The Itô integral

How do you integrate against a Brownian path, which has no slope?

Add up integrand times increment over a fine partition, with the integrand read at the start of each step, before the increment is drawn; for a continuous integrand such as \(W\), the Itô integral is the limit of these sums. When the integrand never looks ahead and has \(\mathbb E\int_0^T H_t^2\,dt \lt \infty\), as in the setup, the integral has mean 0 and is a martingale in its upper limit.

Before you start · 3 quick checks on quadratic variation, martingales and Riemann–Stieltjes sums, about 3 min

Setup: Brownian motion, partitions and the integral

Brownian motion. \(W_t\), \(0 \le t \le T\), is a standard Brownian motion: \(W_0 = 0\), its paths are continuous, and for \(s \lt t\) the increment \(W_t - W_s\) is normal with mean 0 and variance \(t - s\), independent of the path up to time \(s\) (step 9).
Information. \(\mathcal F_t\) is the information in the path up to time \(t\), the filtration. A process \(H_t\) is adapted, or non-anticipating, if each \(H_t\) is determined by \(\mathcal F_t\): it uses no later values of \(W\).
Partition. \(0 = t_0 \lt t_1 \lt \dots \lt t_n = T\) with \(t_k = k\,\Delta t\) and \(\Delta t = T/n\). The increments are \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\), and \(Q_n = \sum_{k=1}^n (\Delta W_k)^2\) is the quadratic variation sum (step 10).

Simple integrands. Suppose \(H_t = \xi_k\) for \(t_{k-1} \lt t \le t_k\), where each \(\xi_k\) is determined by \(\mathcal F_{t_{k-1}}\) and \(\mathbb E[\xi_k^2] \lt \infty\). Then

\[ \int_0^T H_t\,dW_t = \sum_{k=1}^n \xi_k\,\Delta W_k. \]

Each \(\xi_k\) is fixed at the start of its step, before \(\Delta W_k\) is drawn, like a position chosen before a price change.

General integrands, stated without proof. The general integral asks a little more than adapted: \(H\) must be progressively measurable (see the fine print; every adapted process with continuous paths is) and square-integrable, \(\mathbb E\int_0^T H_t^2\,dt \lt \infty\). Then there are simple adapted integrands \(H^{(m)}\) with \(\mathbb E\int_0^T (H_t - H^{(m)}_t)^2\,dt \to 0\), their integrals converge in \(L^2\), that is in mean square, and the limit is \(\int_0^T H_t\,dW_t\). If \(H\) also has continuous paths, the sums \(\sum_k H_{t_{k-1}}\Delta W_k\) converge to \(\int_0^T H_t\,dW_t\) in probability as \(n\) grows.

Fine print: measurability, uniqueness and modes of convergence

Progressively measurable means that for every \(t\), \((s, \omega) \mapsto H_s(\omega)\) on \([0, t]\) is measurable for the product of the Borel sets of \([0, t]\) and \(\mathcal F_t\). It makes \(\int_0^T H_t^2\,dt\) a random variable. Every adapted process with continuous paths, such as \(W_t\) or \(t\), is progressively measurable. The limit \(\int_0^T H_t\,dW_t\) does not depend on which simple integrands \(H^{(m)}\) are used, up to a set of paths of probability 0.

\(X_n \to X\) in \(L^2\) means \(\mathbb E[(X_n - X)^2] \to 0\); it implies convergence in probability. The page simulates paths with partitions that double. Along such partitions the sums on this page also converge on almost every single path, which is why one simulated path already shows the limit.

Paths come from a seeded random number generator, so every visit shows the same paths. Each path is sampled at 8193 equally spaced times, which gives every partition with up to 4096 steps its end points and midpoints exactly.

1 See

Three Riemann sums on one path

Fix \(T = 1\) and split \([0, T]\) into \(n\) equal steps. On each step, multiply the increment \(\Delta W_k\) by a value of \(W\) from that step, and add up:

\[ L_n = \sum_{k=1}^n W_{t_{k-1}}\Delta W_k, \qquad M_n = \sum_{k=1}^n W_{m_k}\Delta W_k, \qquad R_n = \sum_{k=1}^n W_{t_k}\Delta W_k, \]

where \(m_k = \tfrac12(t_{k-1} + t_k)\) is the midpoint in time of step \(k\). These are the left, midpoint and right Riemann–Stieltjes sums for \(\int_0^T W_t\,dW_t\). Make your prediction first; the explorer opens after it.

The sums on this path, rounded to 4 decimals, and the values they settle at.
Sum	Sum at n	Limit as n grows	Limit on this path
Left, \(L_n\)		\(\tfrac12 W_T^2 - \tfrac12 T\)
Midpoint, \(M_n\)		\(\tfrac12 W_T^2\)
Right, \(R_n\)		\(\tfrac12 W_T^2 + \tfrac12 T\)
Average of endpoints, \(A_n = \sum \tfrac12(W_{t_{k-1}} + W_{t_k})\Delta W_k\)		\(\tfrac12 W_T^2\), equal at every \(n\)
Right minus left, \(Q_n = \sum (\Delta W_k)^2\)		\(T\)

The average of the endpoint values telescopes, because on each step

\[ \tfrac12(W_{t_{k-1}} + W_{t_k})\Delta W_k = \tfrac12(W_{t_k}^2 - W_{t_{k-1}}^2), \]

so \(A_n\) is \(\tfrac12 W_T^2\) at every \(n\), up to rounding in the last digit. The time midpoint gives a different sum, \(M_n\), which only tends to \(\tfrac12 W_T^2\).

Averages over many paths


Sum	Average	Standard error	Exact mean at every n
Left, \(L_n\)
Midpoint, \(M_n\)
Right, \(R_n\)
Average of endpoints, \(A_n\)

Only the left sum has mean 0, and it has mean 0 at every \(n\), not only in the limit.

What the left endpoint gives

Mean 0. \(W_{t_{k-1}}\) is determined by the path up to \(t_{k-1}\), and \(\Delta W_k\) is independent of that past with mean 0, so \(\mathbb E[W_{t_{k-1}}\Delta W_k] = \mathbb E[W_{t_{k-1}}]\,\mathbb E[\Delta W_k] = 0\). The right value contains the increment itself: \(\mathbb E[W_{t_k}\Delta W_k] = \mathbb E[W_{t_{k-1}}\Delta W_k] + \mathbb E[(\Delta W_k)^2] = \Delta t\), so \(\mathbb E[R_n] = n\,\Delta t = T\). The midpoint value contains the first half of the increment, and \(\mathbb E[W_{m_k}\Delta W_k] = \tfrac12\Delta t\).

Martingale. Stop the left sum at a partition time: \(L(t_j) = \sum_{k \le j} W_{t_{k-1}}\Delta W_k\). Each later term has \(\mathbb E[W_{t_{k-1}}\Delta W_k \mid \mathcal F_{t_{k-1}}] = W_{t_{k-1}}\,\mathbb E[\Delta W_k] = 0\), so \(\mathbb E[L(t_j) \mid \mathcal F_{t_i}] = L(t_i)\) for \(i \le j\). In the limit, \(\int_0^t W_s\,dW_s = \tfrac12 W_t^2 - \tfrac12 t\) is a martingale in \(t\). The right-sum limit \(\tfrac12 W_t^2 + \tfrac12 t\) has mean \(t\), and the midpoint limit \(\tfrac12 W_t^2\) has mean \(\tfrac12 t\). Both means grow with \(t\), so neither is a martingale.

The conditions. The Itô integral of a general integrand \(H\) is defined, as in the setup, through simple integrands that are constant on each step, and this needs two things. \(H\) must be adapted, or non-anticipating, so that the value on each step is known before \(\Delta W_k\) is drawn (for general \(H\), progressively measurable, as in the setup). And \(\mathbb E\int_0^T H_t^2\,dt \lt \infty\), so that the integrals have finite mean squares and a limit in \(L^2\). Left sums \(\sum_k H_{t_{k-1}}\Delta W_k\) built from the values of \(H\) itself converge to that integral when \(H\) also has continuous paths; without continuity they can fail, since \(H\) is then not determined by its values at the partition times. Under the two conditions \(\int_0^t H_s\,dW_s\) has mean 0 and is a martingale in \(t\) for \(0 \le t \le T\). For \(H = W\) both hold: \(W_t\) is known at time \(t\), and \(\mathbb E\int_0^T W_t^2\,dt\) is finite because \(\mathbb E[W_t^2] = t\).

Key point On a Brownian path the point where the integrand is read changes the limit. Right minus left is \(\sum(\Delta W_k)^2\), which tends to \(T\), not 0. The left endpoint is the only choice that never looks into the step it multiplies, and the only one with mean 0.

2 Try

Spread, fair games and other evaluation points

The questions below use the paths of stage 1 with \(T = 1\).

Move the evaluation point

Read \(W\) at \(t_{k-1} + \lambda\,\Delta t\) on each step, with \(\lambda\) from 0, the left endpoint, to 1, the right endpoint, in steps of \(\tfrac18\). The page computes this sum on path 1 with \(n = 512\), and its average over 2000 fresh paths with \(n = 64\).

Why the limit moves by λT

Split step \(k\) at the evaluation point: \(a_k = W_{t_{k-1} + \lambda\Delta t} - W_{t_{k-1}}\) and \(b_k = W_{t_k} - W_{t_{k-1} + \lambda\Delta t}\), so \(\Delta W_k = a_k + b_k\). The sum minus the left sum is \(\sum a_k(a_k + b_k) = \sum a_k^2 + \sum a_k b_k\).

\(\sum a_k^2\) has mean \(n\lambda\Delta t = \lambda T\) and variance \(2n(\lambda\Delta t)^2 = 2\lambda^2T^2/n\). \(\sum a_k b_k\) has mean 0 and variance \(n\lambda(1-\lambda)(\Delta t)^2 = \lambda(1-\lambda)T^2/n\). Both variances tend to 0, so the sum tends in \(L^2\) to \(\tfrac12 W_T^2 - \tfrac12 T + \lambda T = \tfrac12 W_T^2 + (\lambda - \tfrac12)T\), and its mean is \(\lambda T\) at every \(n\).

3 Prove

An exact identity, then a limit

Theorem (left sums for the integral of W dW)

Assumptions.

\(W\) is a standard Brownian motion with \(W_0 = 0\). \(T \gt 0\), \(n \ge 1\), \(t_k = kT/n\) and \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\).
\(L_n = \sum_{k=1}^n W_{t_{k-1}}\Delta W_k\) and \(Q_n = \sum_{k=1}^n (\Delta W_k)^2\).
Quadratic variation fact (step 10): \(\mathbb E[Q_n] = T\) and \(\mathbb E\big[(Q_n - T)^2\big] = 2T^2/n\).

Conclusion. For every \(n \ge 1\) and every path, \(L_n = \tfrac12 W_T^2 - \tfrac12 Q_n\). Hence \(\mathbb E\big[(L_n - (\tfrac12 W_T^2 - \tfrac12 T))^2\big] = T^2/(2n)\), so \(L_n \to \tfrac12 W_T^2 - \tfrac12 T\) in \(L^2\), and

\[ \int_0^T W_t\,dW_t = \tfrac12 W_T^2 - \tfrac12 T. \]

Scratch work, not part of the proof.

The same algebra with \(W_{t_k}\) in place of \(W_{t_{k-1}}\) gives \(R_n = \tfrac12 W_T^2 + \tfrac12 Q_n\), which tends to \(\tfrac12 W_T^2 + \tfrac12 T\). The Stratonovich integral \(\int_0^T W_t \circ dW_t\) is built from the average of the endpoint values, so it equals \(\tfrac12 W_T^2\) and follows the ordinary chain rule, but its approximating sums use the value of \(W\) at the end of each step, and it is not a martingale.

Key point The left sum equals \(\tfrac12 W_T^2 - \tfrac12 Q_n\) exactly, at every \(n\). Everything the ordinary chain rule misses sits in \(Q_n\), and quadratic variation turns it into the constant \(\tfrac12 T\) in the limit. The next step, Itô's lemma, makes the same correction for smooth functions of \(W\) other than \(W^2\).

4 Apply

Problems

\(W\) is a standard Brownian motion. Sampled increments in the problems are given numbers, not draws from the page.