Categories: Mathematics, Stochastic analysis.

# Itō process

Given two stochastic processes $F_t$ and $G_t$, consider the following random variable $X_t$, where $B_t$ is the Wiener process, i.e. Brownian motion:

\begin{aligned} X_t = X_0 + \int_0^t F_s \dd{s} + \int_0^t G_s \dd{B_s} \end{aligned}

Where the latter is an Itō integral, assuming $G_t$ is Itō-integrable. We call $X_t$ an Itō process if $F_t$ is locally integrable, and the initial condition $X_0$ is known, i.e. $X_0$ is $\mathcal{F}_0$-measurable, where $\mathcal{F}_t$ is the filtration to which $F_t$, $G_t$ and $B_t$ are adapted. The above definition of $X_t$ is often abbreviated as follows, where $X_0$ is implicit:

\begin{aligned} \dd{X_t} = F_t \dd{t} + G_t \dd{B_t} \end{aligned}

Typically, $F_t$ is referred to as the drift of $X_t$, and $G_t$ as its intensity. Because the Itō integral of $G_t$ is a martingale, it does not contribute to the mean of $X_t$:

\begin{aligned} \mathbf{E}[X_t] = \int_0^t \mathbf{E}[F_s] \dd{s} \end{aligned}

Now, consider the following Itō stochastic differential equation (SDE), where $\xi_t = \idv{B_t}{t}$ is white noise, informally treated as the $t$-derivative of $B_t$:

\begin{aligned} \dv{X_t}{t} = f(X_t, t) + g(X_t, t) \: \xi_t \end{aligned}

An Itō process $X_t$ is said to satisfy this equation if $f(X_t, t) = F_t$ and $g(X_t, t) = G_t$, in which case $X_t$ is also called an Itō diffusion. All Itō diffusions are Markov processes, since only the current value of $X_t$ determines the future, and $B_t$ is also a Markov process.

## Itō’s lemma

Classically, given $y \equiv h(x(t), t)$, the chain rule of differentiation states that:

\begin{aligned} \dd{y} = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{x} \end{aligned}

However, for a stochastic process $Y_t \equiv h(X_t, t)$, where $X_t$ is an Itō process, the chain rule is modified to the following, known as Itō’s lemma:

\begin{aligned} \boxed{ \dd{Y_t} = \bigg( \pdv{h}{t} + \pdv{h}{x} F_t + \frac{1}{2} \pdvn{2}{h}{x} G_t^2 \bigg) \dd{t} + \pdv{h}{x} G_t \dd{B_t} } \end{aligned}

We start by applying the classical chain rule, but we go to second order in $x$. This is also valid classically, but there we would neglect all higher-order infinitesimals:

\begin{aligned} \dd{Y_t} = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{X_t} + \frac{1}{2} \pdvn{2}{h}{x} \dd{X_t}^2 \end{aligned}

But here we cannot neglect $\dd{X_t}^2$. We insert the definition of an Itō process:

\begin{aligned} \dd{Y_t} &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big) + \frac{1}{2} \pdvn{2}{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big)^2 \\ &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big) + \frac{1}{2} \pdvn{2}{h}{x} \Big( F_t^2 \dd{t}^2 + 2 F_t G_t \dd{t} \dd{B_t} + G_t^2 \dd{B_t}^2 \Big) \end{aligned}

In the limit of small $\dd{t}$, we can neglect $\dd{t}^2$, and as it turns out, $\dd{t} \dd{B_t}$ too:

\begin{aligned} \dd{t} \dd{B_t} &= (B_{t + \dd{t}} - B_t) \dd{t} \sim \dd{t} \mathcal{N}(0, \dd{t}) \sim \mathcal{N}(0, \dd{t}^3) \longrightarrow 0 \end{aligned}

However, due to the scaling property of $B_t$, we cannot ignore $\dd{B_t}^2$, which has order $\dd{t}$:

\begin{aligned} \dd{B_t}^2 &= (B_{t + \dd{t}} - B_t)^2 \sim \big( \mathcal{N}(0, \dd{t}) \big)^2 \sim \chi^2_1(\dd{t}) \longrightarrow \dd{t} \end{aligned}

Where $\chi_1^2(\dd{t})$ is the generalized chi-squared distribution with one term of variance $\dd{t}$.

The most important application of Itō’s lemma is to perform coordinate transformations, to make the solution of a given Itō SDE easier.

## Coordinate transformations

The simplest coordinate transformation is a scaling of the time axis. Defining $s \equiv \alpha t$, the goal is to keep the Itō process. We know how to scale $B_t$, be setting $W_s \equiv \sqrt{\alpha} B_{s / \alpha}$. Let $Y_s \equiv X_t$ be the new variable on the rescaled axis, then:

\begin{aligned} \dd{Y_s} = \dd{X_t} &= f(X_t) \dd{t} + g(X_t) \dd{B_t} \\ &= \frac{1}{\alpha} f(Y_s) \dd{s} + \frac{1}{\sqrt{\alpha}} g(Y_s) \dd{W_s} \end{aligned}

$W_s$ is a valid Wiener process, and the other changes are small, so this is still an Itō process.

To solve SDEs analytically, it is usually best to have additive noise, i.e. $g = 1$. This can be achieved using the Lamperti transform: define $Y_t \equiv h(X_t)$, where $h$ is given by:

\begin{aligned} \boxed{ h(x) = \int_{x_0}^x \frac{1}{g(y)} \dd{y} } \end{aligned}

Then, using Itō’s lemma, it is straightforward to show that the intensity becomes $1$. Note that the lower integration limit $x_0$ does not enter:

\begin{aligned} \dd{Y_t} &= \bigg( f(X_t) \: h'(X_t) + \frac{1}{2} g^2(X_t) \: h''(X_t) \bigg) \dd{t} + g(X_t) \: h'(X_t) \dd{B_t} \\ &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g^2(X_t) \frac{g'(X_t)}{g^2(X_t)} \bigg) \dd{t} + \frac{g(X_t)}{g(X_t)} \dd{B_t} \\ &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g'(X_t) \bigg) \dd{t} + \dd{B_t} \end{aligned}

Similarly, we can eliminate the drift $f = 0$, thereby making the Itō process a martingale. This is done by defining $Y_t \equiv h(X_t)$, with $h(x)$ given by:

\begin{aligned} \boxed{ h(x) = \int_{x_0}^x \exp\!\bigg( \!-\!\! \int_{x_1}^y \frac{2 f(z)}{g^2(z)} \dd{z} \bigg) \dd{y} } \end{aligned}

The goal is to make the parenthesized first term (see above) of Itō’s lemma disappear, which this $h(x)$ does indeed do. Note that $x_0$ and $x_1$ do not enter:

\begin{aligned} 0 &= f(x) \: h'(x) + \frac{1}{2} g^2(x) \: h''(x) \\ &= \Big( f(x) - \frac{1}{2} g^2(x) \frac{2 f(x)}{g^2(x)} \Big) \exp\!\bigg( \!-\!\! \int_{x_1}^x \frac{2 f(y)}{g^2(y)} \dd{y} \bigg) \end{aligned}

## Existence and uniqueness

It is worth knowing under what condition a solution to a given SDE exists, in the sense that it is finite on the entire time axis. Suppose the drift $f$ and intensity $g$ satisfy these inequalities, for some known constant $K$ and for all $x$:

\begin{aligned} x f(x) \le K (1 + x^2) \qquad \quad g^2(x) \le K (1 + x^2) \end{aligned}

When this is satisfied, we can find the following upper bound on an Itō process $X_t$, which clearly implies that $X_t$ is finite for all $t$:

\begin{aligned} \boxed{ \mathbf{E}[X_t^2] \le \big(X_0^2 + 3 K t\big) \exp\!\big(3 K t\big) } \end{aligned}

If we define $Y_t \equiv X_t^2$, then Itō’s lemma tells us that the following holds:

\begin{aligned} \dd{Y_t} = \big( 2 X_t \: f(X_t) + g^2(X_t) \big) \dd{t} + 2 X_t \: g(X_t) \dd{B_t} \end{aligned}

Integrating and taking the expectation value removes the Wiener term, leaving:

\begin{aligned} \mathbf{E}[Y_t] = Y_0 + \mathbf{E}\! \int_0^t 2 X_s f(X_s) + g^2(X_s) \dd{s} \end{aligned}

Given that $K (1 \!+\! x^2)$ is an upper bound of $x f(x)$ and $g^2(x)$, we get an inequality:

\begin{aligned} \mathbf{E}[Y_t] &\le Y_0 + \mathbf{E}\! \int_0^t 2 K (1 \!+\! X_s^2) + K (1 \!+\! X_s^2) \dd{s} \\ &\le Y_0 + \int_0^t 3 K (1 + \mathbf{E}[Y_s]) \dd{s} \\ &\le Y_0 + 3 K t + \int_0^t 3 K \big( \mathbf{E}[Y_s] \big) \dd{s} \end{aligned}

We then apply the Grönwall-Bellman inequality, noting that $(Y_0 \!+\! 3 K t)$ does not decrease with time, leading us to:

\begin{aligned} \mathbf{E}[Y_t] &\le (Y_0 + 3 K t) \exp\!\bigg( \int_0^t 3 K \dd{s} \bigg) \\ &\le (Y_0 + 3 K t) \exp\!\big(3 K t\big) \end{aligned}

If a solution exists, it is also worth knowing whether it is unique. Suppose that $f$ and $g$ satisfy the following inequalities, for some constant $K$ and for all $x$ and $y$:

\begin{aligned} \big| f(x) - f(y) \big| \le K \big| x - y \big| \qquad \quad \big| g(x) - g(y) \big| \le K \big| x - y \big| \end{aligned}

Let $X_t$ and $Y_t$ both be solutions to a given SDE, but the initial conditions need not be the same, such that the difference is initially $X_0 \!-\! Y_0$. Then the difference $X_t \!-\! Y_t$ is bounded by:

\begin{aligned} \boxed{ \mathbf{E}\big[ (X_t - Y_t)^2 \big] \le (X_0 - Y_0)^2 \exp\!\Big( \big(2 K \!+\! K^2 \big) t \Big) } \end{aligned}

We define $D_t \equiv X_t \!-\! Y_t$ and $Z_t \equiv D_t^2 \ge 0$, together with $F_t \equiv f(X_t) \!-\! f(Y_t)$ and $G_t \equiv g(X_t) \!-\! g(Y_t)$, such that Itō’s lemma states:

\begin{aligned} \dd{Z_t} = \big( 2 D_t F_t + G_t^2 \big) \dd{t} + 2 D_t G_t \dd{B_t} \end{aligned}

Integrating and taking the expectation value removes the Wiener term, leaving:

\begin{aligned} \mathbf{E}[Z_t] = Z_0 + \mathbf{E}\! \int_0^t 2 D_s F_s + G_s^2 \dd{s} \end{aligned}

The Cauchy-Schwarz inequality states that $|D_s F_s| \le |D_s| |F_s|$, and then the given fact that $F_s$ and $G_s$ satisfy $|F_s| \le K |D_s|$ and $|G_s| \le K |D_s|$ gives:

\begin{aligned} \mathbf{E}[Z_t] &\le Z_0 + \mathbf{E}\! \int_0^t 2 K D_s^2 + K^2 D_s^2 \dd{s} \\ &\le Z_0 + \int_0^t (2 K \!+\! K^2) \: \mathbf{E}[Z_s] \dd{s} \end{aligned}

Where we have implicitly used that $D_s F_s = |D_s F_s|$ because $Z_t$ is positive for all $G_s^2$, and that $|D_s|^2 = D_s^2$ because $D_s$ is real. We then apply the Grönwall-Bellman inequality, recognizing that $Z_0$ does not decrease with time (since it is constant):

\begin{aligned} \mathbf{E}[Z_t] &\le Z_0 \exp\!\bigg( \int_0^t 2 K \!+\! K^2 \dd{s} \bigg) \\ &\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big) \end{aligned}

Using these properties, it can then be shown that if all of the above conditions are satisfied, then the SDE has a unique solution, which is $\mathcal{F}_t$-adapted, continuous, and exists for all times.

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