Chapter 2 Itô Integral and Itô’s Lemma
2.1 The Challenge of Stochastic Integration
In regular calculus, we integrate smooth functions:
\[\int_0^t f(s) \, ds = \lim_{\Delta s \to 0} \sum_{i} f(s_i) \Delta s\]
For smooth functions \(f\), it doesn’t matter whether we evaluate at the left endpoint, right endpoint, or midpoint of each interval - we get the same answer in the limit.
But what about:
\[\int_0^t f(s) \, dB(s) \quad ?\]
This means integrating \(f\) with respect to Brownian motion. Instead of accumulating tiny areas \(f(s_i) \Delta s\), we’re accumulating random contributions \(f(s_i) \Delta B_i\) where \(\Delta B_i = B(s_{i+1}) - B(s_i)\).
2.1.1 The Problem
Because Brownian motion is so rough (nowhere differentiable), the choice of evaluation point matters critically! Different choices give different answers, even in the limit.
This isn’t just a technical nuisance - it reflects a fundamental ambiguity about how to define integration with respect to a rough, random path.
2.2 The Itô Integral
Kiyoshi Itô made a choice that turned out to be mathematically natural and practically useful: always evaluate at the left endpoint.
2.2.1 Definition
The Itô integral is defined as:
\[\int_0^t f(s) \, dB(s) = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(s_i) [B(s_{i+1}) - B(s_i)]\]
where \(0 = s_0 < s_1 < \cdots < s_n = t\) is a partition that gets progressively finer, and we use \(f(s_i)\) from the left of each interval.
2.2.2 Why the Left Endpoint?
This choice makes \(f(s_i)\) independent of \(\Delta B_i = B(s_{i+1}) - B(s_i)\), because \(f(s_i)\) depends only on information up to time \(s_i\), while \(\Delta B_i\) is the future increment.
This independence gives us beautiful properties:
- Martingale property: \(\int_0^t f(s) \, dB(s)\) is a martingale (under appropriate conditions on \(f\))
- Zero expectation: \(E\left[\int_0^t f(s) \, dB(s)\right] = 0\)
- Isometry: \(E\left[\left(\int_0^t f(s) \, dB(s)\right)^2\right] = E\left[\int_0^t f^2(s) \, ds\right]\)
The third property is called the Itô isometry and is fundamental for proving convergence.
2.3 A Crucial Example: \(\int_0^t B(s) \, dB(s)\)
Let’s compute this integral using the definition. We want:
\[\lim_{n \to \infty} \sum_{i=0}^{n-1} B(s_i) [B(s_{i+1}) - B(s_i)]\]
Here’s a clever algebraic trick. Note that:
\[B^2(s_{i+1}) - B^2(s_i) = [B(s_{i+1}) - B(s_i)]^2 + 2B(s_i)[B(s_{i+1}) - B(s_i)]\]
Rearranging:
\[B(s_i)[B(s_{i+1}) - B(s_i)] = \frac{1}{2}[B^2(s_{i+1}) - B^2(s_i)] - \frac{1}{2}[B(s_{i+1}) - B(s_i)]^2\]
Summing over all intervals:
\[\sum_{i=0}^{n-1} B(s_i) \Delta B_i = \frac{1}{2}[B^2(t) - B^2(0)] - \frac{1}{2}\sum_{i=0}^{n-1} (\Delta B_i)^2\]
The first term telescopes to give \(\frac{1}{2}B^2(t)\). But what about that second sum?
2.3.1 The Quadratic Variation Miracle
Each \((\Delta B_i)^2\) has expected value \(E[(\Delta B_i)^2] = s_{i+1} - s_i = \Delta t_i\).
As the partition gets finer, by the law of large numbers:
\[\sum_{i=0}^{n-1} (\Delta B_i)^2 \to t\]
This is the quadratic variation we mentioned earlier. Remarkably, it converges to the deterministic value \(t\)!
Therefore:
\[\boxed{\int_0^t B(s) \, dB(s) = \frac{1}{2}B^2(t) - \frac{1}{2}t}\]
2.4 Quadratic Variation and the Formal Rule \((dB)^2 = dt\)
The key insight is:
\[\sum_{i=0}^{n-1} (\Delta B_i)^2 \to t \quad \text{as } n \to \infty\]
In differential notation, we write this as the formal rule:
\[(dB)^2 = dt\]
This is not literally true (both sides are zero!), but it’s a mnemonic for the limiting behavior.
2.4.1 Multiplication Rules
From \((dB)^2 = dt\), we can derive multiplication rules for stochastic differentials:
| Term | Value | Reason |
|---|---|---|
| \(dt \cdot dt\) | \(0\) | Second order in deterministic time |
| \(dt \cdot dB\) | \(0\) | Deterministic times random of order \(\sqrt{dt}\) |
| \(dB \cdot dB\) | \(dt\) | Quadratic variation! |
These rules are essential for applying Itô’s lemma.
2.5 Itô’s Lemma: The Fundamental Theorem
Now we come to the crown jewel - the chain rule for stochastic calculus.
In regular calculus, if \(x(t)\) is differentiable and \(f\) is smooth:
\[\frac{d}{dt}f(x(t)) = f'(x(t)) \cdot \frac{dx}{dt}\]
Or in differential form: \(df = f'(x) \, dx\)
Question: If \(B(t)\) is Brownian motion and \(f\) is smooth, what is \(df(B(t))\)?
Naive guess: \(df = f'(B) \, dB\)? Wrong!
2.5.1 The Derivation
Use Taylor expansion for small changes:
\[f(B(t + dt)) - f(B(t)) = f'(B) \cdot dB + \frac{1}{2}f''(B) \cdot (dB)^2 + \frac{1}{6}f'''(B) \cdot (dB)^3 + \cdots\]
In regular calculus, \((dx)^2\) and higher terms vanish as \(dt \to 0\). But \((dB)^2 = dt\)!
So the second-order term becomes:
\[\frac{1}{2}f''(B) \cdot (dB)^2 = \frac{1}{2}f''(B) \cdot dt\]
This term survives! The higher terms \((dB)^3, (dB)^4, \ldots\) do vanish (they’re of order \(dt^{3/2}, dt^2, \ldots\)).
2.6 Verification: Recovering Our Earlier Result
Let’s verify with \(f(x) = x^2\), so \(f'(x) = 2x\) and \(f''(x) = 2\).
Itô’s lemma gives:
\[d(B^2) = 2B \, dB + \frac{1}{2} \cdot 2 \, dt = 2B \, dB + dt\]
Integrating from 0 to \(t\):
\[B^2(t) - B^2(0) = 2\int_0^t B(s) \, dB(s) + t\]
Since \(B(0) = 0\):
\[B^2(t) = 2\int_0^t B(s) \, dB(s) + t\]
Therefore:
\[\int_0^t B(s) \, dB(s) = \frac{1}{2}B^2(t) - \frac{1}{2}t \quad \checkmark\]
Perfect match!
2.7 Example: The Exponential
Take \(f(x) = e^x\), so \(f'(x) = f''(x) = e^x\).
Itô’s lemma:
\[d(e^B) = e^B \, dB + \frac{1}{2}e^B \, dt = e^B \left(dB + \frac{1}{2}dt\right)\]
Integrating:
\[e^{B(t)} = 1 + \int_0^t e^{B(s)} \, dB(s) + \frac{1}{2}\int_0^t e^{B(s)} \, ds\]
This is an integral equation for the exponential of Brownian motion.
Taking expectations (the stochastic integral has zero mean):
\[E[e^{B(t)}] = 1 + \frac{1}{2}\int_0^t E[e^{B(s)}] \, ds\]
Let \(m(t) = E[e^{B(t)}]\). Then \(m'(t) = \frac{1}{2}m(t)\) with \(m(0) = 1\).
Solution: \(E[e^{B(t)}] = e^{t/2}\)
You can verify this directly: since \(B(t) \sim N(0,t)\), the moment generating function gives \(E[e^{B(t)}] = e^{t/2}\). ✓
2.8 General Itô’s Lemma
For a function \(f(t, x)\) of both time and space, and a process \(X(t)\):
\[\boxed{df(t, X) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x}dX + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(dX)^2}\]
This is the form we’ll use most often in applications.
2.8.1 The Pattern
To apply Itô’s lemma:
- Identify the function \(f\) and the stochastic process \(X\)
- Compute the partial derivatives: \(\frac{\partial f}{\partial t}\), \(\frac{\partial f}{\partial x}\), \(\frac{\partial^2 f}{\partial x^2}\)
- Compute \((dX)^2\) using the multiplication rules
- Substitute everything into the formula
This becomes second nature with practice!
2.9 Why Itô’s Choice?
Other choices for the evaluation point lead to other stochastic calculi:
- Stratonovich integral: Uses the midpoint, gives different rules
- Backward integral: Uses the right endpoint
Itô’s calculus won out because:
- Martingale property: Itô integrals are martingales
- Natural for finance: Fits the “no-anticipation” principle
- Connection to PDEs: Leads naturally to Feynman-Kac
- Computational advantage: Easier to simulate
The Itô convention is now standard in probability, finance, and most applications.