Part 3 - Itô Calculus

Introduction

Recall that we study SDEs of the form,

dx(t)=f(x(t),t) dtdrift+L(x(t),t) dβ(t)diffusion,x(t)x(t0)=t0tf(x(s),s) dsmean square Riemann Integral+t0tL(x(s),s) dβ(s)Itoˆ Integral,\begin{align*} dx(t) & = \underbrace{f(x(t), t) \ dt}_{\text{drift}} + \underbrace{L(x(t), t) \ d\beta(t)}_{\text{diffusion}}, \newline x(t) - x(t_0) & = \underbrace{\int_{t_0}^{t} f(x(s), s) \ ds}_{\text{mean square Riemann Integral}} + \underbrace{\int_{t_0}^{t} L(x(s), s) \ d\beta(s)}_{\text{Itô Integral}}, \end{align*}

where x(t)x(t) is a stochastic process and β(t)\beta(t) is a Brownian motion.

We define the Itô integral (of the diffusion term) as,

t0tL(x(s),s) dβ(s)=l.i.mnP0i=0n1L(x(ti),ti)(β(ti+1)β(ti)).\int_{t_0}^{t} L(x(s), s) \ d\beta(s) = \underset{\substack{n \to \infty \newline |P| \to 0}}{\text{l.i.m}} \sum_{i=0}^{n-1} L(x(t_i), t_i) (\beta(t_{i + 1}) - \beta(t_i)).

Under suitable regularity conditions (e.g., ff and LL Lipschitz 1), the above SDE has a solution x(t)x(t).

Note (Properties of SDEs)
  • The solution x(t)x(t) is a unique and continuous stochastic process, in the mean square sense.
  • In short, x(t)x(t) inherits properties of the Brownian motion and is generally not mean square differentiable.

The Classical Chain Rule

Suppose x(t)x(t) satisfies the ordinary differential equation (ODE),

dx(t)dt=f(x(t)),\frac{dx(t)}{dt} = f(x(t)),
Definition 1 (Definition: The Chain Rule)

If ϕ(x)\phi(x) is a differentiable function, then,

dϕ(x(t))dt=dϕ(x(t))dxϕ(x(t))dx(t)dtx(t)=ϕ(x)f(x(t)).\frac{d \phi(x(t))}{dt} = \underbrace{\frac{d \phi(x(t))}{dx}}_{\phi^{\prime}(x(t))} \underbrace{\frac{d x(t)}{dt}}_{x^{\prime}(t)} = \phi^{\prime}(x) f(x(t)).

Integral Form of the Chain Rule

Suppose x(t)x(t) satisfies the (same) ODE,

dx(t)=f(x(t)) dt,x(t)x(t0)=t0tf(x(s)) ds.\begin{align*} dx(t) & = f(x(t)) \ dt, \newline x(t) - x(t_0) & = \int_{t_0}^{t} f(x(s)) \ ds. \end{align*}
Definition 2 (Definition: The Chain Rule)

If ϕ(x)\phi(x) is a differentiable function, then,

ϕ(x(t))ϕ(x(0))=0tdϕ(x(s))ds ds=0tϕ(x(s))dx(s)=0tϕ(x(s))f(x(s)) dsdϕ(x(t))=ϕ(x(t)) dx(t)=ϕ(x(t))f(x(t)) dt.\begin{align*} \phi(x(t)) - \phi(x(0)) & = \int_{0}^{t} \frac{d \phi(x(s))}{ds} \ ds = \int_{0}^{t} \phi^{\prime}(x(s)) dx(s) \newline & = \int_{0}^{t} \phi^{\prime}(x(s)) f(x(s)) \ ds \newline d \phi(x(t)) & = \phi^{\prime}(x(t)) \ dx(t) = \phi^{\prime}(x(t)) f(x(t)) \ dt. \end{align*}

Derivation of the Integral Form

Consider a partition, a=t0<t1<<tn=ba = t_0 < t_1 < \ldots < t_n = b such that ti+1ti=Δt=bant_{i+1} - t_i = \Delta t = \frac{b - a}{n} for i=1,2,,ni = 1, 2, \ldots, n.

It follows that,

ϕ(x(b))ϕ(x(a))=i=0n1ϕ(x(ti+1))ϕ(x(ti)).\phi(x(b)) - \phi(x(a)) = \sum_{i=0}^{n-1} \phi(x(t_{i + 1})) - \phi(x(t_i)).

We have nn terms and Δt=banO(Δt2)\Delta t = \frac{b - a}{n} \Rightarrow \mathcal{O}(\Delta t^2) can be ignored as nn \to \infty.

Using dx(t)=f(x(t))dtdx(t) = f(x(t)) dt (the ODE) and its Taylor expansion, we get,

x(ti+1)=x(ti)+f(x(ti))Δt+O(Δt2),ϕ(x(ti+1))=ϕ(x(ti))+ϕ(x(ti))(x(ti+1)x(ti))+O(Δt2).\begin{align*} x(t_{i + 1}) & = x(t_i) + f(x(t_i)) \Delta t + \mathcal{O}(\Delta t^2), \newline \phi(x(t_{i + 1})) & = \phi(x(t_i)) + \phi^{\prime}(x(t_i))(x(t_{i + 1}) - x(t_i)) + \mathcal{O}(\Delta t^2). \end{align*}

Thus, we conclude that,

ϕ(x(b))ϕ(x(a))=i=0n1ϕ(x(ti))f(x(ti))Δt=abϕ(x(t))f(x(t)) dt.\phi(x(b)) - \phi(x(a)) = \sum_{i=0}^{n-1} \phi^{\prime}(x(t_i)) f(x(t_i)) \Delta t = \int_{a}^{b} \phi^{\prime}(x(t)) f(x(t)) \ dt.

Recall that the ODE dx(t)=f(x(t)) dtdx(t) = f(x(t)) \ dt and that all O(Δt2)\mathcal{O}(\Delta t^2) terms can be ignored, as nn \to \infty,

dϕ(x(t))=ϕ(x(t)) dx(t)+12ϕ(x(t))(dx(t))2+O((dx(t))3)=ϕ(x(t)) dx(t)=ϕ(x(t))f(x(t)) dt.\begin{align*} d\phi(x(t)) & = \phi^{\prime}(x(t)) \ dx(t) + \frac{1}{2} \phi^{\prime \prime}(x(t)) (dx(t))^2 + \mathcal{O}((dx(t))^3) \newline & = \phi^{\prime}(x(t)) \ dx(t) \newline & = \phi^{\prime}(x(t)) f(x(t)) \ dt. \end{align*}

Solving ODEs

The chain rule is useful in various contexts, but we note that it yields a new ODE,

dx(t)=f(x(t)) dt,dϕ(x(t))=ϕ(x(t)) dx(t)=ϕ(x(t))f(x(t)) dt.\begin{align*} dx(t) & = f(x(t)) \ dt, \newline d\phi(x(t)) & = \phi^{\prime}(x(t)) \ dx(t) \newline & = \phi^{\prime}(x(t)) f(x(t)) \ dt. \end{align*}

We can use this to solve ODEs.

Note (Example)

Solving dx(t)=cx(t) dtdx(t) = -cx(t) \ dt where x(0)=1x(0) = 1.

We introduce ϕ(x(t))=logx(t)\phi(x(t)) = \log x(t). The chain rule implies,

dϕ(x(t))=dx(t)x(t)=c dt.d\phi(x(t)) = \frac{dx(t)}{x(t)} = -c \ dt.

This implies that ϕ(x(t))=ct\phi(x(t)) = -ct and hence,

x(t)=exp(ct).x(t) = \exp(-ct).

Itô Stochastic Calculus

Suppose x(t)x(t) is described (in the Itô sense) by the SDE,

dx(t)=f(x(t),t) dt+L(x(t),t) dβ(t).dx(t) = f(x(t), t) \ dt + L(x(t), t) \ d\beta(t).

The “chain rule” for ϕ(x(t))\phi(x(t)) is called the Itô formula.

To present a short derivation, we recall that,

abg(t) (dβ(t))2=abg(t) dt.\int_{a}^{b} g(t) \ (d\beta(t))^2 = \int_{a}^{b} g(t) \ dt.

It is also easy to verify that all other higher-order terms vanish, e.g.,

abg(t) dt dβ(t)=0.\int_{a}^{b} g(t) \ dt \ d\beta(t) = 0.

The Itô Formula

The Itô formula can thus be formally derived as,

dϕ(x(t))=ϕ(x(t)) dx(t)+12ϕ(x(t))(dx(t))2+O((dx(t))3)=ϕ(x(t))(f(x(t),t) dt+L(x(t),t) dβ(t))+12ϕ(x(t))(L(x(t),t))2dt\begin{align*} d \phi(x(t)) & = \phi^{\prime}(x(t)) \ dx(t) + \frac{1}{2} \phi^{\prime \prime}(x(t)) (dx(t))^2 + \mathcal{O}((dx(t))^3) \newline & = \phi^{\prime}(x(t)) \left( f(x(t), t) \ dt + L(x(t), t) \ d\beta(t) \right) + \frac{1}{2} \phi^{\prime \prime}(x(t)) (L(x(t), t))^2 dt \end{align*}

The Itô formula can be generalized to vector-valued states and ϕ,f\phi, f and LL may all be time-dependent.

Let’s present a scalar version where ϕ\phi depends on state and time, i.e., ϕ(x(t),t)\phi(x(t), t).

Definition 3 (Definition: The Itô formula)

Suppose x(t)x(t) is a scalar Itô process that obeys the SDE,

dx(t)=f(x(t),t) dt+L(x(t),t) dβ(t).dx(t) = f(x(t), t) \ dt + L(x(t), t) \ d\beta(t).

The scalar function ϕ(x(t),t)\phi(x(t), t) can then be described by the SDE,

dϕ(x(t),t)=ϕtdt+ϕxdx(t)+122ϕx2L(x(t))2dt.d \phi(x(t), t) = \frac{\partial \phi}{\partial t} dt + \frac{\partial \phi}{\partial x} dx(t) + \frac{1}{2} \frac{\partial^2 \phi}{\partial x^2} L(x(t))^2 dt.

We can use this to solve SDEs.

Note (Example)

Solving dx(t)=cx(t) dt+x(t) dβ(t)dx(t) = -cx(t) \ dt + x(t) \ d\beta(t) where x(0)=1x(0) = 1.

We introduce ϕ(x(t),t)=logx(t)+ct\phi(x(t), t) = \log x(t) + ct. The Itô formula implies,

dϕ(x(t),t)=dx(t)x(t)(dx(t))22(x(t))2=c dt+dβ(t)12dt.d \phi(x(t), t) = \frac{dx(t)}{x(t)} - \frac{(dx(t))^2}{2 (x(t))^2} = -c \ dt + d\beta(t) - \frac{1}{2} dt.

This implies that ϕ(x(t),t)=(c+12)t+β(t)\phi(x(t), t) = -(c + \frac{1}{2}) t + \beta(t) and hence,

x(t)=exp((c+12)t+β(t)).x(t) = \exp\left(-\left(c + \frac{1}{2}\right) t + \beta(t)\right).

A Toy Example

Note (Example)

Squared Brownian Motion

Suppose x(0) = 0 and dx(t)=dβ(t)dx(t) = d\beta(t) such that x(t)=β(t)x(t) = \beta(t). For ϕ(x(t))=x2(t)\phi(x(t)) = x^2(t), the Itô formula specifies that,

dϕ(x(t))=2x(t) dx(t)+(dx(t))2=2β(t)dβ(t)+dt.d \phi(x(t)) = 2 x(t) \ dx(t) + (dx(t))^2 = 2 \beta(t) d\beta(t) + dt.0tdϕ(x(s))=β2(t)=0t2β(s)dβ(s)+t.\Rightarrow \int_{0}^{t} d \phi(x(s)) = \beta^2(t) = \int_{0}^{t} 2 \beta(s) d\beta(s) + t.

Note, for differentiable functions we get,

0tf(s)df(s)=0tf(s)f(s)ds=120tdf2(s)dsds=f2(t)f2(0)2,\int_{0}^{t} f(s) df(s) = \int_{0}^{t} f(s) f^{\prime}(s) ds = \frac{1}{2} \int_{0}^{t} \frac{d f^2(s)}{ds} ds = \frac{f^2(t) - f^2(0)}{2},

whereas the Itô formula yields and extra tt term.

Starting from the SDE (x(t)=β(t))(x(t) = \beta(t)),

dx(t)=dβ(t),dx(t) = d\beta(t),

we have derived an SDE for ϕ(x(t))=β2(t)\phi(x(t)) = \beta^2(t),

d(B2(t))=2β(t)dβ(t)+t.d(\Beta^2(t)) = 2 \beta(t) d\beta(t) + t.
Brownian motion and its square.
Brownian motion and its square.

To verify the result, let us use the definition of the Itô integrals,

Note (Example: Squared Brownian Motion)

The Itô formula implies that,

abdβ2(t)=ab2β(t)dβ(t)+abdt.\begin{equation} \int_{a}^{b} d \beta^2(t) = \int_{a}^{b} 2 \beta(t) d\beta(t) + \int_{a}^{b} dt. \end{equation}

Using a partition, a=t0<t1<<tn=ba = t_0 < t_1 < \ldots < t_n = b, the first integral is,

i=0n1(β2(ti+1)β2(ti))a2b2=i=0n1(β(ti+1)+β(ti))(β(ti+1)β(ti))=i=0n12β(ti)(β(ti+1)β(ti))+(β(ti+1)β(ti))2,\begin{align*} \sum_{i=0}^{n-1} \underbrace{(\beta^2(t_{i + 1}) - \beta^2(t_i))}_{a^2 - b^2} & = \sum_{i=0}^{n-1} (\beta(t_{i + 1}) + \beta(t_i)) (\beta(t_{i + 1}) - \beta(t_i)) \newline & = \sum_{i=0}^{n-1} 2 \beta(t_i) (\beta(t_{i + 1}) - \beta(t_i)) + (\beta(t_{i + 1}) - \beta(t_i))^2, \end{align*}

This converges (in mean square) to the right-hand side of Equation (1) as nn \to \infty.

We can also visualize the nonlinear function ϕ(x(t))\phi(x(t)) as a function of x(t)x(t).

Note (Example: Squared Brownian Motion (cont’d))

We visualize β2(t)\beta^2(t) as a function of β(t)\beta(t).

Sum of the previous two integrals is the second-order Taylor expansion with respect to β(ti+1)\beta(t_{i + 1}).

The quadratic term converges in mean square to ba=tb - a = t as nn \to \infty.

Squared Brownian motion as a function of Brownian motion.
Notice how the quadratic term ensures that the Taylor expansion converges to the point heightwise.
Squared Brownian motion as a function of Brownian motion. Notice how the quadratic term ensures that the Taylor expansion converges to the point heightwise.

Vector Version

Consider a multi-dimensional SDE,

dx(t)=f(x(t),t) dt+L(x(t),t)dβ(t),d\mathbf{x}(t) = \mathbf{f}(\mathbf{x}(t), t) \ dt + \mathbf{L}(\mathbf{x}(t), t) d\beta(t),

where x\mathbf{x} and f\mathbf{f} are n×1n \times 1, L\mathbf{L} is n×mn \times m and β(t)\beta(t) is m×1m \times 1 with independent Brownian elements.

It follows that,

dϕ(x(t),t)=ϕ(x(t),t)tdt+iϕ(x(t),t)xi(t)dxi(t)+12i,j2ϕ(x(t),t)xi(t)xj(t)dxi(t)dxj(t).d \phi(\mathbf{x}(t), t) = \frac{\partial \phi(\mathbf{x}(t), t)}{\partial t} dt + \sum_i \frac{\partial \phi(\mathbf{x}(t), t)}{\partial x_i(t)} dx_i(t) \newline + \frac{1}{2} \sum_{i,j} \frac{\partial^2 \phi(\mathbf{x}(t), t)}{\partial x_i(t) \partial x_j(t)} dx_i(t) d x_j(t).

All higher-order terms vanish expect (dβ(t))2=dt(d\beta(t))^2 = dt.

We can express the relation on a more compact form,

dϕ(x(t),t)=ϕ(x(t),t)tdt+iϕ(x(t),t)xi(t)dxi(t)+12i,j2ϕ(x(t),t)xi(t)xj(t)dxi(t)dxj(t)=ϕtdt+(xϕ)Tdx(t)+12dx(t)TxxTϕ dx(t).\begin{align*} d \phi(\mathbf{x}(t), t) & = \frac{\partial \phi(\mathbf{x}(t), t)}{\partial t} dt + \sum_i \frac{\partial \phi(\mathbf{x}(t), t)}{\partial x_i(t)} dx_i(t) \newline & + \frac{1}{2} \sum_{i,j} \frac{\partial^2 \phi(\mathbf{x}(t), t)}{\partial x_i(t) \partial x_j(t)} dx_i(t) d x_j(t) \newline & = \frac{\partial \phi}{\partial t} dt + (\nabla_{\mathbf{x}} \phi)^T d\mathbf{x}(t) + \frac{1}{2} d\mathbf{x}(t)^T \nabla_{\mathbf{x}} \nabla_{\mathbf{x}}^T \phi \ d\mathbf{x}(t). \end{align*}

Using the expression for the SDE,

dϕ(x(t),t)=ϕtdt+(xϕ)Tdx(t)+12dβTLTxxTϕL dβ=ϕtdt+(xϕ)Tdx(t)+12tr{LTxxTϕL}dt\begin{align*} d \phi(\mathbf{x}(t), t) & = \frac{\partial \phi}{\partial t} dt + (\nabla_{\mathbf{x}} \phi)^T d \mathbf{x}(t) + \frac{1}{2} d \beta^T \mathbf{L}^T \nabla_{\mathbf{x}} \nabla_{\mathbf{x}}^T \phi \mathbf{L} \ d\beta \newline & = \frac{\partial \phi}{\partial t} dt + (\nabla_{\mathbf{x}} \phi)^T d \mathbf{x}(t) + \frac{1}{2} tr \{\mathbf{L}^T \nabla_{\mathbf{x}} \nabla_{\mathbf{x}}^T \phi \mathbf{L} \} dt \end{align*}
Definition 4 (Definition: The Itô formula for vector states)

Consider a multi-dimensional SDE,

dx(t)=f(x(t),t) dt+L(x(t),t) dβ(t).d\mathbf{x}(t) = \mathbf{f}(\mathbf{x}(t), t) \ dt + \mathbf{L}(\mathbf{x}(t), t) \ d\beta(t).

where x\mathbf{x} and f\mathbf{f} are n×1n \times 1, L\mathbf{L} is n×mn \times m and β(t)\beta(t) is m×1m \times 1 with independent Brownian elements. Under suitable regularity, the scalar function ϕ(x(t),t)\phi(\mathbf{x}(t), t) satisfies,

dϕ(x(t),t)=ϕtdt+(xϕ)Tdx(t)+12tr{LTxxTϕL}dt.d \phi(\mathbf{x}(t), t) = \frac{\partial \phi}{\partial t} dt + (\nabla_{\mathbf{x}} \phi)^T d \mathbf{x}(t) + \frac{1}{2} tr \{\mathbf{L}^T \nabla_{\mathbf{x}} \nabla_{\mathbf{x}}^T \phi \mathbf{L} \} dt.

Stratonovich SDEs

Most SDEs are described using Itô integrals,

abL(x(t)) dβ(t)l.i.mni=0n1L(x(ti))(β(ti+1)β(ti)),\int_{a}^{b} L(x(t)) \ d\beta(t) \triangleq \underset{n \to \infty}{\text{l.i.m}} \sum_{i=0}^{n-1} L(x(t_i)) (\beta(t_{i + 1}) - \beta(t_i)),

but it is not the only option.

Definition 5 (Definition: The Stratonovich Integral)

The stochastic integral of Stratonovich is defined as,

abL(x(t))dβ(t)=l.i.mni=0n1L(x(ti+1))+L(x(ti))2(β(ti+1)β(ti)).\int_{a}^{b} L(x(t)) \circ d\beta(t) = \underset{n \to \infty}{\text{l.i.m}} \sum_{i=0}^{n-1} \frac{L(x(t_{i + 1})) + L(x(t_i))}{2} (\beta(t_{i + 1}) - \beta(t_i)).

One advantage of Stratonovich is that the chain rule holds!

Stratonovich VS. Itô

Note (Converting SDEs)

The Stratonovich SDE,

dx(t)=f(x(t)) dt+L(x(t))dβ(t),dx(t) = f(x(t)) \ dt + L(x(t)) \circ d\beta(t),

is equivalent to the Itô SDE,

dx(t)=(f(x(t))+12L(x(t))L(x(t)))dt+L(x(t)) dβ(t).dx(t) = \left( f(x(t)) + \frac{1}{2} L^{\prime}(x(t)) L(x(t)) \right) dt + L(x(t)) \ d\beta(t).

Hence, we can use results from Itô SDEs (e.g., Euler-Maruyama or the Itô formula) also on Stratonovich SDEs. We will always assume Itô SDEs, but for completeness, one can write, e.g., “in the Itô sense” or “in the Stratonovich sense”.

Sketch of Proof

Consider the Stratonovich SDE,

dx(t)=f(x(t)) dt+L(x(t))dβ(t).dx(t) = f(x(t)) \ dt + L(x(t)) \circ d\beta(t).

The conversion to an Itô SDE follows from (Δβiβ(ti+1β(ti))(\Delta \beta_i \triangleq \beta(t_{i + 1} - \beta(t_i)),

L(x(t))dβ(t)=l.i.mni=0n1L(x(ti+1))+L(x(ti))2Δβi{Taylor Expansion}=l.i.mni=0n1(L(x(ti))+12L(x(ti))(x(ti+1)x(ti)))Δβi=l.i.mni=0n1L(x(ti))Δβi+12L(x(ti))L(x(ti))Δβi2\begin{align*} \int L(x(t)) \circ d\beta(t) & = \underset{n \to \infty}{\text{l.i.m}} \sum_{i=0}^{n-1} \frac{L(x(t_{i + 1})) + L(x(t_i))}{2} \Delta \beta_i \newline \text{\{Taylor Expansion\}} & = \underset{n \to \infty}{\text{l.i.m}} \sum_{i=0}^{n-1} \left( L(x(t_i)) + \frac{1}{2} L^{\prime}(x(t_i))(x(t_{i + 1}) - x(t_i)) \right) \Delta \beta_i \newline & = \underset{n \to \infty}{\text{l.i.m}} \sum_{i=0}^{n-1} L(x(t_i)) \Delta \beta_i + \frac{1}{2} L^{\prime}(x(t_i)) L(x(t_i)) \Delta \beta_i^2 \end{align*}

Summary

Definition 6 (Definition: The Itô formula)

Suppose x(t)x(t) is a scalar Itô process that obeys the SDE,

dx(t)=f(x(t),t) dt+L(x(t),t) dβ(t).dx(t) = f(x(t), t) \ dt + L(x(t), t) \ d\beta(t).

The scalar function ϕ(x(t),t)\phi(x(t), t) can then be described by the SDE,

dϕ(x(t),t)=ϕtdt+ϕxdx(t)+122ϕx2L(x(t))2dt.d \phi(x(t), t) = \frac{\partial \phi}{\partial t} dt + \frac{\partial \phi}{\partial x} dx(t) + \frac{1}{2} \frac{\partial^2 \phi}{\partial x^2} L(x(t))^2 dt.

The Itô formula is essential for understanding and working with SDEs.

Footnotes

  1. Wikipedia, Lipschitz function