Worked Compensated Poisson Integrals

From Jump Sums to Compensated Integrals

Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carry a Poisson counting process \(N_t\) of rate \(\lambda>0\). We assume \(N_0=0\), and that for \(0\le s<t\), \(N_t-N_s\) is independent of \(\mathcal{F}_s\) and has a Poisson distribution with mean \(\lambda(t-s)\). The process \(N_t-\lambda t\) is its compensated martingale. The deterministic term \(\lambda t\) is the compensator: it removes the average rate of jumps.

For a predictable integrand \(H\), the compensated integral over \([0,T]\) is written \(\int_0^T H_s\,d(N_s-\lambda s)\). Here we use integrands that are progressively measurable and satisfy \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); in defining the Poisson integral, we take a predictable version. A simple predictable integrand has the form \(H_s=\sum_{k=0}^{m-1}\xi_k\mathbf{1}_{(t_k,t_{k+1}]}(s)\), where \(0=t_0<\cdots<t_m=T\) and each \(\xi_k\) is \(\mathcal{F}_{t_k}\)-measurable. Its integral is the jump sum minus the compensator contribution:

Each coefficient is chosen using information available at the beginning of its interval. This predictability matters: it ensures that the coefficient cannot use a jump that occurs later in that interval.

Proof. First take a simple integrand and write \(\Delta N_k=N_{t_{k+1}}-N_{t_k}\) and \(\Delta t_k=t_{k+1}-t_k\). Independence of the increment from \(\mathcal{F}_{t_k}\) gives \(\mathbb{E}[\Delta N_k-\lambda\Delta t_k\mid\mathcal{F}_{t_k}]=0\) and \(\mathbb{E}[(\Delta N_k-\lambda\Delta t_k)^2\mid\mathcal{F}_{t_k}]=\lambda\Delta t_k\). Therefore each summand \(X_k=\xi_k(\Delta N_k-\lambda\Delta t_k)\) has mean zero and \(\mathbb{E}[X_k^2]=\lambda\Delta t_k\,\mathbb{E}[\xi_k^2]\). If \(i<j\), then \(X_i\) is \(\mathcal{F}_{t_j}\)-measurable, so \(\mathbb{E}[X_iX_j]=\mathbb{E}[X_i\mathbb{E}[X_j\mid\mathcal{F}_{t_j}]]=0\). Summing the second moments gives \(\mathbb{E}[(\sum_kX_k)^2]=\lambda\sum_k\mathbb{E}[\xi_k^2]\Delta t_k=\lambda\mathbb{E}\int_0^T H_s^2\,ds\). For general predictable \(H\) in the stated class, approximate it in this integral norm by simple predictable integrands. The compensated Poisson isometry, established for such approximations, gives convergence of their integrals in \(L^2(\mathbb{P})\). Expectations and second moments pass to the limit under \(L^2\) convergence, giving the displayed identities. The zero integrand also satisfies them, so no nonzero-integrand condition is needed. \(\square\)

The result says more than that the expected jump sum equals its compensator. It also quantifies the residual randomness. It does not say that an integral with a random integrand has a Poisson distribution: its distribution can depend on how the integrand is chosen from past information.

Deterministic Integrands and Their Distributions

For a deterministic step function \(h\) with value \(c_j\) on \((s_j,s_{j+1}]\), set \(\Delta_j=s_{j+1}-s_j\) and \(P_j=N_{s_{j+1}}-N_{s_j}\). Then the integral is \(\sum_jc_j(P_j-\lambda\Delta_j)\). The \(P_j\) are independent Poisson random variables with means \(\lambda\Delta_j\). The expectation is zero and the variance is \(\lambda\sum_jc_j^2\Delta_j\).

The characteristic function (the expectation of \(e^{iuX}\) for real \(u\)) records the full distribution. For a Poisson random variable \(P\) with mean \(a\), summing its probabilities gives \(\mathbb{E}[e^{ivP}]=\exp(a(e^{iv}-1))\). Applying this to each independent interval gives the following formula.

Proof. Independence lets us multiply the characteristic functions of \(c_j(P_j-\lambda\Delta_j)\). For a single interval, the result is \(\exp(\lambda\Delta_j(e^{iuc_j}-1))\exp(-iu\lambda\Delta_jc_j)\). Multiplying these factors and collecting exponents yields the stated expression. The mean and variance also follow directly by adding the means and variances of the independent centered terms. \(\square\)

Simple Random Integrands

For a simple random integrand, the coefficients may depend on the past. The moment result still applies because each coefficient is fixed before the next Poisson increment is observed. However, the unconditional distribution need not be the distribution obtained by treating those coefficients as deterministic constants. Conditioning on the past is often the most direct way to identify it.

Why the Compensator and Predictability Matter

The compensator subtracts \(\lambda\int_0^T H_s\,ds\) from the raw jump sum. For a deterministic integrand, that subtraction centers the integral. For a predictable random integrand, the same conclusion follows by conditioning each interval’s increment on the information available at its left endpoint. The square-integrability condition ensures that these centered terms have finite second moments and that the integral is a genuine square-integrable martingale, not merely a local martingale.

A common error is to omit the compensator. For the constant-integrand example, the raw jump sum is \(2N_2\), whose expectation is \(2\cdot6=12\), not zero. Subtracting \(2\cdot3\cdot2=12\) produces \(2(N_2-6)\), with mean zero and variance \(24\). Another error is to use a coefficient selected after observing an interval’s jump and then treat it as predictable. The conditional-centering argument above no longer applies: the coefficient may be correlated with that interval’s increment.

When a distribution is needed, first write the integral as a sum of centered Poisson increments. For deterministic step functions, independent increments give an explicit characteristic function. For simple random integrands, condition on the information determining each coefficient; the result may be a mixture of distributions, as in the last example. In either case, checking that the compensator is subtracted and that the coefficients use only past information is essential.