Construction of the Stochastic Integral

Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.

Begin with a continuous martingale \(X\), with \(X_0=0\) and \(\sup_{t\geq0}\mathbb{E}X_t^2<\infty\). For Brownian motion on a fixed interval \([0,T]\), use the stopped integrator \(X_t=B_{t\wedge T}\). The aim is to define

\[\mathcal I_t=\int_0^t H_s\,\mathrm dX_s.\tag{1}\]

The coefficient multiplying an increment must be chosen from information available before that increment. Predictability expresses this requirement and gives the measurable class on which the integral will be constructed.

The predictable sigma-field \(\mathcal P\) on \([0,\infty)\times\Omega\) is the smallest sigma-field making every bounded left-continuous adapted real process measurable. Equivalently,

\[\mathcal P=\sigma\{\{(t,\omega):Y_t(\omega)\in C\}:Y\text{ bounded, left-continuous and adapted},\ C\in\mathscr B(\mathbb R)\}.\]

A process is predictable when it is \(\mathcal P\)-measurable.

Remark 1. Continuous adapted processes generate the same sigma-field. One inclusion follows because continuity implies left continuity. For the other, extend a bounded left-continuous adapted process \(Y\) to negative times by \(Y_s=Y_0\), and set

\[Y_t^n=n\int_{t-1/n}^tY_s\,\mathrm ds.\]

These processes are continuous and adapted, and converge to \(Y_t\) at every \(t\), including zero. For an unbounded process first use bounded truncations. This proves the other inclusion.

Remark 2. A useful set of generators is

\[\{(s,t]\times A:0\leq s<t, A\in\mathcal F_s\}\ \cup\ \{\{0\}\times A:A\in\mathcal F_0\}.\]

Their indicators are left-continuous and adapted. Conversely, a left-continuous adapted process is the pointwise limit of

\[Y_t^n=Y_0\mathbb1_{\{t=0\}}+ \sum_{k\geq1}Y_{(k-1)/n}\mathbb1_{((k-1)/n,k/n]}(t).\tag{2}\]

Each coefficient is measurable at the left endpoint of its interval, so each approximation is measurable for the sigma-field generated by these rectangles.

The construction of (1) has two parts. First define the integral for elementary predictable processes and prove a second-moment identity. Then use that identity to extend the integral by approximation. We begin with a single time interval.

\[H_s=K\mathbb1_{(a,b]}(s),\qquad K\text{ bounded and }\mathcal F_a\text{-measurable}.\tag{3}\]

Define \(\mathcal I_t=K(X_{t\wedge b}-X_{t\wedge a})\). Its terminal value is \(\mathcal I_\infty=K(X_b-X_a)\). The three time ranges give

\[\mathbb{E}[\mathcal I_\infty\mid\mathcal F_t]= \begin{cases}0,&t\leq a,\\K(X_t-X_a),&a<t<b,\\K(X_b-X_a),&t\geq b. \end{cases}\]

For \(t\leq a\), use the tower property through \(\mathcal F_a\); for the middle range use the martingale identity directly. Thus \(\mathcal I\) is a continuous square-integrable martingale. Writing \(A=\langle X\rangle\), the conditional second-moment identity gives

\[\mathbb{E}\mathcal I_\infty^2=\mathbb{E}[K^2(X_b-X_a)^2] =\mathbb{E}[K^2(A_b-A_a)].\]

It also identifies the bracket. For \(a\leq t\leq b\), expand \(X_b-X_a=(X_b-X_t)+(X_t-X_a)\). The mixed term has conditional mean zero, and

\[\mathbb{E}[(X_b-X_a)^2-(A_b-A_a)\mid\mathcal F_t] =(X_t-X_a)^2-(A_t-A_a).\]

The cases before \(a\) and after \(b\) follow by the same conditioning as above. Therefore

\[\mathcal I_t^2-K^2(A_{t\wedge b}-A_{t\wedge a}) \quad\text{is a martingale},\qquad \langle\mathcal I\rangle_t=\int_0^tH_s^2\,\mathrm dA_s.\]

For a finite sum of elementary integrands, refine all endpoints into one deterministic partition. We may then write

\[H_s=\sum_{j=1}^J K_j\mathbb1_{(t_{j-1},t_j]}(s),\qquad \mathcal I_t=\sum_{j=1}^J K_j(X_{t\wedge t_j}-X_{t\wedge t_{j-1}}),\]

where \(K_j\) is bounded and \(\mathcal F_{t_{j-1}}\)-measurable. Disjoint intervals are essential for the orthogonality calculation: for \(i<j\), all factors in the earlier increment and \(K_j\) are measurable at \(t_{j-1}\), while the conditional expectation of the later increment is zero. Consequently

\[\mathbb{E}\mathcal I_\infty^2 =\sum_{j=1}^J\mathbb{E}[K_j^2(X_{t_j}-X_{t_{j-1}})^2] =\mathbb{E}\int_0^\infty H_s^2\,\mathrm dA_s.\]

The bracket calculation also passes from one interval to the next. On an interval, expanding the square of the accumulated integral adds a term twice its left-endpoint value times the new martingale increment, and the centered square term already handled above. Both have conditional mean zero. It follows that

\[\mathcal I_t^2-\int_0^tH_s^2\,\mathrm dA_s\quad\text{is a martingale}.\]

The second-moment identity now specifies the admissible class. Require \(H\) to be predictable and

\[\|H\|_{L^2(A)}^2:=\mathbb{E}\int_0^\infty H_s^2\,\mathrm dA_s<\infty.\tag{4}\]

The measure \(\mu(E)=\mathbb{E}\int\mathbb1_E(s,\omega)\,\mathrm dA_s\) on the predictable sigma-field is finite, because \(\mathbb{E}A_\infty=\mathbb{E}X_\infty^2<\infty\). Its mass at time zero is zero. The rectangle generators and the monotone-class theorem show that finite elementary sums are dense in \(L^2(\mu)\): first approximate bounded indicators using the generated sigma-field, then bounded simple functions, and finally truncate a general square-integrable function. Choose elementary \(H^n\) with \(\|H^n-H\|_{L^2(A)}\to0\).

Their integrals \(I^n\) have terminal values Cauchy in \(L^2\). Let \(I_\infty\) be the limit and \(I_t=\mathbb{E}[I_\infty\mid\mathcal F_t]\). Doob’s inequality and the isometry give

\[\mathbb{E}\sup_{t\geq0}|I_t^n-I_t^m|^2 \leq4\mathbb{E}|I_\infty^n-I_\infty^m|^2 =4\mathbb{E}\int_0^\infty|H_s^n-H_s^m|^2\,\mathrm dA_s.\]

The same bound with \(I\) in place of \(I^m\) follows by taking the limit. A subsequence converges uniformly almost surely, and its limit supplies a continuous version of \(I\). Completeness of the filtration preserves adaptation on the exceptional null set. The martingale identity passes to this limit in \(L^2\); for example,

\[\left|\mathbb{E}[(I_t^n-I_t)\mathbb1_E]\right|\leq\|I_t^n-I_t\|_2\longrightarrow0, \qquad E\in\mathcal F_s, s\leq t.\]

Two approximating sequences give the same limit by the same estimate. The isometry passes to the limit, and so does the centered-square identity. Indeed,

\[\mathbb{E}\int_0^\infty|(H_s^n)^2-H_s^2|\,\mathrm dA_s \leq\|H^n-H\|_{L^2(A)}\bigl(\|H^n\|_{L^2(A)}+\|H\|_{L^2(A)}\bigr)\to0.\]

Also \((I_t^n)^2\to I_t^2\) in \(L^1\), by Cauchy–Schwarz. Conditional expectation is continuous in \(L^1\), so the limit of the centered-square martingales is a martingale. We have therefore constructed

\[I_t=\int_0^tH_s\,\mathrm dX_s,\qquad \mathbb{E}I_t^2=\mathbb{E}\int_0^tH_s^2\,\mathrm d\langle X\rangle_s, \qquad \langle I\rangle_t=\int_0^tH_s^2\,\mathrm d\langle X\rangle_s.\]

Remark 3. On a finite horizon the integral is a linear map from bounded predictable processes to random variables modulo almost-sure equality. On elementary processes it is specified by

\[\int_0^t\xi_s\,\mathrm dX_s =\sum_{k=1}^nK_k(X_{t\wedge t_k}-X_{t\wedge s_k}), \qquad \xi_s=\sum_{k=1}^nK_k\mathbb1_{(s_k,t_k]}(s).\tag{5}\]

If predictable \(\xi^n\to\xi\) pointwise and \(|\xi^n|\leq K\) for one deterministic constant, dominated convergence in the measure \(\mu\) and the isometry imply \(\int_0^t\xi_s^n\,\mathrm dX_s\to\int_0^t\xi_s\,\mathrm dX_s\) in \(L^2\), hence in probability. For a continuous local martingale this conclusion follows by stopping so that its bracket and initial value are controlled.

Remark 4. Linearity, the elementary formula, and bounded pointwise continuity in probability characterize the integral uniquely. The class of bounded predictable integrands on which two such maps agree is a vector space closed under bounded pointwise limits. It contains the rectangle indicators (and integrals of time-zero indicators are zero). The functional monotone-class theorem extends agreement to every bounded predictable integrand.