From Taylor to Itô¶

This page derives Itô's formula mechanically from a second-order Taylor expansion. The key tool is the Itô multiplication table, which evaluates each second-order differential product term by term, reducing the five-term expansion to Itô's formula.

1. Second-Order Taylor Expansion¶

Let \(f(t, x)\) be \(C^{1,2}([0,\infty)\times\mathbb{R})\). A second-order Taylor expansion in the increments \(dt\) and \(dW_t\) gives

\[ f(t+dt,\, W_t + dW_t) - f(t,\, W_t) = f_t \, dt + f_x \, dW_t + \tfrac{1}{2} f_{tt} (dt)^2 + f_{tx} \, dt \, dW_t + \tfrac{1}{2} f_{xx} (dW_t)^2 \]

where all partial derivatives are evaluated at \((t, W_t)\), and

\[ f_t = \frac{\partial f}{\partial t}, \quad f_x = \frac{\partial f}{\partial x}, \quad f_{tt} = \frac{\partial^2 f}{\partial t^2}, \quad f_{tx} = \frac{\partial^2 f}{\partial t \, \partial x}, \quad f_{xx} = \frac{\partial^2 f}{\partial x^2} \]

This expansion has two first-order terms (\(f_t\,dt\) and \(f_x\,dW_t\)) and three second-order terms. The next step is to decide which of the second-order terms survive.

2. The Itô Multiplication Table¶

Each second-order term involves a product of two differentials. The following table gives the limiting value of each product:

\[ \begin{array}{c|cc} \times & dt & dW_t \\ \hline dt & 0 & 0 \\ dW_t & 0 & dt \\ \end{array} \]

That is:

\[ (dt)^2 = 0, \qquad dt \, dW_t = 0, \qquad (dW_t)^2 = dt \]

Why each rule holds:

Product	Order	Verdict
\((dt)^2\)	\(O((\Delta t)^2)\)	vanishes — higher-order infinitesimal
\(dt \, dW_t\)	\(O((\Delta t)^{3/2})\)	vanishes — since \(3/2 > 1\), this is \(o(dt)\)
\((dW_t)^2\)	\(O(\Delta t)\)	survives — this is quadratic variation

The survival of \((dW_t)^2\) is the central fact. A smooth deterministic path satisfies \(\Delta x = O(\Delta t)\), so \((\Delta x)^2 = O((\Delta t)^2)\) vanishes. Brownian motion instead satisfies \(\Delta W_t = O(\sqrt{\Delta t})\), so \((\Delta W_t)^2 = O(\Delta t)\) — the same order as \(dt\) itself. The rigorous statement is that the quadratic variation of Brownian motion equals \(t\); see Quadratic Variation of Brownian Motion.

The key asymmetry

A smooth deterministic path satisfies \(\Delta x = O(\Delta t)\), so \((\Delta x)^2 = O((\Delta t)^2)\) — it vanishes.
Brownian motion satisfies \(\Delta W = O(\sqrt{\Delta t})\), so \((\Delta W)^2 = O(\Delta t)\) — it survives.
This single difference is the source of the Itô correction term that separates stochastic calculus from ordinary calculus.

3. Applying the Table Term by Term¶

Return to the five-term expansion and apply the multiplication table to each second-order product:

\[ f(t+dt,\, W_t+dW_t) - f(t,\, W_t) = f_t \, dt + f_x \, dW_t + \tfrac{1}{2} f_{tt} \underbrace{(dt)^2}_{=\,0} + f_{tx} \underbrace{dt \, dW_t}_{=\,0} + \tfrac{1}{2} f_{xx} \underbrace{(dW_t)^2}_{=\,dt} \]

Dropping the zero terms and substituting \((dW_t)^2 = dt\):

\[ \boxed{ df(t, W_t) = f_t \, dt + f_x \, dW_t + \tfrac{1}{2} f_{xx} \, dt } \]

This is Itô's formula. The term \(\tfrac{1}{2} f_{xx} \, dt\) is the Itô correction — it has no counterpart in ordinary calculus and arises entirely from the survival of \((dW_t)^2\).

\(\square\)

4. Worked Example¶

Let \(f(x) = x^2\). We set \(f(t,x) = x^2\), so \(f_t = f_{tt} = f_{tx} = 0\), \(f_x = 2x\), and \(f_{xx} = 2\).

Step 1. Write all five terms in the same order as Section 1, substituting the zero coefficients explicitly:

\[ f(t, W_t + dW_t) - f(t, W_t) = \underbrace{0 \cdot dt}_{f_t\,dt\,=\,0} + f_x \, dW_t + \tfrac{1}{2} f_{tt} \underbrace{(dt)^2}_{=\,0} + f_{tx} \underbrace{dt \, dW_t}_{=\,0} + \tfrac{1}{2} f_{xx} (dW_t)^2 \]

Step 2. Drop the zero terms and substitute \(f_x = 2x\big|_{x=W_t} = 2W_t\) and \(f_{xx} = 2\):

\[ = 2W_t \, dW_t + \tfrac{1}{2} \cdot 2 \cdot (dW_t)^2 \]

Step 3. Apply \((dW_t)^2 = dt\) and simplify \(\tfrac{1}{2} \cdot 2 = 1\):

\[ d(W_t^2) = 2W_t \, dW_t + dt \]

In ordinary calculus one would expect only \(d(x^2) = 2x\,dx\). The extra \(dt\) term is the Itô correction: the positive curvature of \(x^2\) converts symmetric Brownian fluctuations into a positive drift.

\(\square\)

Extension to general Itô processes

The same procedure extends to a general Itô process \(dX_t = \mu_t\,dt + \sigma_t\,dW_t\). The key step is that \((dX_t)^2 = \sigma_t^2\,dt\), because only the \(dW_t\) component survives squaring: \((\mu_t\,dt)^2 = 0\), \(\mu_t\,dt\cdot\sigma_t\,dW_t = 0\), and \((\sigma_t\,dW_t)^2 = \sigma_t^2\,dt\). Substituting this into the second-order Taylor term yields the general Itô formula; see Itô's Lemma.

Exercises¶

Exercise 1. Using the Itô multiplication table, evaluate each of the following products:

(a) \((3\,dt)(2\,dW_t)\)

(b) \((dW_t)(5\,dW_t)\)

(c) \((\mu\,dt + \sigma\,dW_t)^2\) for constants \(\mu\) and \(\sigma\)

Solution to Exercise 1

(a) \((3\,dt)(2\,dW_t) = 6\,dt\,dW_t = 6 \cdot 0 = 0\) since \(dt\,dW_t = 0\) by the multiplication table.

(b) \((dW_t)(5\,dW_t) = 5(dW_t)^2 = 5\,dt\) since \((dW_t)^2 = dt\).

(c) Expanding:

\[ (\mu\,dt + \sigma\,dW_t)^2 = \mu^2(dt)^2 + 2\mu\sigma\,dt\,dW_t + \sigma^2(dW_t)^2 \]

Applying the multiplication table: \((dt)^2 = 0\), \(dt\,dW_t = 0\), \((dW_t)^2 = dt\). Therefore

\[ (\mu\,dt + \sigma\,dW_t)^2 = \sigma^2\,dt \]

Only the Brownian component survives when squaring an Itô differential.

Exercise 2. Let \(f(x) = e^x\). Write out all five terms of the second-order Taylor expansion for \(f(t, W_t + dW_t) - f(t, W_t)\), apply the multiplication table term by term, and derive Itô's formula for \(d(e^{W_t})\).

Solution to Exercise 2

For \(f(x) = e^x\) (no explicit time dependence), we have \(f_t = 0\), \(f_{tt} = 0\), \(f_{tx} = 0\), and \(f_x = e^x\), \(f_{xx} = e^x\). The five-term expansion is

\[ f(t, W_t + dW_t) - f(t, W_t) = 0 \cdot dt + e^{W_t}\,dW_t + \frac{1}{2}(0)(dt)^2 + 0 \cdot dt\,dW_t + \frac{1}{2}e^{W_t}(dW_t)^2 \]

Applying the multiplication table: \((dt)^2 = 0\), \(dt\,dW_t = 0\), \((dW_t)^2 = dt\):

\[ d(e^{W_t}) = e^{W_t}\,dW_t + \frac{1}{2}e^{W_t}\,dt \]

The Itô correction \(\frac{1}{2}e^{W_t}\,dt\) arises from the positive curvature of \(e^x\) (since \(f''(x) = e^x > 0\)).

Exercise 3. Explain why \((dt \cdot dW_t) = 0\) by analyzing the order of magnitude. Specifically, if \(dW_t = O(\sqrt{dt})\), show that \(dt \cdot dW_t = O((dt)^{3/2})\), and explain why this vanishes faster than \(dt\).

Solution to Exercise 3

If \(dW_t = O(\sqrt{dt})\), then

\[ dt \cdot dW_t = O(dt \cdot \sqrt{dt}) = O((dt)^{3/2}) \]

Since \(3/2 > 1\), this product vanishes faster than \(dt\) as \(dt \to 0\). Formally, \(\frac{dt \cdot dW_t}{dt} = O((dt)^{1/2}) \to 0\), so \(dt \cdot dW_t = o(dt)\). In the Taylor expansion, any term that is \(o(dt)\) makes zero contribution in the infinitesimal limit and is therefore set to zero in the Itô multiplication table.

Exercise 4. Apply the derivation of Section 3 to the function \(f(t, x) = \sin(x)\). Compute \(f_t\), \(f_x\), \(f_{xx}\), substitute into the five-term Taylor expansion, apply the multiplication table, and write the resulting Itô formula for \(d(\sin(W_t))\).

Solution to Exercise 4

For \(f(t, x) = \sin(x)\): \(f_t = 0\), \(f_x = \cos(x)\), \(f_{tt} = 0\), \(f_{tx} = 0\), \(f_{xx} = -\sin(x)\). The five-term expansion gives

\[ d(\sin(W_t)) = 0 \cdot dt + \cos(W_t)\,dW_t + \frac{1}{2}(0)(dt)^2 + 0 \cdot dt\,dW_t + \frac{1}{2}(-\sin(W_t))(dW_t)^2 \]

Applying the multiplication table (only \((dW_t)^2 = dt\) survives):

\[ d(\sin(W_t)) = \cos(W_t)\,dW_t - \frac{1}{2}\sin(W_t)\,dt \]

The Itô correction \(-\frac{1}{2}\sin(W_t)\,dt\) arises from the negative curvature of \(\sin(x)\) at points where \(\sin(x) > 0\).

Exercise 5. Consider a general Itô process \(dX_t = \mu_t\,dt + \sigma_t\,dW_t\).

(a) Expand \((dX_t)^2 = (\mu_t\,dt + \sigma_t\,dW_t)^2\) into four terms.

(b) Apply the multiplication table to each term and show that \((dX_t)^2 = \sigma_t^2\,dt\).

(c) Use this to write the second-order Taylor term \(\frac{1}{2}f_{xx}(dX_t)^2\) for a general \(C^2\) function \(f\).

Solution to Exercise 5

(a) Expanding \((dX_t)^2 = (\mu_t\,dt + \sigma_t\,dW_t)^2\):

\[ (dX_t)^2 = \mu_t^2(dt)^2 + 2\mu_t\sigma_t\,dt\,dW_t + \sigma_t^2(dW_t)^2 \]

(b) Applying the multiplication table term by term:

\(\mu_t^2(dt)^2 = 0\)
\(2\mu_t\sigma_t\,dt\,dW_t = 0\)
\(\sigma_t^2(dW_t)^2 = \sigma_t^2\,dt\)

Therefore \((dX_t)^2 = \sigma_t^2\,dt\).

(c) The second-order Taylor term becomes

\[ \frac{1}{2}f_{xx}(dX_t)^2 = \frac{1}{2}f''(X_t)\sigma_t^2\,dt \]

This is the Itô correction for a general Itô process: it depends only on the diffusion coefficient \(\sigma_t\) and the second derivative of \(f\).

Exercise 6. Let \(f(x) = x^3\). Derive \(d(W_t^3)\) by writing out the full five-term Taylor expansion and applying the multiplication table. Verify that your result matches what you would obtain from applying Itô's formula directly.

Solution to Exercise 6

For \(f(x) = x^3\): \(f_t = 0\), \(f_x = 3x^2\), \(f_{xx} = 6x\), \(f_{tt} = f_{tx} = 0\). The five-term expansion:

\[ d(W_t^3) = 0 \cdot dt + 3W_t^2\,dW_t + 0 + 0 + \frac{1}{2}(6W_t)(dW_t)^2 \]

Applying \((dW_t)^2 = dt\):

\[ d(W_t^3) = 3W_t^2\,dW_t + 3W_t\,dt \]

Verification via Itô's formula: \(df(B_t) = f'(B_t)\,dB_t + \frac{1}{2}f''(B_t)\,dt = 3W_t^2\,dW_t + \frac{1}{2}(6W_t)\,dt = 3W_t^2\,dW_t + 3W_t\,dt\). The results match.

Exercise 7. Suppose two independent Brownian motions \(W_t^1\) and \(W_t^2\) are given. Using the multiplication table extended to two independent Brownian motions (where \(dW_t^1 \cdot dW_t^2 = 0\)), compute \((dW_t^1 + dW_t^2)^2\) and interpret the result.

Solution to Exercise 7

Expanding \((dW_t^1 + dW_t^2)^2\):

\[ (dW_t^1 + dW_t^2)^2 = (dW_t^1)^2 + 2\,dW_t^1\,dW_t^2 + (dW_t^2)^2 \]

Since \(W_t^1\) and \(W_t^2\) are independent: \((dW_t^1)^2 = dt\), \((dW_t^2)^2 = dt\), and \(dW_t^1 \cdot dW_t^2 = 0\). Therefore

\[ (dW_t^1 + dW_t^2)^2 = dt + 0 + dt = 2\,dt \]

Interpretation: the process \(W_t^1 + W_t^2\) has quadratic variation \(2t\), which is twice that of a single Brownian motion. This is consistent with the fact that \(W_t^1 + W_t^2\) has variance \(2t\) (the variances add for independent processes). Equivalently, \(\frac{1}{\sqrt{2}}(W_t^1 + W_t^2)\) is a standard Brownian motion with quadratic variation \(t\).