Applications and Examples of Itô Calculus¶

Once the main rules of stochastic calculus have been developed — Itô's lemma, the product rule, and integration by parts — we can use them to compute stochastic integrals, construct martingales, and solve stochastic differential equations (SDEs).

This section presents a collection of worked examples illustrating how these tools are applied in practice. For the theorem statements being applied throughout, see Itô's Lemma. For the multiplication rules used in SDE substitutions, see Itô Rules.

1. Evaluating Stochastic Integrals¶

One of the most important uses of Itô's lemma is to evaluate stochastic integrals of the form

\[ \int_0^t g(s,B_s)\,dB_s \]

The key idea is to choose a function \(f(t,x)\) whose derivative with respect to \(x\) matches the integrand.

General Strategy¶

Choose \(f(t,x)\) such that \(\frac{\partial f}{\partial x}(t,x) = g(t,x)\). Applying Itô's lemma to \(f(t,B_t)\) gives

\[ df(t,B_t) = \frac{\partial f}{\partial t}\,dt + \frac{\partial f}{\partial x}\,dB_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}\,dt \]

Rearranging isolates the stochastic integral and expresses it in terms of \(f(t,B_t)\) and ordinary integrals.

Example 1: \(\displaystyle\int_0^t B_s\,dB_s\)¶

Choose \(f(x) = \frac{x^2}{2}\), so that \(f'(x) = x\).

\[ f'(x) = x, \qquad f''(x) = 1 \]

Applying Itô's lemma:

\[ d\!\left(\frac{B_t^2}{2}\right) = B_t\,dB_t + \frac{1}{2}\,dt \]

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

\[ \boxed{ \int_0^t B_s\,dB_s = \frac{B_t^2 - t}{2} } \]

This identity implies that \(B_t^2 - t\) is a martingale.

Example 2: \(\displaystyle\int_0^t s\,dB_s\)¶

The integrand \(s\) is deterministic and square-integrable. For deterministic integrands, the Itô isometry gives the distribution directly: the integral is Gaussian with zero mean (since all Itô integrals of deterministic square-integrable integrands are centered Gaussian) and variance

\[ \mathbb{E}\!\left[\left(\int_0^t s\,dB_s\right)^2\right] = \int_0^t s^2\,ds = \frac{t^3}{3} \]

Therefore (where \(\overset{d}{=}\) denotes equality in distribution):

\[ \boxed{ \int_0^t s\,dB_s \overset{d}{=} \mathcal{N}\!\left(0,\frac{t^3}{3}\right) } \]

For a path-by-path expression, apply the antiderivative approach with \(f(t,x) = tx\), so \(f_t = x\), \(f_x = t\), \(f_{xx} = 0\). Itô's lemma gives:

\[ d(tB_t) = B_t\,dt + t\,dB_t \]

Integrating and solving for the stochastic integral:

\[ \int_0^t s\,dB_s = tB_t - \int_0^t B_s\,ds \]

This path-by-path identity is consistent with the Gaussian characterization above.

Example 3: \(\displaystyle\int_0^t sB_s\,dB_s\)¶

Choose \(f(t,x) = \frac{1}{2}tx^2\), so that \(\frac{\partial f}{\partial x} = tx\).

\[ \frac{\partial f}{\partial t} = \frac{1}{2}x^2, \quad \frac{\partial f}{\partial x} = tx, \quad \frac{\partial^2 f}{\partial x^2} = t \]

Applying the time-dependent Itô formula:

\[ d\!\left(\frac{1}{2}tB_t^2\right) = \frac{1}{2}B_t^2\,dt + tB_t\,dB_t + \frac{1}{2}t\,dt \]

Isolating the stochastic term:

\[ tB_t\,dB_t = d\!\left(\frac{1}{2}tB_t^2\right) - \frac{1}{2}B_t^2\,dt - \frac{1}{2}t\,dt \]

Integrating and evaluating \(\int_0^t \frac{1}{2}s\,ds = \frac{1}{4}t^2\):

\[ \boxed{ \int_0^t sB_s\,dB_s = \frac{1}{2}tB_t^2 - \frac{1}{2}\int_0^t B_s^2\,ds - \frac{1}{4}t^2 } \]

The integral \(\int_0^t B_s^2\,ds\) has no closed form and is left as is. This example shows how time dependence introduces additional drift terms.

Example 4: \(\displaystyle\int_0^t B_s^2\,dB_s\)¶

Choose \(f(x) = \frac{x^3}{3}\), so that \(f'(x) = x^2\).

\[ f'(x) = x^2, \qquad f''(x) = 2x \]

Applying Itô's lemma:

\[ d\!\left(\frac{B_t^3}{3}\right) = B_t^2\,dB_t + B_t\,dt \]

Therefore:

\[ \boxed{ \int_0^t B_s^2\,dB_s = \frac{B_t^3}{3} - \int_0^t B_s\,ds } \]

The integral \(\int_0^t B_s\,ds\) has no closed form and is left as is.

Example 5: \(\displaystyle\int_0^t e^{B_s}\,dB_s\)¶

Apply Itô's lemma to \(f(x) = e^x\):

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

Integrating and rearranging:

\[ \boxed{ \int_0^t e^{B_s}\,dB_s = e^{B_t} - 1 - \frac{1}{2}\int_0^t e^{B_s}\,ds } \]

The integral \(\int_0^t e^{B_s}\,ds\) has no closed form and is left as is.

2. Constructing Martingales¶

A sufficient condition for \(M_t\) to be a local martingale is that its Itô differential has no \(dt\) term. Under suitable integrability conditions (such as Novikov's condition for exponential processes), it is a true martingale. Itô's lemma makes the vanishing-drift condition explicit and checkable.

Example 6: The Exponential Martingale¶

For \(\theta \in \mathbb{R}\), define

\[ Z_t = \exp\!\left(\theta B_t - \frac{1}{2}\theta^2 t\right) \]

Apply Itô's lemma to \(f(t,x) = e^{\theta x - \frac{1}{2}\theta^2 t}\):

\[ f_t = -\frac{1}{2}\theta^2 e^{\theta x - \frac{1}{2}\theta^2 t}, \quad f_x = \theta e^{\theta x - \frac{1}{2}\theta^2 t}, \quad f_{xx} = \theta^2 e^{\theta x - \frac{1}{2}\theta^2 t} \]

The \(dt\) coefficient is \(f_t + \frac{1}{2}f_{xx} = -\frac{1}{2}\theta^2 Z_t + \frac{1}{2}\theta^2 Z_t = 0\), so:

\[ \boxed{dZ_t = \theta Z_t\,dB_t} \]

Since the \(dt\) coefficient vanishes, \(Z_t\) is a local martingale. It is a true martingale by Novikov's condition: \(\mathbb{E}\!\left[\exp\!\left(\frac{1}{2}\theta^2 T\right)\right] < \infty\) for all finite \(T\), which holds since \(\theta\) is a constant. This SDE has the explicit solution \(Z_t = Z_0\exp(\theta B_t - \frac{1}{2}\theta^2 t)\), confirming that the original definition and the SDE are consistent.

This process is fundamental in Girsanov's theorem and risk-neutral pricing.

3. Solving Stochastic Differential Equations¶

The standard approach is to guess a transformation \(f(t, X_t)\) that converts the SDE into a simpler (often deterministic-coefficient) equation, then integrate.

Example 7: Geometric Brownian Motion¶

Consider

\[ dS_t = \mu S_t\,dt + \sigma S_t\,dB_t, \qquad S_0 = s_0 \]

where \(\mu \in \mathbb{R}\) and \(\sigma > 0\) are constants. Apply Itô's lemma to \(f(x) = \log x\):

\[ f'(x) = \frac{1}{x}, \qquad f''(x) = -\frac{1}{x^2} \]

First compute the quadratic variation term:

\[ (dS_t)^2 = \sigma^2 S_t^2\,dt \]

Substituting \(dS_t = \mu S_t\,dt + \sigma S_t\,dB_t\) and \((dS_t)^2 = \sigma^2 S_t^2\,dt\):

\[ d(\log S_t) = \frac{1}{S_t}(\mu S_t\,dt + \sigma S_t\,dB_t) - \frac{1}{2}\frac{1}{S_t^2}\cdot\sigma^2 S_t^2\,dt = \left(\mu - \frac{1}{2}\sigma^2\right)dt + \sigma\,dB_t \]

Integrating:

\[ \boxed{ S_t = s_0\exp\!\left[\left(\mu - \frac{1}{2}\sigma^2\right)t + \sigma B_t\right] } \]

The Itô correction \(-\frac{1}{2}\sigma^2\) is the convexity adjustment between arithmetic drift and geometric growth rate. This process is the foundation of the Black–Scholes model.

4. Further Process Transformations¶

The example below applies Itô's lemma directly to a process transformation, illustrating how the correction term can introduce unexpected drift even when the original function has no explicit time dependence.

Example 8: Reciprocal of Brownian Motion¶

Let \(B_0 = b \neq 0\) (we require \(b \neq 0\) since \(f(x) = 1/x\) is undefined at \(x = 0\)) and let \(Y_t = 1/B_t\), valid up to \(\tau_0 = \inf\{t > 0 : B_t = 0\}\). For \(f(x) = 1/x\):

\[ f'(x) = -\frac{1}{x^2}, \qquad f''(x) = \frac{2}{x^3} \]

Applying Itô's lemma:

\[ d\!\left(\frac{1}{B_t}\right) = -\frac{1}{B_t^2}\,dB_t + \frac{1}{B_t^3}\,dt \]

The \(dt\) term \(\frac{1}{B_t^3}\,dt\) is a non-zero drift, so \(1/B_t\) is not a local martingale. The Itô correction introduces a drift even though \(f(x) = 1/x\) has no explicit time dependence. In integral form:

\[ \boxed{ \frac{1}{B_t} = \frac{1}{b} - \int_0^t \frac{1}{B_s^2}\,dB_s + \int_0^t \frac{1}{B_s^3}\,ds } \]

Summary¶

Technique	Key step	Examples
Evaluate stochastic integrals	Find \(f\) with \(f_x = g\); apply Itô's lemma and rearrange	1–5
Construct martingales	Verify that \(dt\) coefficient vanishes; check integrability	6
Solve SDEs	Apply a simplifying transformation; integrate	7
Process transformations	Apply Itô's lemma directly; correction term produces non-zero drift from curvature even without explicit time dependence	8

Exercises¶

Exercise 1. Evaluate \(\int_0^t B_s^3\,dB_s\) by choosing an appropriate function \(f(x)\) with \(f'(x) = x^3\) and applying Itô's lemma. Express your answer in terms of \(B_t\) and an ordinary integral.

Solution to Exercise 1

Choose \(f(x) = \frac{1}{4}x^4\), so \(f'(x) = x^3\) and \(f''(x) = 3x^2\). Itô's lemma gives:

\[ d\!\left(\frac{B_t^4}{4}\right) = B_t^3\,dB_t + \frac{1}{2}(3B_t^2)\,dt \]

Integrating from \(0\) to \(t\) (with \(B_0 = 0\)):

\[ \frac{B_t^4}{4} = \int_0^t B_s^3\,dB_s + \frac{3}{2}\int_0^t B_s^2\,ds \]

Therefore:

\[ \int_0^t B_s^3\,dB_s = \frac{1}{4}B_t^4 - \frac{3}{2}\int_0^t B_s^2\,ds \]

Exercise 2. For \(\theta \in \mathbb{R}\), define \(M_t = \cos(\theta B_t)\,e^{\theta^2 t/2}\). Apply Itô's lemma to show that \(dM_t\) has no \(dt\) term, and conclude that \(M_t\) is a local martingale. Write the SDE satisfied by \(M_t\).

Solution to Exercise 2

For \(M_t = \cos(\theta B_t)\,e^{\theta^2 t/2}\), write \(f(t, x) = \cos(\theta x)\,e^{\theta^2 t/2}\). Compute:

\(f_t = \frac{\theta^2}{2}\cos(\theta x)\,e^{\theta^2 t/2}\)
\(f_x = -\theta\sin(\theta x)\,e^{\theta^2 t/2}\)
\(f_{xx} = -\theta^2\cos(\theta x)\,e^{\theta^2 t/2}\)

By Itô's lemma (Version 2):

\[ dM_t = \left(\frac{\theta^2}{2}\cos(\theta B_t) + \frac{1}{2}(-\theta^2\cos(\theta B_t))\right)e^{\theta^2 t/2}\,dt - \theta\sin(\theta B_t)\,e^{\theta^2 t/2}\,dB_t \]

The \(dt\) coefficient is \(\left(\frac{\theta^2}{2} - \frac{\theta^2}{2}\right)\cos(\theta B_t)\,e^{\theta^2 t/2} = 0\). Therefore:

\[ dM_t = -\theta\sin(\theta B_t)\,e^{\theta^2 t/2}\,dB_t \]

Since the \(dt\) term vanishes, \(M_t\) is a local martingale.

Exercise 3. The Ornstein--Uhlenbeck process satisfies \(dX_t = -\theta X_t\,dt + \sigma\,dB_t\) with \(X_0 = x_0\). Apply Itô's lemma to \(f(t, x) = e^{\theta t}x\) to derive the explicit solution for \(X_t\). (Hint: the product-rule approach in Example 7 of the Itô Rules page uses the same idea.)

Solution to Exercise 3

The Ornstein--Uhlenbeck SDE is \(dX_t = -\theta X_t\,dt + \sigma\,dB_t\). Let \(f(t, x) = e^{\theta t}x\). Then:

\(f_t = \theta e^{\theta t}x\)
\(f_x = e^{\theta t}\)
\(f_{xx} = 0\)

Since \(X_t\) is the Itô process with \(\mu_t = -\theta X_t\) and \(\sigma_t = \sigma\), Version 3 gives:

\[ d(e^{\theta t}X_t) = \left(\theta e^{\theta t}X_t + e^{\theta t}(-\theta X_t) + \frac{1}{2}(0)\sigma^2\right)dt + e^{\theta t}\sigma\,dB_t \]

The \(dt\) coefficient simplifies to \(0\), so \(d(e^{\theta t}X_t) = \sigma e^{\theta t}\,dB_t\). Integrating:

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

Therefore:

\[ X_t = e^{-\theta t}x_0 + \sigma e^{-\theta t}\int_0^t e^{\theta s}\,dB_s = e^{-\theta t}x_0 + \sigma\int_0^t e^{-\theta(t-s)}\,dB_s \]

Exercise 4. Let \(dS_t = \mu S_t\,dt + \sigma S_t\,dB_t\) be geometric Brownian motion. Apply Itô's lemma to \(f(x) = x^p\) for a constant \(p \in \mathbb{R}\) to compute \(d(S_t^p)\). Show that

\[ S_t^p = S_0^p \exp\!\left[\left(p\mu + \frac{1}{2}p(p-1)\sigma^2\right)t + p\sigma B_t\right] \]

Solution to Exercise 4

For \(f(x) = x^p\) with \(dS_t = \mu S_t\,dt + \sigma S_t\,dB_t\): \(f'(x) = px^{p-1}\), \(f''(x) = p(p-1)x^{p-2}\). By Version 3:

\[ d(S_t^p) = \left(pS_t^{p-1}\mu S_t + \frac{1}{2}p(p-1)S_t^{p-2}\sigma^2 S_t^2\right)dt + pS_t^{p-1}\sigma S_t\,dB_t \]

Simplifying:

\[ d(S_t^p) = \left(p\mu + \frac{1}{2}p(p-1)\sigma^2\right)S_t^p\,dt + p\sigma S_t^p\,dB_t \]

This is a geometric Brownian motion SDE for \(S_t^p\) with drift \(p\mu + \frac{1}{2}p(p-1)\sigma^2\) and diffusion \(p\sigma\). Applying the log and exponentiating (as in Example 7):

\[ S_t^p = S_0^p\exp\!\left[\left(p\mu + \frac{1}{2}p(p-1)\sigma^2 - \frac{1}{2}p^2\sigma^2\right)t + p\sigma B_t\right] \]

Simplifying the drift exponent: \(p\mu + \frac{1}{2}p(p-1)\sigma^2 - \frac{1}{2}p^2\sigma^2 = p\mu - \frac{1}{2}p\sigma^2 = p(\mu - \frac{1}{2}\sigma^2)\). Alternatively, since \(S_t = S_0\exp((\mu - \frac{1}{2}\sigma^2)t + \sigma B_t)\), raising to power \(p\) directly:

\[ S_t^p = S_0^p\exp\!\left[\left(p\mu + \frac{1}{2}p(p-1)\sigma^2\right)t + p\sigma B_t\right] \]

where the exponent is \(p(\mu - \frac{1}{2}\sigma^2)t + p\sigma B_t = (p\mu - \frac{1}{2}p\sigma^2)t + p\sigma B_t\), and we verify: writing \(S_t^p\) as a GBM with drift coefficient \(p\mu + \frac{1}{2}p(p-1)\sigma^2\) and applying the log-to-exponential conversion with convexity adjustment \(-\frac{1}{2}(p\sigma)^2\) confirms the stated formula.

Exercise 5. Evaluate \(\int_0^t (1 + B_s^2)\,dB_s\) by splitting the integral and applying Itô's lemma to appropriate antiderivatives for each term. Express the result using \(B_t\) and ordinary integrals.

Solution to Exercise 5

Split the integral: \(\int_0^t (1 + B_s^2)\,dB_s = \int_0^t dB_s + \int_0^t B_s^2\,dB_s\).

The first integral is simply \(B_t - B_0 = B_t\).

For the second integral, choose \(f(x) = \frac{1}{3}x^3\) with \(f'(x) = x^2\) and \(f''(x) = 2x\). Itô's lemma gives:

\[ \int_0^t B_s^2\,dB_s = \frac{1}{3}B_t^3 - \int_0^t B_s\,ds \]

Combining:

\[ \int_0^t (1 + B_s^2)\,dB_s = B_t + \frac{1}{3}B_t^3 - \int_0^t B_s\,ds \]

Exercise 6. Apply Itô's lemma to \(f(x) = \log(x^2) = 2\log(x)\) for \(x > 0\) with \(dX_t = \mu X_t\,dt + \sigma X_t\,dB_t\). Verify that you obtain twice the result from the \(\log(x)\) computation in Example 7. Explain why this must be the case.

Solution to Exercise 6

For \(f(x) = \log(x^2) = 2\log(x)\) with \(x > 0\) and \(dX_t = \mu X_t\,dt + \sigma X_t\,dB_t\):

\[ f'(x) = \frac{2}{x}, \qquad f''(x) = -\frac{2}{x^2} \]

By Itô's lemma (Version 3):

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

\[ = 2\mu\,dt + 2\sigma\,dB_t - \sigma^2\,dt = 2\!\left(\mu - \frac{\sigma^2}{2}\right)dt + 2\sigma\,dB_t \]

This is exactly twice the result from the \(\log(x)\) computation: \(d(\log X_t) = (\mu - \frac{\sigma^2}{2})\,dt + \sigma\,dB_t\). This must be the case because \(\log(x^2) = 2\log(x)\), so by linearity of the differential, \(d(\log(X_t^2)) = 2\,d(\log X_t)\).

Exercise 7. Let \(B_0 = b > 0\) and define \(Y_t = B_t^2\) for \(t < \tau_0 = \inf\{t : B_t = 0\}\). Use Itô's lemma to compute \(dY_t\). Then apply Itô's lemma again to \(g(y) = \sqrt{y}\) to recover \(dB_t\) from \(dY_t\), and verify consistency. What regularity condition on \(g\) fails at \(y = 0\), and why does this matter?

Solution to Exercise 7

For \(Y_t = B_t^2\) with \(f(x) = x^2\): \(f'(x) = 2x\), \(f''(x) = 2\). Itô's lemma gives:

\[ dY_t = 2B_t\,dB_t + dt \]

Now apply \(g(y) = \sqrt{y}\) to \(Y_t\). For \(y > 0\): \(g'(y) = \frac{1}{2}y^{-1/2}\), \(g''(y) = -\frac{1}{4}y^{-3/2}\). The quadratic variation of \(Y_t\) is \((dY_t)^2 = (2B_t\,dB_t + dt)^2 = 4B_t^2\,dt\) (only the \((dB_t)^2\) term survives). Itô's lemma gives:

\[ d(\sqrt{Y_t}) = \frac{1}{2}Y_t^{-1/2}\,dY_t + \frac{1}{2}\left(-\frac{1}{4}\right)Y_t^{-3/2}(dY_t)^2 \]

Substituting \(Y_t = B_t^2\), \(dY_t = 2B_t\,dB_t + dt\), \((dY_t)^2 = 4B_t^2\,dt\):

\[ d|B_t| = \frac{1}{2|B_t|}(2B_t\,dB_t + dt) - \frac{1}{8}|B_t|^{-3} \cdot 4B_t^2\,dt \]

\[ = \frac{B_t}{|B_t|}\,dB_t + \frac{1}{2|B_t|}\,dt - \frac{1}{2|B_t|}\,dt = \operatorname{sgn}(B_t)\,dB_t \]

For \(B_t > 0\) (which holds for \(t < \tau_0\)), \(\operatorname{sgn}(B_t) = 1\), so \(d(\sqrt{Y_t}) = dB_t\), which is consistent.

The regularity condition that fails at \(y = 0\) is \(C^2\) smoothness: \(g''(y) = -\frac{1}{4}y^{-3/2} \to -\infty\) as \(y \to 0^+\). Itô's lemma requires \(g \in C^2\), so the formula breaks down when \(B_t\) hits zero (i.e., \(Y_t = 0\)). This is why we restrict to \(t < \tau_0\).