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<div class="block-container"><p><a class="link-foreground" href="/t/5"><strong>Theorem 1.1.1.</strong></a><span> </span> Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space and $\bracs{B_t|t \ge 0}$ be a $\real^{d}$-valued, $\bracs{\cf_t}$-progressively measurable process such that $B_{0} = 0$ almost surely, then the following are equivalent:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>$\bracs{B_t|t \ge 0}$ is a Brownian motion.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For each $\xi \in \real^{d}$, the process</p><div class="">\[X^{\theta}_{t} = \exp\braks{i \dpn{B_t, \xi}{\real^d} + \frac{1}{2}\norm{\xi}_{\real^d}^2t}\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(3) &#x27;"><p>For each $\phi \in BC^{2}(\real^{d})$, the process</p><div class="">\[Y^{\phi}_{t} = \phi(B_{t}) - \frac{1}{2}\int_{0}^{t} \Delta f(B_{s})ds\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li></ol></div><div class="proof"><p><i class="">Proof <span class="">[<span class="">Theorem 4.1.1</span>, <a class="link-primary" href="/b/1">SV97</a>]</span>. </i>(1) $\Rightarrow$ (2): For each $1 \le s \le t$ and $\xi \in \real^{d}$, </p><div class="" id="">\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&#x26;= \ev\braks{e^{i \dpn{B_s, \xi}{\real^d}}e^{i \dpn{B_t - B_s, \xi}{\real^d}} \big| \cf_s}\\&#x26;= e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}</div><p>Therefore </p><div class="">\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d} + t\norm{\xi}^2/2} \big| \cf_s}= e^{i \dpn{B_s, \xi}{\real^d} + s\norm{\xi}_{\real^d}^2/2}\]</div><p>(2) $\Rightarrow$ (3): For each $0 \le s &#x3C; t$ and $\xi \in \real^{d}$, </p><div class="" id="">\begin{align*}\frac{d}{dt}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}</div><p>Therefore for any $0 \le s &#x3C; t$, </p><div class="" id="">\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} e^{i \dpn{B_s, \xi}{\real^d}}e^{-(r - s)\norm{\xi}_{\real^d}^2/2}dr \\&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} \ev\braks{e^{i \dpn{B_r, \xi}{\real^d}} \big | \cf_{s}}dr\end{align*}</div><p>If $\phi_{\xi}(x) = e^{-i \dpn{x, \xi}{\real^d}}$, then </p><div class="">\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}= \ev\braks{\int_s^t \frac{1}{2}(\Delta \phi_\xi)(B_r)dd \bigg | \cf_s}\]</div><p>so the process </p><div class="">\[Z^{\xi}_{t} = e^{i \dpn{B_t, \xi}{\real^d}}+ \int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>Let $\phi \in \mathcal{S}(\real^{d})$, then by the Fourier Inversion Theorem, there exists $\psi \in \mathcal{S}(\real^{d})$ such that for any $x \in \real^{d}$, </p><div class="">\[\phi(x) = \int e^{i \dpn{\xi, x}{\real^d}}\psi(\xi)d\xi\]</div><p>In which case, for any $t \ge 0$, </p><div class="">\[\phi(B_{t}) + \int_{0}^{t} \frac{1}{2}(\Delta \phi)(B_{r})dr = \int_{\real^d}e^{i \dpn{\xi, B_t}{\real^d}}\psi(\xi)d\xi + \int_{\real^d}\int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r}) \psi(\xi) dr d\xi\]</div><p>By the Dominated Convergence Theorem, the above also holds for any $\phi \in BC^{2}(\real^{d})$. Therefore the process $\bracsn{Y^\phi_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale as well.<span class="qed">$\square$</span></p></div>

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Brownian Motion Let ( , _t|t 0 , ) be a filtered probability space and B_t|t 0 be a d - valued , _t - progressively measurable process such that B 0 = 0 almost surely , then the following are equivalent : (1) B_t|t 0 is a Brownian motion . (2) For each d , the process X t = i B_t , ^d + 1 2 _ ^d ^2t is a F _t - martingale . (3) For each B C 2 ( d ) , the process Y t = ( B t ) - 1 2 0 t f ( B s ) d s is a F _t - martingale . ( 1 ) ( 2 ) : For each 1 s t and d , e^ i B_t , ^d | _s & = e^ i B_s , ^d e^ i B_t - B_s , ^d | _s & = e i B_s , ^d e - ( t - s ) _ ^d ^2 / 2 Therefore e^ i B_t , ^d + t ^2 / 2 | _s = e i B_s , ^d + s _ ^d ^2 / 2 ( 2 ) ( 3 ) : For each 0 s < t and d , d d t e^ i B_t , ^d | _s & = - ^d 2 2 e i B_s , ^d e - ( t - s ) _ ^d ^2 / 2 Therefore for any 0 s < t , e^ i B_t , ^d - e^ i B_s , ^d | _s & = - ^d 2 2 s t e i B_s , ^d e - ( r - s ) _ ^d ^2 / 2 d r & = - ^d 2 2 s t e^ i B_r , ^d | _ s d r If ( x ) = e - i x , ^d , then e^ i B_t , ^d - e^ i B_s , ^d | _s = _s^t 1 2 ( _ ) ( B_r ) dd | _s so the process Z t = e i B_t , ^d + 0 t 1 2 ( ) ( B r ) d r is a F _t - martingale . Let S ( d ) , then by the Fourier Inversion Theorem , there exists S ( d ) such that for any x d , ( x ) = e i , x ^d ( ) d In which case , for any t 0 , ( B t ) + 0 t 1 2 ( ) ( B r ) d r = ^d e i , B_t ^d ( ) d + ^d 0 t 1 2 ( ) ( B r ) ( ) d r d By the Dominated Convergence Theorem , the above also holds for any B C 2 ( d ) . Therefore the process Y^ _t|t 0 is a F _t - martingale as well . Proof Theorem 4.1 .1 Diffusion .

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<div class="block-container"><p><a class="link-foreground" href="/t/5"><strong>Theorem 1.1.1.</strong></a><span> </span> Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space and $\bracs{B_t|t \ge 0}$ be a $\real^{d}$-valued, $\bracs{\cf_t}$-progressively measurable process such that $B_{0} = 0$ almost surely, then the following are equivalent:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>$\bracs{B_t|t \ge 0}$ is a Brownian motion.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For each $\xi \in \real^{d}$, the process</p><div class="">\[X^{\theta}_{t} = \exp\braks{i \dpn{B_t, \xi}{\real^d} + \frac{1}{2}\norm{\xi}_{\real^d}^2t}\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(3) &#x27;"><p>For each $\phi \in BC^{2}(\real^{d})$, the process</p><div class="">\[Y^{\phi}_{t} = \phi(B_{t}) - \frac{1}{2}\int_{0}^{t} \Delta f(B_{s})ds\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li></ol></div><div class="proof"><p><i class="">Proof <span class="">[<span class="">Theorem 4.1.1</span>, <a class="link-primary" href="/b/1">SV97</a>]</span>. </i>(1) $\Rightarrow$ (2): For each $1 \le s \le t$ and $\xi \in \real^{d}$, </p><div class="" id="">\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&#x26;= \ev\braks{e^{i \dpn{B_s, \xi}{\real^d}}e^{i \dpn{B_t - B_s, \xi}{\real^d}} \big| \cf_s}\\&#x26;= e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}</div><p>Therefore </p><div class="">\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d} + t\norm{\xi}^2/2} \big| \cf_s}= e^{i \dpn{B_s, \xi}{\real^d} + s\norm{\xi}_{\real^d}^2/2}\]</div><p>(2) $\Rightarrow$ (3): For each $0 \le s &#x3C; t$ and $\xi \in \real^{d}$, </p><div class="" id="">\begin{align*}\frac{d}{dt}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}</div><p>Therefore for any $0 \le s &#x3C; t$, </p><div class="" id="">\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} e^{i \dpn{B_s, \xi}{\real^d}}e^{-(r - s)\norm{\xi}_{\real^d}^2/2}dr \\&#x26;= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} \ev\braks{e^{i \dpn{B_r, \xi}{\real^d}} \big | \cf_{s}}dr\end{align*}</div><p>If $\phi_{\xi}(x) = e^{-i \dpn{x, \xi}{\real^d}}$, then </p><div class="">\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}= \ev\braks{\int_s^t \frac{1}{2}(\Delta \phi_\xi)(B_r)dd \bigg | \cf_s}\]</div><p>so the process </p><div class="">\[Z^{\xi}_{t} = e^{i \dpn{B_t, \xi}{\real^d}}+ \int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>Let $\phi \in \mathcal{S}(\real^{d})$, then by the Fourier Inversion Theorem, there exists $\psi \in \mathcal{S}(\real^{d})$ such that for any $x \in \real^{d}$, </p><div class="">\[\phi(x) = \int e^{i \dpn{\xi, x}{\real^d}}\psi(\xi)d\xi\]</div><p>In which case, for any $t \ge 0$, </p><div class="">\[\phi(B_{t}) + \int_{0}^{t} \frac{1}{2}(\Delta \phi)(B_{r})dr = \int_{\real^d}e^{i \dpn{\xi, B_t}{\real^d}}\psi(\xi)d\xi + \int_{\real^d}\int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r}) \psi(\xi) dr d\xi\]</div><p>By the Dominated Convergence Theorem, the above also holds for any $\phi \in BC^{2}(\real^{d})$. Therefore the process $\bracsn{Y^\phi_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale as well.<span class="qed">$\square$</span></p></div>

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Let ( , _t|t 0 , ) be a filtered probability space and B_t|t 0 be a d - valued , _t - progressively measurable process such that B 0 = 0 almost surely , then the following are equivalent : (1) B_t|t 0 is a Brownian motion . (2) For each d , the process X t = i B_t , ^d + 1 2 _ ^d ^2t is a F _t - martingale . (3) For each B C 2 ( d ) , the process Y t = ( B t ) - 1 2 0 t f ( B s ) d s is a F _t - martingale .

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<div class="block-container"><p><a class="link-foreground" href="/t/7"><strong>Definition 1.2.1 (Random Differential Operator).</strong></a><span> </span> 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 </p><div class="">\[[L_{t}(\omega)f](x) = \sum_{\abs{\alpha} \le k}a_{\alpha}(t, \omega) \partial^{\alpha} f(x)\]</div><p>then $L = \sum_{|\alpha| \le k}a_{\alpha} \partial^{\alpha}$ is a <b class="textbf">random differential operator</b> of order $k$ with coefficients $\bracs{a_\alpha}_{|\alpha| \le k}$.</p></div><div class="block-container"><p><a class="link-foreground" href="/t/8"><strong>Definition 1.2.2 (Progressively Measurable Random Differential Operator).</strong></a><span> </span> 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 <b class="textbf">progressively measurable</b> 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 $\bracs{\mathcal{F}_t}$.</p></div><div class="block-container"><p><a class="link-foreground" href="/t/9"><strong>Lemma 1.2.3.</strong></a><span> </span> 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 </p><div class="">\[Y_{t}(\omega) = (L_{t}(\omega)f)(X_{t}(\omega))\]</div><p>is a progressively measurable process.</p></div><div class="block-container"><p><a class="link-foreground" href="/t/C"><strong>Theorem 1.2.4 (Integration by Parts).</strong></a><span> </span> 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:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>For every $\omega \in \Omega$, $\phi(\cdot, \omega) \in BV_{\text{loc}}([0, \infty))$.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For all $t \ge 0$,</p><div class="">\[\ev\braks{\sup_{0 \le s \le t}|X_s|(|\phi|(s) + |\phi|(t))}&#x3C; \infty\]</div></li></ol><p>then the process </p><div class="">\[Y_{t} = X_{t} \phi(t) - \int_{0}^{t} X_{r} \phi(dr)\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></div><div class="block-container"><p><a class="link-foreground" href="/t/A"><strong>Theorem 1.2.5 (Equivalence of Martingales).</strong></a><span> </span> Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, </p><div class="">\[L_{t}(\omega)u = \frac{1}{2}\dpn{A_t(\omega), u}{\real^{d \times d}}+ \dpn{B_t(\omega), u}{\real^d}\]</div><p>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:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>For every $f \in C_{c}^{\infty}(\real^{d})$, the process</p><div class="">\[E_{t} = f(X_{t}) - \int_{0}^{t} (L_{r}f)(X_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For every $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,</p><div class="">\[H_{t} = f(t, X_{t}) - \int_{0}^{t} [(\partial_{r} + L_{r})f](r, X_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(3) &#x27;"><p>For each uniformly positive $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,</p><div class="">\[X_{t} = f(t, X_{t})\exp\braks{-\int_0^t \frac{(\partial_{r} + L_{r})f}{f}(r, X_r)dr}\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(4) &#x27;"><p>For each $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$, the process</p><div class="" id="">\begin{align*}Y_{t}^{x, g}&#x26;= \exp\braks{\dpn{x, X_t - X_0 - \int_0^t b(r)dr}{\real^d} + g(t, X_t)}\\&#x26;\cdot \exp\braks{-\frac{1}{2}\int_0^t \dpn{x + Dg, A(r)(x + Dg)}{\real^d}(r, X_r)dr}\\&#x26;\cdot \exp\braks{\int_0^t [(\partial_r + L_r)g](r, X_r)}dr\end{align*}</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(5) &#x27;"><p>For each $\theta \in \real^{d}$,</p></li></ol></div><div class="proof"><p><i class="">Proof <span class="">[<span class="">Theorem 4.2.1</span>, <a class="link-primary" href="/b/1">SV97</a>]</span>. </i>(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, </p><div class="" id="">\begin{align*}f(t, X_{t}) - f(s, X_{s})&#x26;= f(t, X_{t}) - f(s, X_{t}) + f(s, X_{t}) - f(s, X_{s}) \\&#x26;= \int_{s}^{t} \partial_{r}f(r, X_{t})dr + f(s, X_{t}) - f(s, X_{s})\end{align*}</div><p>where by (1), </p><div class="">\[\ev\braks{f(s, X_t) - f(s, X_s) |\cf_s}= \ev\braks{\int_s^t (L_rf)(s, X_r)dr \bigg | \cf_s}\]</div><p>By the Fundamental Theorem of Calculus, </p><div class="" id="">\begin{align*}\int_{s}^{t} (\partial_{r}f)(r, X_{t})dr&#x26;= \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 \\&#x26;= \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*}</div><p>By (1) applied to $(\partial_{r} f)$, </p><div class="">\[\ev\braks{\int_s^t (\partial_rf)(r, X_t)dr \bigg | \cf_s}= \braks{\int_s^t (\partial_r f)(r, X_r)dr + \int_s^t \int_r^t (\partial_r f)(r, X_u)du dr \bigg | \cf_s}\]</div><p>Similarly, by the Fundamental Theorem of Calculus, </p><div class="" id="">\begin{align*}\int_{s}^{t} (L_{r}f)(s, X_{r})dr&#x26;= \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 \\&#x26;= \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*}</div><p>Since the two iterated integrals are over the same region, </p><div class="">\[\ev\braks{f(t, X_t) - f(s, X_s) | \cf_s}= \ev\braks{\int_s^t (\partial_r + L_rf)(r, X_r)dr \bigg | \cf_s}\]</div><p>Therefore $\bracs{E_t}$ is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>(2) $\Rightarrow$ (3): Let $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive. Define </p><div class="">\[Z_{t} = f(t, X_{t}) - \int_{0}^{t} (\partial_{r}f + L_{r}f)(r, X_{r})dr\]</div><p>and </p><div class="">\[\phi_{t} = \exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\]</div><p>then using the by parts formula for Riemann-Stieltjes integrals, </p><div class="" id="">\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&#x26;= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} Z(ds) \\&#x26;= 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*}</div><p>Since </p><div class="">\[\partial_{s} \phi_{s} = -\phi_{s} \frac{\partial_{s} f + L_{s}f}{f}(s, X_{s})\]</div><p>Using the by parts formula again,</p><div class="" id="">\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&#x26;= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} f(ds, X_{ds}) + \int_{0}^{t} f(s, X_{s})\phi(ds) \\&#x26;= Z_{0}\phi_{0} + f(t, X_{t})\phi_{t} - f(0, X_{0})\phi_{0} = f(t, X_{t})\phi_{t} \\&#x26;= f(t, X_{t})\exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\end{align*}</div><p>Therefore <a href="/t/C" class="link-primary">Theorem 1.2.4</a> implies that the above process is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>(3) $\Rightarrow$ (4): Let $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$. Define </p><div class="">\[f: [0, \infty) \times \real^{d} \to \real \quad (t, y) \mapsto \exp(\dpn{x, y}{\real^d}+ g(t, x))\]</div><p>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.</p><p>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 </p><div class="">\[\tau_{n} = \inf\bracs{t \ge 0 \bigg | \sup_{0 \le s \le t}|X_t(u)| \ge n}\wedge n\]</div><p>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, </p><div class="">\[(Y_{t \wedge \tau_n}^{x, g})^{2} \le Y_{t \wedge \tau_n}^{2x, 2g}\cdot \exp\braks{\int_0^{t \wedge \tau_n} \dpn{x + Dg, A_s(x + Dg) }{\real^d}(s, X_s)ds}\le CY_{t \wedge \tau_n}^{2x, 2g}\]</div><p>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 $L^{1}$.<span class="qed">$\square$</span></p></div>

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Equivalence of Certain Martingales Let ( , , ) be a filtered probability space , s 0 , and k . For each multi - index 0 d with k , let a : [ s , ) be measurable function . For each f C k ( d ) and t s , let [ L t ( ) f ] ( x ) = k a ( t , ) f ( x ) then L = | | k a is a random differential operator of order k with coefficients a_ | | k . Random Differential Operator Let ( , , ) be a filtered probability space and L = | | k a be a random differential operator on , then L is progressively measurable with respect to F _t if for each 0 with | | k , a : [ s , ) is progressively measurable with respect to F _t . Progressively Measurable Random Differential Operator Let ( , _t|t 0 , ) be a filtered probability space , s 0 , k , L = | | k a be a F _t - progressively measurable random differential operator , f C k ( d ) , and X_t|t 0 be a d - valued progressively measurable process , then Y t ( ) = ( L t ( ) f ) ( X t ( ) ) is a progressively measurable process . Let ( , _t|t 0 , ) be a filtered probability space , X_t be a F _t - martingale , and : [ 0 , ) be a continuous , progressively measurable function . If : (1) For every , ( , ) B V loc ( [ 0 , ) ) . (2) For all t 0 , _ 0 s t |X_s| ( | | ( s ) + | | ( t ) ) < then the process Y t = X t ( t ) - 0 t X r ( d r ) is a F _t - martingale . Integration by Parts Let ( , _t|t 0 , ) be a filtered probability space , L t ( ) u = 1 2 A_t ( ) , u ^ d d + B_t ( ) , u ^d be a second - order F _t - progressively measurable random differential operator , and X_t|t 0 be a F _t - progressively measurable process with continuous sample paths , then the following are equivalent : (1) For every f C c ( d ) , the process E t = f ( X t ) - 0 t ( L r f ) ( X r ) d r is a F _t - martingale . (2) For every f C b 1 , 2 ( [ 0 , ) d ) , H t = f ( t , X t ) - 0 t [ ( r + L r ) f ] ( r , X r ) d r is a F _t - martingale . (3) For each uniformly positive f C b 1 , 2 ( [ 0 , ) d ) , X t = f ( t , X t ) - _0^t ( r + L r ) f f ( r , X_r ) dr is a F _t - martingale . (4) For each x d and g C b 1 , 2 ( [ 0 , ) d ) , the process Y t x , g & = x , X_t - X_0 - _0^t b ( r ) dr ^d + g ( t , X_t ) & - 1 2 _0^t x + Dg , A ( r ) ( x + Dg ) ^d ( r , X_r ) dr & _0^t [ ( _r + L_r ) g ] ( r , X_r ) d r is a F _t - martingale . (5) For each d , Equivalence of Martingales ( 1 ) ( 2 ) : Let 0 s t and f C c ( [ 0 , ) d ) , then by the Fundamental Theorem of Calculus , f ( t , X t ) - f ( s , X s ) & = f ( t , X t ) - f ( s , X t ) + f ( s , X t ) - f ( s , X s ) & = s t r f ( r , X t ) d r + f ( s , X t ) - f ( s , X s ) where by ( 1 ) , f ( s , X_t ) - f ( s , X_s ) | _s = _s^t ( L_rf ) ( s , X_r ) dr | _s By the Fundamental Theorem of Calculus , s t ( r f ) ( r , X t ) d r & = s t ( r f ) ( r , X r ) d r + s t ( r f ) ( r , X t ) - ( r f ) ( r , X r ) d r & = s t ( r f ) ( r , X r ) d r + s t ( r f ) ( r , X t ) - ( r f ) ( r , X r ) d r By ( 1 ) applied to ( r f ) , _s^t ( _rf ) ( r , X_t ) dr | _s = _s^t ( _r f ) ( r , X_r ) dr + _s^t _r^t ( _r f ) ( r , X_u ) du dr | _s Similarly , by the Fundamental Theorem of Calculus , s t ( L r f ) ( s , X r ) d r & = s t ( L r f ) ( r , X r ) d r + s t ( L r f ) ( s , X r ) - ( L r f ) ( r , X r ) d r & = s t ( L r f ) ( r , X r ) d r - s t s r ( r L r f ) ( u , X r ) d u d r Since the two iterated integrals are over the same region , f ( t , X_t ) - f ( s , X_s ) | _s = _s^t ( _r + L_rf ) ( r , X_r ) dr | _s Therefore E_t is a F _t - martingale . ( 2 ) ( 3 ) : Let f C b 1 , 2 ( [ 0 , ) d ) be uniformly positive . Define Z t = f ( t , X t ) - 0 t ( r f + L r f ) ( r , X r ) d r and t = - _0^t r f + L r f f ( r , X_r ) dr then using the by parts formula for Riemann - Stieltjes integrals , Z t t - 0 t Z s ( d s ) & = Z 0 0 + 0 t s Z ( d s ) & = Z 0 0 + 0 t s f ( d s , X ds ) + 0 t s ( s f L s f ) ( s , X s ) d s Since s s = - s s f + L s f f ( s , X s ) Using the by parts formula again , Z t t - 0 t Z s ( d s ) & = Z 0 0 + 0 t s f ( d s , X ds ) + 0 t f ( s , X s ) ( d s ) & = Z 0 0 + f ( t , X t ) t - f ( 0 , X 0 ) 0 = f ( t , X t ) t & = f ( t , X t ) - _0^t r f + L r f f ( r , X_r ) dr Therefore theorem : martingale - ibp implies that the above process is a F _t - martingale . ( 3 ) ( 4 ) : Let x d and g C b 1 , 2 ( [ 0 , ) d ) . Define f : [ 0 , ) d ( t , y ) ( x , y ^d + g ( t , x ) ) then ( 3 ) applied to f yields that Y_t|t 0 is a F _t - martingale . However , since f is unbounded and not uniformly positive , a direct application is ineffective . Let f_n C b 1 , 2 ( [ 0 , ) d ) be uniformly positive such that f n | |y| n = f . Define n = t 0 | _ 0 s t |X_t ( u ) | n n then Y_ t _n ^ x , g is a F _t - martingale with Y t _n x , g Y t x , g almost surely as n . In addition , ( Y t _n x , g ) 2 Y t _n 2x , 2g _0^ t _n x + Dg , A_s ( x + Dg ) ^d ( s , X_s ) ds C Y t _n 2x , 2g where Y_ t _n ^ 2x , 2g = f ( s , X s ) 2 ( 2 g u ) . Therefore Y_ t _n ^ x , g is bounded in L 2 , uniformly integrable in L 1 , and converges to Y t x , g in L 1 . Proof Theorem 4.2 .1 Diffusion .

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<div class="block-container"><p><a class="link-foreground" href="/t/7"><strong>Definition 1.2.1 (Random Differential Operator).</strong></a><span> </span> 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 </p><div class="">\[[L_{t}(\omega)f](x) = \sum_{\abs{\alpha} \le k}a_{\alpha}(t, \omega) \partial^{\alpha} f(x)\]</div><p>then $L = \sum_{|\alpha| \le k}a_{\alpha} \partial^{\alpha}$ is a <b class="textbf">random differential operator</b> of order $k$ with coefficients $\bracs{a_\alpha}_{|\alpha| \le k}$.</p></div>

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Random Differential Operator Let ( , , ) be a filtered probability space , s 0 , and k . For each multi - index 0 d with k , let a : [ s , ) be measurable function . For each f C k ( d ) and t s , let [ L t ( ) f ] ( x ) = k a ( t , ) f ( x ) then L = | | k a is a random differential operator of order k with coefficients a_ | | k . Random Differential Operator

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<div class="block-container"><p><a class="link-foreground" href="/t/8"><strong>Definition 1.2.2 (Progressively Measurable Random Differential Operator).</strong></a><span> </span> 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 <b class="textbf">progressively measurable</b> 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 $\bracs{\mathcal{F}_t}$.</p></div>

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Progressively Measurable Random Differential Operator Let ( , , ) be a filtered probability space and L = | | k a be a random differential operator on , then L is progressively measurable with respect to F _t if for each 0 with | | k , a : [ s , ) is progressively measurable with respect to F _t . Progressively Measurable Random Differential Operator

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<div class="block-container"><p><a class="link-foreground" href="/t/9"><strong>Lemma 1.2.3.</strong></a><span> </span> 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 </p><div class="">\[Y_{t}(\omega) = (L_{t}(\omega)f)(X_{t}(\omega))\]</div><p>is a progressively measurable process.</p></div>

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Let ( , _t|t 0 , ) be a filtered probability space , s 0 , k , L = | | k a be a F _t - progressively measurable random differential operator , f C k ( d ) , and X_t|t 0 be a d - valued progressively measurable process , then Y t ( ) = ( L t ( ) f ) ( X t ( ) ) is a progressively measurable process .

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<div class="block-container"><p><a class="link-foreground" href="/t/A"><strong>Theorem 1.2.5 (Equivalence of Martingales).</strong></a><span> </span> Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, </p><div class="">\[L_{t}(\omega)u = \frac{1}{2}\dpn{A_t(\omega), u}{\real^{d \times d}}+ \dpn{B_t(\omega), u}{\real^d}\]</div><p>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:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>For every $f \in C_{c}^{\infty}(\real^{d})$, the process</p><div class="">\[E_{t} = f(X_{t}) - \int_{0}^{t} (L_{r}f)(X_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For every $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,</p><div class="">\[H_{t} = f(t, X_{t}) - \int_{0}^{t} [(\partial_{r} + L_{r})f](r, X_{r})dr\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(3) &#x27;"><p>For each uniformly positive $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,</p><div class="">\[X_{t} = f(t, X_{t})\exp\braks{-\int_0^t \frac{(\partial_{r} + L_{r})f}{f}(r, X_r)dr}\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(4) &#x27;"><p>For each $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$, the process</p><div class="" id="">\begin{align*}Y_{t}^{x, g}&#x26;= \exp\braks{\dpn{x, X_t - X_0 - \int_0^t b(r)dr}{\real^d} + g(t, X_t)}\\&#x26;\cdot \exp\braks{-\frac{1}{2}\int_0^t \dpn{x + Dg, A(r)(x + Dg)}{\real^d}(r, X_r)dr}\\&#x26;\cdot \exp\braks{\int_0^t [(\partial_r + L_r)g](r, X_r)}dr\end{align*}</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></li><li style="list-style-type: &#x27;(5) &#x27;"><p>For each $\theta \in \real^{d}$,</p></li></ol></div><div class="proof"><p><i class="">Proof <span class="">[<span class="">Theorem 4.2.1</span>, <a class="link-primary" href="/b/1">SV97</a>]</span>. </i>(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, </p><div class="" id="">\begin{align*}f(t, X_{t}) - f(s, X_{s})&#x26;= f(t, X_{t}) - f(s, X_{t}) + f(s, X_{t}) - f(s, X_{s}) \\&#x26;= \int_{s}^{t} \partial_{r}f(r, X_{t})dr + f(s, X_{t}) - f(s, X_{s})\end{align*}</div><p>where by (1), </p><div class="">\[\ev\braks{f(s, X_t) - f(s, X_s) |\cf_s}= \ev\braks{\int_s^t (L_rf)(s, X_r)dr \bigg | \cf_s}\]</div><p>By the Fundamental Theorem of Calculus, </p><div class="" id="">\begin{align*}\int_{s}^{t} (\partial_{r}f)(r, X_{t})dr&#x26;= \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 \\&#x26;= \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*}</div><p>By (1) applied to $(\partial_{r} f)$, </p><div class="">\[\ev\braks{\int_s^t (\partial_rf)(r, X_t)dr \bigg | \cf_s}= \braks{\int_s^t (\partial_r f)(r, X_r)dr + \int_s^t \int_r^t (\partial_r f)(r, X_u)du dr \bigg | \cf_s}\]</div><p>Similarly, by the Fundamental Theorem of Calculus, </p><div class="" id="">\begin{align*}\int_{s}^{t} (L_{r}f)(s, X_{r})dr&#x26;= \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 \\&#x26;= \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*}</div><p>Since the two iterated integrals are over the same region, </p><div class="">\[\ev\braks{f(t, X_t) - f(s, X_s) | \cf_s}= \ev\braks{\int_s^t (\partial_r + L_rf)(r, X_r)dr \bigg | \cf_s}\]</div><p>Therefore $\bracs{E_t}$ is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>(2) $\Rightarrow$ (3): Let $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive. Define </p><div class="">\[Z_{t} = f(t, X_{t}) - \int_{0}^{t} (\partial_{r}f + L_{r}f)(r, X_{r})dr\]</div><p>and </p><div class="">\[\phi_{t} = \exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\]</div><p>then using the by parts formula for Riemann-Stieltjes integrals, </p><div class="" id="">\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&#x26;= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} Z(ds) \\&#x26;= 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*}</div><p>Since </p><div class="">\[\partial_{s} \phi_{s} = -\phi_{s} \frac{\partial_{s} f + L_{s}f}{f}(s, X_{s})\]</div><p>Using the by parts formula again,</p><div class="" id="">\begin{align*}Z_{t}\phi_{t} - \int_{0}^{t} Z_{s} \phi(ds)&#x26;= Z_{0}\phi_{0} + \int_{0}^{t} \phi_{s} f(ds, X_{ds}) + \int_{0}^{t} f(s, X_{s})\phi(ds) \\&#x26;= Z_{0}\phi_{0} + f(t, X_{t})\phi_{t} - f(0, X_{0})\phi_{0} = f(t, X_{t})\phi_{t} \\&#x26;= f(t, X_{t})\exp\braks{-\int_0^t \frac{\partial_{r}f + L_{r}f}{f}(r, X_r)dr}\end{align*}</div><p>Therefore <a href="/t/C" class="link-primary">Theorem 1.2.4</a> implies that the above process is a $\bracs{\mathcal{F}_t}$-martingale.</p><p>(3) $\Rightarrow$ (4): Let $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$. Define </p><div class="">\[f: [0, \infty) \times \real^{d} \to \real \quad (t, y) \mapsto \exp(\dpn{x, y}{\real^d}+ g(t, x))\]</div><p>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.</p><p>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 </p><div class="">\[\tau_{n} = \inf\bracs{t \ge 0 \bigg | \sup_{0 \le s \le t}|X_t(u)| \ge n}\wedge n\]</div><p>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, </p><div class="">\[(Y_{t \wedge \tau_n}^{x, g})^{2} \le Y_{t \wedge \tau_n}^{2x, 2g}\cdot \exp\braks{\int_0^{t \wedge \tau_n} \dpn{x + Dg, A_s(x + Dg) }{\real^d}(s, X_s)ds}\le CY_{t \wedge \tau_n}^{2x, 2g}\]</div><p>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 $L^{1}$.<span class="qed">$\square$</span></p></div>

contentText:
Equivalence of Martingales Let ( , _t|t 0 , ) be a filtered probability space , L t ( ) u = 1 2 A_t ( ) , u ^ d d + B_t ( ) , u ^d be a second - order F _t - progressively measurable random differential operator , and X_t|t 0 be a F _t - progressively measurable process with continuous sample paths , then the following are equivalent : (1) For every f C c ( d ) , the process E t = f ( X t ) - 0 t ( L r f ) ( X r ) d r is a F _t - martingale . (2) For every f C b 1 , 2 ( [ 0 , ) d ) , H t = f ( t , X t ) - 0 t [ ( r + L r ) f ] ( r , X r ) d r is a F _t - martingale . (3) For each uniformly positive f C b 1 , 2 ( [ 0 , ) d ) , X t = f ( t , X t ) - _0^t ( r + L r ) f f ( r , X_r ) dr is a F _t - martingale . (4) For each x d and g C b 1 , 2 ( [ 0 , ) d ) , the process Y t x , g & = x , X_t - X_0 - _0^t b ( r ) dr ^d + g ( t , X_t ) & - 1 2 _0^t x + Dg , A ( r ) ( x + Dg ) ^d ( r , X_r ) dr & _0^t [ ( _r + L_r ) g ] ( r , X_r ) d r is a F _t - martingale . (5) For each d , Equivalence of Martingales

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<div class="block-container"><p><a class="link-foreground" href="/t/C"><strong>Theorem 1.2.4 (Integration by Parts).</strong></a><span> </span> 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:</p><ol class="enumerate"><li style="list-style-type: &#x27;(1) &#x27;"><p>For every $\omega \in \Omega$, $\phi(\cdot, \omega) \in BV_{\text{loc}}([0, \infty))$.</p></li><li style="list-style-type: &#x27;(2) &#x27;"><p>For all $t \ge 0$,</p><div class="">\[\ev\braks{\sup_{0 \le s \le t}|X_s|(|\phi|(s) + |\phi|(t))}&#x3C; \infty\]</div></li></ol><p>then the process </p><div class="">\[Y_{t} = X_{t} \phi(t) - \int_{0}^{t} X_{r} \phi(dr)\]</div><p>is a $\bracs{\mathcal{F}_t}$-martingale.</p></div>

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Integration by Parts Let ( , _t|t 0 , ) be a filtered probability space , X_t be a F _t - martingale , and : [ 0 , ) be a continuous , progressively measurable function . If : (1) For every , ( , ) B V loc ( [ 0 , ) ) . (2) For all t 0 , _ 0 s t |X_s| ( | | ( s ) + | | ( t ) ) < then the process Y t = X t ( t ) - 0 t X r ( d r ) is a F _t - martingale . Integration by Parts

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# bib

key: Diffusion
type: book
author: Stroock, D.W. and Varadhan, S.R.S.
title: Multidimensional Diffusion Processes
year: 1997
publisher: Springer Berlin Heidelberg
url: https://books.google.ca/books?id=DuDsmoyqCy4C
aux: {"isbn":"9783540903536","lccn":"96027093","series":"Grundlehren der mathematischen Wissenschaften"}
bibtex:
@book{Diffusion,
    title={Multidimensional Diffusion Processes}
    author={Stroock, D.W. and Varadhan, S.R.S.}
    isbn={9783540903536}
    lccn={96027093}
    series={Grundlehren der mathematischen Wissenschaften}
    url={https://books.google.ca/books?id=DuDsmoyqCy4C}
    year={1997}
    publisher={Springer Berlin Heidelberg}
}


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