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Let $N$ be a Poisson point process with intensity measure $\mu$, and let $g$ be a real-valued measurable function. For the ordinary Poisson sum $S_g=\int gdN$, the condition $\int |g|d\mu <\infty$ givesIf also $\int g^{2}d\mu <\infty$, thenExpand the square: equal-index terms contribute $\int g^{2}d\mu$, while the Poisson factorial moment measure makes distinct-index terms contribute $(\int gd\mu {)}^2$. Subtracting the squared expected value proves the variance formula. The Campbell first-moment formula also holds for general point processes; this variance identity uses the Poisson assumption.^[ref]

Square integrability alone is insufficient for the ordinary sum on an infinite-intensity space. Take intensity $dx$ on $[1,\infty )$ and $g(x)=1/x$. Then $\int g^{2}dx=1$, but the sum is infinite almost surely: the ordered points satisfy $Z_i/i\to 1$ by the strong law for Poisson arrival times, so comparison with the harmonic series proves divergence.

For this same $g$, the compensated Poisson integral exists as a mean-zero limit under convergence in Lp with $p=2$ and has variance $1$. Its definition is the L2 construction of compensated Poisson integrals, rather than subtraction of two infinite ordinary integrals. When the total intensity is finite, Cauchy-Schwarz inequality makes L2 imply L1, explaining why the shorter variance statement works in that special case.^[ref]

For a Poisson random measure $N$ with sigma-finite intensity measure $\mu$, a real deterministic simple integrand $f={\sum }_i{a}_i{1}_{A_i}$ on disjoint sets of finite intensity has compensated Poisson integralIts expected value is zero. Independence of disjoint counts and the Poisson variance formula give $\mathbb{E}[I(f{)}^2]=\int f^{2}d\mu$.

This isometry extends the integral by completion to deterministic $f\in L^2(\mu )$. If $f$ is not also integrable, writing two separate finite integrals $\int fdN-\int fd\mu$ can be invalid; the compensated integral is defined through the joint centered limit. For a constant-rate Poisson process, taking $f=1_{[0,t]}$ gives $N_t-\lambda t$ with variance $\lambda t$.^[ref]

The Lévy-Itô decomposition builds a real Lévy process from drift, continuous Gaussian noise and Poisson jumps. For the triplet $(a,b,K)$ in the Lévy-Khintchine formula with small-jump cutoff one, take independent standard Brownian motion $B$ and Poisson random measure $N$ of intensity $dsK(dx)$. ThenHere $b\ge 0$, $K$ is a Lévy measure and $\tilde{N}=N-dsK(dx)$ denotes compensation. The large-jump term has finitely many jumps on each bounded time interval because $K(|x|>1)<\infty$.

The small-jump integral is defined by limits, rather than by subtracting two potentially infinite sums. For $\epsilon >0$, let $J_t^{\epsilon }={\int }_0^t{\int }_{\epsilon <|x|\le 1}x\tilde{N}(ds,dx)$. These are square-integrable martingales. The compensated Poisson isometry givesas $0<\delta <\epsilon \downarrow 0$. Doob's L2 maximal inequality also controls the supremum on $[0,T]$, so the limit exists locally uniformly in probability, indeed in the L2 norm of that supremum. This uses exactly the small-jump second-moment condition of a Lévy measure.^[ref]

The components have independent increments; multiplying their characteristic functions recovers the Lévy-Khintchine exponent. The decomposition separates continuous fluctuations from large jumps and possibly infinitely many small jumps. A finite small-jump Second moment does not require a finite first absolute jump moment, which explains why compensation and its limiting construction are essential.

For the small-jump cutoff convention in the Lévy-Khintchine formula, let $(a,b,K)$ be the triplet of a real Lévy process. Its paths are continuous almost surely exactly when $K=0$. They are everywhere differentiable almost surely exactly when $K=0$ and $b=0$, giving the deterministic path $X_t=at$. A compound Poisson path can have a derivative between its jumps without being an everywhere differentiable path.

finite variation on compact time intervals has the different criterionUnder these conditions the uncompensated Lévy-Khintchine formula gives a drift plus an absolutely summable small-jump process and finitely many large jumps. A nonzero Brownian component precludes finite variation, while an infinite absolute small-jump sum already forces infinite variation.

Integrability requires only the large-jump first moment:The small compensated jumps are square integrable and hence integrable. When this condition holds, $\mathbb{E}{X}_t=t[a+{\int }_{|x|>1}xK(dx)]$; the small-jump compensation convention is why no small-jump mean appears.

Likewise a finite second moment is equivalent to $\int x^{2}K(dx)<\infty$, and thenThe Lévy-Itô decomposition proves these moment statements by independent Brownian, compensated-small-jump and compound-Poisson-large-jump parts. The large-jump one-event case also shows why the tail moment conditions are necessary.