Categories: Mathematics, Stochastic analysis.

Itō integral

The Itō integral offers a way to integrate a given stochastic process $G_t$ with respect to a 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.

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}

Proof

Proof. 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):

\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}

Furthermore, Itō integrals are martingales, meaning that the average noise contribution is zero, which makes intuitive sense, since true white noise cannot be biased.

Proof

Proof. 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 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.

References

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