$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ # Solving our first SDE Let $b$ be a bounded measurable function $\RR_+\times \RR$, which is bounded temporally by an $L^2$ function; there exists some $g\in L^2(\RR_+, \BBB(\RR_+), dt)$ such that $\abs{b(t,x)}\leq g(t)$ identically. Notably, if $b$ is bounded on some $[0,A]\times \RR_+$ and vanishes on $t > A$ then we're chilling. Now, let $B$ be an $(\FFF_t)$-BM. Consider the CLM $L_t = \int_0^t b(s, B_s)dB_s$ and its exponential martingale $D_t = \EEE(L)_t$. Then, $\braket{L,L}_\infty = \int_0^t b(s,B_s)^2 ds < C < \infty$ for some constant $C$, and thus $\EE\left[\exp \frac12 \braket{L,L}_\infty\right] < \infty$. Hence, by Novikov's criterion, Girsanov's theorem applies and $D$ is a UI martingale. We now set $\QQ = D_\infty\cdot \PP$, such that $\beta_t = B_t - \braket{B_t, L} = B_t - \int_0^t b(s,B_s)ds$is an $(\FFF_t)$-BM under $\QQ$. Flipping this around, we have obtained a solution $X = B$ to the SDE $dX_t = d\beta_t + b(t, X_t)dt$. This is really cool, because unlike the main sequence of our study of [[Stochastic Differential Equations Homepage|SDEs]], we are making no assumptions on the regularity of $b$; being bounded is enough. # Cameron-Martin Formula Suppose that $b(t,x) = g(t)$ is independent of $x$ now. Then, $h(t) = \int_0^t g(s)ds$ is just some (deterministic) function. We call the space of all such functions ==**Cameron-Martin space**==; this is just a regularity condition which says you "have a derivative in $L^2$." We then call $\dot h = g$. >[!idea] >This derivative is in the sense of distributions. I should figure out what distributions are at some point. Under $\QQ = D_\infty\cdot \PP$ as before, $\beta_t = B_t - h(t)$ is a BM. But this means for every non-negative measurable $\Phi$ on the space of continuous functions $C(\RR_+, \RR)$,$\EE_\PP\left[D_\infty\Phi(B)\right] = \EE_\QQ[\Phi(B)] = \EE_\QQ[\Phi(\beta + h)] = \EE_\PP[\Phi(B + h)].$ >[!idea]- Do I understand the last equality? >The last equality is because the value of $\EE_\PP[\Phi(B+h)]$ is a function solely of the pushforward measure $\PP\circ (B+h)^{-1}$ on $C(\RR_+,\RR)$, which is described completely by its finite-dimensional marginals, which are prescribed by the definition of Brownian motion, and are hence the same for $\QQ\circ (\beta + h)^{-1}$. To rewrite it with all the integrals: > [!theorem] Cameron-Martin Formula > Let $W(dw)$ be the Wiener measure on $C(\RR_+, \RR)$, and let $h$ be a function in CM space. Then, for every non-negative measurable $\Phi: \CC(\RR_+,\RR)\to \RR_+$, > $ > \int W(dw)\Phi(w + h) = \int W(dw) \exp\left(\int_0^\infty \dot{h}(s)dw(s) - \frac12 \int_0^\infty \dot h(s)^2 ds\right)\Phi(w). > $ The integrals shown are with respect to a BM, so they can be interpreted as either Ito or Wiener integrals.