Itô Product Rule and Derived Identities¶

1. Concept Definition¶

The Itô product rule describes how the product of two Itô processes evolves over time. It is the stochastic analogue of the classical product rule, with one additional term:

\[ \boxed{ d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t } \]

where \([X,Y]_t\) is the quadratic covariation of \(X_t\) and \(Y_t\).

The extra term \(d[X,Y]_t\) has no classical counterpart. It arises because Brownian increments satisfy \((dW_t)^2 = dt\), so products of stochastic differentials do not vanish. The multiplication rules behind this are established in From Taylor to Itô.

From the product rule two further identities follow directly:

Stochastic integration by parts
Itô quotient rule

2. Why the Classical Product Rule Gains an Extra Term¶

In classical calculus, \(dX\,dY = 0\) because increments scale like \(O(dt)\), making cross terms second-order and negligible. For Itô processes driven by Brownian motion, \(dX_t\) contains a \(dW_t\) component with \(dW_t = O(\sqrt{dt})\), so

\[ dW_t \cdot dW_t = dt \neq 0 \]

Products of stochastic differentials therefore contribute at first order and cannot be dropped.

	Classical calculus	Stochastic calculus
Product rule	\(d(XY) = X\,dY + Y\,dX\)	\(d(XY) = X\,dY + Y\,dX + d[X,Y]\)
Cross term	\(dX\,dY = 0\)	\(d[X,Y]_t\) survives
Reason	\(dX = O(dt)\)	\(dX_t\) has a \(dW_t\) component: \(O(\sqrt{dt})\)

3. Itô Product Rule¶

Let \(X_t\) and \(Y_t\) be scalar Itô processes:

\[ dX_t = a_t\,dt + b_t\,dW_t, \qquad dY_t = c_t\,dt + e_t\,dW_t \]

The quadratic covariation is computed by expanding \(dX_t\,dY_t\) and applying the multiplication table \((dW_t)^2 = dt\), \(dt\,dW_t = 0\), \((dt)^2 = 0\):

\[ dX_t\,dY_t = \underbrace{a_t c_t (dt)^2}_{=\,0} + \underbrace{a_t e_t\,dt\,dW_t}_{=\,0} + \underbrace{b_t c_t\,dW_t\,dt}_{=\,0} + \underbrace{b_t e_t\,(dW_t)^2}_{=\,b_t e_t\,dt} \]

\[ = b_t e_t\,dt \]

so \(d[X,Y]_t = b_t e_t\,dt\). Substituting \(dX_t\), \(dY_t\), and \(d[X,Y]_t\) into \(X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t\):

\[ d(X_t Y_t) = X_t(c_t\,dt + e_t\,dW_t) + Y_t(a_t\,dt + b_t\,dW_t) + b_t e_t\,dt \]

Collecting \(dt\) and \(dW_t\) terms:

\[ d(X_t Y_t) = (X_t c_t + Y_t a_t + b_t e_t)\,dt + (X_t e_t + Y_t b_t)\,dW_t \]

When the correction term vanishes

\(d[X,Y]_t = 0\) when one process is deterministic (so \(b_t = 0\) or \(e_t = 0\)), or when the two processes are driven by independent Brownian motions (separate driving noises \(dW^X\) and \(dW^Y\) that are uncorrelated). For the multi-noise case, see Multidimensional Itô's Lemma.

4. Derived Rules¶

4.1 Stochastic Integration by Parts¶

Integrate the product rule \(d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t\) from \(0\) to \(t\):

\[ X_t Y_t - X_0 Y_0 = \int_0^t X_s\,dY_s + \int_0^t Y_s\,dX_s + [X,Y]_t \]

Rearranging for \(\int_0^t X_s\,dY_s\):

\[ \boxed{ \int_0^t X_s\,dY_s = X_t Y_t - X_0 Y_0 - \int_0^t Y_s\,dX_s - [X,Y]_t } \]

This is the stochastic analogue of classical integration by parts.

4.2 Itô Quotient Rule¶

Write \(X_t / Y_t = X_t \cdot Y_t^{-1}\) and derive in two steps.

Step 1. Apply Itô's lemma (see Itô's Lemma) to \(g(y) = y^{-1}\) with \(g'(y) = -y^{-2}\) and \(g''(y) = 2y^{-3}\):

\[ d(Y_t^{-1}) = -\frac{1}{Y_t^2}\,dY_t + \frac{1}{2} \cdot \frac{2}{Y_t^3}\,(dY_t)^2 = -\frac{dY_t}{Y_t^2} + \frac{e_t^2}{Y_t^3}\,dt \]

where \((dY_t)^2 = e_t^2\,dt\) from the multiplication table. This shows that \(Y_t^{-1}\) is itself an Itô process with diffusion coefficient \(-e_t/Y_t^2\).

Step 2. Apply the product rule to \(X_t \cdot Y_t^{-1}\). Since \(Y_t^{-1}\) has diffusion coefficient \(-e_t/Y_t^2\) (from Step 1), the quadratic covariation between \(X_t\) (diffusion coefficient \(b_t\)) and \(Y_t^{-1}\) is:

\[ d[X, Y^{-1}]_t = b_t \cdot \left(-\frac{e_t}{Y_t^2}\right)dt = -\frac{b_t e_t}{Y_t^2}\,dt \]

Applying the product rule \(d(X_t \cdot Y_t^{-1}) = Y_t^{-1}\,dX_t + X_t\,d(Y_t^{-1}) + d[X, Y^{-1}]_t\) and substituting:

\[ \boxed{ d\!\left(\frac{X_t}{Y_t}\right) = \frac{dX_t}{Y_t} - \frac{X_t\,dY_t}{Y_t^2} - \frac{b_t e_t}{Y_t^2}\,dt + \frac{X_t e_t^2}{Y_t^3}\,dt } \]

where \(b_t\) is the diffusion coefficient of \(X_t\) and \(e_t\) is the diffusion coefficient of \(Y_t\). The first two terms mirror the classical quotient rule; the last two are the Itô corrections arising from the non-zero quadratic variation of \(Y_t\).

5. Worked Examples¶

Example 1 — Computing \(\int_0^t s\,dW_s\)¶

Let \(X_t = t\) (deterministic) and \(Y_t = W_t\), so \(dX_t = dt\) and \(dY_t = dW_t\). Since \(X_t\) is deterministic, its quadratic covariation with any Itô process is zero, so \([X, Y]_t = 0\). Note also that \(X_0 = 0\).

Applying the integration by parts formula with \(X_s = s\), \(Y_s = W_s\):

\[ \int_0^t s\,dW_s = \underbrace{tW_t}_{X_t Y_t} - \underbrace{0}_{X_0 Y_0\,=\,0} - \int_0^t W_s\,ds - \underbrace{0}_{[X,Y]_t} \]

so

\[ \boxed{ \int_0^t s\,dW_s = tW_t - \int_0^t W_s\,ds } \]

Example 2 — Solving the Ornstein–Uhlenbeck SDE¶

Consider the SDE with initial condition \(X_0 = x_0\):

\[ dX_t = -\theta X_t\,dt + \sigma\,dW_t, \qquad X_0 = x_0 \]

Multiply through by the integrating factor \(e^{\theta t}\) and let \(Y_t = e^{\theta t} X_t\).

Apply the product rule to \(Y_t = e^{\theta t} X_t\). Since \(e^{\theta t}\) is deterministic, its quadratic covariation with \(X_t\) is zero and the correction term vanishes:

\[ dY_t = e^{\theta t}\,dX_t + \theta e^{\theta t} X_t\,dt \]

Substituting the SDE:

\[ dY_t = e^{\theta t}(-\theta X_t\,dt + \sigma\,dW_t) + \theta e^{\theta t} X_t\,dt = \sigma e^{\theta t}\,dW_t \]

Integrating from \(0\) to \(t\):

\[ e^{\theta t} X_t = x_0 + \sigma \int_0^t e^{\theta s}\,dW_s \]

Therefore

\[ \boxed{ X_t = e^{-\theta t} x_0 + \sigma e^{-\theta t} \int_0^t e^{\theta s}\,dW_s } \]

6. Summary¶

Rule	Formula
Product rule	\(d(XY) = X\,dY + Y\,dX + d[X,Y]\)
Integration by parts	\(\int_0^t X\,dY = X_t Y_t - X_0 Y_0 - \int_0^t Y\,dX - [X,Y]_t\)
Quotient rule	\(d(X/Y) = Y^{-1}\,dX - XY^{-2}\,dY - b_t e_t Y^{-2}\,dt + X e_t^2 Y^{-3}\,dt\)

Here \(b_t\) and \(e_t\) denote the diffusion coefficients of \(X_t\) and \(Y_t\) respectively (i.e., the coefficients of \(dW_t\) in their SDEs).

The logical dependency is:

flowchart TD
    A["Ito Multiplication Table"] --> B["Ito Lemma"]
    B --> C["Product Rule"]
    C --> D["Integration by Parts"]
    C --> E["Quotient Rule"]

The diagram reflects the logical dependency across the section: the multiplication table (established in From Taylor to Itô) feeds into Itô's lemma (see Itô's Lemma), which is then used in Section 4.2 to derive the quotient rule. Readers encountering this page before ito_lemma.md can treat the lemma as a black box and return to the diagram after reading that page.

\(\square\)

Exercises¶

Exercise 1. Let \(X_t = t\) and \(Y_t = W_t^2\). Compute \(dY_t\) using Itô's lemma, then apply the product rule to compute \(d(tW_t^2)\). Write the result in the form \((\cdots)\,dt + (\cdots)\,dW_t\).

Solution to Exercise 1

First compute \(dY_t\) for \(Y_t = W_t^2\) using Itô's lemma: \(dY_t = 2W_t\,dW_t + dt\).

Now apply the product rule to \(Z_t = tW_t^2 = X_t Y_t\) with \(X_t = t\) (deterministic, \(dX_t = dt\)). Since \(X_t\) is deterministic, \(d[X, Y]_t = 0\):

\[ d(tW_t^2) = t\,dY_t + W_t^2\,dX_t + 0 \]

Substituting \(dY_t = 2W_t\,dW_t + dt\) and \(dX_t = dt\):

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

Exercise 2. Let \(X_t\) and \(Y_t\) both satisfy \(dX_t = dY_t = \sigma\,dW_t\) (pure diffusion, no drift), with \(X_0 = x_0\) and \(Y_0 = y_0\). Use the product rule to compute \(d(X_t Y_t)\). Identify the quadratic covariation term \(d[X, Y]_t\) and verify that it equals \(\sigma^2\,dt\).

Solution to Exercise 2

With \(dX_t = dY_t = \sigma\,dW_t\), the diffusion coefficients are \(b_t = e_t = \sigma\). The quadratic covariation is:

\[ d[X, Y]_t = b_t e_t\,dt = \sigma^2\,dt \]

Applying the product rule:

\[ d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t = X_t\sigma\,dW_t + Y_t\sigma\,dW_t + \sigma^2\,dt \]

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

The covariation \(d[X,Y]_t = \sigma^2\,dt\) is verified: both processes have the same diffusion coefficient \(\sigma\) and are driven by the same Brownian motion.

Exercise 3. Derive the stochastic integration by parts formula for \(\int_0^t W_s^2\,dW_s\) by choosing \(X_s = W_s\) and \(Y_s = W_s\), applying the product rule to \(d(W_s \cdot W_s)\), and solving for the integral. Verify that your result matches Example 4 from the Applications page.

Solution to Exercise 3

First, apply the product rule with \(X_s = Y_s = W_s\) to get \(d(W_t^2) = 2W_t\,dW_t + dt\), which gives \(\int_0^t W_s\,dW_s = \frac{1}{2}(W_t^2 - t)\).

To derive \(\int_0^t W_s^2\,dW_s\), apply integration by parts with \(X_s = W_s\) and \(Y_s = W_s^2\). From Itô's lemma, \(dY_s = 2W_s\,dW_s + ds\), and \(dX_s = dW_s\). The diffusion coefficient of \(X\) is \(1\) and the diffusion coefficient of \(Y\) is \(2W_s\), so the covariation is \(d[X, Y]_s = 2W_s\,dt\).

The integration by parts formula gives:

\[ \int_0^t W_s\,dY_s = W_t \cdot W_t^2 - 0 - \int_0^t W_s^2\,dW_s - [X, Y]_t \]

The left side expands as \(\int_0^t W_s(2W_s\,dW_s + ds) = 2\int_0^t W_s^2\,dW_s + \int_0^t W_s\,ds\), and \([X,Y]_t = \int_0^t 2W_s\,ds\). Substituting:

\[ 2\int_0^t W_s^2\,dW_s + \int_0^t W_s\,ds = W_t^3 - \int_0^t W_s^2\,dW_s - 2\int_0^t W_s\,ds \]

\[ 3\int_0^t W_s^2\,dW_s = W_t^3 - 3\int_0^t W_s\,ds \]

\[ \int_0^t W_s^2\,dW_s = \frac{1}{3}W_t^3 - \int_0^t W_s\,ds \]

This matches Example 4 from the Applications page.

Exercise 4. Let \(dX_t = \mu X_t\,dt + \sigma X_t\,dW_t\) (geometric Brownian motion). Use the quotient rule to compute \(d(1/X_t)\), and show that \(1/X_t\) also follows a geometric Brownian motion SDE. Identify its drift and diffusion coefficients.

Solution to Exercise 4

For \(dX_t = \mu X_t\,dt + \sigma X_t\,dW_t\), use the quotient rule with \(X_t\) in the numerator replaced by the constant \(1\) (i.e., compute \(d(1/X_t)\)). Equivalently, apply Itô's lemma to \(g(x) = 1/x\):

\[ g'(x) = -x^{-2}, \qquad g''(x) = 2x^{-3} \]

With \(\mu_t = \mu X_t\) and \(\sigma_t = \sigma X_t\):

\[ d(X_t^{-1}) = \left(-X_t^{-2}\mu X_t + \frac{1}{2}(2X_t^{-3})\sigma^2 X_t^2\right)dt + (-X_t^{-2})\sigma X_t\,dW_t \]

\[ = (-\mu + \sigma^2)X_t^{-1}\,dt - \sigma X_t^{-1}\,dW_t \]

So \(1/X_t\) follows a geometric Brownian motion SDE with drift \(-\mu + \sigma^2\) and diffusion \(-\sigma\). The Itô correction \(+\sigma^2\) in the drift arises from the positive curvature of \(1/x\).

Exercise 5. Let \(X_t = e^{W_t}\) and \(Y_t = e^{-W_t}\). Compute \(dX_t\) and \(dY_t\) using Itô's lemma. Then use the product rule to compute \(d(X_t Y_t)\). Since \(X_t Y_t = 1\) for all \(t\), verify that \(d(X_t Y_t) = 0\) and confirm that the classical terms and the correction term cancel exactly.

Solution to Exercise 5

For \(X_t = e^{W_t}\), Itô's lemma gives \(dX_t = e^{W_t}\,dW_t + \frac{1}{2}e^{W_t}\,dt\).

For \(Y_t = e^{-W_t}\), Itô's lemma gives \(dY_t = -e^{-W_t}\,dW_t + \frac{1}{2}e^{-W_t}\,dt\).

The diffusion coefficients are \(b_t = e^{W_t}\) and \(e_t = -e^{-W_t}\). The covariation is:

\[ d[X, Y]_t = e^{W_t} \cdot (-e^{-W_t})\,dt = -dt \]

Product rule:

\[ d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t \]

\[ = e^{W_t}\!\left(-e^{-W_t}\,dW_t + \frac{1}{2}e^{-W_t}\,dt\right) + e^{-W_t}\!\left(e^{W_t}\,dW_t + \frac{1}{2}e^{W_t}\,dt\right) - dt \]

\[ = -dW_t + \frac{1}{2}\,dt + dW_t + \frac{1}{2}\,dt - dt \]

\[ = 0 \]

The \(dW_t\) terms cancel (\(-dW_t + dW_t = 0\)), and the \(dt\) terms cancel (\(\frac{1}{2} + \frac{1}{2} - 1 = 0\)). This confirms \(d(X_tY_t) = 0\), consistent with \(X_tY_t = e^{W_t}e^{-W_t} = 1\).

Exercise 6. Consider the Ornstein--Uhlenbeck SDE: \(dX_t = -\theta X_t\,dt + \sigma\,dW_t\). Let \(Y_t = e^{\theta t}\) (deterministic). Apply the product rule to \(Z_t = X_t Y_t\) and show that \(dZ_t = \sigma e^{\theta t}\,dW_t\). Explain why the quadratic covariation \(d[X, Y]_t\) vanishes.

Solution to Exercise 6

For \(Z_t = X_t Y_t = X_t e^{\theta t}\) with \(Y_t = e^{\theta t}\) deterministic (\(dY_t = \theta e^{\theta t}\,dt\)):

The product rule gives \(d(X_tY_t) = Y_t\,dX_t + X_t\,dY_t + d[X,Y]_t\).

Since \(Y_t\) is deterministic, its diffusion coefficient is zero, so \(d[X,Y]_t = 0\). Substituting \(dX_t = -\theta X_t\,dt + \sigma\,dW_t\) and \(dY_t = \theta e^{\theta t}\,dt\):

\[ dZ_t = e^{\theta t}(-\theta X_t\,dt + \sigma\,dW_t) + X_t \theta e^{\theta t}\,dt \]

\[ = -\theta X_t e^{\theta t}\,dt + \sigma e^{\theta t}\,dW_t + \theta X_t e^{\theta t}\,dt = \sigma e^{\theta t}\,dW_t \]

The drift terms cancel, leaving \(dZ_t = \sigma e^{\theta t}\,dW_t\). The quadratic covariation \(d[X,Y]_t\) vanishes because \(Y_t = e^{\theta t}\) is a deterministic function of time with no stochastic component (its diffusion coefficient is zero).

Exercise 7. Let \(dX_t = a_t\,dt + b_t\,dW_t\) and \(dY_t = c_t\,dt + e_t\,dW_t\). Starting from the product rule \(d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + d[X,Y]_t\), derive the formula for \(d(X_t^2)\) by setting \(Y_t = X_t\). Show that

\[ d(X_t^2) = 2X_t\,dX_t + b_t^2\,dt \]

and interpret the term \(b_t^2\,dt\) as the Itô correction.

Solution to Exercise 7

Setting \(Y_t = X_t\) in the product rule: \(dX_t = a_t\,dt + b_t\,dW_t\), so the diffusion coefficients for both "copies" are \(b_t\). The covariation is \(d[X, X]_t = b_t^2\,dt\):

\[ d(X_t^2) = X_t\,dX_t + X_t\,dX_t + d[X,X]_t = 2X_t\,dX_t + b_t^2\,dt \]

The term \(b_t^2\,dt\) is the Itô correction. It arises because \((dX_t)^2 = b_t^2\,dt \neq 0\): the squared diffusion coefficient contributes a deterministic drift to \(X_t^2\) that has no classical counterpart. Geometrically, \(x^2\) is convex (\(f'' = 2 > 0\)), so symmetric random fluctuations of size \(b_t\,dW_t\) produce a net positive contribution \(b_t^2\,dt\) to the expected change in \(X_t^2\).