The Compensated Poisson Isometry

Why Compensation Gives an Isometry

A Poisson count has nonnegative mean; its mean is positive when its rate is positive. Thus raw increments of a Poisson process are generally not centered. Subtracting their expected increase produces increments with conditional mean zero. That centering is what makes increments from different time intervals orthogonal in the second-moment calculation, even when the integrand on a later interval depends on the earlier history.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) and a Poisson counting process \(N_t\) of constant rate \(\lambda\ge0\). The Brownian motion is not used here. Assume \(N_0=0\), and that for \(0\le s<t\), \(N_t-N_s\) is independent of \(\mathcal{F}_s\) and has the Poisson distribution with parameter \(\lambda(t-s)\). In particular, the filtration contains the observed history of \(N\). Define the compensated process \(M_t=N_t-\lambda t\).

Such an integrand is adapted and progressively measurable, and it satisfies \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). More generally, the notation \(\mathcal{L}^2\) refers to progressively measurable integrands with this finite expected time integral of the square. The result below is proved for simple predictable integrands; it does not assume the existence of an integral for every member of \(\mathcal{L}^2\).

Conditional Moments of a Compensated Increment

Write \(\Delta t_i=t_{i+1}-t_i\) and \(\Delta M_i=M_{t_{i+1}}-M_{t_i}=(N_{t_{i+1}}-N_{t_i})-\lambda\Delta t_i\). Conditional on \(\mathcal{F}_{t_i}\), the count increment is Poisson with parameter \(\lambda\Delta t_i\). A Poisson random variable with parameter \(a\ge0\) has mean \(a\) and variance \(a\); when \(a=0\), it equals zero almost surely. It follows that, almost surely,

These identities also verify that \(M_t\) is a true martingale on every finite time interval. Indeed, \(M_t\) is integrable because \(N_t\) has finite mean, and for \(s\le t\), independence and the Poisson mean give \(\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\). Therefore \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). The second conditional-moment identity supplies the variance calculation needed for the isometry.

Proof. The second-moment formula for the increment implies \(\mathbb{E}[(\Delta M_i)^2]=\lambda\Delta t_i<\infty\). By Cauchy–Schwarz, \(H_i\Delta M_i\) is integrable. Conditioning on \(\mathcal{F}_{t_i}\) and using that \(H_i\) is measurable there gives \(\mathbb{E}[H_i\Delta M_i]=\mathbb{E}[H_i\mathbb{E}[\Delta M_i\mid\mathcal{F}_{t_i}]]=0\). For the second identity, condition on the same sigma-field and use the conditional second moment: \(\mathbb{E}[H_i^2(\Delta M_i)^2]=\mathbb{E}[H_i^2\mathbb{E}[(\Delta M_i)^2\mid\mathcal{F}_{t_i}]]=\lambda\Delta t_i\,\mathbb{E}[H_i^2]\). The conditional-expectation step is valid for nonnegative \(H_i^2(\Delta M_i)^2\); the resulting expectation is finite by the assumed square integrability of \(H_i\). If \(\lambda\Delta t_i=0\), the increment is zero almost surely and both identities still hold. \(\square\)

The Compensated Poisson Isometry

Proof. Set \(X_i=H_i\Delta M_i\). Each \(X_i\) is square-integrable: by the Proposition, \(\mathbb{E}[X_i^2]=\lambda\Delta t_i\,\mathbb{E}[H_i^2]<\infty\). Expanding the square of the finite sum gives diagonal terms \(\mathbb{E}[X_i^2]\) and cross terms \(\mathbb{E}[X_iX_j]\) for \(i<j\). The diagonal terms sum to \(\lambda\sum_i\Delta t_i\,\mathbb{E}[H_i^2]\).

For a cross term with \(i<j\), both \(H_i\Delta M_i\) and \(H_j\Delta M_j\) are square-integrable by the diagonal identity. Hence their product is integrable by Cauchy–Schwarz; this establishes integrability of the full cross term, including its later compensated increment. Also \(X_i\) is \(\mathcal{F}_{t_j}\)-measurable, since its interval ends at or before \(t_j\). The coefficient \(H_j\) is \(\mathcal{F}_{t_j}\)-measurable, and \(\Delta M_j\) is independent of \(\mathcal{F}_{t_j}\) with mean zero. Thus \(\mathbb{E}[X_iH_j\Delta M_j]=0\). More explicitly, for an unbounded \(\mathcal{F}_{t_j}\)-measurable multiplier \(X_iH_j\), truncate it to bounded multipliers, use independence and centering for each truncation, and pass to the limit using the integrability of \(X_iH_j\Delta M_j\). Every cross term therefore vanishes. Finally, the definition of \(H\) gives \(\mathbb{E}\int_0^T H_t^2\,dt=\sum_i\Delta t_i\,\mathbb{E}[H_i^2]\), proving the identity. \(\square\)

The conditional-expectation argument matters because a later coefficient need not be independent of earlier increments. It may depend on them. What matters is that the later compensated increment is centered conditionally on the information available when its coefficient is chosen.

Worked Calculations

Why Compensation Is Essential

A frequent mistake is to apply the isometry to the raw jump sum \(\int_0^T H_t\,dN_t\), forgetting the compensator \(\lambda\,dt\). For a concrete check, take \(\lambda=2\), \(T=1\), and \(H_t=1\). The raw jump sum is \(N_1\), and \(\mathbb{E}[N_1^2]=\operatorname{Var}(N_1)+(\mathbb{E}N_1)^2=2+2^2=6\). Substituting into the compensated isometry as though it were raw would incorrectly predict \(6=2\mathbb{E}\int_0^1 1\,dt=2\). The correction is to center the count: the compensated integral is \(N_1-2\), whose second moment is its variance, \(\mathbb{E}[(N_1-2)^2]=2\), exactly the isometry's right-hand side.

The theorem is useful because it converts a random integral's second moment into an ordinary expected time integral. Its proof relies on the simple predictable form and the conditional Poisson moments. In particular, this tutorial establishes the identity for finite step integrands; it does not silently assume a construction or an isometry for a larger class of integrands.