---
title: "Itō integral"
sort_title: "Ito integral" # sic
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.