Centering a Poisson Random Measure
A Poisson random measure records random points while its intensity measure describes their average distribution. Integrating a deterministic function against the points gives a random sum. Subtracting the corresponding intensity integral centers that sum: the result has mean zero, and its variance is determined by the squared size of the function integrated against the intensity. These facts make it possible to calculate uncertainty without finding the full distribution of the random sum.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \((E,\mathcal{E})\) be a measurable space, and let \(\nu\) be a sigma-finite measure on it. A Poisson random measure \(N\) with intensity \(\nu\) assigns a count \(N(A)\) to each \(A\in\mathcal{E}\). For any finite collection of disjoint sets of finite \(\nu\)-measure, their counts are independent, and \(N(A)\) has a Poisson distribution with mean \(\nu(A)\). In particular, \(\mathbb{E}N(A)=\operatorname{Var}(N(A))=\nu(A)\) when \(\nu(A)<\infty\).
For a deterministic measurable function \(f:E\to\mathbb{R}\), the random integral \(\int f\,dN\), when it is defined, sums the values of \(f\) over the points counted by \(N\), with multiplicity. The deterministic integral \(\int f\,d\nu\) is its centering term when that integral is finite.
Simple functions supported on sets of finite measure are dense in \(L^2(\nu)\). One way to obtain approximations is to restrict \(f\) to an increasing sequence of finite-measure sets whose union is \(E\), truncate its values to bounded ranges, and then approximate those values by a finite grid. The resulting simple functions converge to \(f\) in squared-integral norm. The following result shows why the mean-square limit in the definition exists and gives its moments.
Proof. First take a simple function \(f=\sum_{\ell=1}^r a_\ell\mathbf{1}_{A_\ell}\) as in the definition. Each centered count \(N(A_\ell)-\nu(A_\ell)\) has mean zero and variance \(\nu(A_\ell)\). The centered counts are independent because the sets are disjoint, so their pairwise covariances are zero. Therefore $$ \mathbb{E}[I(f)]=\sum_{\ell=1}^r a_\ell\bigl(\mathbb{E}N(A_\ell)-\nu(A_\ell)\bigr)=0 $$ and $$ \mathbb{E}[I(f)^2]=\operatorname{Var}(I(f)) =\sum_{\ell=1}^r a_\ell^2\nu(A_\ell)=\int f^2\,d\nu. $$ Applying this calculation to the simple function \(f-g\) gives \(\mathbb{E}[(I(f)-I(g))^2]=\int(f-g)^2\,d\nu\) for simple \(f,g\).
Now let \(f\in L^2(\nu)\), and choose simple \(f_m\) with \(\int(f_m-f)^2\,d\nu\to0\). The simple-function identity gives $$ \mathbb{E}[(I(f_m)-I(f_n))^2]=\int(f_m-f_n)^2\,d\nu\longrightarrow0 $$ as \(m,n\to\infty\), because \(f_m\) converges in \(L^2(\nu)\). Thus \(I(f_m)\) is Cauchy in mean square and has a mean-square limit, which defines \(I(f)\). Mean-square convergence implies convergence in mean: \(\mathbb{E}|I(f_m)-I(f)|\leq\sqrt{\mathbb{E}[(I(f_m)-I(f))^2]}\). Hence \(\mathbb{E}I(f)=0\). Also, the simple-function identities give \(\mathbb{E}[I(f_m)^2]=\int f_m^2\,d\nu\); taking limits in mean square on the left and in \(L^2(\nu)\) on the right yields \(\mathbb{E}[I(f)^2]=\int f^2\,d\nu\). Since the mean is zero, this is its variance. Finally, apply the variance identity to \(f+g\) and expand both squares: $$ \mathbb{E}[I(f)I(g)] =\frac{1}{2}\left(\int(f+g)^2\,d\nu-\int f^2\,d\nu-\int g^2\,d\nu\right) =\int fg\,d\nu. $$ All the integrals on the right are finite by the Cauchy–Schwarz inequality. \(\square\)
The theorem says that the map \(f\mapsto I(f)\) preserves squared-integral distance in mean square. In particular, the variance depends on \(f^2\), not on the sign of \(f\). If \(f\) is also integrable, then the compensated integral agrees with the usual centered random sum \(\int f\,dN-\int f\,d\nu\). For nonnegative \(f\), the expectation formula for \(\int f\,dN\) follows first for simple functions and then by monotone convergence. For signed integrable \(f\), apply that argument to its positive and negative parts.
Worked Example: A Centered Count on One Set
Suppose \(A\) is measurable with \(\nu(A)=2\), and take \(f=3\mathbf{1}_A\). This is a simple function supported on a finite-measure set, and \(I(f)=3(N(A)-2)\). The set count has distribution \(\operatorname{Poisson}(2)\), so \(\mathbb{E}N(A)=2\) and \(\operatorname{Var}(N(A))=2\). Thus $$ \mathbb{E}[I(f)]=3(2-2)=0,\qquad \operatorname{Var}(I(f))=3^2\operatorname{Var}(N(A))=9\cdot2=18. $$ The isometry gives the same variance directly: \(\int f^2\,d\nu=9\nu(A)=9\cdot2=18\). The two calculations agree.
Worked Example: A Signed Step Function
Take \(E=[0,3]\) with Lebesgue intensity \(\nu\), and let \(A_1=[0,1)\) and \(A_2=[1,3]\). These disjoint sets have measures \(1\) and \(2\); the inclusion of the endpoint \(3\) does not change either measure. Set \(f=2\mathbf{1}_{A_1}-\mathbf{1}_{A_2}\). Then $$ I(f)=2(N(A_1)-1)-(N(A_2)-2). $$ The two counts are independent Poisson random variables with respective means \(1\) and \(2\). Therefore the displayed expression has mean \(2(1-1)-(2-2)=0\), and its variance is $$ 2^2\cdot1+(-1)^2\cdot2=4+2=6. $$ Alternatively, the isometry gives \(\int f^2\,d\nu=4\cdot1+1\cdot2=6\). This calculation also shows why the two pieces contribute positively to the variance even though the integrand is negative on \(A_2\).
Worked Example: A Nonconstant Integrand
Let \(N\) have Lebesgue intensity on \([0,1]\), and take \(f(x)=x\). Since \(\int_0^1x^2\,dx=\tfrac13<\infty\), the compensated integral is defined. The centering term is \(\int_0^1x\,dx=\tfrac12\), so $$ I(f)=\int_0^1x\,N(dx)-\tfrac12,\qquad \mathbb{E}[I(f)]=0,\qquad \operatorname{Var}(I(f))=\int_0^1x^2\,dx=\tfrac13. $$ For an independent check, the total count \(N([0,1])\) is Poisson with mean \(1\); conditional on that count being \(n\), the \(n\) points are independent uniform random variables on \([0,1]\). Write \(S=\int_0^1x\,N(dx)\) for their sum. A uniform random variable \(U\) on \([0,1]\) has \(\mathbb{E}U=\tfrac12\) and \(\mathbb{E}U^2=\tfrac13\). Conditioning on the count, or using the compound-Poisson second-moment calculation, gives $$ \mathbb{E}S=\tfrac12,\qquad \mathbb{E}S^2=1\cdot\tfrac13+1^2\cdot\left(\tfrac12\right)^2 =\tfrac13+\tfrac14=\tfrac7{12}. $$ Consequently \(\operatorname{Var}(I(f))=\operatorname{Var}(S)=\tfrac7{12}-(\tfrac12)^2=\tfrac7{12}-\tfrac3{12}=\tfrac13\), agreeing with the isometry.
When the Raw Centering Formula Applies
The \(L^2\) definition does not require \(\int |f|\,d\nu<\infty\), but writing the integral as a raw random sum minus a deterministic integral does require care. If \(f\in L^1(\nu)\cap L^2(\nu)\), then \(\int f\,dN\) is integrable, its expectation is \(\int f\,d\nu\), and \(I(f)=\int f\,dN-\int f\,d\nu\). The expectation statement follows by approximation for nonnegative functions and by separating positive and negative parts for signed functions. If only \(f\in L^2(\nu)\), the centered integral still exists in mean square, but either raw term in that expression may fail to be finite.
Square integrability is the condition that makes the variance finite, not merely a convenient way to do the calculation. For example, on \([0,1]\) with Lebesgue intensity, let \(f(x)=x^{-3/4}\) for \(x>0\). Then \(\int_0^1 f(x)\,dx=4\), but \(\int_0^1 f(x)^2\,dx=\int_0^1x^{-3/2}\,dx=\infty\). The random sum \(S=\int f\,dN\) is finite almost surely: the total number of points on \([0,1]\) is finite almost surely, and each point lies strictly above \(0\) almost surely. Its expectation is \(4\), but its second moment is infinite. Indeed, for \(f_m(x)=f(x)\mathbf{1}_{[1/m,1]}(x)\), the compound-Poisson second-moment formula gives \(\mathbb{E}[(\int f_m\,dN)^2]=(\int f_m\,d\nu)^2+\int f_m^2\,d\nu\), and the second term tends to infinity. Since these nonnegative sums increase to \(S\), monotone convergence gives \(\mathbb{E}S^2=\infty\). Thus the centered raw sum has no finite variance in this example.
A Common Centering Mistake
A deterministic centering term changes the mean but not the variance. For any square-integrable random variable \(X\) and deterministic constant \(c\), $$ \operatorname{Var}(X-c)=\operatorname{Var}(X). $$ More generally, \(\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)-2\operatorname{Cov}(X,Y)\). When \(Y=c\) is deterministic, \(\operatorname{Var}(Y)=0\) and \(\operatorname{Cov}(X,Y)=0\), giving the stated identity.
For a concrete check, let \(N(A)\) have a Poisson distribution with mean \(4\). Then \(X=N(A)-4\) is centered, and $$ \operatorname{Var}(N(A)-4)=\operatorname{Var}(N(A))=4. $$ The expression \(\operatorname{Var}(N(A))-\operatorname{Var}(4)\) also equals \(4-0=4\); it is not unequal to the variance. The correct general formula includes the covariance term, and here subtracting a deterministic number leaves the variance unchanged.
Keep the distinction between the random count and its deterministic compensator explicit. For \(f=\mathbf{1}_A\), the compensated integral is \(N(A)-\nu(A)\), with mean zero and variance \(\nu(A)\). For a general deterministic \(f\in L^2(\nu)\), the corresponding rule is that the variance is \(\int f^2\,d\nu\), not \((\int f\,d\nu)^2\). The latter is the square of the centering term and is not the variance formula.
Check Your Understanding
For each question, state the assumptions you use and check any variance by applying the isometry.
- If \(\nu(A)=5\), find the mean and variance of \(I(2\mathbf{1}_A)\).
- For disjoint finite-measure sets \(A\) and \(B\), write \(I(a\mathbf{1}_A+b\mathbf{1}_B)\) and its variance.
- On \([0,1]\) with Lebesgue intensity, find the mean and variance of \(I(f)\) for \(f(x)=1-x\).
- Explain why mean-square convergence of simple compensated integrals is enough to define \(I(f)\) for \(f\in L^2(\nu)\).
- If \(X\) is square-integrable and \(c\) is deterministic, compute \(\operatorname{Var}(X-c)\) using the covariance formula.