Predictable Integrands and Poisson Jump Sums

Sampling Before a Jump

A Poisson counting process records jumps at random times. To integrate against it, a coefficient attached to a jump must be determined using information available before that jump. A simple predictable process makes this restriction explicit: it holds a value fixed over each time interval, with that value chosen using the information at the interval’s left endpoint. Its integral against the count process is therefore a finite sum of those values times the numbers of jumps in the corresponding intervals.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) and an adapted Poisson counting process \(N_t\) of rate \(\lambda>0\). The Brownian motion is not used in the integrals below. Assume that \(N_0=0\), its jumps have size one, and \(\widetilde N_t=N_t-\lambda t\) is a martingale in the given filtration. No independence between \(B\) and \(N\) is needed. We work on a fixed finite time interval \([0,T]\).

At a deterministic partition time \(t_i\), the process as defined above takes the value from the interval ending at \(t_i\), not from the one beginning there. This convention assigns a jump at \(t_i\) to the interval \((t_{i-1},t_i]\). A Poisson process of finite rate has probability zero of jumping at any specified deterministic time, so endpoint conventions do not affect the integral almost surely. Keeping a consistent convention is still useful when writing pathwise sums.

Integrating Against the Counting Process

For a simple predictable \(H\), define its integral against \(N\) by multiplying each interval’s coefficient by the count increment on that interval. Each integral is a finite random sum because \(N_T\) is finite almost surely. The square-integrability condition above is the class used here; it also ensures that the jump sum is integrable, by the compensator identity for nonnegative predictable integrands.

The first expression counts jumps with their predictable weights. The second subtracts, on each interval, the rate \(\lambda\) times its length, multiplied by the same weight. That subtraction is the compensator term. In differential shorthand, \(d\widetilde N_t=dN_t-\lambda\,dt\), but the finite-sum definitions specify exactly what the shorthand means for these integrands.

Proof. Fix a path on which \(N_T\) is finite, as happens almost surely. For each partition interval \((t_i,t_{i+1}]\), the increment \(N_{t_{i+1}}-N_{t_i}\) is the sum of \(\Delta N_s\) over the jumps in that interval. The integrand equals \(\xi_i\) throughout that interval, including its right endpoint, so \(\xi_i(N_{t_{i+1}}-N_{t_i})=\sum_{t_i<s\le t_{i+1}}\xi_i\Delta N_s=\sum_{t_i<s\le t_{i+1}}H_s\Delta N_s\). Summing over \(i\) counts each jump in \((0,T]\) exactly once. At a jump strictly inside an interval, \(H_s=\xi_i\) was chosen at the earlier time \(t_i\); at a partition endpoint, the stated interval convention assigns the jump to the interval ending there. Thus the expression is the claimed pre-jump-weighted sum. \(\square\)

The formula uses the value of the predictable process at a jump time; for these step processes, that value is the coefficient already selected at the left endpoint of the interval. It does not mean that the coefficient can use the jump’s occurrence or size to decide its own weight.

Compensation as a Time-Integral Correction

Proof. On each interval \((t_i,t_{i+1}]\), the definition gives the contribution \(\xi_i\big((N_{t_{i+1}}-N_{t_i})-\lambda(t_{i+1}-t_i)\big)\). Summing the count terms gives \(\int_0^T H_s\,dN_s\) by its definition. Since \(H_s=\xi_i\) on that interval, its ordinary time integral is \(\int_{t_i}^{t_{i+1}}H_s\,ds=\xi_i(t_{i+1}-t_i)\). Summing these time integrals gives exactly the subtracted term \(\lambda\int_0^T H_s\,ds\). Combining the two sums proves the identity. All terms are well-defined: there are finitely many intervals, the coefficients are finite almost surely, and the square-integrability condition implies \(\mathbb{E}\int_0^T|H_s|\,ds<\infty\) by Cauchy–Schwarz. \(\square\)

This identity is a pathwise algebraic relation for simple processes; it does not itself prove a martingale property or an isometry. Those require additional arguments. In particular, an integral against \(N\) records actual jumps, whereas the compensating time integral records the predictable rate-weighted contribution.

Reading the Integral Jump by Jump

The interval formula can be expanded into individual jumps. This is especially useful when coefficients depend on the count observed at earlier partition times. The coefficient stays fixed between partition points, even if new jumps occur before the next partition point. Refining the partition may change the integrand itself, so it is not legitimate to replace an interval’s fixed coefficient with a value updated after every jump unless that updated process is separately defined as an admissible predictable integrand.

A Common Predictability Error

A frequent mistake is to use information revealed by a jump to set the coefficient multiplying that same jump. For example, on a path with two jumps in \((0,1]\), a post-jump rule that assigns the current count \(N_s\) to each jump would assign weights \(1\) and \(2\), giving \(1+2=3\). Those weights use the jump being counted: the process \(N_s\) is not the left-endpoint coefficient of a simple predictable process.

For a predictable step process with coefficient chosen at time \(0\), the weight is fixed throughout \((0,1]\). If that coefficient is \(N_0=0\), the integral over the interval is \(N_0(N_1-N_0)=0\), not \(3\). If instead the intended integrand is the left-limit process \(N_{s-}\), then its weights on the same two jumps are \(0\) and \(1\), giving \(0+1=1\); that is a different integrand, not the fixed step process \(N_0\) on the whole interval. The distinction is precisely why the predictable sampling rule must be stated before calculating a jump sum.

For simple integrands, the key procedure is concrete: identify the deterministic intervals, determine each coefficient using only information at its left endpoint, multiply by the count increment on that interval, and subtract \(\lambda\) times the coefficient-weighted interval length for the compensated integral. The jump-sum representation is the same calculation with each interval’s count increment expanded into its individual jumps.