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First and second moments of a Poisson random integral

Statement

Let η be a Poisson point process with locally finite intensity measure ν. For every nonnegative Borel function f, E[η(f)]=ν(f),E[η(f)2]=ν(f2)+ν(f)2, as identities in [0,∞]. Here η(f)=∫f dη and ν(f)=∫f dν.

Facts & Assumptions

Given: A Poisson point process η with locally finite intensity measure ν and a nonnegative Borel function f.

[F1]

Counts on disjoint Borel sets are independent and have Poisson laws with their intensity means; an infinite-mean count equals +∞ almost surely. (Poisson and Cox point processes)

[F2]

For N∼Pois⁡(a) with finite a, EN=a and EN2=a+a2. (The scalar Poisson count law, Normalization, moments, and generating function of the Poisson count law)

[F3]

Nonnegative Borel functions admit increasing simple approximations, and increasing limits pass through nonnegative integrals. (Every nonnegative measurable function is the increasing limit of simple measurable functions, Monotone convergence for the integral)

Proof

technique · direct
1.1givenF1F2algebra

First let s=∑j=1mcj1Bj be a nonnegative simple function with disjoint Borel sets Bj and coefficients cj>0, after discarding zero coefficients. Write aj=ν(Bj). If some aj=∞, [F1] gives η(s)=∞ almost surely and both displayed right-hand quantities are infinite. If all aj<∞, [F1]-[F2] give Eη(s)=∑jcjaj=ν(s) and Eη(s)2=∑jcj2(aj+aj2)+2∑i<jcicjaiaj=ν(s2)+ν(s)2. The zero simple function also satisfies both identities.

2.1givenstep 1.1F3∎

For arbitrary f≥0, choose simple sn↑f by [F3]. Then η(sn)↑η(f), η(sn)2↑η(f)2, ν(sn)↑ν(f), and ν(sn2)↑ν(f2). Applying monotone convergence [F3] to the two expectation integrals and to the two measure integrals, and using step 1.1 for each sn, yields both asserted identities, including the extended-infinite cases.

Depends on

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Sources