---
title: "Itō integral"
date: 2021-11-06
categories:
- Mathematics
- Stochastic analysis
layout: "concept"
---

The **Itō integral** offers a way to integrate
a given [stochastic process](/know/concept/stochastic-process/) $G_t$
with respect to a [Wiener process](/know/concept/wiener-process/) $B_t$,
which is also a stochastic process.
The Itō integral $I_t$ of $G_t$ is defined as follows:

$$\begin{aligned}
    \boxed{
        I_t
        \equiv \int_a^b G_t \dd{B_t}
        \equiv \lim_{h \to 0} \sum_{t = a}^{t = b} G_t \big(B_{t + h} - B_t\big)
    }
\end{aligned}$$

Where have partitioned the time interval $[a, b]$ into steps of size $h$.
The above integral exists if $G_t$ and $B_t$ are adapted
to a common filtration $\mathcal{F}_t$,
and $\mathbf{E}[G_t^2]$ is integrable for $t \in [a, b]$.
If $I_t$ exists, $G_t$ is said to be **Itō-integrable** with respect to $B_t$.


## Motivation

Consider the following simple first-order differential equation for $X_t$,
for some function $f$:

$$\begin{aligned}
    \dv{X_t}{t}
    = f(X_t)
\end{aligned}$$

This can be solved numerically using the explicit Euler scheme
by discretizing it with step size $h$,
which can be applied recursively, leading to:

$$\begin{aligned}
    X_{t+h}
    \approx X_{t} + f(X_t) \: h
    \quad \implies \quad
    X_t
    \approx X_0 + \sum_{s = 0}^{s = t} f(X_s) \: h
\end{aligned}$$

In the limit $h \to 0$, this leads to the following unsurprising integral for $X_t$:

$$\begin{aligned}
    \int_0^t f(X_s) \dd{s}
    = \lim_{h \to 0} \sum_{s = 0}^{s = t} f(X_s) \: h
\end{aligned}$$

In contrast, consider the *stochastic differential equation* below,
where $\xi_t$ represents white noise,
which is informally the $t$-derivative
of the Wiener process $\xi_t = \idv{B_t}{t}$:

$$\begin{aligned}
    \dv{X_t}{t}
    = g(X_t) \: \xi_t
\end{aligned}$$

Now $X_t$ is not deterministic,
since $\xi_t$ is derived from a random variable $B_t$.
If $g = 1$, we expect $X_t = X_0 + B_t$.
With this in mind, we introduce the **Euler-Maruyama scheme**:

$$\begin{aligned}
    X_{t+h}
    &= X_t + g(X_t) \: (\xi_{t+h} - \xi_t) \: h
    \\
    &= X_t + g(X_t) \: (B_{t+h} - B_t)
\end{aligned}$$

We would like to turn this into an integral for $X_t$, as we did above.
Therefore, we state:

$$\begin{aligned}
    X_t
    = X_0 + \int_0^t g(X_s) \dd{B_s}
\end{aligned}$$

This integral is *defined* as below,
analogously to the first, but with $h$ replaced by
the increment $B_{t+h} \!-\! B_t$ of a Wiener process.
This is an Itō integral:

$$\begin{aligned}
    \int_0^t g(X_s) \dd{B_s}
    \equiv \lim_{h \to 0} \sum_{s = 0}^{s = t} g(X_s) \big(B_{s + h} - B_s\big)
\end{aligned}$$

For more information about applying the Itō integral in this way,
see the [Itō calculus](/know/concept/ito-calculus/).


## Properties

Since $G_t$ and $B_t$ must be known (i.e. $\mathcal{F}_t$-adapted)
in order to evaluate the Itō integral $I_t$ at any given $t$,
it logically follows that $I_t$ is also $\mathcal{F}_t$-adapted.

Because the Itō integral is defined as the limit of a sum of linear terms,
it inherits this linearity.
Consider two Itō-integrable processes $G_t$ and $H_t$,
and two constants $v, w \in \mathbb{R}$:

$$\begin{aligned}
    \int_a^b v G_t + w H_t \dd{B_t}
    = v\! \int_a^b G_t \dd{B_t} +\: w\! \int_a^b H_t \dd{B_t}
\end{aligned}$$

By adding multiple summations,
the Itō integral clearly satisfies, for $a < b < c$:

$$\begin{aligned}
    \int_a^c G_t \dd{B_t}
    = \int_a^b G_t \dd{B_t} + \int_b^c G_t \dd{B_t}
\end{aligned}$$

A more interesting property is the **Itō isometry**,
which expresses the expectation of the square of an Itō integral of $G_t$
as a simpler "ordinary" integral of the expectation of $G_t^2$
(which exists by the definition of Itō-integrability):

$$\begin{aligned}
    \boxed{
        \mathbf{E} \bigg( \int_a^b G_t \dd{B_t} \bigg)^2
        = \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-isometry"/>
<label for="proof-isometry">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-isometry">Proof.</label>
We write out the left-hand side of the Itō isometry,
where eventually $h \to 0$:

$$\begin{aligned}
    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
    &= \sum_{t = a}^{t = b} \sum_{s = a}^{s = b} \mathbf{E} \bigg[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \bigg]
\end{aligned}$$

In the particular case $t \ge s \!+\! h$,
a given term of this summation can be rewritten
as follows using the *law of total expectation*
(see [conditional expectation](/know/concept/conditional-expectation/)):

$$\begin{aligned}
    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
    = \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$

Recall that $G_t$ and $B_t$ are adapted to $\mathcal{F}_t$:
at time $t$, we have information $\mathcal{F}_t$,
which includes knowledge of the realized values $G_t$ and $B_t$.
Since $t \ge s \!+\! h$ by assumption, we can simply factor out the known quantities:

$$\begin{aligned}
    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
    = \mathbf{E} \bigg[ G_t G_s (B_{s + h} \!-\! B_s) \: \mathbf{E} \Big[ (B_{t + h} \!-\! B_t) \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$

However, $\mathcal{F}_t$ says nothing about
the increment $(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h)$,
meaning that the conditional expectation is zero:

$$\begin{aligned}
    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
    = 0
    \qquad \mathrm{for}\; t \ge s + h
\end{aligned}$$

By swapping $s$ and $t$, the exact same result can be obtained for $s \ge t \!+\! h$:

$$\begin{aligned}
    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
    = 0
    \qquad \mathrm{for}\; s \ge t + h
\end{aligned}$$

This leaves only one case which can be nonzero: $[t, t\!+\!h] = [s, s\!+\!h]$.
Applying the law of total expectation again yields:

$$\begin{aligned}
    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
    &= \sum_{t = a}^{t = b} \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big]
    \\
    &= \sum_{t = a}^{t = b} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$

We know $G_t$, and the expectation value of $(B_{t+h} \!-\! B_t)^2$,
since the increment is normally distributed, is simply the variance $h$:

$$\begin{aligned}
    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
    &= \sum_{t = a}^{t = b} \mathbf{E} \big[ G_t^2 \big] h
    \longrightarrow
    \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
\end{aligned}$$
</div>
</div>

Furthermore, Itō integrals are [martingales](/know/concept/martingale/),
meaning that the average noise contribution is zero,
which makes intuitive sense,
since true white noise cannot be biased.

<div class="accordion">
<input type="checkbox" id="proof-martingale"/>
<label for="proof-martingale">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-martingale">Proof.</label>
We will prove that an arbitrary Itō integral $I_t$ is a martingale.
Using additivity, we know that the increment $I_t \!-\! I_s$
is as follows, given information $\mathcal{F}_s$:

$$\begin{aligned}
    \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
    = \mathbf{E} \bigg[ \int_s^t G_u \dd{B_u} \bigg| \mathcal{F}_s \bigg]
    = \lim_{h \to 0} \sum_{u = s}^{u = t} \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_s \Big]
\end{aligned}$$

We rewrite this [conditional expectation](/know/concept/conditional-expectation/)
using the *tower property* for some $\mathcal{F}_u \supset \mathcal{F}_s$,
such that $G_u$ and $B_u$ are known, but $B_{u+h} \!-\! B_u$ is not:

$$\begin{aligned}
    \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
    &= \lim_{h \to 0} \sum_{u = s}^{u = t}
    \mathbf{E} \bigg[ \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_u \Big] \bigg| \mathcal{F}_s \bigg]
    = 0
\end{aligned}$$

We now have everything we need to calculate $\mathbf{E} [ I_t | \mathcal{F_s} ]$,
giving the martingale property:

$$\begin{aligned}
    \mathbf{E} \big[ I_t | \mathcal{F}_s \big]
    = \mathbf{E} \big[ I_s | \mathcal{F}_s \big] + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
    = I_s + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
    = I_s
\end{aligned}$$

For the existence of $I_t$,
we need $\mathbf{E}[G_t^2]$ to be integrable over the target interval,
so from the Itō isometry we have $\mathbf{E}[I]^2 < \infty$,
and therefore $\mathbf{E}[I] < \infty$,
so $I_t$ has all the properties of a Martingale,
since it is trivially $\mathcal{F}_t$-adapted.
</div>
</div>



## References
1.  U.H. Thygesen,
    *Lecture notes on diffusions and stochastic differential equations*,
    2021, Polyteknisk Kompendie.