From Taylor to Itô¶
This page is the canonical home of the Itô multiplication table and the mechanical derivation of Itô's formula from a second-order Taylor expansion. Other pages in this section recall the rule \((dB_t)^2 = dt\) and link back here rather than rederive it.
Recall (see § Quadratic Approximation): the second-order Taylor expansion already contains the term \(\tfrac12 f_{xx}(\Delta x)^2\). The novelty here is purely in the scaling of \((\Delta B_t)^2\) — the rest is bookkeeping.
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
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:
That is:
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. The scaling argument behind each rule — why deterministic squares vanish but Brownian squares survive — is developed in Quadratic Taylor Expansion and Quadratic Variation of Brownian Motion. Here we simply apply the table mechanically.
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 \((dW_t)^2 = dt\):
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 \(f_{xx} = 2\):
Step 3. Apply \((dW_t)^2 = dt\) and simplify \(\tfrac{1}{2} \cdot 2 = 1\):
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.
5. Role in the System¶
This page is the origin of Itô calculus: a single mechanism (Brownian scaling) generates a single rule \((dW_t)^2 = dt\), which forces a single new identity (Itô's formula). Everything downstream is a corollary:
flowchart LR
A[Taylor Expansion] --> B["(dW)² = dt"]
B --> C[Itô's Lemma]
C --> D[Generator]
D --> E[Dynkin Formula]
Subsequent pages apply the multiplication table rather than rederive it: Itô's Lemma states the transformation rule, Itô Rules develop the derived identities (product, quotient, integration by parts), and Applications puts the lemma to work.
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:
Applying the multiplication table: \((dt)^2 = 0\), \(dt\,dW_t = 0\), \((dW_t)^2 = dt\). Therefore
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
Applying the multiplication table: \((dt)^2 = 0\), \(dt\,dW_t = 0\), \((dW_t)^2 = dt\):
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
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
Applying the multiplication table (only \((dW_t)^2 = dt\) survives):
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 \((dX_t)^2 = (\mu_t\,dt + \sigma_t\,dW_t)^2\):
(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
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:
Applying \((dW_t)^2 = dt\):
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 \((dW_t^1 + dW_t^2)^2\):
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
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 \(t\).