \to 0$ weakly, $\max_{1 \le k \le N}\angles{h_n, x_k^*}_E \to 0$ as $n \to \infty$. As $\eps > 0$ is arbitrary, $\norm{h_n}_E \to 0$ as $n \to \infty$.
\end{proof}
\begin{proposition}
Let $(H, E, \wien)$ be an abstract Wiener space. If $E$ is infinite-dimensional, then $\wien(H) = 0$.
\end{proposition}
\begin{proof}
Let $\seq{x_n^*} \subset E^*$ such that $\seq{\iota^* x_n^*}$ is an orthonormal basis for $H$. For each $n \in \nat$, let $X_n(x) = \dpb{x, x_n^*}{E}$, then $\seq{X_n}$ is a family of independent Gaussian random variables with variance $1$. Thus $\sum_{n \in \nat} X_n^2 = \infty$ almost surely. By (2) of \autoref{lem:banach-hilbert}, $\wien(H) = 0$.
\end{proof}
\begin{theorem}
\label{thm:paley-wiener}
Let $(H, E, \wien)$ be an abstract Wiener space, then
\begin{enumerate}
\item There exists a unique linear isometry $\ci: H \to L^2(\wien; \real)$ such that $\ci(\iota^*x^*) = \dpb{\cdot, x^*}{E}$ for all $x^* \in E^*$.
\item $\bracs{\ci(h): h \in H}$ is a Gaussian family.
\end{enumerate}
The map in (1) is the \textbf{Paley-Wiener map}.
\end{theorem}
\begin{proof}
Let $x^* \in E^*$, then $\dpb{\cdot, x^*}{E}$ is a centred Gaussian random variable with variance $\norm{\iota^*x^*}_H^2$. By (3) of \autoref{lem:banach-hilbert}, $\iota^*(E^*) \subset H$ is dense. Thus $\ci$ extends uniquely into a continuous linear isometry.
\end{proof}
\begin{theorem}[Cameron-Martin, {{\cite[Theorem 8.2.7]{stroock}}}]
\label{thm:cameron-martin}
Let $(H, E, \wien)$ be an abstract Wiener space and $g \in E$. Denote $\tau_g: E \to E$ with $x \mapsto x + g$ as the translation map, then:
\begin{enumerate}
\item If $g \in H$, then $(\tau_g)_* \ll \wien$ and
\[
\frac{d(\tau_g)_*\wien}{d\wien}(x) = e^{\ci (g) - \norm{g}_H^2/2}
\]
\item If $g \not\in H$, then $(\tau_g)_* \wien \perp \wien$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1): Let $\mu = (\tau_g)_* \wien$, then for any $x^* \in E^*$,
\[
\wh \mu(x^*) = \ev^\wien\braks{e^{i\angles{x + g, x^*}_E}} = e^{i\dpb{g, x^*}{E} - \norm{\iota^*x^*}_H^2/2}
\]
On the other hand, for any $x, y \in H$, $\ci(x)$ and $\ci(y)$ are jointly Gaussian, so for any $\xi, \eta \in \complex$,
\[
\ev^\wien\braks{e^{\xi \ci(x) + \eta \ci(y)}} = \exp\braks{\frac{1}{2}\paren{\xi^2\norm{x}_H^2 + 2\xi \eta \dpb{x, y}{H} + \eta^2\norm{y}_H^2}}
\]
In particular, if $\xi = 1$, $\eta = i$, $x = g$, and $y = x^*$, then
\[
\ev^\wien\braks{e^{\ci (g) + ix^*}} = e^{i\dpb{g, x^*}{E} - \norm{\iota^*x^*}_H^2/2} = \wh \mu(x^*)
\]
(2): Let $x^* \in E^*$ with $\norm{\iota^*x^*}_H = 1$, and $\mathcal F = \sigma(x^*)$, then
\[
\frac{d\mu|_{\cf}}{d\wien_\cf}(x) = e^{\dpb{g, x^*}{E}\dpb{x, x^*}{E} - \dpb{g, x^*}{E}^2/2}
\]
If $\mu = \lambda + Fd\wien$ is the Lebesgue decomposition of $\mu$ with respect to $\wien$, then
\[
\frac{d\mu|_{\cf}}{d\wien_\cf} \ge \ev^\wien[F|\cf]
\]
Given that $\norm{\iota^*x^*}_H = 1$,
\[
e^{\dpb{g, x^*}{E}^2}= \ev^\wien\braks{\paren{\frac{d\mu|_{\cf}}{d\wien_\cf}}^2} \ge \ev^\wien(F^2)
\]
where $\ev^\wien(F^2)$ does not depend on $x^*$. If $F \ne 0$, then the above bound implies that
\[
\sup_{\substack{x^* \in E^* \\ ||\iota^* x^*||_H = 1}}|\dpb{g, x^*}{E}^2| < \infty
\]
Thus $g \in H$.
\end{proof}

135
src/aws/donsker.tex Normal file
View File

@@ -0,0 +1,135 @@
\section{Donsker's Theorem}
\label{section:donsker}
In this section, equip $\Omega = C(\real_{\ge 0}; \real)$ with the topology of uniform convergence on compact sets.
For every $\delta > 0$ and $N \in \nat$, denote
\[
V_{\delta, n}(\omega) = \sup\bracs{\abs{\pi_s(\omega) - \pi_t(\omega)}| s, t \in [0, N], \abs{s - t} \le \delta}
\]
then
\begin{theorem}[Arzela-Ascoli]
\label{thm:arzela-ascoli}
Let $K \subset \Omega$, then $K$ is compact if and only if the following holds:
\begin{enumerate}
\item $\pi_0(K)$ is precompact.
\item For each $N \in \nat$,
\[
\lim_{\delta \to 0}\sup_{\omega \in K}V_{\delta, N}(\omega) = 0
\]
\end{enumerate}
\end{theorem}
\begin{theorem}[Prokhorov]
\label{thm:prokhorov}
Let $X$ be a Polish space and $\mathcal{P} \subset \mathbf{M}_1(X, \cb(X))$, then the following are equivalent:
\begin{enumerate}
\item For each $\alpha \in (0, 1)$, there exists $K \subset X$ compact such that
\[
\inf_{\mu \in \mathcal{P}}\mu(K) \ge \alpha
\]
\item $\mathcal{P}$ is weakly precompact.
\end{enumerate}
\end{theorem}
\begin{proposition}
\label{prop:weakly-precompact-theta}
Let $\seq{\mu_n} \subset \mathbf{M}_1(\Omega)$, then $\seq{\mu_n}$ is weakly precompact if and only if the following holds:
\begin{enumerate}
\item $\seq{(\pi_0)_*\mu_n}$ is tight.
\item For any $N \in \nat$ and $\eta > 0$,
\[
\lim_{\delta \to 0}\sup_{n \in \nat}\mu_n(\bracs{V_{\delta, N} > \eta}) = 0
\]
\end{enumerate}
\end{proposition}
\begin{proof}
Using Prokhorov's theorem, it is sufficiently to show that $\seq{\mu_n}$ is tight.
To this end, let $\eps > 0$, then there exists $K \subset \real$ compact such that
\[
\sup_{n \in \nat}\mu_n(\bracs{\pi_0 \not\in K}) \le \eps
\]
Fix $N \in \nat$, then for each $k \in \nat$, there exists $\delta(N, k) > 0$ such that
\[
\sup_{n \in \nat}\mu_n\paren{\bracs{V_{\delta(N, k), N} > \frac{1}{k}}} \le 2^{-N-k} \cdot \eps
\]
In which case,
\[
\sup_{n \in \nat}\mu_n\paren{\bracs{\omega \in \Omega \bigg | \forall k \in \nat, V_{\delta(N, k), N}(\omega) \le \frac{1}{k}}^c} \ge 1 - 2^{-N}\eps
\]
Let $K'$ be the intersection of $\bracs{\pi_0 \in K}$ and the above compact set, then $K'$ is precompact by the Arzela-Ascoli theorem, and $\sup_{n \in \nat}\mu_n(\Omega \setminus K') \le 2\eps$.
\end{proof}
\begin{definition}
\label{define:remainder}
Let $t \ge 0$ and $n \in \nat$, then
\[
\floor{t}_n = \max\bracs{\frac{k}{n} \bigg |k \in \nat_0, \frac{k}{n} \le t}
\]
is the \textbf{$n$-adic floor} of $t$, respectively, and
\[
r_n(t) = t - \fl{t}_n
\]
is the \textbf{$n$-adic remainder} of $t$.
\end{definition}
\begin{definition}
\label{define:linearisation}
Let $f: \bracs{k/n|k \in \nat_0} \to \real$, then the function
\[
L_nf: \real_{\ge 0} \to \real \quad t \mapsto (1 - nr_n(t))f(\fl{t}_n) + nr_n(t)f(\fl{t}_n + n^{-1})
\]
is the \textbf{$n$-adic linearisation} of $f$.
\end{definition}
\begin{theorem}[Donsker]
\label{thm:donsker}
Let $\seq{S_n}$ be the symmetric random walk on $\integer$. For each $n \in \nat$, let $S^n_t = L_nP_n$ be the $n$-adic linearisation of
\[
P_n: \bracs{k/n|k \in \nat_0} \to \real \quad \frac{k}{n} \mapsto \frac{S_k}{\sqrt{n}}
\]
Let $B$ be the standard Brownian motion, then $S^n \to B$ in distribution as $n \to \infty$.
\end{theorem}
\begin{proof}
For any $0 \le s < t \le <\infty$,
\[
S^n_t - S^n_s = \frac{S_{n\fl{t}_n} - S_{n\fl{s}_n}}{\sqrt{n}} + \frac{R_{n, s, t}}{\sqrt{n}}
\]
where $\sup_{n \in \nat}\norm{R_{n, s, t}}_{L^\infty} < \infty$\footnote{I believe that 3 is an easy bound, but I cannot be bothered to check.}, so $R_{n, s, t}/\sqrt{n} \to 0$ in $L^\infty$ as $n \to \infty$.
Suppose that $S^n$ \textit{converges} in distribution. By the above calculation, it is sufficient to consider the convergence in distribution of $(S_{n\floor{t}_n} - S_{n\floor{s}_n})/\sqrt{n}$.
Since $\sqrt{n}(\fl{t}_n - \fl{s}_n)/\sqrt{n} \to (t - s)$ as $n \to \infty$, for any $0 \le t_1 < \cdots < t_k < \infty$,
\[
\frac{1}{\sqrt{n}}\paren{S_{n\fl{t_2}_n} - S_{n\fl{t_1}_n}, \cdots, S_{n\fl{t_k}_n} - S_{n\fl{t_{k-1}}_n}} \to (X_1, \cdots, X_k)
\]
in distribution by the central limit theorem, where $\bracs{X_j}_1^k$ are independent random variables with $X_j \sim \gamma_{0, t_{j+1} - t_{j}}$. Therefore $S^n \to B$ in distribution.
To see that $S^n$ does converge in distribution, it is sufficient to show that it is tight. Given that $S^n_0 = 0$ for all $n \in \nat$, it is sufficient to check (2) of \autoref{prop:weakly-precompact-theta}.
Let $\lambda > 0$. To achieve the appropriate order of bound, apply Doob's maximal inequality to $S_n^4$.
\begin{align*}
\bp\bracs{\max_{1 \le k \le n}|S_k| > \lambda \sqrt{n}} &= \bp\bracs{\max_{1 \le k \le n}|S_k^4| > \lambda^4n^2} \le \frac{1}{\lambda^4n^2}\ev\braks{|S_n^4|} \\
&= \frac{1}{\lambda^4 n^2}\braks{n + {n \choose 2}{4 \choose 2}} \\
&= \frac{3n^2 - 2n}{\lambda^4n^2} \le \frac{3}{\lambda^4}
\end{align*}
For any $\delta > 0$ and $\eps > 0$, if $3/\sqrt{N} < \eps$, then for any $0 \le s < t$ with $t - s < \delta$, and $n \ge N$,
\begin{align*}
\bp\bracs{|S_t^n - S^n_s| \ge 2\eps} &\le \bp\bracs{|S_{n\floor{t}_n} - S_{n\floor{s}_n}| \ge \frac{\eps}{\sqrt{\delta}}\sqrt{n \delta}}\\
&\le \frac{3 \delta^2}{\eps^4}
\end{align*}
In particular, for any $\eta > 0$, a choice of $\delta = \eta \eps^4/3$ yields
\[
\bp\bracs{|S_t^n - S^n_s| \ge 2\eps} \le \frac{3 \delta^2}{\eps^4} = \frac{\eta^2\eps^4}{3} = \eta \delta
\]
Partition the interval $[0, N]$ into $\lceil N/\delta \rceil$ pieces of of intervals of length $\delta$, then
\[
\bp\bracs{\sup_{\substack{0 \le s < t \\ t - s < \delta}}|S_t^n - S_s^n| \ge 2\eps} \le (1 + N/\delta)\eta \delta = \eta \delta + N\eta
\]
\end{proof}

97
src/aws/gaussian.tex Normal file
View File

@@ -0,0 +1,97 @@
\section{Gaussian Measures}
\label{sec:gaussian}
\begin{definition}
\label{define:gaussian}
Let $E$ be a separable Banach space over $\real$ and $\wien$ be a Borel measure on $E$, then $\wien$ is a \textbf{Gaussian measure} if $x \mapsto \angles{x, x^*}_E$ is a centred Gaussian random variable for each $x^* \in E^*$. If each $x \mapsto \angles{x, x^*}_E$ is non-degenerate, then $\wien$ is \textbf{non-degenerate}.
\end{definition}
\begin{lemma}
\label{lem:shift}
Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X, X': \Omega \to E$ be i.i.d Gaussian random variables, then $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ have the same distribution as $(X, X')$.
\end{lemma}
\begin{proof}
Let $x^*, y^* \in E^*$, $Y(\omega) = \angles{X(\omega), x^*}_E$ and $Y'(\omega) = \anglesn{X'(\omega), x^*}_E$, then $Y$ and $Y'$ are independent Gaussian random variables, so
\begin{align*}
\ev^\bp\braks{e^{i\angles{X, x^*}_E + i\anglesn{X', y^*}_E}} &= \ev^\bp\braks{e^{i\angles{X, x^*}_E}}\ev^\bp\braks{e^{i\anglesn{X', y^*}_E}} \\
&= \exp\braks{-\frac{1}{2}\paren{\ev^\bp\braks{\angles{X, x^*}_E^2} + \ev^\bp\braks{\angles{X', y^*}_E^2}}}
\end{align*}
On the other hand,
\begin{align*}
&\ev^\bp\braks{e^{i\anglesn{(X + X')/\sqrt{2}, x^*}_E + i\anglesn{(X - X')/\sqrt{2}, y^*}_E}} \\
&= \ev^\bp\braks{e^{i\anglesn{X/\sqrt{2}, x^* + y^*}_E}}\ev^\bp\braks{e^{i\anglesn{X'/\sqrt{2}, x^* - y^*}_E}} \\
&= \exp\braks{-\frac{1}{4}\paren{\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2}}}
\end{align*}
where
\begin{align*}
\angles{X, x^* + y^*}_E^2 &= \angles{X, x^*}_E^2 + \angles{X, y^*}_E^2 + 2\angles{X, x^*}_E\angles{X, y^*}_E \\
\angles{X', x^* - y^*}_E^2 &= \angles{X', x^*}_E^2 + \angles{X', y^*}_E^2 - 2\angles{X', x^*}_E\angles{X', y^*}_E
\end{align*}
Since $X$ and $X'$ are identically distributed,
\begin{align*}
&\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2} \\
&= \ev^\bp\braks{\angles{X, x^*}_E^2 + \angles{X', x^*}_E^2} + \ev^\bp\braks{\angles{X, y^*}_E^2 + \angles{X', y^*}_E^2} \\
&+2\ev^\bp\braks{\angles{X, x^*}_E\angles{X, y^*}_E} - \ev^\bp\braks{2\angles{X', x^*}_E\angles{X', y^*}_E} \\
&= 2\ev^\bp\braks{\angles{X, x^*}_E^2} + 2\ev^\bp\braks{\angles{X', y^*}_E^2}
\end{align*}
Thus the characteristic functions of $(X, X')$ and $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ are equal. By \autoref{def:charfunc}, their distributions are equal.
\end{proof}
\begin{theorem}[{{Fernique, \cite[Theorem 8.2.1]{stroock}}}]
\label{thm:fernique}
Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X: \Omega \to E$ be a centred Gaussian random variable. Let $R > 0$ such that $\bp\bracs{\norm{X}_E \le R} \ge 9/10$, then
\[
\ev^\bp\braks{e^{\frac{\norm{X}_E^2}{18R^2}}} \le e^{1/2} + \sum_{n = 0}^\infty \paren{\frac{e}{3}}^{2^n}
\]
\end{theorem}
\begin{proof}
By enlarging the sample space, let $X': \Omega \to E$ be independent from $X$ and with the same distribution as $X$. For any $0 \le s \le t < \infty$,
\begin{align*}
&\bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t} \\
&= \bp\bracs{\norm{X}_E \le s}\bp\bracsn{\normn{X'}_E \ge t} = \bp\bracs{\norm{X}_E \le s, \normn{X'}_E \ge t} \\
&= \bp\bracs{\norm{X - X'}_E \le \sqrt{2}s, \normn{X + X'}_E \ge \sqrt{2}t} \\
&\le \bp\bracs{\abs{\norm{X}_E - \normn{X'}_E} \le \sqrt{2}s, \norm{X}_E + \normn{X'}_E \ge \sqrt{2}t}
\end{align*}
Here, if $\abs{\norm{X}_E - \normn{X'}_E} \le \sqrt{2}s$, then
\[
\sqrt{2}t \le \norm{X}_E + \norm{X'}_E \le 2(\norm{X}_E \wedge \norm{X'}_E) + \sqrt{2}s
\]
Thus $\norm{X}_E + \norm{X'}_E \ge \sqrt{2}(t - s)$, so
\begin{align*}
\bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t} &\le \bp\bracs{\norm{X}_E \wedge \norm{X'}_E \ge (t - s)/\sqrt{2}} \\
&= \bp\bracs{\norm{X}_E \ge (t - s)/\sqrt{2}}^2
\end{align*}
Let $t_0 = R$, and $t_n = R + \sqrt{2}t_{n - 1}$ for all $n \in \nat$. From here, for each $n \in \nat$,
\begin{align*}
\bp\bracs{\norm{X}_E \le R}\bp\bracs{\norm{X}_E \ge t_n} &\le \bp\bracs{\norm{X}_E \ge (t_n - R)/\sqrt{2}}^2 \\
&= \bp\bracs{\norm{X}_E \ge t_{n-1}}^2 \\
\frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}} &\le \braks{\frac{\bp\bracs{\norm{X}_E \ge t_{n - 1}}}{\bp\bracs{\norm{X}_E \le R}}}^2
\end{align*}
Thus
\begin{align*}
\frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}} &\le \braks{\frac{\bp\bracs{\norm{X}_E \ge R}}{\bp\bracs{\norm{X}_E \le R}}}^{2^n} \le 9^{-2^n} \\
\bp\bracs{\norm{X}_E \ge t_n} &\le 9^{-2^{n}} \le 3^{-2^{n+1}}
\end{align*}
so for any $K > 0$,
\begin{align*}
\ev^\bp\braks{e^{\norm{X}_E^2/K}} &\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K}\bp\bracs{\norm{X}_E \ge t_n} \\
&\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K} \cdot 9^{-2^n}\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K} \cdot 3^{-2^{n+1}}
\end{align*}
Now, since
\begin{align*}
t_N &= R + \sqrt{2}t_{N - 1} = R + \sqrt{2}(R + \sqrt{2}t_{N - 2}) \\
&= R + \sqrt{2}R + \sqrt{2}^2(R + \sqrt{2}t_{N - 3}) = R\sum_{n = 0}^{N}\sqrt{2}^n = \frac{R(1 - \sqrt{2}^{N+1})}{1 - \sqrt{2}} \\
&\le 3R\sqrt{2}^{N+1} = 3R \cdot 2^{(N+1)/2}
\end{align*}
so if $K = 18R^2$,
\[
\frac{t^2_{n+1}}{K} \le \frac{9R^2 \cdot 2^{(N+2)}}{18R^2} = 2^{N+1}
\]
Thus
\[
\ev^\bp\braks{e^{\norm{X}_E^2/K}} \le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{2^{n+1}} \cdot 3^{-2^{n+1}} \le e^{1/2} + \sum_{n = 0}^\infty \paren{\frac{e}{3}}^{2^n}
\]
\end{proof}

10
src/aws/index.tex Normal file
View File

@@ -0,0 +1,10 @@
\chapter{Abstract Wiener Spaces}
\label{chap:aws}
\input{./classical.tex}
\input{./gaussian.tex}
\input{./aws.tex}
\input{./cm.tex}
\input{./wiener.tex}
\input{./donsker.tex}

57
src/aws/wiener.tex Normal file
View File

@@ -0,0 +1,57 @@
\section{Wiener's Construction}
\label{section:wiener-construction}
\begin{theorem}[{{\cite[Theorem 8.3.1]{stroock}}}]
\label{thm:wiener-consurtcion}
Let $H$ be an infinite-dimensional, separable Hilbert space over $\real$, $E$ be a Banach space with $H$ continuously embedded as a dense subspace.
Let $\seq{h_n} \subset H$ be an orthonormal basis and $\seq{X_n}$ be mutually independent $\gamma_{0, 1}$-distributed random variables on a probability space $(\Omega, \cf, \bp)$. If
\begin{enumerate}
\item $\sum_{n = 1}^\infty X_n h_n$ converges in $E$ almost surely.
\item $S: \Omega \to E$ is given by
\[
S(\omega) = \begin{cases}
\sum_{n = 1}^\infty X_n(\omega) h_n &\text{The series converges in } $E$ \\
0 &\text{Otherwise}
\end{cases}
\]
\item $\wien = S_* \bp$.
\end{enumerate}
then $(H, E, \wien)$ is an abstract Wiener space.
\end{theorem}
\begin{proof}
Let $S_n = \sum_{k = 1}^n X_kh_k$, then
\[
\wh \wien(x^*) = \limv{n}\ev^\bp\braks{e^{\dpb{S_n, x^*}{E}}} = \limv_{n}\prod_{k = 1}^n e^{-\dpb{\iota^*x^*, h_k}{H}^2/2} = e^{-\norm{\iota^*x^*}_H^2/2}
\]
\end{proof}
\begin{theorem}
\label{thm:wiener-structure}
Let $(H, E, \wien)$ be an abstract Wiener space and $\seq{h_n}$ be an orthonormal sequence, then
\begin{enumerate}
\item For any $p \in [1, \infty)$,
\[
\ev^\wien\braks{\sup_{n \in \nat}\norm{\sum_{k = 1}^n \ci(h_k)h_k}_E^p} < \infty
\]
\item For $\wien$-almost every $x \in E$,
\[
\sum_{n = 1}^\infty \ci(h_k)(x)h_k = \ev^\wien\braks{x|\sigma(\bracs{\ci(h_n): n \in \nat})}
\]
\item $\sum_{n = 1}^\infty \ci(h_n)h_n$ is $\wien$-independent of $x - \sum_{n = 1}^\infty \ci(h_n)h_m$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $n \in \nat$, $\cf_n = \sigma(\bracs{\ci(h_k): 1 \le k \le n})$, and $S_n = \sum_{k = 1}^n \ci(h_k)h_k$, then for any $x^* \in E^*$, $\dpb{x - S_n(x), x^*}{E} \perp h_k$ in $L^2(\wien; E)$ for all $k \le n$. Since $\bracs{\ci(h_n): n \in \nat}$ is a Gaussian family, $x - S_n(x)$ is independent of $\seqf{h_k}$ and $\cf_n$.
(1): By \autoref{thm:fernique}, $x \in L^p(\wien; E)$ for all $p \in [1, \infty)$, so (1) holds.
(2): $S_n = \ev^\wien[x|\cf_n]$, and $S_n \to \ev^\wien[x|\sigma(\bracs{\ci(h_n): n \in \nat})]$ $\wien$-almost surely.
(3): Since $x - S_n(x)$ is independent of $\seq{h_k}$ for each $n \in \nat$, $x - S$ is independent of $S$.
\end{proof}

1
src/index.tex
View File

@@ -1,6 +1,7 @@
\part{Diffusion Processes}
\label{part:diffusion}
\input{./aws/index.tex}
\input{./operator/index.tex}
\input{./diffusion/index.tex}
\input{./calculus/index.tex}