$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ # Initial Interface We assume a complete filtration. >[!theorem] Quadratic Variation Uniqueness and Existence >Let $M$ be a CLM. There exists an **increasing process** called the ==**Quadratic Variation**== $\braket{M, M}$, unique up to indistinguishability, such that $M_t^2 - \braket{M,M}_t$ is a CLM. >[!theorem] QV is Variance additivity along a mesh a.s. >For every $t > 0$, pick a mesh. > $\braket{M,M}_t = \lim\sum \Delta M^2$ >>[!idea]- Expand >>$\braket{M,M}_t = \lim_{n\to \infty} \sum_{i = 1}^{p_n} \left(M_{t^n_i} - M_{t^n_{i-1}}\right)^2$ >>[[types of convergence|in probability]]. >[!example] Brownian Motion >$\braket{B,B}_t = t$. Uniqueness is clear because FVP $\cap$ CLM = $\{0\}$. The proof of existence and the characterization follow from elementary but somewhat involved proofs. Here are the main steps: 1. Consider a bounded $M$ and find a mesh on $[0,K]$. Then, define bounded martingale $X^n_t = \sum_1^{p_n} M_{t^n_{i-1}} \left(M_{t^n_i\land t} - M_{t^n_{i-1}\land t}\right)$. 2. Prove this lemma (big bash): $\lim_{n,m\to \infty} \EE\left[\left(X^n_K - X^m_K\right)^2\right] = 0$. No external lemmas are used. 3. By Doob's inequality in $L^2$, $\lim_{n,m\to \infty} \EE\left[ \left(\sup_{0\leq t\leq K} (X^n_t - X^m_t)\right)^2\right]\to 0$. We then obtain a pointwise AS-convergent sequence, and extract AS-continuous sample paths via uniform convergence. 4. Finish. # Basic Properties - Let $M$ be a CLM and $T$ a stopping time. Then a.s. for all $t\geq 0$, $\braket{M^T, M^T}_t = \braket{M,M}_{t\land T}$. - If $M$ is a CLM and $M_0 = 0$, then $\braket{M, M} = 0$ iff $M = 0$. Indeed, if $M_t^2$ is a nonnegative CLM then [[Continuous Local Martingales|it is a supermartingale]], thus $\EE[M_t^2] \leq \EE[M_0^2] = 0$ and $M_t = 0$ a.s. for all $t$. >[!idea] $L^2$ is strong >The following bounds are really good, and occur a lot in practice. The moment you shield $\braket{M,M}$, everything becomes regular. > [!claim] $L^2$ bounds > Suppose $M$ is a CLM with $M_0 \in L^2$. > - $M$ is a true martingale bounded in $L^2$ (we say $M\in \Hh$) iff $\EE[\braket{M,M}_\infty] < \infty$. This also implies $M^2 - \braket{M,M}$ is a UI martingale. > - $M$ is a true martingale such that $M_t \in L^2$ for all $t\geq 0$ iff $\EE[\braket{M,M}_t] < \infty$ for every $t\geq 0$. This also implies $M^2 - \braket{M, M}$ is a martingale. > [!proof]- Forward, part 1 > Suppose $M$ is a true martingale bounded in $L^2$, i.e. $\sup_{t\geq 0} \EE[M_t^2] = \frac{C}{4} < \infty$. By [[Discrete Doob's Lp Inequalities|Doob's Lp inequality]], $\EE[\sup_{t\geq 0} M_t^2] \leq C$ which is really good. > >>[!idea] >>Now, if $M^2 - \braket{M,M}$ were a martingale, we would be done, because >>$\EE\left[\braket{M,M}_t\right] = \EE[M^2_t] \leq \EE[\sup_s M_s^2] \leq C$ and we can use monotone convergence because $\braket{M,M}$ is IP. > > So we shield with $S_n = \inf\{t\geq 0: \braket{M,M}_t\geq n\}$. The CLM $(M^2)^{S_n} - \braket{M,M}^{S_n}$ is dominated by $\sup M_s^2 + n$, thus$\EE[\braket{M,M}^{S_n}_t] \leq C$as desired. If we take $n$ to $\infty$, this recovers $\EE[\braket{M,M}_t]\leq C$, and if we take $t\to \infty$ we conclude by monotone convergence. The other directions are more of the same flavor of proof, omitted. >[!problem] Lol >Prove part 2 using part 1. (No calculations!) >[!solution]- >Consider $M^t_s = M_{s\land t}$. A problem from your 676 midterm: >[!claim] Exact characterization of $M_\infty$ existence >Let $M$ be a CLM. The following two sets are AS equal: >$ > \left\{\lim_{t\to\infty} M_t \text{ exists and is finite}\right\},\qquad \{\braket{M,M}_\infty < \infty\} >$ # Important Lemma This lemma is so important for calculations; it's what makes our [[Riemann Summation Shorthand]] intuition tick. I will write out the full proof because I was confused about it for a long time. > [!claim] The measures actually work > Give me a mesh. Let $\mu_n$ be the [[Random Measure]] on $[0,t]$ which assigns weight $\left(X_{t_i} - X_{t_{i-1}}\right)^2$ to $\{t_{i-1}\}$, i.e. $\mu_n = \sum_i \left(X_{t_i} - X_{t_{i-1}}\right)^2 \delta_{\{t_{i-1}\}}.$ > The $\mu_n$ converge [[Random Measure|weakly in probability]] to $\mu$, the measure corresponding to $\braket{X,X}$. >[!proof] Diagonalization > Start by letting $D_n$ be the dyadics such that $D = \bigcup D_n$ is countable. By definition, for any particular $s\in D$, $\mu_n\left([0,s]\right)\to \braket{X,X}_s$ in probability. Often, we will desire that this holds for all $s$ at once. This is precisely what we're going to show here. > > By diagonalization, we can extract a subsequence such that, for all $s\in D$, $\PP\left(\mu_n([0,s])\to \braket{X,X}_s\right) = 1$. This follows because any $\PP$-convergent sequence admits an AS-convergent subsequence. > > However, then, $\PP\left( \bigcap_{s\in D} \left\{\mu_n([0,s])\to \braket{X,X}_s\right\}\right) = 1$. >[!proof] Some quick analysis > Hence, let $F_n$ be the CDF of $\mu_n$. $\PP\left(\bigcap_{s\in D} F_n(s)\to \braket{X,X}_s\right) = 1$. Because $F(s) = \braket{X,X}_s$ is path-continuous, on almost-sure $\omega$, for any $\eps > 0$, we can always find $\alpha < s < \beta$ inside $D$ such that $F(s) - \eps/2 < F(\alpha) \leq F(s) \leq F(\beta) < F(s) + \eps/2$. Then, there will exist an $n$ such that on all $m\geq n$, $F(s) - \eps < F_m(\alpha) < F_m(s) < F_m(\beta) < F(s) + \eps$. > >>[!idea] >>This could still have worked if $F(s)$ was merely taken to be right-continuous; the definition of weak convergence only requires this convergence to hold on every $s$ such that $F(s)$ is left-continuous. >[!proof] Using the [[types of convergence|$\PP$ to AS boostrap]] > Hence, we've shown that there is a subsequence of our original mesh along which $\mu_n\to \mu$ weakly, almost surely. > > But nothing about our original mesh was special. This is true for any subsequence of our original mesh, thus we obtain that $\mu_n\to \mu$ weakly in probability. >[!idea] Reminding myself how to use this > Along our subsequence, for almost all $\omega$, for continuous (and hence bounded on compact $[0,t]$) $g(s)\in C_b$, $\int g(s) d\mu_n(s)\to \int g(s) d\mu(s)$. Notably, for any path-continuous random processes $g_s(\omega)$, $\int g(s) d\mu_n(s)\to \int g(s)d\mu(s)$ almost surely. This $Z_n = \int g(s)d\mu_n(s)$ is just some random variable. > > Hence, for any path-continuous random process $g_s(\omega)$, along any subsequence of our mesh, there is some AS-convergent subsequence of $Z_ns. Thus, for all path-continuous random processes $g_s(\omega)$, $\int g(s)d\mu_n(s)\to \int g(s)d\mu(s)$ in probability.