Ito Calculus

Continuation from Moments of the OU Process. We want to calculate higher order moments.

The idea is that we use a “stochastic” chain rule for computing $d(f(x))$ where $d$ is the differential operator.

Taylor Series (Big O notation)

Let $f:\mathbb{R}\to \mathbb{R}$ be smooth. Then

$$f(a+\epsilon )=f(a)+\underset{O(\epsilon )}{\underbrace{\epsilon f^{\prime }(a)}}+\underset{O({\epsilon }^2)}{\underbrace{\frac{1}{2}{\epsilon }^2{f}^{\prime \prime }(a)}}+\cdots$$

This comes from Taylor’s Theorem. Similarly, we have First-Order Taylor Expansion using Big O Notation. We rewrite it here to motivate the following lemma.

Ito’s Lemma

Consider the 1D SDE for $X(t)\in \mathbb{R}$:

$$dX=u(X)dt+\sigma (X)dB$$

Stochastic calculus differs from standard calculus. Recall that $dB\sim \sqrt{dt}\cdot Z\in O(\sqrt{dt})$ from the formulation and $\mathbb{E}[d{B}^2]=dt⟹d{B}^2\in O(dt)$ (from the variance). This implies $dX\in O(\sqrt{dt})$ because the noise term dominates the drift term as $dt\to 0$.

The goal for Ito’s Lemma is, given function $f(X)$ and $dX$, to find an expression for $df$. The strategy is to use the Taylor expansion of $f$ up to $O(dt)$ and treating $d{B}^2=dt$ (since the variance of $d{B}^2$ vanishes faster than its mean as $dt\to 0$).

Consider the transformation of $f(X)$:

$$df(X)=f(X+dX)-f(X)=f^{\prime }(X)dX+\frac{1}{2}{f}^{\prime \prime }(X)(dX{)}^2+\dots$$

Next, we find $dX$ and $(dX{)}^2$ to $O(dt)$:

$$\begin{array}{rl}dX& =u(X)dt+\sigma (X)dB\\ (dX{)}^2& =(udt+\sigma dB{)}^2\\ & =u^2(dt{)}^2+2u\sigma (dtdB)+{\sigma }^2(dB{)}^2\\ & =O(d{t}^2)+O(d{t}^{3/2})+O(dt)\\ & ={\sigma }^{2}dt+O(d{t}^{3/2})\end{array}$$

We substitute these back into the expansion, keeping only terms up to $O(dt)$:

$$\begin{array}{rl}df(X)& =f^{\prime }(X)[udt+\sigma dB]+\frac{1}{2}{f}^{\prime \prime }(X)[{\sigma }^{2}dt+O(d{t}^{3/2})]+O(d{t}^{3/2})\\ & =(u(X)f^{\prime }(X)+\frac{1}{2}\sigma (X{)}^2{f}^{\prime \prime }(X))dt+\sigma (X)f^{\prime }(X)dB\end{array}$$

Note that the $O(d{t}^{3/2})$ terms are omitted because as $dt\to 0$, they vanish significantly faster than the $dt$ and $dB$ terms. The final line is Ito’s Lemma.

Remarks (Ito’s Lemma)

As previously mentioned, this is the stochastic version of the chain rule:

• If $\sigma =0$ (i.e. no noise, is deterministic), then $dX=u(X)dt$, such that $df=u(X)f^{\prime }(X)dt=f^{\prime }(X)dX$.
• This works because $(dX{)}^2=(udt{)}^2=O(d{t}^2)$ which is small enough to ignore. See the derivation of $(dX{)}^2$ above.

In the stochastic case, $df\ne f^{\prime }(X)dX$ at $O(dt)$ in general because $dX\sim O\left(\sqrt{dt}\right)$, so we need to work $O(d{X}^2)$ to get expression correct to $O(dt)$.

But what about the $(\sigma (X)dB{)}^2$ term?

Discretization Dilemma

ODE Case

In the ODE case,

$$\frac{dx}{dt}=u(x)x(0)=a$$

Let $\delta t>0$ and set $t_i=i\delta t$. Then via Taylor expansion,

$$\begin{array}{rl}x(t_{i+1})& =x(t_i)+\delta t\frac{dx}{dt}{|}_{t=t_i}+O\left(d{t}^2\right)\\ & =x(t_i)+u(x(t_i))\delta t+O\left(\delta t^2\right)\end{array}$$

But then

$$\begin{array}{rl}u(x(t_{i+1}))& =u(x(t_i)+O(\delta t))\\ & =u(x(t_i))+O(\delta t)\end{array}$$

implying it is equal at $O(\delta t)$. Implying that

$$x(t_{i+1})=x(t_i)+u(x(t_{i+1}))\delta t+O\left(\delta t^2\right)$$

We get equivalent discretizations of $x(t)$, where

$$\begin{array}{rlrl}{x}_{i+1}& =x_i+u(x_i)\delta t,& & x_0=a\\ {\tilde{x}}_{i+1}& ={\tilde{x}}_i+u({\tilde{x}}_{i+1})\delta t,& & {\tilde{x}}_0=a\end{array}$$

both $x_{i+1},{\tilde{x}}_{i+1}\approx x(t_i)$. Te reason they are “equal at $O(\delta t)$” is because $x(t)$ is differentiable and smooth (no noise).

This effect is similar to how the Riemann-Stieltjes Integral evaluated on an interval from left to right via Riemann Sums gives us the same value.

Note that “equal at $O(\delta t)$” is with respect to the Taylor expansion.

SDE Case (Ito-SDE)

Consider the SDE:

$$dX=u(X)dt+\sigma (X)dBX(0)=x_0$$

Set $\delta t>0$ and discretize up to $O(\delta t)$ only.

$$X_{i+1}=X_i+u(X_i)\delta t+\sigma (X_i)\sqrt{\delta t}\cdot Z_i{X}_0=x_0$$

where $Z_i\sim \mathcal{N}(0,1)$, the standard normal, independent of $X_i$. This is different from

$$X_{i+1}=X_i+u(X_{i+1})\delta t+\sigma (X_{i+1})\sqrt{\delta t}\cdot Z_i$$

The first discretization is an example of an Ito-SDE. The noise $Z_i$ is independent of the current state $X_i$, and the expectation is $0$, so it does not push the particle in some specific direction.

This is the most common type of SDE, and Ito Calculus holds for Ito-SDEs.

Example 1 (Discrete-Time Step with Ito’s Lemma)

We can derive with Ito’s Lemma.

$$\begin{array}{rl}f(X_{i+1})& =f\left(X_i+u(X_i)\delta t+\sigma (X_i)\sqrt{\delta t}{Z}_i\right)\\ & =f(X_i)+f^{\prime }(X_i)\left[u(X_i)\delta t+\sigma (X_i)\sqrt{\delta t}{Z}_i\right]+\frac{1}{2}{f}^{\prime \prime }(X_i)\left[\sigma (X_i{)}^2\delta t{Z}_i^2\right]+O(\delta t^{3/2})\\ & =f(X_i)+\left(f^{\prime }(X_i)u(X_i)+f^{\prime \prime }(X_i)\cdot \frac{1}{2}\sigma (X_i^2)\right)\delta t+\underset{\text{only depends on Xi}}{\underbrace{f^{\prime }(X_i)\sigma (X_i)\sqrt{\delta t}{Z}_i}}+O(\delta t^{3/2})\end{array}$$

giving us a discrete expression for $df\approx f(X_{i+1})-f(X_i)$. The continuum limit gives Ito Calculus.

Example 2 (Simple Ito Lemma)

We can do a much easier example. What is $df$ for $f(x)=x^2$? Using Ito’s Lemma,

$$\begin{array}{rl}df& =(-\alpha X{f}^{\prime }+D{f}^{\prime \prime })dt+f^{\prime }\sqrt{2D}dB\\ & =(-2\alpha X^2+2D)dt+2X\sqrt{2D}dB\end{array}$$

where $\alpha$ represents a constant to replace $u(X)$.

Because stochastic processes are nowhere differentiable, we cannot describe a simple derivative of a function. Instead, we discuss how it updates deterministically and how multiplicative noise affects it.

Backwards Ito SDE

The first SDE of SDE Case (Ito-SDE) is covered by Examples 1 and 2. We can show the derivation for the second rule, where diffusion depends on the future state. This is the continuum limit of

$$X_{i+1}=X_i+u(X_{i+1})\delta t+\sigma (X_{i+1})\sqrt{\delta t}\cdot Z_i$$

for standard normal $Z_i$. Let $X_i,Z_i$ be independent and $\delta t>0$. The SDE of this discretized form is

$$dX=u(X)dt+\sigma (X)\bullet dB$$

This is

• Anticipatory: Diffusion term depends on future state.
• $\mathbb{E}\left[dX\right]\ne \mathbb{E}\left[u(X)\right]dt$ because the correlation between $\sigma (X_{i+1})$ and $dB$ creates additional drift.
• The bullet $\bullet$ is used to represent the backwards Ito. This represents the anticipatory nature of the SDE.

Not sure if continuum limit gives the correct SDE.

Stratanovich SDE

We take the continuum limit of

$$X_{i+1}=X_i+u\left(\frac{X_i+X_{i+1}}{2}\right)\delta t+\sigma \left(\frac{X_i+X_{i+1}}{2}\right)\sqrt{\delta t}{Z}_i$$

where $Z_i\sim \mathcal{N}(0,1)$ and $X_i,Z_i$ are independent with $\delta t>0$. The SDE is

$$dX=u(X)dt+\sigma (X)dB$$

Notation can vary.

Remarks (Ito v.s Ito Backwards)

The Ito-SDE can sometimes be written

$$dX=u(X,t)dt+\sigma (X,t)\ast dB$$

where the asterisk $\ast$ represents non-anticipatory updates.

We consider Ito-SDEs most common since

$$\mathbb{E}\left[dX\right]=\mathbb{E}\left[u(X,t)\right]dt$$

Different SDEs can be transformed into each other. Suppose $X(t)$ satisfies

$$d(X)=u(X)dt+\sigma (X)\bullet dB$$

or the backwards Ito. Then

$$dX=\left(u(X)+\sigma (X){\sigma }^{\prime }(X)\right)dt+\sigma (X)dB$$

represents the Ito SDE. In general, different SDEs have different chain rules.

Example 3 (Apply OU on Ito SDE)

We can analyze an OU-process with Ito’s Lemma. Recall that OU-processes are of the following form. Let $X(t)\in \mathbb{R}$ with

$$dX=-\alpha Xdt+\sqrt{2D}dBX(0)=x_0$$

where $\alpha ,D$ are constants and $D>0$. We can take the expectation to get

$$d\mathbb{E}\left[X\right]=-\alpha \mathbb{E}\left[X\right]dt$$

Set $y(t)=\mathbb{E}\left[X(t)\right]$. Then

$$\frac{dy}{dt}=-\alpha y⟹y(t)=x_0\exp (-\alpha t)$$

giving us the general form for the expectation. Recall from example 2 that

$$d(X^2)=(-2\alpha X^2+2D)dt+2X\sqrt{2D}dB$$

On average, where is the particle spending time in? This means we need to take the expectation:

$$\begin{array}{rl}d\left(\mathbb{E}\left[X^2\right]\right)& =(-2\alpha \mathbb{E}\left[X^2\right]+2D)dt+\underset{0\text{ since Ito SDE}}{\underbrace{\mathbb{E}\left[2X\sqrt{2D}dB\right]}}\\ \frac{d}{dt}\left(\mathbb{E}\left[X^2\right]\right)& =-2\alpha \left(\mathbb{E}\left[X^2\right]\right)+2D\end{array}$$

Letting $y(t)=\mathbb{E}\left[X^2\right]$, this amounts to solving

$$\frac{dy}{dt}=-2\alpha y+2D$$

where $y(0)=x_0^2$. Now we solve like a typical ODE. Write

$$y(t)=y_h(t)+y_p(t)$$

and solve for the homogeneous solution:

$$\frac{d{y}_h}{dt}=-\alpha y_h⟹y_h(t)=A{e}^{-2\alpha t}$$

and the particular solution:

$$0=-2\alpha y_p+2D⟹y_p=\frac{D}{\alpha }$$

This implies that

$$y(t)=\mathbb{E}\left[X^2\right]=A{e}^{-2\alpha t}+\frac{D}{\alpha }$$

We can also solve via separating the variables:

$$\begin{array}{rl}\int \frac{1}{2D-2\alpha y}dy& =\int dt\\ -\frac{1}{2\alpha }\log (2D-2\alpha y)=t+C& \end{array}$$

The rest is trivial.

Example 4 (Geometric BM)

Let $X(t)\in \mathbb{R}$. Consider Geometric Brownian Motion:

$$dX=\mu Xdt+\sigma XdBX(0)=x_0$$

The expectation gives

$$\begin{array}{rl}d(\mathbb{E}\left[X\right])& =\mu \mathbb{E}\left[X\right]dt+\underset{0\text{ since Ito SDE}}{\underbrace{\mathbb{E}\left[\sigma XdB\right]}}\\ \frac{d}{dt}\mathbb{E}\left[X\right]& =\mu \mathbb{E}\left[X\right]\\ \mathbb{E}\left[X\right]& =x_0{e}^{\mu t}\end{array}$$

We can again apply this to $f(X)=X^2$ with Ito’s Lemma to find the second moment:

$$\begin{array}{rl}df& =\left(\mu X{f}^{\prime }+\frac{1}{2}{\sigma }^2{X}^2{f}^{\prime \prime }\right)dt+f^{\prime }\sigma XdB\\ d(X^2)& =\left(2\mu X^2+{\sigma }^2{X}^2\right)dt+2\sigma X^{2}dB\\ d\left(\mathbb{E}\left[X^2\right]\right)& =(2\mu +{\sigma }^2)\left(\mathbb{E}\left[X^2\right]\right)dt+0\\ \frac{d}{dt}\left(\mathbb{E}\left[X^2\right]\right)& =(2\mu +{\sigma }^2)\cdot \mathbb{E}\left[X^2\right]\\ \mathbb{E}\left[X^2\right]& =x_0^2\exp (2\mu t)\end{array}$$

or the second moment.

Context: GBM is the standard model for stock prices because the noise is multiplicative ($XdB$). This ensures that returns $\frac{dX}{X}$ are independent of the absolute price level, and it prevents the price from ever becoming negative.

Example 5 (Geometric BM with Log Expectation)

To solve the SDE exactly, we use Ito’s Lemma on $f(X)=\log X$. So,

$$\begin{array}{rl}d(\log X)& =\left(\mu X\frac{1}{X}+\frac{1}{2}{\sigma }^2{X}^2\left(-\frac{1}{X^2}\right)\right)dt+\frac{1}{\lambda }\sigma XdB\\ & =\underset{\text{Brownian Motion with Drift!}}{\underbrace{\left(\mu -\frac{1}{2}{\sigma }^2\right)dt+\sigma dB}}\end{array}$$

The result is Brownian Motion with Drift where the coefficients are constants. This means that $\log X$ is Normal.

Indeed,

$$\begin{array}{rl}\mathbb{E}\left[\log X\right]& =\left(\mu -\frac{1}{2}{\sigma }^2\right)t+x_0\\ \text{Var}(\log X)& ={\sigma }^{2}t\\ \log X(t)& \sim \mathcal{N}\left[\left(\mu -\frac{1}{2}{\sigma }^2\right)t+x_0,{\sigma }^{2}t\right]\end{array}$$

Insight: The $-\frac{1}{2}{\sigma }^2$ term is the “Ito Correction.” It shows that volatility actually drags down the median growth of a stock, even if the average growth remains $\mu$.

Example 6 (Arithmetic BM with Drift)

Consider an SDE with with constant additive noise and drift. Let $X(t)\in \mathbb{R}$ and

$$dX=udt+\sqrt{2D}dBX(0)=x_0$$

where $u,D$ are constants and $D>0$. The expectation gives the linear trend:

$$\begin{array}{rl}d\mathbb{E}\left[X\right]& =udt\\ \frac{d}{dt}\mathbb{E}\left[X\right]& =u\\ \mathbb{E}\left[X(t)\right]& =ut+x_0\end{array}$$

Using Ito’s Lemma with $f(X)=X^2$,

$$\begin{array}{rl}df& =\left(2uX+2D\right)dt+2X\sqrt{2D}dB\\ d\mathbb{E}\left[X^2\right]& =(2u\mathbb{E}\left[X\right]+2D)dt\\ \frac{d}{dt}\mathbb{E}\left[X^2\right]& =(2u^{2}t+2u{x}_0)+2D\\ \mathbb{E}\left[X^2\right]& =u^2{t}^2+2u{x}_{0}t+2Dt+x_0^2\end{array}$$

This also gives us the variance.

$$\text{Var}X(t)=\mathbb{E}\left[X^2\right]-{\left(\mathbb{E}\left[X\right]\right)}^2=2Dt$$

We also know the full distribution:

$$X(t)={\int }_0^{t}dX={\int }_0^{t}ud{t}^{\prime }+{\int }_0^t\sqrt{2D}dB+x_0=ut+x_0+\sqrt{2D}B(t)$$

implying that $X(t)\sim \mathcal{N}(ut+x_0,2Dt)$, since $B(t)\sim \mathcal{N}(0,t)$ by definition.

Comparison: Unlike GBM (where variance grows exponentially), Arithmetic BM has a variance that grows linearly with time ($2Dt$). This is used for physical diffusion (e.g., a particle in a flow) rather than finance, as it allows for negative values.