The Ordinary Time Integral: An Intuitive Introduction¶
1. Concept Definition¶
This page introduces time integrals of stochastic processes and builds intuition for why these ordinary integrals behave differently from stochastic integrals such as the Itô integral.
The central distinction is the integrator. When integrating with respect to time,
the integrator \(ds\) is deterministic and smooth. When integrating with respect to Brownian motion,
the integrator \(dB_s\) is random and irregular. This single difference changes everything.
Pathwise construction¶
The time integral is constructed in three steps.
- Fix a sample path \(\omega\).
- Treat \(s \mapsto f(s, B_s(\omega))\) as an ordinary deterministic function of time.
- Compute the usual Lebesgue integral.
For each fixed \(\omega\), the function \(s \mapsto f(s, B_s(\omega))\) is an ordinary time-dependent function. If it is integrable on \([0,t]\), then
is defined as a standard Lebesgue integral. Under suitable measurability conditions, the resulting pathwise integral is a random variable.
Discrete approximation¶
The integral can be approximated using Riemann sums:
where \(\Pi = \{0 = t_0 < t_1 < \cdots < t_n = t\}\) is a partition of the interval.
Because the increment \(ds\) is deterministic, the only randomness comes from the function \(f(s,B_s)\). Any evaluation point (left, midpoint, right) gives the same limit—unlike the stochastic case where the choice of evaluation point matters.
2. Intuition and Financial Interpretation¶
Think of \(f(s, B_s)\) as a rate of accumulation per unit time. Then \(f(s,B_s)\,ds\) is the small accumulated amount over a short interval, and
is the total accumulated quantity over \([0,t]\).
The randomness changes the shape of the curve \(f(s,B_s)\), but the mechanism of accumulation—multiplying a rate by a time duration and summing—is the same as in ordinary calculus. This contrasts with stochastic integrals where the increment \(dB_s\) is itself random.
Financial examples of time integrals:
- Cumulative cashflow: \(\int_0^t c_s\, ds\)
- Accumulated short-rate discount: \(\int_0^t r_s\, ds\)
- Time-weighted position: \(\int_0^t H_s\, ds\)
In each case the rate \(c_s\), \(r_s\), or \(H_s\) may be random, but the accumulation mechanism is deterministic.
3. Discrete Approximation from Coin Flips¶
To build intuition, we approximate Brownian motion by a scaled random walk.
Divide \([0,1]\) into \(n\) equal steps of size \(\Delta t = 1/n\). At each step the Brownian increment is approximated by \(\Delta B_k = \pm \sqrt{\Delta t}\) with equal probability. This scaling ensures the correct variance: after \(k\) steps, \(\operatorname{Var}(B_{t_k}) = t_k\).
As the time step shrinks, the scaled random walk converges to Brownian motion (Donsker's theorem). The corresponding discrete sums converge to the continuous-time integral:
Each term is the area of a thin rectangle with random height \(f(t_k,B_{t_k})\) and deterministic width \(\Delta t\).
4. Worked Examples¶
We illustrate using the coin sequence \(H,H,T,H,T,T,H,H,H,T\) with \(n=10\), giving increments \(\pm 1/\sqrt{10}\).
Example 1: Integrating \(B_s\) with respect to time¶
For this path the discrete sum gives approximately \(0.411\).
Mean and variance: Since \(\mathbb{E}[B_s] = 0\) for all \(s\), by linearity and Fubini's theorem:
The variance has a clean closed form. Using \(\mathbb{E}[B_s B_t] = \min(s,t)\):
Computing explicitly:
This double-integral calculation foreshadows the Itô isometry: both results compute a second moment by reducing a stochastic quantity to an ordinary integral.
Example 2: Integrating time¶
When the integrand does not depend on the random path at all, the integral becomes deterministic and is identical for every Brownian path:
The Riemann sum approximation gives \(\sum t_k\,\Delta t \approx 0.45\), converging to the exact value as the grid becomes finer.
Example 3: Integrating \(sB_s\)¶
Now the integrand depends on both time and the Brownian path:
This value depends on the realized path. The mean is zero by the same Fubini argument. The variance follows from \(\mathbb{E}[B_s^2] = s\):
5. Monte Carlo Illustration¶
The following script simulates many Brownian paths and computes several integrals of the form \(\int_0^T f(s, B_s)\, ds\).
```python import matplotlib.pyplot as plt import numpy as np import scipy.stats as stats from dataclasses import dataclass from typing import Optional
@dataclass class BrownianMotionResult: time_steps: np.ndarray time_step_size: float brownian_paths: np.ndarray
class BrownianMotion: DEFAULT_STEPS_PER_YEAR = 252
def __init__(self, maturity_time: float = 1.0, seed: Optional[int] = None):
if maturity_time <= 0:
raise ValueError("maturity_time must be positive")
self.maturity_time = maturity_time
self.rng = np.random.RandomState(seed)
def simulate(self, num_paths: int = 1,
num_steps: Optional[int] = None) -> BrownianMotionResult:
if num_paths <= 0:
raise ValueError("num_paths must be positive")
if num_steps is None:
num_steps = int(self.maturity_time * self.DEFAULT_STEPS_PER_YEAR)
if num_steps <= 0:
raise ValueError("num_steps must be positive")
time_steps = np.linspace(0, self.maturity_time, num_steps + 1)
dt = time_steps[1] - time_steps[0]
increments = self.rng.standard_normal((num_paths, num_steps)) * np.sqrt(dt)
brownian_paths = np.concatenate(
[np.zeros((num_paths, 1)), increments.cumsum(axis=1)], axis=1
)
return BrownianMotionResult(time_steps=time_steps,
time_step_size=dt,
brownian_paths=brownian_paths)
if name == "main": num_paths = 10000 T = 1
bm = BrownianMotion(maturity_time=T, seed=0)
result = bm.simulate(num_paths)
t = result.time_steps
dt = result.time_step_size
b = result.brownian_paths
# Approximate ∫ f(s,B_s) ds using a left-point Riemann sum.
# b[:, :-1] excludes the final time point so each row aligns with dt.
integrands = (
lambda t, b: b[:, :-1], # B_t
lambda t, b: b[:, :-1] ** 2, # B_t^2
lambda t, b: t[:-1] * b[:, :-1], # t B_t
lambda t, b: t[:-1] ** 2 * b[:, :-1], # t^2 B_t
lambda t, b: t[:-1] * b[:, :-1] ** 2, # t B_t^2
lambda t, b: t[:-1] * np.exp(b[:, :-1]) # t e^{B_t}
)
labels = ("B_t", "B_t^2", "t B_t", "t^2 B_t", "t B_t^2", "t e^{B_t}")
fig, axes = plt.subplots(len(integrands), 2, figsize=(12, 3 * len(integrands)))
for i, (integrand, label) in enumerate(zip(integrands, labels)):
ax0, ax1 = axes[i, 0], axes[i, 1]
integral = np.cumsum(integrand(t, b) * dt, axis=1)
integral = np.concatenate((np.zeros((num_paths, 1)), integral), axis=1)
ax0.set_title(f"Time integral with f = {label}")
ax0.plot(t, b[0, :], "--b", label="Brownian Motion")
ax0.plot(t, integral[0, :], "r", label="Integral")
ax0.legend()
ax0.grid(True)
ax1.set_title(r"Distribution of $\int_0^T f(s,B_s)ds$")
ax1.hist(integral[:, -1], bins=70, density=True)
mu = integral[:, -1].mean()
sigma = integral[:, -1].std()
x = np.linspace(-3, 3, 101) * sigma + mu
ax1.plot(x, stats.norm(loc=mu, scale=sigma).pdf(x),
"--r", lw=3, label="Normal curve (matching mean and variance)")
ax1.legend()
ax1.grid(True)
plt.tight_layout()
plt.savefig("./image/lebesgue_integration_intuitive_figure.png", dpi=150)
plt.show()
```
The dashed normal curve matches the simulated mean and variance and serves only as a visual reference. The distribution of these integrals need not be exactly normal in general.
6. Key Observations¶
Four patterns appear consistently in the simulation.
-
Each Brownian path produces a different integral value. The integral \(\int_0^T f(s, B_s)\,ds\) depends on the realized path.
-
Randomness enters through the integrand only. The increment \(ds\) is deterministic; it does not contribute randomness.
-
Deterministic integrands give deterministic integrals. For example, \(\int_0^1 s\,ds = 0.5\) is the same for every path.
-
Random integrands give random integrals. When \(f(s,B_s)\) depends on \(B_s\), the integral is a random variable.
7. Comparison with Itô Integrals¶
Time integrals and Itô integrals differ fundamentally.
| Feature | Time integral \(\int f(s,B_s)\,ds\) | Itô integral \(\int f(s,B_s)\,dB_s\) |
|---|---|---|
| Integrator | deterministic time \(ds\) | Brownian motion \(dB_s\) |
| Definition | pathwise Lebesgue integral | \(L^2\)-limit of left-endpoint sums |
| Randomness | enters through integrand | enters through integrator |
| Mean | depends on integrand | zero (martingale property) |
| Quadratic variation | zero | \(\int_0^t f^2\,ds > 0\) |
| Evaluation point | any point gives same limit | left endpoint essential |
| Interpretation | accumulated quantity over time | accumulation against random fluctuations |
8. Summary¶
The integral \(\int_0^t f(s,B_s)\,ds\) is an ordinary time integral, constructed pathwise as standard Lebesgue integration.
- The increment \(ds\) is deterministic; randomness enters only through \(f(s,B_s)\).
- Standard Riemann sums converge and any evaluation point gives the same limit.
- The integral is a random variable when \(f\) depends on \(B_s\), with mean and variance computable by ordinary calculus and Fubini's theorem.
9. Why Ordinary Calculus Breaks for Brownian Motion¶
Although Brownian motion is continuous, its total variation is infinite on every interval while its quadratic variation is finite. This unusual combination is the fundamental reason stochastic calculus differs from ordinary calculus.
The time integral works because \(ds\) is smooth. Now suppose we try to define
Because Brownian paths have infinite variation, summing absolute increments \(\sum |B_{t_{k+1}} - B_{t_k}|\) diverges as the partition becomes finer. Moreover, unlike the time integral, the limit of discrete sums depends on where the integrand is sampled—left endpoint, midpoint, and right endpoint give different limits. Different conventions lead to different stochastic integrals: most notably the Itô integral (left endpoint) and the Stratonovich integral (midpoint).
To make sense of \(\int_0^t f(s, B_s)\, dB_s\), we must introduce a new definition of integration specifically designed for stochastic processes—the Itô integral studied in the next section.
Exercises¶
Exercise 1. Let \(B_t\) be a standard Brownian motion. Compute the mean and variance of the time integral
Hint: Use Fubini's theorem and \(\mathbb{E}[B_s^2] = s\), \(\mathbb{E}[B_s^4] = 3s^2\).
Solution to Exercise 1
Mean. By Fubini's theorem and \(\mathbb{E}[B_s^2] = s\):
Variance. We need \(\mathbb{E}\!\left[\left(\int_0^T B_s^2\, ds\right)^2\right] - \left(\frac{T^2}{2}\right)^2\). Expand the square of the integral using Fubini:
For jointly Gaussian variables with zero mean, \(\mathbb{E}[B_s^2 B_u^2] = \mathbb{E}[B_s^2]\mathbb{E}[B_u^2] + 2(\mathbb{E}[B_s B_u])^2 = su + 2(\min(s,u))^2\). This follows from the Isserlis (Wick) theorem: \(\mathbb{E}[X^2 Y^2] = (\mathbb{E}[X^2])(\mathbb{E}[Y^2]) + 2(\mathbb{E}[XY])^2\) for zero-mean jointly Gaussian \(X, Y\). Therefore:
The first integral is \(\left(\int_0^T s\, ds\right)^2 = T^4/4\). For the second, by symmetry:
So the second moment is \(T^4/4 + T^4/3\). The variance is:
Exercise 2. Using the coin-flip approximation with \(n = 10\) and the sequence \(T, H, H, T, T, H, T, H, H, H\), compute the discrete Riemann sum approximation to
Compare your answer with the theoretical mean of this integral.
Solution to Exercise 2
With \(n = 10\), we have \(\Delta t = 1/10\) and \(\Delta B_k = \pm 1/\sqrt{10}\). The coin sequence \(T, H, H, T, T, H, T, H, H, H\) gives increments \(-,+,+,-,-,+,-,+,+,+\) (in units of \(1/\sqrt{10}\)).
The cumulative Brownian path values at \(t_k = k/10\) are:
| \(k\) | \(\Delta B_k\) | \(B_{t_k}\) |
|---|---|---|
| 0 | \(0\) | |
| 1 | \(-1/\sqrt{10}\) | \(-1/\sqrt{10}\) |
| 2 | \(+1/\sqrt{10}\) | \(0\) |
| 3 | \(+1/\sqrt{10}\) | \(1/\sqrt{10}\) |
| 4 | \(-1/\sqrt{10}\) | \(0\) |
| 5 | \(-1/\sqrt{10}\) | \(-1/\sqrt{10}\) |
| 6 | \(+1/\sqrt{10}\) | \(0\) |
| 7 | \(-1/\sqrt{10}\) | \(-1/\sqrt{10}\) |
| 8 | \(+1/\sqrt{10}\) | \(0\) |
| 9 | \(+1/\sqrt{10}\) | \(1/\sqrt{10}\) |
| 10 | \(+1/\sqrt{10}\) | \(2/\sqrt{10}\) |
The Riemann sum is:
The theoretical mean is \(\mathbb{E}[\int_0^1 B_s\, ds] = 0\). Our single-path approximation of \(-0.0316\) is close to zero, consistent with the theoretical mean.
Exercise 3. Explain why the Riemann sum approximation to \(\int_0^t f(s, B_s)\, ds\) converges to the same limit regardless of whether the integrand is evaluated at the left endpoint, midpoint, or right endpoint of each subinterval. Why does this property fail for stochastic integrals of the form \(\int_0^t f(s, B_s)\, dB_s\)?
Solution to Exercise 3
For the time integral \(\int_0^t f(s, B_s)\, ds\), the integrator is \(ds\), which is deterministic and smooth (it has bounded variation). For Riemann sums:
where \(t_k^*\) is any evaluation point in \([t_k, t_{k+1}]\). Since \(f(s, B_s(\omega))\) is a continuous function of \(s\) for each fixed \(\omega\), and the integrator \(ds\) has bounded variation, the Riemann-Stieltjes theory guarantees convergence regardless of the evaluation point choice. The difference between evaluations at different points within a subinterval is bounded by the oscillation of \(f\) times \(\Delta t\), which vanishes as the mesh goes to zero.
For the stochastic integral \(\int_0^t f(s, B_s)\, dB_s\), the integrator is \(B_s\), which has infinite variation. The critical difference is that Brownian increments satisfy \((\Delta B_k)^2 \approx \Delta t\), so second-order terms do not vanish. Changing the evaluation point from left to midpoint introduces an additional contribution proportional to \(\frac{1}{2}\sum_k f'(\cdot)(\Delta B_k)^2 \approx \frac{1}{2}\int f'(\cdot)\, ds\), which is the Stratonovich correction. The infinite variation of \(B_s\) means that the integrand's value at slightly different points within the same subinterval leads to a macroscopic difference in the limit.
Exercise 4. Compute the variance of
by evaluating the double integral \(\int_0^T \int_0^T s^2 t^2 \min(s,t)\, ds\, dt\).
Solution to Exercise 4
We need to compute:
By symmetry, this equals:
Exercise 5. In the short-rate model, the discount factor is \(D(0,T) = \exp\!\left(-\int_0^T r_s\, ds\right)\), where \(r_t = r_0 + \sigma B_t\) for constants \(r_0 > 0\) and \(\sigma > 0\). Show that \(\int_0^T r_s\, ds\) is Gaussian and find its mean and variance.
Solution to Exercise 5
Since \(r_t = r_0 + \sigma B_t\):
The random part is \(\sigma \int_0^T B_s\, ds\). Since \(\int_0^T B_s\, ds\) is the integral of a Gaussian process against a deterministic measure, it is Gaussian. (More precisely, it is the \(L^2\) limit of Riemann sums, each of which is a finite linear combination of Gaussian random variables, and hence Gaussian. The limit of Gaussian random variables in \(L^2\) is Gaussian.)
Mean:
Variance: Using the computation from Example 1 in the text (with \(T\) replacing 1):
Therefore:
In summary, \(\int_0^T r_s\, ds \sim \mathcal{N}\!\left(r_0 T,\; \frac{\sigma^2 T^3}{3}\right)\).
Exercise 6. Let \(f(s) = e^{-\alpha s}\) for a constant \(\alpha > 0\). Show that the integral \(\int_0^T f(s) \, ds\) is deterministic and compute its value. Then consider the integral \(\int_0^T f(s) B_s \, ds\) and compute its variance.
Solution to Exercise 6
Deterministic part. Since \(f(s) = e^{-\alpha s}\) does not depend on \(B_s\):
This is the same for every Brownian path.
Variance of \(\int_0^T e^{-\alpha s} B_s\, ds\). Using Fubini and \(\mathbb{E}[B_s B_u] = \min(s,u)\):
By symmetry:
The inner integral \(\int_0^u s e^{-\alpha s}\, ds\) is computed by integration by parts:
So the variance is:
Evaluating each integral:
- \(\int_0^T e^{-\alpha u}\, du = \frac{1}{\alpha}(1 - e^{-\alpha T})\)
- \(\int_0^T e^{-2\alpha u}\, du = \frac{1}{2\alpha}(1 - e^{-2\alpha T})\)
- \(\int_0^T u e^{-2\alpha u}\, du = \frac{1}{4\alpha^2}(1 - e^{-2\alpha T}(1 + 2\alpha T))\)
Substituting:
Exercise 7. Prove that \(\int_0^t B_s \, ds\) has the same distribution as \(\int_0^t (t - s) \, dB_s\) by computing the mean and variance of both sides. Why does the representation \(\int_0^t B_s\, ds = tB_t - \int_0^t s\, dB_s\) (integration by parts) make this plausible?
Solution to Exercise 7
Mean of both sides. The mean of \(\int_0^t B_s\, ds\) is \(\int_0^t \mathbb{E}[B_s]\, ds = 0\). The mean of \(\int_0^t (t-s)\, dB_s\) is \(0\) by the martingale property of the Ito integral.
Variance of \(\int_0^t B_s\, ds\). From the text:
Variance of \(\int_0^t (t-s)\, dB_s\). Since \(H_s = t - s\) is deterministic, the Ito isometry gives:
Both integrals have mean zero and variance \(t^3/3\). Moreover, \(\int_0^t (t-s)\, dB_s\) is Gaussian (deterministic integrand), and \(\int_0^t B_s\, ds\) is also Gaussian (as a continuous linear functional of a Gaussian process). Since a Gaussian distribution is determined by its mean and variance, the two have the same distribution.
Why integration by parts makes this plausible. From the identity \(\int_0^t B_s\, ds = tB_t - \int_0^t s\, dB_s\), we can verify:
using \(B_t = \int_0^t dB_s\) and linearity of the Ito integral. This gives the exact identity \(\int_0^t B_s\, ds = \int_0^t (t-s)\, dB_s\), which is stronger than merely having the same distribution — the two random variables are equal almost surely.