%%% ----------------------------------------------------------------
%\newpage
%\subsection{Lawler 1 temp newpage}
%
%%xxx this is just a transcription of \cite{bib:lawler}.
%
%Let
%\begin{align}
%\label{eqn:law_1_dZ}
%	dZ_t = a(Z_t)\,dt + b(Z_t)\,dW_t.
%\end{align}
%We call $Z_t$ a (time homogeneous) diffusion since $a$ and $b$ depend on $Z_t$
%but not $t$ directly.  Note that $Z_t$ is Markovian.
%
%Let $f(x)$ and $v(x)$ be two functions and let
%\begin{align*}
%	J_t &= \exp\left\{\int_0^t v(Z_s)\,ds \right\}, \\
%	V(t,x) &= \bbE^x[f(Z_t)J_t]
%\end{align*}
%where
%$\bbE^x[Y]$ denotes $\bbE[Y \mid Z_0 = x]$.  We assume that this expectation
%exists for all $t,x$.  If $s<t$ then
%\begin{align}
%\label{eqn:law_1_M_s}
%	\bbE[f(Z_t)J_t\mid \mcF_s] &= J_s \bbE\left[
%		f(Z_t) \exp\left\{ \int_s^t v(Z_r) \,dr \right\}
%		\mid \mcF_s
%		\right] \\
%	&= J_s V(t-s,Z_s).
%\end{align}
%
%Call the left-hand side $M_s$.  We claim that $M_s$ is a martingale for $0 \le
%s < t$.  To see this, suppose $r<s$.  Then
%\begin{align*}
%	\bbE[\bbE[f(Zt) J_t \mid \mcF_s] \mid \mcF_r]
%		&= \bbE[f(Z_t) J_t \mid \mcF_r]
%\end{align*}
%
%Next we obtain our primary result.
%
%\begin{prop}
%The solution to the partial differential equation
%\begin{align*}
%	\frac{\D}{\Dt} V(t,x)
%		&= \frac{1}{2} b^2(x) \frac{\D^2}{\Dx^2} V(t,x)
%		+ a(x) \frac{\D}{\Dx} V(t,x) + v(x) V(t,x)
%\end{align*}
%with initial condition
%$$
%	V(0,x) = f(x)
%$$
%is
%$$
%	V(t,x) = \bbE^x\left[
%		f(Z_t) \exp\left\{ \int_0^t v(Z_s) \,ds \right\}
%	\right],
%$$
%where $Z_t$ satisfies
%$$
%	dZ_t = a(Z_t)\,dt + b(Z_t)\,dW_t.
%$$
%\begin{proof}
%Assuming sufficient differentiability, we can use It\=o's formula and the
%stochastic product rule to write, with $M_s$ as in equation
%\ref{eqn:law_1_M_s},
%\begin{align*}
%	dM_s &= J_s \,dV(t-s,Z_s) + V(t-s, Z_s) \frac{\D}{\Dt} J_s\, ds \\
%	&= J_s [ v(Z_s) V(t-s,Z_s) - \frac{\D}{\Dt}V(t-s,Z_s)
%		+ \frac{\D}{\Dx}V(t-s, Z_s) a(Z_s) \\
%		&+ \frac{1}{2} \frac{\D^2}{\Dx^2}V(t-s, Z_s) b^2(Z_s) ] \,ds \\
%		&+ J_s \frac{\D}{\Dt}V(t-s, Z_s) b(Z_s)\,dW_s.
%\end{align*}
%Since $M_s$ is a martingale, the $ds$ term must be zero.  Putting $s=0$
%and recalling $Z_0=x$, we are done.
%\end{proof}
%\end{prop}