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Bokuan Li
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root = true
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end_of_line = lf

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name: Compile Project
run-name: Compile the project using spec.
on: [push]
jobs:
Compile:
runs-on: ubuntu-latest
steps:
- name: Check out repository code
uses: actions/checkout@v4
- name: Add to PATH
run: echo "/root/.nvm/versions/node/v24.14.0/bin" >> $GITHUB_PATH
- name: Copy source
run: |
mkdir -p /srv/builds/${{ github.event.repository.name }}
cp -r . /srv/builds/${{ github.event.repository.name }}/
- name: Compile Website
run: spec compile --all
working-directory: /srv/builds/${{ github.event.repository.name }}

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%\documentclass{report}
\usepackage{amssymb, amsmath, hyperref}
\usepackage{preamble}
\begin{document}
\input{./src/index}
%\input{./src/process/index}
\bibliographystyle{alpha} % We choose the "plain" reference style
\bibliography{refs.bib} % Entries are in the refs.bib file
\end{document}

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% \NeedsTeXFormat{LaTeX2e}
% \ProvidesPackage{jerrylicious}[Jerry's Tex Mess]
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@book{Diffusion,
title={Multidimensional Diffusion Processes},
author={Stroock, D.W. and Varadhan, S.R.S.},
isbn={9783540903536},
lccn={96027093},
series={Grundlehren der mathematischen Wissenschaften},
url={https://books.google.ca/books?id=DuDsmoyqCy4C},
year={1997},
publisher={Springer Berlin Heidelberg}
}

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database = "spec.db"
document = "document.tex"
siteTitle = "Unnamed Website"
[compiler]
compileAll = false
redoTags = false
indirectReferences = true
[website]
font = "roboto"
fontSize = 16
lineHeight = 1.3
textAlign = "left"
primaryColour = "blue"
neutralColour = "grey"
searchLimit = 16
maxSearchPages = 48
recentChanges = 10
tableOfContentsDepth = 2
advertiseSpec = true

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\section{Brownian Motion}
\label{section:brownian-motion}
\begin{theorem}
\label{theorem:brownian-martingales}
Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space and $\bracs{B_t|t \ge 0}$ be a $\real^d$-valued, $\bracs{\cf_t}$-progressively measurable process such that $B_0 = 0$ almost surely, then the following are equivalent:
\begin{enumerate}
\item $\bracs{B_t|t \ge 0}$ is a Brownian motion.
\item For each $\xi \in \real^d$, the process
\[
X^\theta_t = \exp\braks{i \dpn{B_t, \xi}{\real^d} + \frac{1}{2}\norm{\xi}_{\real^d}^2t}
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\item For each $\phi \in BC^2(\real^d)$, the process
\[
Y^\phi_t = \phi(B_t) - \frac{1}{2}\int_0^t \Delta f(B_s)ds
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 4.1.1]{Diffusion}}}. ]
(1) $\Rightarrow$ (2): For each $1 \le s \le t$ and $\xi \in \real^d$,
\begin{align*}
\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s} &= \ev\braks{e^{i \dpn{B_s, \xi}{\real^d}}e^{i \dpn{B_t - B_s, \xi}{\real^d}} \big| \cf_s} \\
&= e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}
\end{align*}
Therefore
\[
\ev\braks{e^{i \dpn{B_t, \xi}{\real^d} + t\norm{\xi}^2/2} \big| \cf_s} = e^{i \dpn{B_s, \xi}{\real^d} + s\norm{\xi}_{\real^d}^2/2}
\]
(2) $\Rightarrow$ (3): For each $0 \le s < t$ and $\xi \in \real^d$,
\begin{align*}
\frac{d}{dt}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s} &= -\frac{\norm{\xi}_{\real^d}^2}{2} e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}
\end{align*}
Therefore for any $0 \le s < t$,
\begin{align*}
\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s} &= -\frac{\norm{\xi}_{\real^d}^2}{2}\int_s^t e^{i \dpn{B_s, \xi}{\real^d}}e^{-(r - s)\norm{\xi}_{\real^d}^2/2}dr \\
&= -\frac{\norm{\xi}_{\real^d}^2}{2}\int_s^t \ev\braks{e^{i \dpn{B_r, \xi}{\real^d}} \big | \cf_{s}}dr
\end{align*}
If $\phi_\xi(x) = e^{-i \dpn{x, \xi}{\real^d}}$, then
\[
\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s} = \ev\braks{\int_s^t \frac{1}{2}(\Delta \phi_\xi)(B_r)dd \bigg | \cf_s}
\]
so the process
\[
Z^\xi_t = e^{i \dpn{B_t, \xi}{\real^d}} + \int_0^t \frac{1}{2}(\Delta \phi_\xi)(B_r)dr
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
Let $\phi \in \mathcal{S}(\real^d)$, then by the Fourier Inversion Theorem, there exists $\psi \in \mathcal{S}(\real^d)$ such that for any $x \in \real^d$,
\[
\phi(x) = \int e^{i \dpn{\xi, x}{\real^d}}\psi(\xi)d\xi
\]
In which case, for any $t \ge 0$,
\[
\phi(B_t) + \int_0^t \frac{1}{2}(\Delta \phi)(B_r)dr
= \int_{\real^d} e^{i \dpn{\xi, B_t}{\real^d}}\psi(\xi)d\xi + \int_{\real^d}\int_0^t \frac{1}{2}(\Delta \phi_\xi)(B_r) \psi(\xi) dr d\xi
\]
By the Dominated Convergence Theorem, the above also holds for any $\phi \in BC^2(\real^d)$. Therefore the process $\bracsn{Y^\phi_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale as well.
\end{proof}

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\chapter{Stochastic Calculus of Diffusion Processes}
\label{chap:stochastic-calculus}
\input{./brownian.tex}
\input{./martingale.tex}

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\section{Equivalence of Certain Martingales}
\label{section:martingale-equivalence}
% I made this one up.
\begin{definition}[Random Differential Operator]
\label{definition:random-differential-operator}
Let $(\Omega, \cf, \bp)$ be a filtered probability space, $s \ge 0$, and $k \in \natp$. For each multi-index $\alpha \in \nat_0^d$ with $\abs{\alpha} \le k$, let $a_\alpha: [s, \infty) \times \Omega \to \real$ be measurable function. For each $f \in C^k(\real^d)$ and $t \ge s$, let
\[
[L_t(\omega)f](x) = \sum_{\abs{\alpha} \le k} a_\alpha(t, \omega) \partial^\alpha f(x)
\]
then $L = \sum_{|\alpha| \le k}a_\alpha \partial^\alpha$ is a \textbf{random differential operator} of order $k$ with coefficients $\bracs{a_\alpha}_{|\alpha| \le k}$.
\end{definition}
\begin{definition}[Progressively Measurable Random Differential Operator]
\label{definition:random-do-progressively-measurable}
Let $(\Omega, \cf, \bp)$ be a filtered probability space and $L = \sum_{|\alpha| \le k}a_\alpha$ be a random differential operator on $\Omega$, then $L$ is \textbf{progressively measurable} with respect to $\bracs{\mathcal{F}_t}$ if for each $\alpha \in \nat_0$ with $|\alpha| \le k$, $a_\alpha: [s, \infty) \times \Omega \to \real$ is progressively measurable with respect to $\bracs{\mathcal{F}_t}$.
\end{definition}
\begin{lemma}
\label{lemma:random-diff-process}
Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, $s \ge 0$, $k \in \natp$, $L = \sum_{|\alpha| \le k}a_\alpha \partial^\alpha$ be a $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, $f \in C^k(\real^d)$, and $\bracs{X_t|t \ge 0}$ be a $\real^d$-valued progressively measurable process, then
\[
Y_t(\omega) = (L_t(\omega)f)(X_t(\omega))
\]
is a progressively measurable process.
\end{lemma}
\begin{theorem}[Integration by Parts]
\label{theorem:martingale-ibp}
Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, $\bracs{X_t}$ be a $\bracs{\mathcal{F}_t}$-martingale, and $\phi: [0, \infty) \times \Omega \to \complex$ be a continuous, progressively measurable function. If:
\begin{enumerate}
\item For every $\omega \in \Omega$, $\phi(\cdot, \omega) \in BV_{\text{loc}}([0, \infty))$.
\item For all $t \ge 0$,
\[
\ev\braks{\sup_{0 \le s \le t}|X_s|(|\phi|(s) + |\phi|(t))} < \infty
\]
\end{enumerate}
then the process
\[
Y_t = X_t \phi(t) - \int_0^t X_r \phi(dr)
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\end{theorem}
\begin{theorem}[Equivalence of Martingales]
\label{theorem:equivalence-of-martingales}
Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space,
\[
L_t(\omega)u = \frac{1}{2}\dpn{A_t(\omega), u}{\real^{d \times d}} + \dpn{B_t(\omega), u}{\real^d}
\]
be a second-order $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, and $\bracs{X_t|t \ge 0}$ be a $\bracs{\mathcal{F}_t}$-progressively measurable process with continuous sample paths, then the following are equivalent:
\begin{enumerate}
\item For every $f \in C_c^\infty(\real^d)$, the process
\[
E_t = f(X_t) - \int_0^t (L_rf)(X_r)dr
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\item For every $f \in C_b^{1, 2}([0, \infty) \times \real^d)$,
\[
H_t = f(t, X_t) - \int_0^t [(\partial_r + L_r)f](r, X_r)dr
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\item For each uniformly positive $f \in C_b^{1, 2}([0, \infty) \times \real^d)$,
\[
X_t = f(t, X_t)\exp\braks{-\int_0^t \frac{(\partial_r + L_r)f}{f}(r, X_r)dr}
\]
is a $\bracs{\mathcal{F}_t}$-martingale.
\item For each $x \in \real^d$ and $g \in C_b^{1, 2}([0, \infty) \times \real^d)$, the process
\begin{align*}
Y_t^{x, g} &= \exp\braks{\dpn{x, X_t - X_0 - \int_0^t b(r)dr}{\real^d} + g(t, X_t)} \\
&\cdot \exp\braks{-\frac{1}{2}\int_0^t \dpn{x + Dg, A(r)(x + Dg)}{\real^d}(r, X_r)dr} \\
&\cdot \exp\braks{\int_0^t [(\partial_r + L_r)g](r, X_r)}dr
\end{align*}
is a $\bracs{\mathcal{F}_t}$-martingale.
\item For each $\theta \in \real^d$,
\end{enumerate}
\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 4.2.1]{Diffusion}}}. ]
(1) $\Rightarrow$ (2): Let $0 \le s \le t$ and $f \in C_c^\infty([0, \infty) \times \real^d)$, then by the Fundamental Theorem of Calculus,
\begin{align*}
f(t, X_t) - f(s, X_s) &= f(t, X_t) - f(s, X_t) + f(s, X_t) - f(s, X_s) \\
&= \int_s^t \partial_rf(r, X_t)dr + f(s, X_t) - f(s, X_s)
\end{align*}
where by (1),
\[
\ev\braks{f(s, X_t) - f(s, X_s) |\cf_s} = \ev\braks{\int_s^t (L_rf)(s, X_r)dr \bigg | \cf_s}
\]
By the Fundamental Theorem of Calculus,
\begin{align*}
\int_s^t (\partial_rf)(r, X_t)dr &= \int_s^t (\partial_r f)(r, X_r)dr
+ \int_s^t (\partial_r f)(r, X_t) - (\partial_r f)(r, X_r)dr \\
&= \int_s^t (\partial_r f)(r, X_r)dr
+ \int_s^t (\partial_r f)(r, X_t) - (\partial_r f)(r, X_r)dr \\
\end{align*}
By (1) applied to $(\partial_r f)$,
\[
\ev\braks{\int_s^t (\partial_rf)(r, X_t)dr \bigg | \cf_s} =
\braks{\int_s^t (\partial_r f)(r, X_r)dr
+ \int_s^t \int_r^t (\partial_r f)(r, X_u)du dr \bigg | \cf_s}
\]
Similarly, by the Fundamental Theorem of Calculus,
\begin{align*}
\int_s^t (L_rf)(s, X_r)dr &= \int_s^t (L_rf)(r, X_r)dr
+ \int_s^t (L_rf)(s, X_r) - (L_rf)(r, X_r)dr \\
&= \int_s^t(L_rf)(r, X_r)dr - \int_s^t \int_s^r (\partial_rL_rf)(u, X_r)dudr
\end{align*}
Since the two iterated integrals are over the same region,
\[
\ev\braks{f(t, X_t) - f(s, X_s) | \cf_s} = \ev\braks{\int_s^t (\partial_r + L_rf)(r, X_r)dr \bigg | \cf_s}
\]
Therefore $\bracs{E_t}$ is a $\bracs{\mathcal{F}_t}$-martingale.
(2) $\Rightarrow$ (3): Let $f \in C_b^{1, 2}([0, \infty) \times \real^d)$ be uniformly positive. Define
\[
Z_t = f(t, X_t) - \int_0^t (\partial_rf + L_rf)(r, X_r)dr
\]
and
\[
\phi_t = \exp\braks{-\int_0^t \frac{\partial_rf + L_rf}{f}(r, X_r)dr}
\]
then using the by parts formula for Riemann-Stieltjes integrals,
\begin{align*}
Z_t\phi_t - \int_0^t Z_s \phi(ds) &= Z_0\phi_0 + \int_0^t \phi_s Z(ds) \\
&= Z_0\phi_0 + \int_0^t \phi_s f(ds, X_{ds}) + \int_0^t \phi_s \cdot (\partial_s f L_sf)(s, X_s)ds
\end{align*}
Since
\[
\partial_s \phi_s = -\phi_s \frac{\partial_s f + L_sf}{f}(s, X_s)
\]
Using the by parts formula again,
\begin{align*}
Z_t\phi_t - \int_0^t Z_s \phi(ds) &= Z_0\phi_0 + \int_0^t \phi_s f(ds, X_{ds}) + \int_0^t f(s, X_s)\phi(ds) \\
&= Z_0\phi_0 + f(t, X_t)\phi_t - f(0, X_0)\phi_0 = f(t, X_t)\phi_t \\
&= f(t, X_t)\exp\braks{-\int_0^t \frac{\partial_rf + L_rf}{f}(r, X_r)dr}
\end{align*}
Therefore \autoref{theorem:martingale-ibp} implies that the above process is a $\bracs{\mathcal{F}_t}$-martingale.
(3) $\Rightarrow$ (4): Let $x \in \real^d$ and $g \in C_b^{1, 2}([0, \infty) \times \real^d)$. Define
\[
f: [0, \infty) \times \real^d \to \real \quad (t, y) \mapsto \exp(\dpn{x, y}{\real^d} + g(t, x))
\]
then (3) applied to $f$ yields that $\bracs{Y_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale. However, since $f$ is unbounded and not uniformly positive, a direct application is ineffective.
Let $\seq{f_n} \subset C_b^{1, 2}([0, \infty) \times \real^d)$ be uniformly positive such that $f_n|_{\bracs{|y| \le n}} = f$. Define
\[
\tau_n = \inf\bracs{t \ge 0 \bigg | \sup_{0 \le s \le t}|X_t(u)| \ge n} \wedge n
\]
then $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is a $\bracs{\mathcal{F}_t}$-martingale with $Y_{t \wedge \tau_n}^{x, g} \to Y_{t}^{x, g}$ almost surely as $n \to \infty$. In addition,
\[
(Y_{t \wedge \tau_n}^{x, g})^2 \le Y_{t \wedge \tau_n}^{2x, 2g} \cdot \exp\braks{\int_0^{t \wedge \tau_n} \dpn{x + Dg, A_s(x + Dg) }{\real^d}(s, X_s)ds} \le CY_{t \wedge \tau_n}^{2x, 2g}
\]
where $\ev\braks{Y_{t \wedge \tau_n}^{2x, 2g}} = f(s, X_s)^2 \le \exp(2\norm{g}_u)$. Therefore $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is bounded in $L^2$, uniformly integrable in $L^1$, and converges to $Y_{t}^{x, g}$ in $L^1$.
\end{proof}

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\part{Diffusion Processes}
\label{part:diffusion}
\input{./diffusion/index.tex}