> [!definition] > > Let $\bracs{X_t}$ be a one-dimensional $\bracs{\cf_t}$-[[Martingale|martingale]]. $X \in L^2$ if $X_t \in L^2$ for all $t \ge 0$. In which case, $X_t^2$ is a non-negative $\bracs{\cf_t}$-submartingale. > > By the [[Doob-Meyer Decomposition]], there exists a stochastic process $\angles{X}$ such that $X_t^2 - \angles{X}_t$ is a martingale, where $\angles{X}$ is previsible and has non-decreasing RCLL sample paths, known as the **predictable quadratic variation** of $X$. > [!theorem] > > Let $B_t$ be the standard [[Brownian Motion]] on $\real$, then $B_t^2 - t$ is a martingale, so $\angles{B}_t = t$ is the predictable quadratic variation of $B$. > [!theorem] > > If $X \in L^2$ is a one-dimensional square integrable martingale. Suppose that $X_0 = 0$ and $\angles{X}_t = t$, then $X$ is the standard Brownian motion. > [!theorem] > > Let $N_t$ be the [[Simple Poisson Jump Process|simple Poisson jump process]] with parameter $1$, then $N_t - t$ is a martingale. From here, let $s < t$, then > $ > \begin{align*} > \ev\braks{(N_t - t)^2|\cf_s} &= (N_s - s)^2 + (t - s) > \end{align*} > $ > so $(N_t - t)^2 - t$ is a martingale, and $\angles{N}_t = t$ is the predictable quadratic variation of $N_t$. > [!theorem] > > Let $\bracs{X_t}, \bracs{Y_t}$ be $L^2$, $\bracs{\cf_t}$-martingales with [[RCLL]] sample paths, then their linear combinations are also $L^2$ $\bracs{\cf_t}$martingales. Define > $ > \angles{X, Y}_t = \frac{1}{4}\paren{\angles{X + Y}_t - \angles{X - Y}_t} > $ > as their **predictable cross variation**. The mapping $\angles{\cdot, \cdot}$ is symmetric and bilinear on the space of $L^2$ martingales. > 1. $\angles{X, Y}_0 = 0$. > 2. $\angles{X, Y}$ is RCLL. > 3. $\angles{X, Y}$ has locally bounded variation. > 4. $\angles{X, Y}_t$ is $\cf_{t^-}$-measurable. > [!theorem] > > $X_tY_t - \angles{X, Y}_t$ is a martingale. > > *Proof*. > $ > \begin{align*} > \ev\braks{X_tY_t|\cf_s} &= \frac{1}{4}\ev\braks{(X_t + Y_t)^2 - (X_t - Y_t)^2|\cf_s} \\ > &= \frac{1}{4}\ev\braks{(X_t + Y_t)^2 - \angles{X + Y}_t - (X_t - Y_t)^2 + \angles{X - Y}_t|\cf_s} \\ > &+ \angles{X, Y}_t \\ > &= \frac{1}{4}(X_s + Y_s)^2 - \angles{X + Y}_s - (X_s - Y_s)^2 + \angles{X - Y}_s + \angles{X, Y}_t > \end{align*} > $