%% ================================================================
\section{Reference}

Before presenting problem solutions, I will first tabulate some results for
ready reference.  The following are collected from \cite{bib:bernt} and from my
problem solutions, using definitions and notation as in \cite{bib:bernt}.  See
also \cite{bib:gs} and/or \cite{bib:lawler} for more background information on
Brownian motion, martingales, and stochastic differential equations.

%% ----------------------------------------------------------------
\subsection{The \Ito formula in one dimension}

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\begin{prop}
\label{prop:ito_1d}
If
\begin{align}
\label{eqn:dX_edt_fdb}
	dX_t = e(X_t,t)\,dt + f(X_t,t)\,db_t
\end{align}
is an \Ito process
and
$$
	y = g(t,x)
$$
is a $C^2$ map from $[0,+\infty) \times \R$ to $\R$, then
$$
	Y_t = g(t, X_t)
$$
is again an \Ito process with
\begin{align}
\label{eqn:ito_1d_1}
	dY_t = \fracDgt(t,X_t)\,dt + \fracDgx(t,X_t)\,dX_t
		+ \frac{1}{2} \fracDgxx(t,X_t)\,dX_t^2.
\end{align}
\begin{rem*}
Recall that we use the \Ito calculus rules
$$
	db_t^2 = dt, \qquad
	dt^2 = 0, \qquad
	dt\,db_t = 0
$$
when expanding $dX_t^2$.
\end{rem*}
\begin{rem*}
If we expand $dX_t$ (equation \ref{eqn:dX_edt_fdb}) into \ref{eqn:ito_1d_1},
we obtain an alternate expression
\begin{align}
\label{eqn:ito_1d_2}
	dX_t^2 &= e^2\,dt^2 + 2ef\,dt\,db_t + f^2 db_t^2 = f^2 dt \\
	dY_t &= \fracDgt(t,X_t)\,dt + \fracDgx(t,X_t)\,(e\,dt + f\,db_t)
		+ \frac{f^2}{2} \fracDgxx(t,X_t)\,dt \\
	&= \left(\fracDgt(t,X_t)
		+ e\,\fracDgx(t,X_t)
		+ \frac{f^2}{2} \fracDgxx(t,X_t)
		\right)\,dt
		+ f\,\fracDgx(t,X_t)\,db_t.
\end{align}
\end{rem*}
\end{prop}

The following two cases appear more often in the problem solutions than the
general statement of proposition \ref{prop:ito_1d}, so they are worth writing
down to relieve the brain of the task of error-prone on-the-fly relabeling.

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\begin{prop}
\label{prop:ito_1d_b_t}
Special case with $e=0$ and $f=1$: If $x = g(t,u)$ is $C^2$ then
$$
	X_t = g(t,b_t)
$$
is an \Ito process with
\begin{align}
	dX_t &= \fracDgt(t,b_t)\,dt
		+ \fracDgu(t,b_t)\,dX_t
		+ \frac{1}{2} \fracDguu(t,b_t)\,dX_t^2 \\
	&= \left(\fracDgt(t,b_t)
		+ \frac{1}{2} \fracDguu(t,b_t)\right)\,dt
		+ \fracDgu(t,b_t)\,db_t.
\end{align}
\end{prop}

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\begin{prop}
\label{prop:ito_1d_b_no_t}
Special case with $e=0$ and $f=1$ and no explicit time dependence:
If
$$
	X_t = g(b_t)
$$
then
\begin{align}
	dX_t &= \frac{1}{2} \fracDguu(b_t)\,dt + \fracDgu(b_t)\,db_t.
\end{align}
\end{prop}

%% ----------------------------------------------------------------
\subsection{The \Ito formula in higher dimensions}

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\begin{prop}
\label{prop:ito_nd}
Let
$$
	\colvecthree{dX^{(1)}_t}{\vdots}{dX^{(n)}_t}
	=
	\colvecthree{u_1\,dt}{\vdots}{u_n\,dt}
	+
	\prmatrix{
		v_{1,1} & \cdots & v_{1,m} \\
		\vdots  &        & \vdots  \\
		v_{n,1} & \cdots & v_{n,m} \\
	}
	\colvecthree{db^{(1)}_t}{\vdots}{db^{(m)}_t},
$$
i.e. in matrix notation
$$
	d\vecX_t = \vecu\,dt + \vecV \,d\vecb_t.
$$
(This is called am $n$-dimensional \Ito process.)
Let $\vecg:[0,+\infty) \times \R^n \to \R^p$ be a $C^2$ map:
$$
	\colvecthree{y_1}{\vdots}{y_p}
	= \colvecthree
		{g_1(t, \vecx)}
		{\vdots}
		{g_p(t, \vecx)}
	= \colvecthree
		{g_1(t, x_1, \ldots, x_n)}
		{\vdots}
		{g_p(t, x_1, \ldots, x_n)}.
$$
Then
$$
	\vecY_t = \vecg(t, \vecX_t)
$$
is again an \Ito process with $k$th component (for $k = 1, \ldots, p$)
$$
	dY^{(k)}_t
		= \fracDgt(t,\vecX_t)\,dt
		+ \sum_{i=1}^n \frac{\Dg_k}{\Dx_i} (t,\vecX_t)\, dX^{(i)}_t
		+ \frac{1}{2} \sum_{i,j=1}^n \frac{\D^2 g_k}{\Dx_i \Dx_j}
			(t,\vecX_t)\, dX^{(i)}_t \,dX^{(j)}_t.
$$
\begin{rem*}
Recall that we use the \Ito calculus rules
$$
	db^{(i)}_t \, db^{(j)}_t = \delta_{ij}, \qquad
	dt^2 = 0, \qquad
	dt\,db^{(i)}_t = 0
$$
when expanding $dX^{(i)}_t \, dX^{(j)}_t$.
\end{rem*}
\end{prop}

Again, there is a special case when the input \Ito process is simply
$n$-dimensional Brownian motion; again, it is worth writing down as such since
this case arises often.

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\begin{prop}
\label{prop:ito_ndb}

Let $\vecg:\R^+ \times \R^m \to \R^n$ be a $C^2$ map:
$$
	\colvecthree{x_1}{\vdots}{x_n}
	= \colvecthree
		{g_1(t, \vecu)}
		{\vdots}
		{g_n(t, \vecu)}
	= \colvecthree
		{g_1(t, u_1, \ldots, u_m)}
		{\vdots}
		{g_n(t, u_1, \ldots, u_m)}.
$$
Then
$$
	\vecX_t = \vecg(t, \vecb_t)
$$
is again an \Ito process with $k$th component (for $k = 1, \ldots, n$)
$$
	dX^{(k)}_t
		= \fracDgt(t,\vecb_t)\,dt
		+ \sum_{i=1}^m \frac{\Dg_k}{\Du_i} (t,\vecb_t)\,db^{(i)}_t
		+ \frac{1}{2} \sum_{i,j=1}^m \frac{\D^2 g_k}{\Du_i \Du_j}
			(t,\vecb_t)\,db^{(i)}_t \,db^{(j)}_t.
$$
\end{prop}

%% ----------------------------------------------------------------
\subsection{The \Ito formula from two dimensions to one}

\begin{prop}
\label{prop:ito_2_1}
Let $\vecg:\R^+ \times \R^2 \to \R$ be a $C^2$ map
$$
	z = g(t, x, y).
$$
Then
$$
	Z_t = g(t, X_t, Y_t)
$$
is again an \Ito process with
\begin{align*}
	dZ_t &= \fracDgt(t,X_t,Y_t)\,dt
		+ \fracDgx(t,X_t,Y_t)\,dX_t
		+ \fracDgy(t,X_t,Y_t)\,dY_t \\
	&	+ \frac{1}{2} \fracDgxx(t,X_t,Y_t)\,dX_t^2
		+ \frac{1}{2} \fracDgyy(t,X_t,Y_t)\,dY_t^2
		+ \fracDgxy(t,X_t,Y_t)\,dX_t\, dY_t.
\end{align*}
\end{prop}

%% ----------------------------------------------------------------
\subsection{A derivation of the one-dimensional \Ito formula}

This is not a proof (see e.g. \cite{bib:bernt} section 4.1) --- rather, it is a
heuristic discussion intended to remind me why which terms appear where.

\begin{itemize}

\item Start with
	$$
		dX_t = e(X_t, t)\,dt + f(X_t,t)\,db_t.
	$$

\item Consider
	$$
		Y_t = g(X_t,t).
	$$

\item We need to show (using the formulation of equation \ref{eqn:ito_1d_2})
	$$
		dY_t = \left(\fracDgt(t,X_t)
		+ e\,\fracDgx(t,X_t)
		+ \frac{f^2}{2} \fracDgxx(t,X_t)
		\right)\,dt
		+ f\,\fracDgx(t,X_t)\,db_t.
	$$

\item Using the shorthand partial-derivative notation $g_t = \Dg/\Dt$ etc.,
Taylor-expand $dY$ as
	\begin{align*}
		dY &= g_t \,dt + g_x \,dX + \frac{g_{tt}}{2} \,dt^2
			+ \frac{g_{xx}}{2} \,dX^2 + g_{tx} \,dt \,dX
			+ \textrm{higher order terms} \\
		%
		&= g_t \,dt + g_x (e\,dt+f\,db_t) + \frac{g_{tt}}{2} \,dt^2
			+ \frac{g_{xx}}{2} (e\,dt+f\,db_t)^2 + g_{tx} \,dt (e\,dt+f\,db_t)
			+ \ldots.
	\end{align*}

\item Use $dt^2=0$, $dt\,db_t=0$, $db_t^2=dt$:
	\begin{align*}
		dY &= g_t \,dt + g_x (e\,dt+f\,db_t) + \frac{g_{xx}}{2} f^2 \,dt \\
		&= g_t \,dt + e\, g_x\,dt + f^2 \,\frac{g_{xx}}{2} \,dt + f\,g_x\,db_t\\
		&= \left(g_t + e\, g_x + f^2 \,\frac{g_{xx}}{2}\right) \,dt
			+ f\,g_x\,db_t.
	\end{align*}

\end{itemize}

%% ----------------------------------------------------------------
\subsection{The stochastic product rule}

\begin{prop}
\label{prop:stochastic_product_rule}
Let\index{product rule} $X_t$, $Y_t$ be \Ito processes.  Then
$$
	d(X_t\, Y_t) = X_t\, dY_t + Y_t\,dX_t + dX_t\,dY_t.
$$
\begin{proof}
Using the \Ito formula in the form of proposition \ref{prop:ito_2_1} with
$g(x,y) = xy$,
we have
$$
	\fracDgt  = 0, \qquad
	\fracDgx  = y, \qquad
	\fracDgy  = x, \qquad
	\fracDgxx = 0, \qquad
	\fracDgyy = 0, \qqand
	\fracDgxy = 1.
$$
Then
$$
	d(X_t\,Y_t) = Y_t\,dX_t + X_t\,dY_t + dX_t\,dY_t.
$$
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Stochastic integration by parts}

\begin{prop}
\label{prop:stochastic_parts}
For\index{parts}\index{integration by parts} \Ito processes $X_t$ and $Y_t$,
$$
	\int_0^t X_s dY_s = X_t Y_t - X_0 Y_0
		- \int_0^t Y_s \,dX_s - \int_0^t dX_s\,dY_s.
$$
\begin{proof}
Using the stochastic product rule (proposition
\ref{prop:stochastic_product_rule}), we have
\begin{align*}
	d(X_t\, Y_t) &= X_t\, dY_t + Y_t\,dX_t + dX_t\,dY_t \\
	X_t\, dY_t &= d(X_t\, Y_t) - Y_t\,dX_t - dX_t\,dY_t.
\end{align*}
Integrating from $0$ to $t$ gives
\begin{align*}
	\int_0^t X_s\, dY_s &= \int_0^t d(X_s\, Y_s)
		- \int_0^t Y_s\,dX_s - \int_0^t dX_s\,dY_s \\
	&= X_t Y_t - X_0 Y_0 - \int_0^t Y_s\,dX_s - \int_0^t dX_s\,dY_s.
\end{align*}
\end{proof}
\end{prop}

\begin{cor}
\label{prop:stochastic_parts_cor}
For $f_t$ not depending on $b_t$, we have
$$
	\int_0^t f_s\,db_s = f_t b_t - \int_0^t b_s\,df_s.
$$
\begin{proof}
Putting $X_t = f_t$ and $Y_t = b_t$, the proposition gives
$$
	\int_0^t f_s\,db_s = f_t b_t - \int_0^t b_s\,df_s
		- \int_0^t df_s\,db_s
$$
since $b_0=0$.  By the \Ito calculus rule $dt\,db_t = 0$, the last integral
vanishes as long as there is no $db_t$ in $df_t$.
\end{proof}
\end{cor}

%% ----------------------------------------------------------------
\subsection{The \Ito isometry}

\begin{prop}
\label{prop:ito_isometry}
For\index{isometry}\index{It\=o isometry} all $f$ such that the \Ito integral
is defined,
$$
	E\left[ \left(\int_{t_1}^{t_2} f_s\,db_s\right)^2 \right]
	=
	E\left[ \int_{t_1}^{t_2} f_s^2\,ds \right].
$$
\end{prop}

%% ----------------------------------------------------------------
\subsection{Expectation of integrals}

\begin{prop}
\label{prop:property_iii}
For\index{expecation of integrals} all $f$ such that the \Ito integral is
defined,
$$
	E\left[ \int_{t_1}^{t_2} f_s db_s \right] = 0.
$$
\begin{proof}
This is property (iii) on p. 30 of {\O}ksendahl.  The proof is quick
for step functions, since
$$
	E[c\,b_t] =
	cE[\,b_t] =
	0
$$
on each piece; for the general case, one takes limits of step functions.
\end{proof}
\end{prop}
