Itô Formula
1 Itô Process
Let \(B_{t}\) be a 1-dimensional Brownian motion on \((\Omega, \mathcal{F}, \mathbb{P})\). A (1-dimensional) Itô process (or stochastic integral) is a stochastic process \(X_{t}\) on \((\Omega, \mathcal{F}, \mathbb{P})\) of the form \[ X_{t}=x+\int_{0}^{t}{u(s, \omega )}~d{s} + \int_{0}^{t}{v(s,\omega)}~d{B_{s}}, \] where we say that \(u\) is the drift coefficient (controls the growth of the process) and \(v\) is the diffusion coefficient (controls the size of the noise). Sometimes this is written in the shorter differential form \[ dX_{t}=udt+vdB_{t}. \]
2 Itô Formula
See also Itô Integral for the Itô integral and Stochastic Calculus for the wider picture.
Itô’s formula or Itô’s lemma is an identity used in Itô calculus to find the differential of a time-dependent function of a stochastic process. It serves as the stochastic calculus counterpart of the chain rule.
Let \(X_{t}\) be an Itô process given by \[ dX_{t}=udt+vdB_{t}. \] Further, let \(g(t,x)\in C^2([0,\infty)\times \mathbb{R})\) (i.e. twice differentiable on \([0,\infty)\times \mathbb{R}\)). Then \[ Y_{t}=g(t,X_{t}) \] is again an Itô process and \[ dY_{t}=g_{t}(t, X_{t})dt+g_{x}(t, X_{t})dX_{t}+ \frac{1}{2}g_{x x}(t, X_{t})d\left< X,X \right>_{t}. \]
Here \(d\left< X,X \right>_{t}=(dX_{t})\cdot(dX_{t})\) is computed according to the following: \[ dt\cdot dt=dt\cdot dB_{t}=dB_{t}\cdot dt=0,\quad dB_{t}\cdot dB_{t}=dt. \]
Proof. Substituting \(dX_{t}=udt+vdB_{t}\) into the Itô formula we get \[ \begin{align} g(t,X_{t}) & = g(0,X_{0}) + \int_{0}^t\left( \frac{{\partial g}}{\partial s}(s,X_{s})+u_{s}\frac{{\partial g}}{\partial x}(s,X_{s})+ \frac{1}{2} v_{s}^2 \frac{{\partial^2g}}{\partial x^2}(s,X_{s}) \right)ds \\ & + \int_{0}^t v_{s} \frac{{\partial g}}{\partial x}(s,X_{s})dB_{s}. \end{align} \] We may assume that \(g, g_{t}, g_{x}, g_{x x}\) are bounded for each \(n\) and converge uniformly on compact subsets of \([0,\infty)\times \mathbb{R}\) to \(g,g_{t},g_{x},g_{x x}\) respectively. Moreover, we may assume that \(u,v\) are elementary functions. Using Taylor’s theorem we get \[ \begin{align} g(t,X_{t}) & = g(0,X_{0}) + \sum_{j}\Delta g(t_{j}, X_{j}) \\ & = g(0,X_{0})+\sum_{j} \frac{{\partial g}}{\partial t} \Delta t_{j}+\sum_{j} \frac{{\partial g}}{\partial x}\Delta X_{j}+ \frac{1}{2}\sum_{j} \frac{{\partial^2g}}{\partial t^2}(\Delta t_{j}^2) \\ & \qquad + \sum_{j} \frac{{\partial^2g}}{\partial t\partial x}(\Delta t_{j})(\Delta X_{j}) + \frac{1}{2}\sum_{j} \frac{{\partial^2g}}{\partial x^2}(\Delta X_{j})^2 + \sum_{j}R_{j}, \end{align} \] where \(\frac{{\partial g}}{\partial t}, \frac{{\partial g}}{\partial x}\) etc. are evaluated at the points \((t_{j}, X_{t_{j}})\), \[ \begin{align} \Delta t_{j} & =t_{j+1}-t_{j} \\ \Delta X_{j} & = X_{t_{j+1}}-X_{t_{j}} \\ \Delta g(t_{j},X_{j}) & = g(t_{j+1},X_{t_{j+1}})-g(t_{j},X_{j}), \end{align} \] and \(R_{j}=o(\lvert \Delta t_{j} \rvert^2+\lvert \Delta X_{j} \rvert^2)\) for all \(j\). If \(\Delta t_{j}\to 0\) then \[ \begin{align} \sum_{j} \frac{{\partial g}}{\partial t}\Delta t_{j} & =\sum_{j} \frac{{\partial g}}{\partial t}(t_{j}, X_{j})\Delta t_{j} \to \int_{0}^t \frac{{\partial g}}{\partial t}(s,X_{s})~ds \\ \sum_{j} \frac{{\partial^2 g}}{\partial x^2} \Delta X_{j} & = \sum_{j} \frac{{\partial g}}{\partial x}(t_{j}, X_{j})\Delta X_{j} \to \int_{0}^t \frac{{\partial g}}{\partial x}(s,X_{s})dX_{s}. \end{align} \] Moreover, since \(u\) and \(v\) are elementary we get \[ \begin{align} \sum_{j} \frac{{\partial^2 g}}{\partial x^2}(\Delta X_{j})^2 & = \underbrace{\sum_{j} \frac{{\partial^2g}}{\partial x^2}u_{j}^2 (\Delta t_{j})^2}_{\to 0\text{ as }\Delta t_{j}\to 0} + \underbrace{2 \sum_{j} \frac{{\partial^2g}}{\partial x^2} u_{j}v_{j}(\Delta t_{j})(\Delta B_{j})}_{\to 0\text{ as }\Delta t_{j}\to 0} +\sum_{j} \frac{{\partial^2g}}{\partial x^2}v_{j}^2 \cdot (\Delta B_{j})^2, \end{align} \] where \(u_{j}=u(t_{j},\omega)\) and \(v_{j}=v(t_{j},\omega)\) and convergence in \(L_2\). We claim the third term tends to \[ \int_{0}^t \frac{{\partial^2g}}{\partial x^2}v^2~ds. \] To prove this put \(a(t)= \frac{{\partial^2g}}{\partial x^2}(t,X_{t})v^2(t,\omega)\), \(a_{j}=a(t_{j})\) and consider \[ \mathbb{E}\left[ \left( \sum_{j} a_{j}(\Delta B_{j})^2-\sum_{j}a_{j}\Delta t_{j} \right)^2 \right] = \sum_{i,j} \mathbb{E}[a_{i}a_{j}((\Delta B_{i})^2-\Delta t_{i})((\Delta B_{j})^2 - \Delta t_{j})]. \] If \(i<j\) then \(a_{i}a_{j}((\Delta B_{i})^2-\Delta t_{i})\) and \((\Delta B_{j})^2-\Delta t_{j}\) are independent so the terms vanish, and similarly if \(i>j\). This leaves \[ \begin{align} \sum_{j}\mathbb{E}[a_{j}^2((\Delta B_{j})^2-\Delta t_{j})^2] & = \sum_{j}\mathbb{E}[a_{j}^2] \cdot \mathbb{E}[(\Delta B_{j})^4-2(\Delta B_{j})^2\Delta t_{j} + (\Delta t_{j})^2] \\ & = \sum_{j}\mathbb{E}[a_{j}^2]\cdot (3(\Delta t_{j})^2 - 2(\Delta t_{j})^2+(\Delta t_{j})^2) \\ & =2\sum_{j}\mathbb{E}[a_{j}^2]\cdot(\Delta t_{j})^2 \\ & \to 0\quad (\Delta t_{j}\to 0). \end{align} \] Thus we have established that \[ \sum_{j}a_{j}(\Delta B_{j})^2 \to \int_{0}^t a(s)ds, \] which is often expressed as \[ (dB_{t})^2=dt. \] The argument above also proves that \(\sum R_{j}\to 0\) as \(\Delta t_{j}\to_{0}\).
Example (Itô Formula Application I) Consider the integral \[ Y_{t}=\int_{0}^tB_{s}dB_{s}. \] Assuming we can write \(Y_{t}=g(t, B_{t})\) from Itô’s formula we have \[ dY_{t}=\left[ g_{t}(t,B_{t})+\frac{1}{2}g_{BB}(t,B_{t}) \right] dt+g_{B}(t,B_{t})dB_{t}, \] which written in integral form gives \[ Y_{t}-Y_{0}=\int_{0}^t\left[ g_{t}(s,B_{s})+\frac{1}{2}g_{B B}(s,B_{s}) \right]ds+\int_{0}^t g_{B}(s, B_{s})dB_{s}. \] Hence we need some \(g(t,B_{t})\) such that \[ (1)~~g_{t}(t,B_{t})+\frac{1}{2}g_{BB}(t, B_{t})=0\quad \&\quad (2)~~ g_{B}(t,B_{t})=B_{t}. \] Since \(g_{B}=B_{t}\implies g_{BB}=1\) and so \(g_{t}=-\frac{1}{2}\). An obvious candidate therefore is \[ g(t,B_{t})=\frac{1}{2}B_{t}^2 - \frac{1}{2}t, \] which in fact is the result, as can be verified by plugging into Itô’s formula.
Example (Itô Formula Application II) Consider the Itô process \[ g(t,B_{t})=\exp(bt+\sigma B_{t}), \] where \(b,\sigma\) are constants for drift and diffusion. To compute \(dg\) we apply the Itô formula \[ \begin{align} dg(t,B_{t}) & =b\exp(bt+\sigma B_{t})dt+\sigma \exp(bt+\sigma B_{t})dB_{t}+ \frac{1}{2} \sigma^2\exp(bt+\sigma B_{t})d\left< B,B \right>_{t} \\ & = g(t,B_{t})\left\{ \left( b+ \frac{1}{2} \sigma^2 \right)dt+\sigma dB_{t} \right\}. \end{align} \] The process \(X_{t}=g(t,B_{t})\) defines geometric BM and is used to simulate the change of the stock price in the Black-Scholes-Merton Model (\(\mu\) as mean return and \(\sigma\) as volatility).
For a continuous function \(f(s,\omega)\) that is of bounded variation with respect to \(s \in [0,t]\) almost surely we have that \[ \int_{0}^t f(s)dB_{s}=f(t)B_{t}-\int_{0}^t B_{s}df_{s}. \]
3 Multi-dimensional Itô Formula
Considering the situation in higher dimensions we define a \(d\)-dimensional Itô process as the integral with respect to the \(m\)-dimensional BM as \[ X_{t}^i=x_{0}^i+\int_{0}^t\psi_{s}^ids+\sum_{k=1}^m \int_{0}^t \varphi_{s}^{i,k}dB_{s}^k, \] for \(k = 1, \dots, m\) and \(i = 1, \dots, d\) where we note that such an Itô process exists in \(\mathbb{R}^d\) and is constructed with respect to BM \(B_{s}=(B_{s}^1, \dots, B_{s}^m)\). To write this in a more compact form \[ X_{t}=x_{0}+\int_{0}^t \psi_{s}ds+\int_{0}^t\varphi_{s}\cdot dB_{s}, \] where \(X_{t}, x_{0},\psi_{s} \in \mathbb{R}^d\) and \(\varphi_{s}\in \mathbb{R}^{d \times m}\) and here \(\varphi_{s}^{i,k}\) stands for the process \(\varphi\) used to construct the \(i\)-th coordinate of the Itô process as a stochastic integral with respect to the \(k\)-th coordinate of the \(m\)-dimensional BM.
For some vector-valued function \(g:\mathbb{R}_{+}\times \mathbb{R}^d\to \mathbb{R}^p \in C^{1,2}\), then the multi-dimensional Itô formula holds \[ \begin{align}dg^i(t,X_{t}) & =\partial_{t}g^i(t,X_{t})dt+\nabla_{x} g^i(t,X_{t})\cdot dX_{t}+ \frac{1}{2}\mathrm{Tr}(\varphi_{t}^T H \varphi_{t})dt \\ & = \partial_{t}g^i(t,X_{t})dt+\sum_{k=1}^{d}\partial_{x_{k}}g^i(t,X_{t})dX_{t}^k+ \frac{1}{2}\sum_{j,k = 1}^{d}\partial_{x_{j}, x_{k}}g^i(t,X_{t})d\left< X^j, X^k \right> _{t},\end{align} \] where \(\nabla _x f\) denotes the gradient of \(f\) with respect to vector \(x\), \(\varphi_{t}\in \mathbb{R}^{d \times m}\) is a matrix and \(H_{d \times d}\) is the Hessian of \(g\) restricted on its action on \(x \in \mathbb{R}^d\) at \((t, X_{t})\). The bracket \(d\left< X^j, X^k \right>_{t}=\sum_{l=1}^{m}\varphi_{t}^{j,l}\varphi_{t}^{k,l}dt\).
Proof. The structure of the proof is still exactly the same as the 1-dimensional case. We need only verify that \(dB_{t}^pdB_{t}^q=0\) for \(p \neq q\) and \(d\left< X^j,X^k \right>_{t}=\sum_{l=1}^{m}\varphi_{t}^{j,l}\varphi_{t}^{k,l}dt\).