title: Ito Processes
author: Keith A. Lewis
institute: KALX, LLC
fleqn: true
classoption: fleqn
abstract: Stochastic differential equations and integrals
...
\newcommand{\RR}{\boldsymbol{R}}
\newcommand{\Var}{\operatorname{Var}}
\renewcommand{\o}[1]{\bar{#1}}
A large class of stochastic processes can be defined using
standard Brownian motion $(B_t)_{t\ge 0}$.
The stochastic differential equation
$$
dX = \mu,dt + \sigma,dB, X_{t_0} = x_0
$$
is shorthand for
$$
X_t = x_0 + \int_{t_0}^t \mu,ds + \int_{t_0}^t \sigma,dB_s,
$$
which is shorthand for
$$
X_t(\omega) = x_0 + \int_{t_0}^t \mu(s,\omega),ds + \int_{t_0}^t \sigma(s,\omega),dB_s(\omega),
$$
where $\mu,\sigma\colon [0,\infty)\times\Omega\to\RR$ are the drift and
diffusion coefficients, respectively.
The first integral is (pointwise in $\omega$) Riemann and the second is the Ito integral.
Just as an integral is a continuous linear transformation from a vector space of functions to the real
numbers, a stochastic integral is a continuous linear transformation from a vector space of functions
to a vector space of random variables.
Recall that Brownian motion is a stochastic process $B_t\colon\Omega\to\RR$, $t\ge 0$,
where $B_t(\omega) = \omega(t)$ for $\omega\in\Omega = C[0,\infty)$, the space
of continuous function from $[0,\infty)$ to the real numbers $\RR$.
The information available at time $t$ is specified by $\mathcal{F}t$,
the smallest $\sigma$-algebra for which $(B_s){0\le s\le t}$ are measurable.
If $\omega\mapsto\sigma(t, \omega)$ is $\mathcal{F}t$ measurable (only depends on
the information available at time $t$) then
so is the Ito integral $\int_0^t \sigma(s, \omega),dB_s(\omega)$ defined as a limit
$$
\int_0^t \sigma(s, \omega),dB_s(\omega)
= \lim{\Delta T\to 0} \sum_{0\le j < n} \sigma(t_j, \omega),\Delta B_j(\omega)
$$
where $0 = t_0 < \cdots < t_n = t$ is a partition of $[0,t]$ and $\Delta B_j = B_{t_{j+1}} - B_{t_j}$.
We let $T = {t_j}$ denote the partition and define $\Delta T = \max_{0\le j < n}{\Delta t_j}$.
The set of all partitions of $[0,t]$ is a net and
the limit is defined as described therein.
Exercise.
Show $E[\sum{0\le j < n} B_{t_j}\Delta B_j] = 0$ for any partition
$0 = t_0 < \cdots < t_n = t$ of $[0,t]$_.
This shows $E[\int_0^t B_s,dB_s] = 0$. Later we will see $\int_0^t B_s dB_s = (B_t^2 - t)/2$.
Exercise.
Show $E[\sum{0\le j < n} B_{t_{j+1}}\Delta B_j] = t \not= 0$_.
Hint: $E[B_t B_u] = \min{t, u}$. Note the integrand $\omega\mapsto B_{t_{j+1}}(\omega)$
is not $\mathcal{F}_{t_j}$ measurable.
The Riemann integral $\int_0^t f(s),ds = \lim_{\Delta T\to 0} \sum_j f(t_j^)\Delta t_j$ converges
to the same value if $f$ is continuous and $t_j^$ is any point in $[t_j, t_{j+1}]$.
Unlike the Riemann integral, the Ito integral requires the left endpoint be used.
Exercise. Show $E[\sum{0\le j < n} (\Delta B_j)^2] = t$ for any partition_.
Hint: $E[(\Delta B)^2] = \Delta t$.
Exercise. Show $\lim{\Delta T\to 0}\sum_{0\le j < n} (\Delta t_j)^2 = 0$_.
Hint: $\sum_j (\Delta t_j)^2 \le (\max_j \Delta t_j) \sum_j \Delta t_j = (\max_j \Delta t_j)t$.
Shorthand notation for this is $(dt)^2 = 0$.
Exercise Show $\lim{\Delta T\to 0}\Var(\sum_{0\le j < n} (\Delta B_j)^2) = 0$_.
Hint: $E[(\Delta B)^4] = 3(\Delta t)^2$.
Solution
We have $\Var(\sum_{0\le j < n} (\Delta B_j)^2) = \sum_{0\le j < n} 3(\Delta t)^2$
since Brownian motion has independent increments and $E[(\Delta B)^4] 3(\Delta t)^2$.
Although $\sum_{0\le j < n} (\Delta B_j)^2$ is random for any given partition, it always
has expected value $t$ and its variance tends to 0 as the partition gets finer.
Shorthand notation for this is $(dB)^2 = dt$.
Exercise Show $\sum{0\le j < n} \Delta B_j \Delta t_j$ has mean 0 and its variance tends to 0
as $\Delta T\to 0$_.
Shorthand notation for this is $dB,dt = 0$.
The Ito integral can be extended to $m$-dimensional Brownian motion, $B_t = (B_t^1,\ldots,B_t^m)$,
where the $B_t^j$ are independent standard Brownian motions. In this case
$\mu\colon [0,\infty)\times\Omega\to\RR^n$ is vector-valued
and $\sigma\colon [0,\infty)\times\Omega\to\RR^{n,m}$ is matrix-valued.
The $\sigma dB$ term is the matrix-vector product.
The Ito integral can be generalized to stochastic process other than Brownian motion.
If $X_t$ is any stochastic process we can define $dY_t = \sigma,dX_t$, $Y_0 = y$, by
$$
Y_t(\omega) = y + \lim_{\Delta T\to 0}\sum_j \sigma(t_j, \omega) \Delta X_j(\omega)
$$
where $\Delta X_j = X_{t_{j+1}} - X_{t_j}$. For $Y_t$ to be $\mathcal{F}_t$-measurable
we need $\sigma$ to be adapted. Of course there must be restrictions on $X_t$ too in order
to ensure convergence and continuity. If $X_t$ is a martingale that is (almost surely)
continuous from the right and has left limits then a well-behaved stochastic integral can be define.
Note Brownian motion satisfies this since it is almost everywhere continuous.
This can be generalized by adding a stochastic process to the martingale that has bounded variation.
As shown in [@Pro04], this is the most general process
for which a well-behaved stochastic integral can be defined.
An Ito process $X_t:\Omega\to\RR$, $t\ge0$, satisfies the SDE
$$
dX_t(\omega) = \mu(t,\omega),dt + \sigma(t,\omega),dB_t(\omega), X_{t_0} = x_0, t\ge t_0.
$$
where $\mu,\sigma:[0,\infty)\times\Omega\to\RR$ are functions of time and the Brownian
sample path $\omega\in\Omega$.
If $\omega\mapsto \mu(t, \omega)$ is $\mathcal{F}_t$ measurable for all $t$ then so is
the Riemann integral $\omega\mapsto \int_0^t \mu(s, \omega),ds$.
If $\omega\mapsto \sigma(t, \omega)$ is $\mathcal{F}_t$ measurable for all $t$ then so is
the Ito integral $\omega\mapsto \int_0^t \sigma(s, \omega),dB_s(\omega)$.
Exercise. In this case the Ito integral $\int_0^t \sigma dB$ is a martingale.
Hint: $E[B_u - B_t\mid\mathcal{F}_s] = 0$ if $s\le t\le u$.
Exercise. If $dX = \mu,dt + \sigma,dB$ is a martingale then $\mu = 0$.
Hint: Show $\int_t^u E_t[\mu(s,\omega)],ds = 0$ for $t\le u$, $\omega\in\Omega$.
Exercise. The sum of Ito processes is an Ito process.
Hint: If $dX^j = \mu_j,dt + \sigma_j,dB$ are Ito processes what are $\mu$ and $\sigma$
for $X = \sum_j X_j$? Write out the integrals in full.
If $X_t$ and $Y_t$ are Ito processes then so is their product $X_tY_t$.
Exercise. Show $d(XY) = Y,dX + X,dY + dX,dY$.
Hint: Show $X_t Y_t = X_0 Y_0 + \lim_{\Delta T\to 0}\sum_j Y_j\Delta X_j + Y_j\Delta Y_j + \Delta X_j\Delta Y_j$
where where $\Delta X_j = X_{t_{j+1}} - X_{t_j}$, etc., using
$(X_j + \Delta X_j)(Y_j + \Delta Y_j) = X_jY_j + Y_j\Delta X_j + X_j\Delta Y_j + \Delta X_j \Delta Y_j$.
Exercise. If $dX = \mu,dt + \sigma,dB$ and $dY = \nu,dt + \tau,dB$
show $d(XY) = (\mu Y + \nu X + \sigma\tau),dt + (\sigma Y + \tau X),dB$.
Hint: Show
$$
\begin{aligned}
X_t(\omega)Y_t(\omega) = &X_0 Y_0 \
&+ \int_0^t (\mu(s,\omega) Y_s(\omega) + \nu(s,\omega) X_s(\omega) + \sigma(s, \omega)\tau(s,\omega)),ds \
&+ \int_0^t (\sigma(s,\omega) Y_s(\omega) + \tau(s, \omega) X_s(\omega)),dB_s(\omega) \
\end{aligned}
$$
The Ito calculus uses $(dt)^2 = 0$, $dt,dB = 0 = dB,dt$, and $(dB)^2 = dt$ to simplify such calculations.
$$
\begin{aligned}
d(XY) &= Y,dX + X,dY + dX,dY \
&= Y(\mu,dt + \sigma,dB) + X(\nu,dt + \tau,dB) + (\mu,dt + \sigma,dB)(\nu,dt + \tau,dB) \
&= (\mu Y + \nu X + \sigma\tau),dt + (\sigma Y + \tau X),dB.
\end{aligned}
$$
Solution
Use $(XY)_t = X_0Y_0 + \int_0^t d(XY)_s = X_0Y_0 + \int_0^t Y_s\,dX_s X_s + X_s\,dY_s
+ dX_s\,dY_s$.
Using the previous exercises we have $\int_0^t Y_s,dX_s = \int_0^t Y_s(\mu,dt + \sigma,dB_s)$ and
$\int_0^t X_s,dY_s = \int_0^t X_s(\nu,dt + \tau,dB_s)$.
$\int_0^t dX_s,dY_s = \int_0^t (\mu,dt + \sigma,dB_s)(\nu,dt + \tau,dB_s)
= \int_0^t \sigma\tau,dB^2_s = \int_0^t \sigma\tau,ds$.
Exercise. If $dS/S = \mu,dt + \Sigma,dB$ and $\Sigma^2 = (1/t)\int_0^t (dS/S)^2$
show $\Sigma$ is constant.
Hint. Compute $t\Sigma^2$ two ways.
If $X_t$ is an Ito process and $f\colon[0,\infty)\times\RR\to\RR$ then $Y_t = f(t, X_t)$ is
also an Ito process satisfying the SDE
$$
dY_t = f_t(t, X_t),dt + f_x(t, X_t),dX_t + \frac{1}{2} f_{xx}(t, X_t) (dX_t)^2
$$
If $X_t = B_t$ is standard Brownian motion then
$dY_t = (f_t + \frac{1}{2}f_{xx}),dt + f_x,dB$.
Exercise. If $f_t + \frac{1}{2}f{xx} = 0$ then $f(t,B_t)$ is a martingale_.
Exercise. Show $f(t, x) = x^2 - t$ and $f(t, x) = se^{\sigma x - \sigma^2t/2}$
satisfy $f_t + \frac{1}{2}f{xx} = 0$_.
Exercise. Show $\int_0^t B_s,dB_s = B_t^2/2$.
An Ito diffusion $\o{X}_t(\omega)$ satisfies
$$
d\o{X}_t(\omega) = \o{\mu}(t,\o{X}_t(\omega)),dt + \o{\sigma}(t,\o{X}t(\omega)),dB_t(\omega),
\o{X}{t_0} = x, t \ge t_0
$$
where $\o{\mu},\o{\sigma}\colon [0,\infty)\times\RR\to\RR$.
Exercies. Show an Ito diffusion is an Ito process.
The Ito formula holds for diffusions but we have $f(t, X_t)$ is also an Ito diffusion.
Exercise. If $X_t$ and $Y_t$ are Ito diffusions then so are $X_t + Y_t$ and $X_t Y_t$.
Ito diffusions satisfy the Markov property. Loosely speaking,
$E[f(t, \o{X}_t)\mid\mathcal{F}_t] = E[f(t, \o{X}_t)\mid \o{X}_t]$.
The conditional expectation does not depend on
the trajectory $s\mapsto X_s(\omega)$, $0\le s\le t$, it only depends
on the final value $X_t(\omega)$.
We can define vector-valued Ito processes. Let $B_t = (B_t^j)_{j=1}^n$ be
independent Brownian motions. We write
$dX = \mu,dt + \sigma,dB_t$ where $\mu$
takes values in $\RR^n$ and $\sigma$ is
an $n\times n$ matrix.