Why Compensation Changes the Drift
A compensated Poisson integral is a martingale, but applying a nonlinear function to it generally produces drift. The source is the difference between a function’s actual change at a jump and its linear approximation. Compensation removes the linear jump contribution; it does not remove this nonlinear remainder. Keeping those two effects separate is essential, especially when there are infinitely many small jumps.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Poisson random measure \(N(ds,dz)\) with compensator \(ds\,\nu(dz)\), where \(\nu\) is a sigma-finite measure on the mark space. Write \(\widetilde N(ds,dz)=N(ds,dz)-ds\,\nu(dz)\). An integrand \(H(s,z)\) for an integral against \(\widetilde N\) is predictable, and its square-integrable class on \([0,T]\) consists of those satisfying \(\mathbb{E}\int_0^T\int H(s,z)^2\,\nu(dz)\,ds<\infty\). Such an integral is defined as an \(L^2\) limit and has mean zero; it need not be expressible as a jump sum minus a finite-variation compensator.
Itô’s Formula and Its Compensator Drift
For a twice continuously differentiable function \(f\), define the jump remainder \(R_f(y,u)=f(y+u)-f(y)-f'(y)u\). It measures the discrepancy between the true change and its linear part. If \(\lvert f''\rvert\le C\), Taylor’s theorem gives \(\lvert R_f(y,u)\rvert\le \tfrac12 C u^2\). This is a size bound, not a sign claim: the remainder is nonnegative when \(f\) is convex between \(y\) and \(y+u\), and nonpositive when \(f\) is concave there.
Proof. First restrict the marks to sets \(E_n\) of finite \(\nu\)-measure increasing to the mark space, and put \(\gamma_n=\gamma\mathbf{1}_{E_n}\) and \(X^n_t=x+\int_0^t\int\gamma_n\,\widetilde N\). On each finite-activity truncation, the finite-activity jump Itô formula applies: between jumps \(X^n\) has the continuous finite-variation compensating motion, and at a jump of size \(\gamma_n(s,z)\), the exact change is \(f(X^n_{s-}+\gamma_n)-f(X^n_{s-})\). Subtracting its linear part leaves precisely \(R_f(X^n_{s-},\gamma_n)\). This gives the displayed identity for \(X^n\).
To pass to the limit, the compensated Poisson isometry gives \(\mathbb{E}|X^n_t-X_t|^2=\mathbb{E}\int_0^t\int|\gamma_n-\gamma|^2\,\nu(dz)\,ds\to0\). The same convergence holds uniformly in time in probability by Doob’s \(L^2\) maximal inequality, a standard result. Since \(f'\) and \(f''\) are bounded, \(f'\) is Lipschitz and \(\lvert R_f(y,u)\rvert\le C u^2/2\). These bounds, the square-integrability of \(\gamma\), and convergence in measure imply convergence of the linear stochastic integrands in \(L^2\) and convergence of the remainder integrals in \(L^1\). Thus the truncated identities converge in probability to the stated identity. Finally, the compensator identity for absolutely integrable predictable integrands gives \(\int R_f\,N=\int R_f\,\widetilde N+\int R_f\,\nu(dz)\,ds\); the absolute integrability follows from \(\mathbb{E}\int|R_f|\,\nu(dz)\,ds\le(C/2)\mathbb{E}\int\gamma^2\,\nu(dz)\,ds\). This proves both forms. \(\square\)
The second form makes the drift visible. It is not generally the integral of \(f'(X_{s-})\) against a finite-variation compensator for \(X\). Rather, it is the compensator of the nonlinear jump remainder. When \(f'\gamma\) is square-integrable, its compensated integral is a true martingale by the compensated Poisson isometry and the conditional-mean-zero property. The remainder integral is an \(L^1\) martingale because its predictable integrand is absolutely integrable in expectation; it need not be square-integrable.
Worked Calculations
Worked Example: A Compensated Compound Poisson Process
Let \(N_t\) be a Poisson process of rate \(\lambda=3\), and let \(X_t=1+2(N_t-3t)\). Apply the formula to \(f(y)=y^2\), and calculate the compensator drift.
Each jump has size \(2\), while the compensating motion between jumps has rate \(-6\). The derivatives are \(f'(y)=2y\), \(f''(y)=2\), and the jump remainder is \(R_f(y,2)=(y+2)^2-y^2-2y\cdot2=4\). Therefore the drift contributed by the remainder is \(\lambda R_f=3\cdot4=12\) per unit time.
The formula reads $$ X_t^2=1+\int_0^t 4X_{s-}\,\widetilde N(ds)+12t. $$ The stochastic integrand is square-integrable on bounded intervals: \(X\) has finite second moments, and the rate and jump size are finite. Hence its compensated integral is a mean-zero martingale. Taking expectations gives \(\mathbb{E}[X_t^2]=1+12t\).
Check: Directly, \(N_t\) has mean and variance \(3t\), so \(X_t=1+2(N_t-3t)\) has mean \(1\) and variance \(4(3t)=12t\). Thus \(\mathbb{E}[X_t^2]=\operatorname{Var}(X_t)+(\mathbb{E}X_t)^2=12t+1\), as the formula gives.
Worked Example: Infinite Activity Without an Ordinary Compensator
On \(0<z<1\), take \(\nu(dz)=z^{-2}\,dz\) and \(\gamma(s,z)=z\). Then \(\int_0^1 z^2\nu(dz)=\int_0^1 1\,dz=1\), so the compensated integral defining \(X_t\) is square-integrable. But \(\int_0^1 z\,\nu(dz)=\int_0^1 dz/z=\infty\); the compensator cannot be written as a finite-variation term obtained by subtracting \(\int z\,\nu(dz)\,ds\).
Use \(f(y)=y^2\). Here \(f'(y)=2y\), \(f''(y)=2\), and the remainder for a jump of size \(z\) is \((y+z)^2-y^2-2yz=z^2\). Its compensator is \(\int_0^1 z^2\nu(dz)\,ds=ds\). Consequently $$ X_t^2=2\int_0^t\int X_{s-}z\,\widetilde N(ds,dz)+t. $$ Although \(f'\) is unbounded, this application is justified by stopping \(X\) on bounded ranges and then removing the stopping; \(\mathbb{E}\int_0^t\int X_{s-}^2z^2\,\nu(dz)\,ds=\int_0^t\mathbb{E}[X_{s-}^2]\,ds<\infty\), since the isometry gives \(\mathbb{E}[X_s^2]=s\). The stochastic integral is therefore square-integrable.
Check: The isometry also gives \(\mathbb{E}[X_t]=0\) and \(\mathbb{E}[X_t^2]=t\). Taking expectations in the formula gives the same second moment. The calculation uses the compensated integral as an \(L^2\) limit; it never assigns a finite value to the divergent integral \(\int_0^1 z\,\nu(dz)\).
Worked Example: An Exponential and a Direct Moment Check
Let \(N_t\) have rate \(1\), set \(X_t=(\log 2)(N_t-t)\), and take \(f(y)=e^y\). The jump size is \(a=\log 2\). Since \(f'(y)=f''(y)=e^y\), the remainder at a jump is \(e^y(e^a-1-a)=e^y(1-\log 2)\), using \(e^a=2\). Thus $$ e^{X_t}=1+\int_0^t e^{X_{s-}}(e^a-1)\,\widetilde N(ds) +(1-\log 2)\int_0^t e^{X_{s-}}\,ds. $$ Here \(e^{X_{s-}}\) is square-integrable on every bounded interval, since the Poisson variable has finite exponential moments, so the compensated integral is a true square-integrable martingale.
Taking expectations yields \(m(t)=1+(1-\log 2)\int_0^t m(s)\,ds\), where \(m(t)=\mathbb{E}[e^{X_t}]\). Solving this scalar integral equation gives \(m(t)=e^{(1-\log 2)t}\).
Check: Directly, \(e^{X_t}=2^{N_t-t}=2^{-t}2^{N_t}\). For a rate-one Poisson variable, \(\mathbb{E}[2^{N_t}]=\sum_{k=0}^{\infty}2^k e^{-t}t^k/k!=e^{-t}e^{2t}=e^t\). Therefore \(\mathbb{E}[e^{X_t}]=2^{-t}e^t=e^{(1-\log 2)t}\), agreeing with the formula.
Interpretation and Common Pitfalls
When the relevant absolute integrability holds, one may also write the formula in an uncompensated form. For example, if \(\int|\gamma(s,z)|\,\nu(dz)\,ds\) is finite on bounded time intervals, then the linear compensation can be represented as a finite-variation term. But square-integrability alone does not imply this condition. In infinite activity, the compensated integral may only exist as an \(L^2\) limit, while the remainder still has an integrable compensator because it is of quadratic order in the jump size.
A common incorrect step is to replace the remainder by a nonnegative quantity just because its absolute value is bounded by \(Cu^2/2\). For \(f(y)=\sin y\), \(y=0\), and \(u=1\), the remainder is \(\sin 1-1<0\); for \(f(y)=-\sin y\), it is \(1-\sin 1>0\). The bound controls magnitude only. Convexity makes the remainder nonnegative, and concavity makes it nonpositive, on the interval between the pre-jump and post-jump values.
Another error is to use the continuous Itô correction \(\tfrac12 f''(X)\,dt\) as though it described a jump. A jump changes \(f\) by the full amount \(f(X_-+\Delta X)-f(X_-)\); the exact remainder is what remains after subtracting the linear term. For small jumps, Taylor’s theorem explains its quadratic size, but for a finite jump the full remainder—not merely a second-derivative approximation—belongs in the formula.
Check Your Understanding
In each question, assume the integrals mentioned are defined under the stated square-integrability or absolute-integrability conditions.
- For a jump of size \(u\), write the exact remainder for \(f(y)=y^3\). Does its sign have to be the same for all \(y\) and \(u\)?
- For \(f(y)=y^2\), \(\gamma(s,z)=c\), and a Poisson rate \(\lambda\), compute the remainder compensator per unit time.
- Why does \(\int\gamma^2\,d\nu<\infty\) not ensure that the compensation \(\int\gamma\,d\nu\) is finite?
- State what additional property of \(f\) ensures that \(R_f(y,u)\ge0\) for every \(u\) whose segment from \(y\) lies in the domain.
- In the exponential example, compute \(\mathbb{E}[e^{X_t}]\) directly from the Poisson probabilities and compare it with the drift equation’s prediction.