From Elementary Integrands to General Integrands
The compensated Poisson isometry gives a way to define integrals beyond finite sums of predictable step processes. The key requirement is not that the integrand be bounded or that it have only finitely many possible values. It is that its expected squared size, measured against the compensator, be finite. Approximation in this squared-integral norm then gives a well-defined stochastic integral and preserves the isometry.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual conditions. Let \(E\) be a measurable mark space with a sigma-finite measure \(\nu\), and let \(N(dt,dz)\) be a Poisson random measure on \([0,\infty)\times E\) with compensator \(dt\,\nu(dz)\), relative to this filtration. Thus \(N\) counts the Poisson points, and its compensated measure is \(\widetilde N(dt,dz)=N(dt,dz)-dt\,\nu(dz)\). Integrals against \(\widetilde N\) below are Ito integrals.
Predictability means that the value of the integrand at time \(s\) is determined by information available immediately before \(s\). In particular, a process such as \(H_s=\mathbf{1}_{\{N_{s-}=0\}}\) is predictable: it uses the count just before time \(s\), not the count after a possible jump at \(s\). Integrands are identified when they agree outside a set of zero measure for \(\mathbb{P}(d\omega)\,ds\,\nu(dz)\).
For elementary predictable integrands the compensated Poisson isometry states that the integral has mean zero and that its second moment equals the expected integral of the squared integrand. The theorem below extends this construction. The finite-time condition in the definition is essential: it ensures that the approximating integrals converge in \(L^2(\mathbb{P})\), meaning that the expected squared error tends to zero.
Extension Theorem and Martingale Properties
Proof. By density of elementary predictable integrands in the square-integrable predictable class, choose elementary \(H^{(n)}\) with \(\mathbb{E}\int_0^T\int_E|H^{(n)}-H|^2\,\nu(dz)\,ds\to0\). Write \(M^{(n)}=\int H^{(n)}\,d\widetilde N\). The elementary isometry gives, for every \(t\le T\), $$ \mathbb{E}\big[(M_t^{(n)}-M_t^{(m)})^2\big] =\mathbb{E}\int_0^t\int_E|H_s^{(n)}(z)-H_s^{(m)}(z)|^2\,\nu(dz)\,ds. $$ The right side tends to zero as \(m,n\to\infty\). Thus the integrals converge in \(L^2\) at each \(t\), and the isometry shows that their limit does not depend on the chosen approximating sequence. Each \(M_t^{(n)}\) has mean zero, so \(L^2\), hence \(L^1\), convergence gives \(\mathbb{E}[M_t]=0\). Applying the same isometry to \(H^{(n)}\) and passing to the limit gives the asserted second-moment identity.
For \(A\in\mathcal{F}_s\), the integrand \(\mathbf{1}_A\mathbf{1}_{(s,t]}H\) is predictable. Applying the isometry to this integrand, first by elementary approximation and then by \(L^2\) convergence, gives $$ \mathbb{E}\big[\mathbf{1}_A(M_t-M_s)^2\big] =\mathbb{E}\left[\mathbf{1}_A\int_s^t\int_E H_u(z)^2\,\nu(dz)\,du\right]. $$ Since this holds for every \(A\in\mathcal{F}_s\), it is precisely the stated conditional second-moment identity. The same approximation shows that \(M^{(n)}\) converges in \(L^2\) at each time and that \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\), by passing to the limit in the elementary martingale identity. Consequently \(M\) is an integrable martingale; its finite second moments make it square-integrable.
Set \(A_t=\int_0^t\int_E H_s(z)^2\,\nu(dz)\,ds\). The martingale increment has conditional mean zero, so expanding \(M_t^2=(M_s+(M_t-M_s))^2\) and conditioning on \(\mathcal{F}_s\) gives $$ \mathbb{E}[M_t^2-M_s^2\mid\mathcal{F}_s] =\mathbb{E}[(M_t-M_s)^2\mid\mathcal{F}_s] =\mathbb{E}[A_t-A_s\mid\mathcal{F}_s]. $$ Therefore \(M^2-A\) is a martingale, which identifies \(A\) as the predictable bracket. For the final identity, use the standard semimartingale integral rule \([\,\int H\,dX\,]=\int H^2\,d[X]\) and the jump quadratic variation of a compensated Poisson random measure, \([\,\widetilde N\,]=N\). This standard jump-semimartingale rule gives \([M]_t=\int_0^t\int_E H_s(z)^2N(ds,dz)\); its general proof belongs to the theory of quadratic variation for jump processes. The integral is finite almost surely on \([0,T]\), since its expectation is \(\mathbb{E}\int_0^T\int_E H_s(z)^2\,\nu(dz)\,ds<\infty\). \(\square\)
The theorem separates two related quantities. The predictable bracket \(A_t\) averages the squared integrand against the compensator, while \([M]_t\) sums its squared values at the actual Poisson points. They are generally not equal path by path. Their expectations agree: the compensator property gives \(\mathbb{E}[M]_t=\mathbb{E}A_t=\mathbb{E}[M_t^2]\).
Worked Examples
Worked Example: A Constant Integrand
Let \(N_t=N((0,t]\times E)\) be a Poisson counting process of rate \(\lambda=3\), and take \(H_s(z)=1\) on \(0\le s\le2\). Compute the integral at time \(2\), its mean and variance, and its quadratic variation.
The compensator mass on \([0,2]\) is \(\lambda\cdot2=3\cdot2=6\). Hence $$ M_t=\int_0^t\int_E1\,\widetilde N(ds,dz)=N_t-3t, \qquad M_2=N_2-6. $$ The squared-integrability check is \(\mathbb{E}\int_0^2\int_E1\,\nu(dz)\,ds=3\cdot2=6<\infty\). Thus the theorem gives \(\mathbb{E}[M_2]=0\) and \(\mathbb{E}[M_2^2]=6\), so \(\operatorname{Var}(M_2)=6\). Its predictable bracket is \(A_2=6\), while its quadratic variation is \([M]_2=N_2\).
Check: \(N_2\) has a Poisson distribution with parameter \(6\), so its mean and variance are both \(6\). Therefore \(N_2-6\) has mean \(6-6=0\) and variance \(6\), agreeing with the isometry.
Worked Example: A Predictable Integrand Stopped by the First Arrival
Let \(N_t\) have rate \(1\), let \(\tau\) be its first arrival time, and set \(H_s=\mathbf{1}_{\{N_{s-}=0\}}\) on \([0,1]\). Find \(M_1\) and verify its mean and second moment.
The integrand is predictable and equals \(1\) strictly before the first arrival and \(0\) afterward. Its integrability check is $$ \mathbb{E}\int_0^1 H_s^2\,ds =\mathbb{E}[\min(1,\tau)] =\int_0^1\mathbb{P}(\tau>s)\,ds =\int_0^1e^{-s}\,ds =1-e^{-1}. $$ Here \(\mathbb{P}(\tau>s)=e^{-s}\) for a rate-one Poisson process. The jump-sum and compensator parts of the integral are respectively \(\mathbf{1}_{\{\tau\le1\}}\) and \(\min(1,\tau)\). Thus \(M_1=\mathbf{1}_{\{\tau\le1\}}-\min(1,\tau)\).
For the mean, \(\mathbb{E}[\mathbf{1}_{\{\tau\le1\}}]=1-e^{-1}\), and the computation above gives \(\mathbb{E}[\min(1,\tau)]=1-e^{-1}\). Their difference is zero. For the second moment, split according to whether the first arrival occurs by time \(1\): $$ \mathbb{E}[M_1^2] =\int_0^1(1-s)^2e^{-s}\,ds+e^{-1} =1-e^{-1}. $$ In the integral, \(\tau\) has density \(e^{-s}\) on \((0,1)\); on \(\{\tau>1\}\), \(M_1=-1\). The displayed integral evaluates to \(1-2e^{-1}\), so adding \(e^{-1}\) gives \(1-e^{-1}\), exactly the isometry value.
Worked Example: A Time-Dependent Deterministic Integrand
Let \(N_t\) have rate \(2\), take \(H_s=s\) on \([0,1]\), and compute the integral, its second moment and both bracket quantities.
Since \(\int_0^1s\,ds=\tfrac12\), the compensator contribution is \(2\int_0^1s\,ds=1\), and $$ M_1=\int_0^1s\,N(ds)-1 =\sum_{\tau_i\le1}\tau_i-1, $$ where \(\tau_i\) are the arrival times. The squared-integrability condition and the isometry give $$ \mathbb{E}\int_0^1s^2\,2\,ds=2\cdot\frac13=\frac23, \qquad \mathbb{E}[M_1]=0,\qquad \mathbb{E}[M_1^2]=\frac23. $$ The predictable bracket at time \(1\) is \(\int_0^1s^2\,2\,ds=\tfrac23\). The pathwise quadratic variation is \([M]_1=\sum_{\tau_i\le1}\tau_i^2\), whose expectation is \(2\int_0^1s^2\,ds=\tfrac23\).
Check: the compensator identity for deterministic integrands gives \(\mathbb{E}\sum_{\tau_i\le1}\tau_i=2\int_0^1s\,ds=1\). Subtracting the compensator \(1\) verifies the zero mean independently. The second moment and expected quadratic variation both equal \(\tfrac23\), as required.
What the Results Do—and Do Not—Say
The martingale in the theorem is a true square-integrable martingale, not merely a local martingale: for every \(t\le T\), its second moment is finite, and its conditional expectation satisfies the martingale identity. The mean-zero statement is unconditional, while \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) is the stronger statement that records how the integral relates to the filtration.
The bracket \(A_t\) is continuous in time because it is an ordinary time integral. In contrast, \([M]_t\) records the realized squared jumps and is generally random and discontinuous. For the constant-integrand example, \(A_2=6\) on every outcome, whereas \([M]_2=N_2\) varies with the observed number of arrivals. The equality is in expectation, not path by path.
Common mistake. Replacing the compensated integral by its jump sum alone loses the compensator and changes the mean. For the rate-three constant-integrand example, the incorrect calculation \(M_2=N_2\) would give \(\mathbb{E}[M_2]=6\), not zero. The correct calculation is \(M_2=N_2-3\cdot2=N_2-6\), whose mean is \(6-6=0\). The compensator is part of the integral, not an optional correction.
Check Your Understanding
Use the theorem and the definitions to answer the following questions.
- For a rate-\(\lambda\) Poisson process and deterministic \(H_s=h(s)\), state the condition on \(h\) over \([0,T]\) and give \(\mathbb{E}[M_T^2]\).
- Why does \(M_t^2-A_t\) being a martingale identify \(A_t\) as the predictable bracket rather than as the pathwise quadratic variation?
- For \(H_s=\mathbf{1}_{\{N_{s-}=0\}}\) at rate \(\lambda\), write the jump-sum and compensator terms of \(M_t\) for a fixed \(t>0\).
- For the rate-two example with \(H_s=s\) on \([0,1]\), compute the predictable bracket at time \(t\in[0,1]\).
- Explain why finite expected squared integral implies that the random sum defining \([M]_T\) is finite almost surely.