> [!theorem] > > Let $X \in L^2$ be a one-dimensional [[L2 Martingale|square integrable]] $\bracs{\cf_t}$-[[Martingale|martingale]] with continuous sample paths such that > 1. $X_0 = 0$ > 2. $\angles{X}$ has **strictly** increasing sample paths. > 3. $\angles{X}_t \to \infty$ as $t \to \infty$. > > then there exists a [[Filtration|filtration]] $\bracs{\mathcal G_t}$ and a [[Stochastic Process|stochastic process]] $B_t$, progressively measurable with respect to $\bracs{\mathcal G_t}$, such that > 4. $B_t$ is the standard [[Brownian Motion]]. > 5. $X_t = B_{\angles{X}_t}$. > > *Proof*. Since $\angles{X}$ has strictly increasing sample paths, each sample path is invertible. Let $Z(\omega)$ be the inverse of $\angles{Z}(\omega)$, then $Z(\omega)$ is strictly increasing and $Z(\omega)(t) \to \infty$ as $t \to \infty$. Moreover, if $0 \le s \le t$, > $ > \bracs{\omega \in \Omega: Z_s(\omega)\le t} = \bracs{\omega \in \Omega: \angles{X}(\omega)_t \le s} \in \cf_t > $ > So $Z_s$ is a stopping time with respect to $\bracs{\cf_t}$, and $\mathcal G_s = \cf_{Z_s}$ is a filtration. > > Let $\omega \in \Omega$, and define $B_s(\omega) = X_{Z_s(\omega)}(\omega)$, then $B_s$ is $\mathcal G_s$-measurable. Since for each $\omega \in \Omega$, $B_t$ has continuous sample paths and is progressively measurable with respect to $\mathcal G_s$. From here, it's sufficient to check that $B_s$ and $B_s^2 - s$ are martingales. > > Let $t \ge 0$, then > $ > B_s = X_{Z_s} = \limv{T}X_{T \wedge Z_s} > $ > and by the [[Monotone Convergence Theorem for Sequences|monotone convergence theorem]], > $ > \begin{align*} > \ev\braks{\sup_{0 \le t \le Z_s}X_t^2} &= \limv{T}\ev\braks{\sup_{0 \le t \le Z_s \wedge T}X_t^2} \\ > &\le \limv{T}\ev\braks{\sup_{0 \le t \le Z_s}X_{t \wedge T}^2} > \end{align*} > $ > since $X_{t \wedge T}^2$ is a submartingale, > $ > \begin{align*} > \ev\braks{\sup_{0 \le t \le Z_s}X_t^2} &= \limv{T}2\int_0^\infty r\bp\bracs{\sup_{0 \le t \le eZ_s}X_{t \wedge T}^2}dr \\ > &\le \limv{T}4\ev\braks{X_{T \wedge Z_s}^2} = \limv{T}4\ev\braks{\angles{X}_{T \wedge Z_s}} \\ > &\le 4s > \end{align*} > $ > by passing through [[Hunt's Theorem]]. Thus $B_s$ is square integrable, with > $ > \abs{X_{T \wedge Z_s} - B_s}^2 \le C\braks{\sup_{t \in [0, Z_s]}X_t^2 + B_s^2} \in L^1 > $ > By [[Dominated Convergence Theorem]], $X_{T \wedge Z_s} \to B_s$ as $T \to \infty$ in $L^2$. > > Let $A \in \mathcal G_s$, then for any $T \ge 0$, $A \cap \bracs{Z_s \le T} \in \cf_{T \wedge Z_s}$. By Hunt's Theorem, if $r \ge s$, > $ > \begin{align*} > \ev\braks{X_{Z_r \wedge T}| \cf_{Z_s \wedge T}} &= X_{Z_s \wedge T} \\ > \ev\braks{X_{Z_r \wedge T}^2 - \angles{X}_{Z_r \wedge T}| \cf_{Z_s \wedge T}} &= X_{Z_s \wedge T} - \angles{X}_{Z_s \wedge T} > \end{align*} > $ > so > $ > \begin{align*} > \ev\braks{X_{Z_r \wedge T}; A \cap \bracs{Z_s \le T}} &= \ev\braks{X_{Z_s \wedge T}; A \cap \bracs{Z_s \le T}} \\ > \ev\braks{X_{Z_r \wedge T}^2 - \angles{X}_{Z_r \wedge T}; A \cap \bracs{Z_s \le T}} &= \ev\braks{X_{Z_s \wedge T} - \angles{X}_{Z_s \wedge T}; A \cap \bracs{Z_s \le T}} > \end{align*} > $ > Sending $T \to \infty$ over the dominated convergence theorem yields that > $ > \begin{align*} > \ev\braks{B_r; A} &= \ev\braks{B_s ; A} \\ > \ev\braks{X_{Z_r \wedge T}^2 - \angles{X}_{Z_r \wedge T}; A} &= \ev\braks{X_{Z_s \wedge T} - \angles{X}_{Z_s \wedge T}; A} > \end{align*} > $ > which yields the desired conditional expectations. > [!theorem] > > If $X$ satisfies the requirements of the above theorem, then > $ > 1 = \limsup_{t \to \infty}\frac{X_t}{\sqrt{2\angles{X}_t\ln\ln\angles{X_t}}} = - \liminf_{t \to \infty}\frac{X_t}{\sqrt{2\angles{X}_t\ln\ln\angles{X_t}}} > $