> [!definition] > > Let $\phi \in BV_{\text{loc}}([0, \infty))$ and $\psi \in C([0, \infty))$, then $\psi$ is **Riemann-Stieltjes** integrable with respect to $\phi$ if for any finite partition $\mathcal P = \list{t}{L}$ of $[0, t]$, > $ > \sum_{j = 1}^n\psi(t_j^*)[\psi(t_j) - \psi(t_{j - 1})] \to \int_0^t \psi(r)d\phi(r) > $ > converges to the same value as the mesh of the partition goes to $0$, regardless of the choice of $t_j^* \in [t_j - t_{j - 1}]$, and the choice of partitions. The resulting limit is known as the **Riemann-Stieltjes** integral of $\psi$ with respect to $\phi$. > [!theorem] > > Suppose that $\psi$ is integrable with respect to $\phi$, and denote $\mathcal I(\psi, \varphi)_t = \int_0^t \psi(r)d\phi(r)$, then > 1. $t \mapsto \mathcal I(\psi, \varphi)_t$ has locally bounded variation, with $\norm{\mathcal I(\psi, \varphi)}_{\text{var}, [0, t]}$ bounded by $\norm{\psi}_{u, [0, t]}\cdot\norm{\varphi}_{\text{var}, [0, t]}$. > 2. If $\varphi$ is RCLL, so is $\mathcal I(\psi, \varphi)_t$. > 3. If $\varphi$ is continuous, so is $I(\psi, \varphi)_t$. > > Lastly, $\phi$ is integrable with respect to $\psi$, with > $ > \int_0^t \varphi d\psi = \varphi(t)\psi(t) - \varphi(0)\psi(0) - \int_0^t \psi d\varphi > $ > [!definition] > > Let $(\Omega, \cf, \bracs{\cf_t}, \bp)$ be a [[Filtration|filtered]] [[Probability|probability space]], $B_t$ be the standard [[Brownian Motion]], adapted to $\bracs{\cf_t}$, and $\varphi \in BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))$, then for any $\omega \in \Omega$ and $t \ge 0$, the Riemann-Stieltjes integral > $ > \mathcal I(\varphi)_t(\omega) = \int_0^t \varphi(r)dB_r > $ > is well-defined. The process $\mathcal I(\varphi)$ is known as the **Paley-Wiener integral** of $\varphi$. Fix $\omega \in \Omega$, then > 1. $t \mapsto \mathcal I(\varphi)_t(\omega)$ is continuous. > 2. $\mathcal I(\varphi)_0(\omega) = 0$, and $\mathcal I(\varphi)_t = \limv{n}\sum_{m = 1}^{2^n}\varphi((m-1)2^{-n}t)(B_{m2^{-n}t} - B_{(m-1)2^{-n}t})$. > [!theorem] > > As a process, $\mathcal I(\varphi)_t$ is progressively measurable with respect to $\bracs{\cf_t}$. Since $\varphi$ is continuous, $\phi \in L^2_{\text{loc}}([0, \infty))$. $\mathcal I(\varphi)_t$ is a Gaussian process with independent increments, where $\mathcal I(\varphi)_t - \mathcal I(\varphi)_s$ being Gaussian distributed with mean $0$ and variance $\int_{[s, t]}\varphi^2$. > > Thus $\mathcal I(\varphi)_t$ is a [[L2 Martingale|square integrable martingale]] with respect to $\bracs{\cf_t}$. > > *Proof*. > $ > \begin{align*} > \ev\braks{e^{i\xi \mathcal I(\varphi)_t}} &= \limv{n}\ev\braks{e^{i\xi\sum_{m = 1}^{2^n}\varphi((m-1)2^{-n}t)(B_{m2^{-n}t} - B_{(m-1)2^{-n}t})}} \\ > &= \limv{n}\prod_{m = 1}^{2^n}\ev\braks{e^{i\xi\varphi((m-1)2^{-n}t)(B_{m2^{-n}t}-B_{(m-1)2^{-n}t})}} \\ > &= \limv{n}\exp\braks{-\frac{\xi^2}{2}\sum_{m = 1}^{2^n}\varphi((m-1)2^{-n}t)^22^{-n}t} \\ > &= \exp\braks{{-\xi^2/2 \cdot \int_{[0, t]}\varphi^2}} > \end{align*} > $ > Similarly, if $0 \le s \le t$, > $ > \mathcal I(\varphi)_t - \mathcal I(\varphi)_s = \int_s^t \varphi(r) dB_r > $ > Performing the same manipulation yields that $\mathcal I(\varphi)_t -\mathcal I(\varphi)_s$ is independent from $\cf_s$, and is Gaussian distributed with mean $0$ and variance $\int_{[s, t]}\varphi^2$. > [!theorem] > > $\mathcal I(\varphi)_t$ is a Gaussian process with $\ev\braks{\mathcal I(\varphi)_t} = 0$ and > $ > \ev\braks{\mathcal I(\varphi)_t\mathcal I(\varphi)_s} = \int_{[0, s]} \varphi^2 > $ > *Proof*. > $ > \ev\braks{\mathcal I(\varphi)_t\mathcal I(\varphi)_s} = > \ev\braks{(\mathcal I(\varphi)_s + \mathcal I(\varphi)_t - \mathcal I(\varphi)_s)\mathcal I(\varphi)_s} > =\int_{[0, s]} \varphi^2 > $ > [!theorem] > > The set > $ > \bracs{\mathcal I(\varphi)_t: \varphi \in BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))} > $ > is a Gaussian family. That is, for any $\seqf{\varphi_j}$, $\seqf{\mathcal I(\varphi_j)_t}$ is a multivariate Gaussian, where > $ > \cov{\mathcal I(\varphi_1), \mathcal I(\varphi_2)} = \int_{[0, t]}\varphi_1\varphi_2 = \angles{\varphi_1, \varphi_2}_{[0, t]} > $ > > *Proof*. Let $\psi_1, \psi_2$ and $\varphi_1, \varphi_2 \in BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))$, > $ > \begin{align*} > \ev\braks{e^{i\xi_1 \mathcal I(\varphi_1)_t + i\xi_2\mathcal I(\varphi_2)}} &= \exp\braks{-\frac{\xi_1^2}{2}\int_{[0, t]} \varphi_1^2 - \frac{\xi_2^2}{2}\xi_2\int_{[0, t]} \varphi_2^2 - \xi_1\xi_2\int_{[0, t]} \varphi_1\varphi_2} > \end{align*} > $ > [!theorem] > > Let $\varphi \in BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))$, then the predictable quadratic variation > $ > \angles{\mathcal I(\varphi)}_t = \int_{[0, t]}\varphi^2 > $ > *Proof*. > $ > \begin{align*} > \ev\braks{\mathcal I(\varphi)_t^2|\cf_s} &= \ev\braks{(\mathcal I(\varphi)_s + \mathcal I(\varphi)_t - \mathcal I(\varphi)_s)^2|\cf_t} \\ > &= \mathcal I(\varphi)_s^2 - \int_{[0, s]}\varphi^2 + \int_{[0, t]}\varphi^2 > \end{align*} > $ > [!theorem] > > Let $\varphi_1, \varphi_2 \in BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))$, then > $ > \angles{\mathcal I(\varphi_1), \mathcal I(\varphi_2)}_t = \angles{\varphi_1, \varphi_2}_{[0, t]} > $ > [!definition] > > Fix $t \ge 0$. Identifying $BV_{\text{loc}}([0, \infty)) \cap C([0, \infty))$ as a subspace of $L^2([0, t])$, > $ > \mathcal I_t: BV_{\text{loc}}([0, \infty)) \cap C([0, \infty)) \to L^2(\bp) > $ > defined by $\varphi \mapsto I(\varphi)_t$, is an isometry, that is > $ > \angles{\mathcal I(\varphi_1)_t, \mathcal I(\varphi_2)_t}_{L^2(\bp)} = \angles{\varphi_1, \varphi_2}_{L^2([0, t])} > $ > and admits a [[Linear Extension Theorem|unique extension]] to $L^2([0, t])$ as an isometry, known as the **Paley-Wiener map**. > [!theorem] > > The family $\bracs{\mathcal I(\varphi)_t: \varphi \in L^2_{\text{loc}}}$ is Gaussian. > > *Proof*. Let $\varphi,\psi \in L^2_{\text{loc}}([0, \infty))$ with supporting sequence $\seq{\varphi_n}, \seq{\psi_n}$, then for any $0 \le r \le s \le t$ and $\xi_1, \xi_2 \in \real$, assuming a.s. convergence, > $ > \begin{align*} > &\ev\braks{\exp\paren{i\xi_1\mathcal I(\varphi)_r + i\xi_2\mathcal I(\psi)_s}} \\ > &= \limv{n}\ev\braks{\exp\paren{i\xi_1\mathcal I(\varphi_n)_r + i\xi_2\mathcal I(\psi_n)_s}} \\ > &= \exp\braks{-\frac{\xi_1^2}{2}\int_{[0, r]} \varphi^2 - \frac{\xi_2^2}{2}\xi_2\int_{[r, s]} \psi^2 - \xi_1\xi_2\int_{[s, t]} \varphi\psi} > \end{align*} > $ > [!theorem] > > For any $\varphi \in L^2_{\text{loc}}$, $\mathcal I(\varphi)_t$ has continuous sample paths. > > *Proof*. For any $n \ge 1$, $\bracs{\mathcal I(\varphi)_s: s \le t}$ is a martingale. Let $\eps > 0$, then by Doob's maximal inequality, > $ > \begin{align*} > \bp\bracs{\norm{\mathcal I(\varphi_m) - \mathcal I(\varphi_n)}_{u, [0, s]} > \eps} &\le \frac{1}{\eps^2} \norm{\mathcal I(\varphi_n)_t - \mathcal I(\varphi_m)_t}_2^2 > \end{align*} > $ > choosing a subsequence yields the desired uniform convergence. > [!theorem] > > If $\varphi \in L^2([0, \po{\infty)})$, then $\mathcal I(\varphi)$ is a $L^2$-bounded martingale, and converges to its limit in $L^2$ by [[Martingale Convergence Theorems|MCT (III)]]. If $\varphi \in L^2_{\text{loc}} \setminus L^2$, then $\mathcal I(\varphi)$ follows the behaviour of $B_{\int_{[0, t]}\varphi^2}$.