1.2 Equivalence of Certain Martingales

Definition 1.2.1 (Random Differential Operator). Let $(\Omega, \cf, \bp)$ be a filtered probability space, $s \ge 0$, and $k \in \natp$. For each multi-index $\alpha \in \nat_{0}^{d}$ with $\abs{\alpha}\le k$, let $a_{\alpha}: [s, \infty) \times \Omega \to \real$ be measurable function. For each $f \in C^{k}(\real^{d})$ and $t \ge s$, let

\[[L_{t}(\omega)f](x) = \sum_{\abs{\alpha} \le k}a_{\alpha}(t, \omega) \partial^{\alpha} f(x)\]

then $L = \sum_{|\alpha| \le k}a_{\alpha} \partial^{\alpha}$ is a random differential operator of order $k$ with coefficients $\bracs{a_\alpha}_{|\alpha| \le k}$.

Definition 1.2.2 (Progressively Measurable Random Differential Operator). Let $(\Omega, \cf, \bp)$ be a filtered probability space and $L = \sum_{|\alpha| \le k}a_{\alpha}$ be a random differential operator on $\Omega$, then $L$ is progressively measurable with respect to $\bracs{\mathcal{F}_t}$ if for each $\alpha \in \nat_{0}$ with $|\alpha| \le k$, $a_{\alpha}: [s, \infty) \times \Omega \to \real$ is progressively measurable with respect to $\bracs{\mathcal{F}_t}$.

Lemma 1.2.3. Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, $s \ge 0$, $k \in \natp$, $L = \sum_{|\alpha| \le k}a_{\alpha} \partial^{\alpha}$ be a $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, $f \in C^{k}(\real^{d})$, and $\bracs{X_t|t \ge 0}$ be a $\real^{d}$-valued progressively measurable process, then

\[Y_{t}(\omega) = (L_{t}(\omega)f)(X_{t}(\omega))\]

is a progressively measurable process.

Theorem 1.2.4 (Integration by Parts). Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, $\bracs{X_t}$ be a $\bracs{\mathcal{F}_t}$-martingale, and $\phi: [0, \infty) \times \Omega \to \complex$ be a continuous, progressively measurable function. If:

  1. For every $\omega \in \Omega$, $\phi(\cdot, \omega) \in BV_{\text{loc}}([0, \infty))$.

  2. For all $t \ge 0$,

    \[\ev\braks{\sup_{0 \le s \le t}|X_s|(|\phi|(s) + |\phi|(t))}< \infty\]

then the process

\[Y_{t} = X_{t} \phi(t) - \int_{0}^{t} X_{r} \phi(dr)\]

is a $\bracs{\mathcal{F}_t}$-martingale.

Theorem 1.2.5 (Equivalence of Martingales). Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space,

\[L_{t}(\omega)u = \frac{1}{2}\dpn{A_t(\omega), u}{\real^{d \times d}}+ \dpn{B_t(\omega), u}{\real^d}\]

be a second-order $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, and $\bracs{X_t|t \ge 0}$ be a $\bracs{\mathcal{F}_t}$-progressively measurable process with continuous sample paths, then the following are equivalent:

  1. For every $f \in C_{c}^{\infty}(\real^{d})$, the process

    \[E_{t} = f(X_{t}) - \int_{0}^{t} (L_{r}f)(X_{r})dr\]

    is a $\bracs{\mathcal{F}_t}$-martingale.

  2. For every $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,

    \[H_{t} = f(t, X_{t}) - \int_{0}^{t} [(\partial_{r} + L_{r})f](r, X_{r})dr\]

    is a $\bracs{\mathcal{F}_t}$-martingale.

  3. For each uniformly positive $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,

    \[X_{t} = f(t, X_{t})\exp\braks{-\int_0^t \frac{(\partial_{r} + L_{r})f}{f}(r, X_r)dr}\]

    is a $\bracs{\mathcal{F}_t}$-martingale.

  4. For each $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$, the process

    \begin{align*}Y_{t}^{x, g}&= \exp\braks{\dpn{x, X_t - X_0 - \int_0^t b(r)dr}{\real^d} + g(t, X_t)}\\&\cdot \exp\braks{-\frac{1}{2}\int_0^t \dpn{x + Dg, A(r)(x + Dg)}{\real^d}(r, X_r)dr}\\&\cdot \exp\braks{\int_0^t [(\partial_r + L_r)g](r, X_r)}dr\end{align*}

    is a $\bracs{\mathcal{F}_t}$-martingale.

  5. For each $\theta \in \real^{d}$,

Proof [Theorem 4.2.1, SV97]. (1) $\Rightarrow$ (2): Let $0 \le s \le t$ and $f \in C_{c}^{\infty}([0, \infty) \times \real^{d})$, then by the Fundamental Theorem of Calculus,

\begin{align*}f(t, X_{t}) - f(s, X_{s})&= f(t, X_{t}) - f(s, X_{t}) + f(s, X_{t}) - f(s, X_{s}) \\&= \int_{s}^{t} \partial_{r}f(r, X_{t})dr + f(s, X_{t}) - f(s, X_{s})\end{align*}

where by (1),

\[\ev\braks{f(s, X_t) - f(s, X_s) |\cf_s}= \ev\braks{\int_s^t (L_rf)(s, X_r)dr \bigg | \cf_s}\]

By the Fundamental Theorem of Calculus,

\begin{align*}\int_{s}^{t} (\partial_{r}f)(r, X_{t})dr&= \int_{s}^{t} (\partial_{r} f)(r, X_{r})dr + \int_{s}^{t} (\partial_{r} f)(r, X_{t}) - (\partial_{r} f)(r, X_{r})dr \\&= \int_{s}^{t} (\partial_{r} f)(r, X_{r})dr + \int_{s}^{t} (\partial_{r} f)(r, X_{t}) - (\partial_{r} f)(r, X_{r})dr \\\end{align*}

By (1) applied to $(\partial_{r} f)$,

\[\ev\braks{\int_s^t (\partial_rf)(r, X_t)dr \bigg | \cf_s}= \braks{\int_s^t (\partial_r f)(r, X_r)dr + \int_s^t \int_r^t (\partial_r f)(r, X_u)du dr \bigg | \cf_s}\]

Similarly, by the Fundamental Theorem of Calculus,

\begin{align*}\int_{s}^{t} (L_{r}f)(s, X_{r})dr&= \int_{s}^{t} (L_{r}f)(r, X_{r})dr + \int_{s}^{t} (L_{r}f)(s, X_{r}) - (L_{r}f)(r, X_{r})dr \\&= \int_{s}^{t}(L_{r}f)(r, X_{r})dr - \int_{s}^{t} \int_{s}^{r} (\partial_{r}L_{r}f)(u, X_{r})dudr\end{align*}

Since the two iterated integrals are over the same region,

\[\ev\braks{f(t, X_t) - f(s, X_s) | \cf_s}= \ev\braks{\int_s^t (\partial_r + L_rf)(r, X_r)dr \bigg | \cf_s}\]

Therefore $\bracs{E_t}$ is a $\bracs{\mathcal{F}_t}$-martingale.

(2) $\Rightarrow$ (3): Let $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive. Define

\[Z_{t} = f(t, X_{t}) - \int_{0}^{t} (\partial_{r}f + L_{r}f)(r, X_{r})dr\]

and

\[\phi_{t} = \exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\]

then using the by parts formula for Riemann-Stieltjes integrals,

\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} Z(ds) \\&= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} f(ds, X_{ds}) + \int_{0}^{t} \phi_{s} \cdot (\partial_{s} f L_{s}f)(s, X_{s})ds\end{align*}

Since

\[\partial_{s} \phi_{s} = -\phi_{s} \frac{\partial_{s} f + L_{s}f}{f}(s, X_{s})\]

Using the by parts formula again,

\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} f(ds, X_{ds}) + \int_{0}^{t} f(s, X_{s})\phi(ds) \\&= Z_{0}\phi_{0} + f(t, X_{t})\phi_{t} - f(0, X_{0})\phi_{0} = f(t, X_{t})\phi_{t} \\&= f(t, X_{t})\exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\end{align*}

Therefore Theorem 1.2.4 implies that the above process is a $\bracs{\mathcal{F}_t}$-martingale.

(3) $\Rightarrow$ (4): Let $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$. Define

\[f: [0, \infty) \times \real^{d} \to \real \quad (t, y) \mapsto \exp(\dpn{x, y}{\real^d}+ g(t, x))\]

then (3) applied to $f$ yields that $\bracs{Y_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale. However, since $f$ is unbounded and not uniformly positive, a direct application is ineffective.

Let $\seq{f_n}\subset C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive such that $f_{n}|_{\bracs{|y| \le n}}= f$. Define

\[\tau_{n} = \inf\bracs{t \ge 0 \bigg | \sup_{0 \le s \le t}|X_t(u)| \ge n}\wedge n\]

then $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is a $\bracs{\mathcal{F}_t}$-martingale with $Y_{t \wedge \tau_n}^{x, g}\to Y_{t}^{x, g}$ almost surely as $n \to \infty$. In addition,

\[(Y_{t \wedge \tau_n}^{x, g})^{2} \le Y_{t \wedge \tau_n}^{2x, 2g}\cdot \exp\braks{\int_0^{t \wedge \tau_n} \dpn{x + Dg, A_s(x + Dg) }{\real^d}(s, X_s)ds}\le CY_{t \wedge \tau_n}^{2x, 2g}\]

where $\ev\braks{Y_{t \wedge \tau_n}^{2x, 2g}}= f(s, X_{s})^{2} \le \exp(2\norm{g}_{u})$. Therefore $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is bounded in $L^{2}$, uniformly integrable in $L^{1}$, and converges to $Y_{t}^{x, g}$ in $L^{1}$.$\square$