> [!definition] > > Let $(\Omega, \cf, \bracs{\cf_t}, \bp)$ be a [[Filtration|filtered]] [[Probability|probability]] space, and $B_t$ be the standard [[Brownian Motion]], adapted to $\bracs{\cf_t}$. Define > - $\Phi^2$ as the family of [[Stochastic Process|stochastic processes]] on $\Omega$ with [[RCLL]] sample paths, progressively measurable with respect to $\bracs{\cf_t}$, and $\ev\braks{\int_{[0, \infty]}\varphi^2_rdr} < \infty$. > - $S\Phi^2$ be the simple process in $\Phi^2$, where there exists $n \ge 1$ such that $\varphi \in S\Phi^2$ if and only if > - For all $\omega \in \Omega$, $\varphi_t(\omega) = \varphi_{\fl{t}_n}(\omega)$ for all $t \ge 0$, where $\fl{t}_n$ is the $n$-th dyadic floor. > - $\varphi_t$ is uniformly bounded $\sup_{\omega \in \Omega}\varphi(\omega)< \infty$. > > > then $S\Phi^2$ is dense in $\Phi^2$ under > $ > \norm{\varphi}_2^2 = \ev\braks{\int\varphi_r^2dr} > $ > *Proof*. Let $\varphi \in \Phi^2$, $k > 0$, and $M \ge 0$. Define > $ > \varphi_t^{k, M}= \begin{cases} > 0 & t > M \text{ or } \abs{\varphi_t} > k \\ > \varphi_t &\text{otherwise} > \end{cases} > $ > so $\varphi^{k, M} \to \varphi$ as $k, M \to \infty$. For each $n \in \nat$, define $\varphi_t^{n} = \varphi_{\fl{t}_n}$, then $\varphi_{t}^n \to \varphi_{t}$ if $t$ is a continuity point (which happens a.e.). Therefore $(\varphi^{k, M})^{n} \to \varphi^{k, M}$ as $n \to \infty$. # Simple Processes > [!theorem] > > Let $\varphi \in S\Phi^2$. Since $\varphi$ is piecewise constant, $\varphi \in BV_{\text{loc}}$, then > $ > \mathcal I(\varphi)_t(\omega) = \int_0^t \varphi_r(\omega) dB_r(\omega) > $ > is a well-defined Riemann-Stieltjes integral, and $\mathcal I(\varphi)$ is known as Itô's integral. Firstly, $\mathcal I(\varphi)_0 = 0$. Moreover, there exists $n \ge 0$ such that $\varphi_t = \varphi_{\fl{t}_n}$, so if $t \in [(m-1)2^{-n}, m2^{-n})$, then > $ > \mathcal I(\varphi)_t = \sum_{m = 1}^{m-2}\varphi_{j2^{-n}}(B_{(j+1)2^{-n}} - B_{j2^{-n}}) + \varphi_{(m-1)2^{-n}}(B_t - B_{(m-1)2^{-n}}) > $ > so $\mathcal I(\varphi)$ has continuous sample paths, and progressively measurable with respect to $\bracs{\cf_t}$. > [!theorem] > > Let > $ > E\varphi_t = \exp\braks{\mathcal I(\varphi)_t - \frac{1}{2}\int_0^t\varphi_r^2dr} > $ > then $E\varphi$ is a progressively measurable [[Martingale|martingale]]. > > *Proof*. Let $n \in \nat$ such that $\varphi_t = \varphi_{\fl{t}_t}$, $m \in \nat$, and $0 \le s \le t$ such that $s, t \in [(m-1)2^{-n}, m2^{-n})$, then > $ > \begin{align*} > \ev\braks{\exp\paren{\mathcal I(\varphi)_t}|\cf_s} &= \exp\braks{\mathcal I(\varphi)_{\fl{s}_n} - \varphi_{\fl{s}_n}(B_s - B_{\fl{s}_n})|\cf_s} \\ > &\times \ev\braks{\varphi_{\fl{s}_n}(B_t - B_{\fl{s}_n})|\cf_s} \\ > &= \exp\braks{\mathcal I(\varphi)_{\fl{s}_n} - \varphi_{\fl{s}_n}(B_s - B_{\fl{s}_n})|\cf_s} \\ > &\times \ev\braks{\varphi_{\fl{s}_n}(B_t - B_s + B_s - B_{\fl{s}_n})|\cf_s} \\ > &= \exp\braks{\mathcal I(\varphi)_{\fl{s}_n}} \cdot \exp\paren{\frac{1}{2}\varphi^2_{\fl{s}_n}(t - s)} \\ > &= \exp\braks{\mathcal I(\varphi)_{\fl{s}_n}} \cdot \exp\paren{\frac{1}{2}\int_s^t\varphi^2_rdr} \\ > \ev\braks{E\varphi_t|\cf_t} &= E\varphi_s > \end{align*} > $ > [!theorem] > > The process > $ > \mathcal I^2(\varphi)_t - \int_0^t \varphi_r^{2}dr > $ > is a martingale, so $\angles{\mathcal I(\varphi)}_t = \int_0^t \varphi_r^2dr$. For any $\mathcal I(\varphi), \mathcal I(\psi)$, $\int_0^t \varphi \psi dr$. # All Processes > [!definition] > > For any $\varphi \in S\Phi^2$, by [[Martingale Convergence Theorems|MCT (III)]], there exists $\mathcal I(\varphi)_\infty$ where $\mathcal I(\varphi)_t \to \mathcal I(\varphi)_\infty$ almost surely and in $L^2$. The mapping $\varphi \mapsto \mathcal I(\varphi)_\infty$ is an isometry where > $ > \angles{\mathcal I(\varphi)_t, \mathcal I(\psi)_t} = \angles{\varphi, \psi}_{[0, t]} = \ev\braks{\angles{\varphi(\omega), \psi(\omega)}_{[0, t]}} > $ > which admits a [[Linear Extension Theorem|unique extension]] to $\Phi^2$. Thus $\mathcal I(\varphi)_t$ exists for all $\varphi \in \Phi^2$ and $t \ge 0$, and is known as Itô's integral of $\varphi$. > [!theorem] > > For any $\varphi \in \Phi^2$, $\bracs{\mathcal I(\varphi)_t}$ and $\{\mathcal I(\varphi)_t^2 - \int_{[0, t]}\varphi^2\}$ are both martingales. > > *Proof*. Let $\{\varphi^{(n)}\}_1^\infty \subset S\Phi^2$ such that $\varphi^{(n)} \to \varphi$. Assume without loss of generality that $\mathcal I(\varphi^{(n)})_t \to \mathcal I(\varphi)_t$ and $\mathcal I(\varphi^{(n)})^2 \to \mathcal I(\varphi)_t$ [[Almost Everywhere|a.s.]] and in $L^1$, and $\int_{[0, t]}[\varphi^{(n)}]^2 \to \int_{[0, t]}\varphi^2$ almost surely. > > Note that > $ > \ev\braks{\int_{[0, t]}[\varphi^{(n)}]^2} \to \ev\braks{\int_{[0, t]}\varphi^2} > $ > since the random variables are non-negative, the convergence is in $L^1$. > > Thus all components of the processes converge in $L^1$, which implies the convergence of conditional expectations. > [!theorem] > > For any $\varphi \in \Phi^2$, $\bracs{\mathcal I(\varphi)_t}$ has continuous sample paths. > > *Proof*. Let $\{\varphi^{(n)}\}_1^\infty \subset S\Phi^2$ such that $\varphi^{(n)} \to \varphi$. For each $m, n \in \nat$, > $ > \bracs{\mathcal I(\varphi^{(m)})_t - \mathcal I(\varphi^{(n)})_t: t \ge 0} > $ > is a martingale. By [[Doob's Maximal Inequality]], > $ > \begin{align*} > &\bp\bracs{\norm{\mathcal I(\varphi^{(m)}) - \mathcal I(\varphi^{(n)})}_{u, [0, T]} > \eps} \\ > &\le \frac{1}{\eps^2}\ev\braks{\norm{\mathcal I(\varphi^{(m)})_T - \mathcal I(\varphi^{(n)})_T}_2^2} \\ > &\le \frac{1}{\eps^2} \ev\braks{\norm{\mathcal I(\varphi^{(m)} - \varphi^{(n)})_T}_2^2} \\ > &\le \frac{1}{\eps^2}\norm{\varphi^{(m)} - \varphi^{(n)}}_{\Phi^2}^2 > \end{align*} > $ > So there exists a subsequence such that $\norm{\varphi^{(n_{k+1})} - \varphi^{(n_k)}}_{\Phi^2}^2 \le 2^{-k}$. Choose $\eps = 2^{-k/4}$ yields the desired uniform convergence. > [!theorem] > > Let $\varphi \in \Phi^2$ and > $ > E\varphi_t = \exp\braks{\mathcal I(\varphi)_t - \frac{1}{2}\int_0^t\varphi_r^2dr} > $ > then $\{E\varphi_t\}$ is a supermartingale. If for every $t \ge 0$, there exists $C_{\varphi, t} \ge 0$ such that $\int_0^t \varphi^2(\omega) \le C_{\varphi, t}$ for all $\omega \in \Omega$, then $E\varphi_t$ is a martingale. > > *Proof*. Let $\{\varphi^{(n)}\}_1^\infty \subset S\Phi^2$ such that $\varphi^{(n)} \to \varphi$, then $\mathcal I(\varphi^{(n)})_t \to \mathcal I(\varphi)_t$ and $\angles{\mathcal I(\varphi^{(n)})}_t \to \angles{\mathcal I(\varphi)}_t$ a.s. and in $L^1$. Therefore $E\varphi^{(n)}_t \to E\varphi_t$ in probability. > > Let $0 \le s \le t$, and assume without loss of generality that $E\varphi_s^{(n)} \to E\varphi_s$ and $E\varphi_t^{(n)} \to E\varphi_t$ almost surely as well. > $ > \ev\braks{E\varphi_t|\cf_s} = \ev\braks{\limv{n}E\varphi^{(n)}_t|\cf_s} \le \liminf_{n \to \infty }\ev\braks{E\varphi^{(n)}_t|\cf_s} = E\varphi_s > $ > > Suppose that there exists $C_{\varphi} \ge 0$ such that $\abs{\varphi_r(\omega)} \le C_\varphi$ for all $r \ge 0$ and $\omega \in \Omega$. Under the same conditions as before, > $ > \begin{align*} > \ev\braks{(E\varphi^{(n)}_t)^2} &\le \ev\braks{\exp\braks{2\mathcal I(\varphi)_t - \int_0^t(\varphi^{(n)})^2}} \\ > &= \ev\braks{E(2\varphi^{(n)})_t \cdot \exp\braks{\int_{[0, t]}(\varphi^{(n)})^2}} \\ > &\le e^{tC_\varphi}\ev(E(2\varphi)) = e^{tC_\varphi^2} > \end{align*} > $ > Since $\bracs{E\varphi_t^{(n)}: k \ge 1}$ is uniformly integrable, $E\varphi_t^{(n)} \to E\varphi_t$ in $L^1$. A similar argument holds for $E\varphi_s^{(n)} \to E\varphi_s$ in $L^1$. > > Lastly, let $C_{\varphi, t}, C_{\varphi, s} \ge 0$. Define $\varphi_t^M = \one_{\bracs{\abs{\varphi_t} \le M}}$, then $\{E\varphi^{M}_t\}$ is a martingale. Since $\int_{[0, t]}(\varphi^M)^2 \upto \int_{[0, t]}\varphi^2$ by the monotone convergence theorem, and $\mathcal I(\varphi^M)_t \to \mathcal I(\varphi)_t$ in $L^2$. From here, $E\varphi^M \to E\varphi$ in probability as $M \to \infty$. Assume without loss of generality that $E\varphi^M \to E\varphi$ almost surely by taking a subsequence. From here, using the same steps as before and applying the bound yields the desired result.