From a Lévy Triplet to Its Random Components
A Lévy process can have continuous random fluctuations and jumps of many sizes. The Lévy–Itô decomposition separates these sources: a deterministic drift, a Brownian component, compensated jumps of size at most one, and a sum of jumps larger than one. The distinction between the two jump terms is essential. There can be infinitely many small jumps in a finite time interval, so their sum is defined by a limit; there are almost surely only finitely many large jumps on such an interval.
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 random measure \(N(dt,dx)\) on \((0,\infty)\times(\mathbb{R}\setminus\{0\})\). Assume that the filtration is the usual augmentation of the filtration generated by \(B\) and \(N\), and that \(B\) and \(N\) are independent. In particular, the future increments of \(N\) after time \(s\) are independent of \(\mathcal{F}_s\). This filtration assumption is needed for the conditional-expectation martingale statements below.
The Lévy-measure condition implies \(\nu(\{|x|>1\})<\infty\), so large jumps have finite intensity. It also implies \(\int_{|x|\le1}x^2\,\nu(dx)<\infty\), which is the condition that makes compensated small-jump integrals converge in mean square. It does not require \(\int_{|x|\le1}|x|\,\nu(dx)<\infty\); in fact that integral may be infinite.
The Decomposition and Its Small-Jump Limit
The full representation theorem, which identifies the jump measure of a given Lévy process and constructs its continuous Brownian component, is a standard deep result; that construction is not proved here. The cutoff convergence and exponent calculations that explain the form of the decomposition are proved below.
For every cutoff \(\varepsilon>0\), the small-jump integrand \(H_s(x)=x\mathbf{1}_{\{\varepsilon<|x|\le1\}}\) is deterministic, hence predictable, and satisfies $$ \mathbb{E}\int_0^T\int_{\mathbb{R}\setminus\{0\}}H_s(x)^2\,\nu(dx)\,ds =T\int_{\varepsilon<|x|\le1}x^2\,\nu(dx)<\infty. $$ Thus its compensated Poisson integral is square-integrable. For the Brownian integral, the integrand \(\sigma\) is adapted and progressively measurable and satisfies \(\mathbb{E}\int_0^T\sigma^2\,ds=\sigma^2T<\infty\), so it belongs to the Brownian \(L^2\) integrand class.
Proof. For \(0<\delta<\varepsilon\), the difference \(M_T^\delta-M_T^\varepsilon\) is the compensated integral over \(\delta<|x|\le\varepsilon\). The compensated Poisson isometry gives $$ \mathbb{E}\big[(M_T^\delta-M_T^\varepsilon)^2\big] =T\int_{\delta<|x|\le\varepsilon}x^2\,\nu(dx) \le T\int_{0<|x|\le\varepsilon}x^2\,\nu(dx). $$ The last integral tends to zero as \(\varepsilon\downarrow0\), since \(x^2\) is integrable on \(\{0<|x|\le1\}\) with respect to \(\nu\). Therefore the cutoff integrals are Cauchy in \(L^2\), which is complete, and have an \(L^2\) limit \(M_T\).
For each cutoff, \(M^\varepsilon\) is a martingale: its integrand is square-integrable, and the compensated Poisson integral over \((s,t]\) has conditional expectation zero given \(\mathcal{F}_s\). This conditional statement follows here because the future increments of \(N\) are independent of \(\mathcal{F}_s\), by the filtration assumption. For \(s\le t\), \(L^2\) convergence at both times implies \(L^1\) convergence. Conditional expectation is a contraction in \(L^1\), so passing to the limit in \(\mathbb{E}[M_t^\varepsilon\mid\mathcal{F}_s]=M_s^\varepsilon\) gives \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). Also \(M_t\) is square-integrable as an \(L^2\) limit. Thus the limit is a true square-integrable martingale, not merely a local martingale. \(\square\)
The isometry also gives \(\mathbb{E}[M_t^2]=t\int_{0<|x|\le1}x^2\,\nu(dx)\). Uniform convergence in probability on \([0,T]\) follows by applying Doob's \(L^2\) maximal inequality to cutoff differences: their expected squared suprema are at most \(4T\int_{0<|x|\le\varepsilon}x^2\,\nu(dx)\), which tends to zero.
Why These Terms Give the Lévy–Khintchine Exponent
The large-jump integral is an uncompensated sum over the atoms of \(N\). On \([0,T]\), the number of atoms with \(|x|>1\) is Poisson with mean \(T\nu(\{|x|>1\})<\infty\); hence it is finite almost surely. No integrability of \(|x|\) over the large-jump region is needed to define this finite sum. Its expectation, however, need not exist.
Proof of the exponent calculation. Fix \(u\in\mathbb{R}\) and first retain only jumps with \(|x|>\varepsilon\). The independent-increment property and the characteristic function of a Poisson sum give the exponent per unit time $$ i u\gamma-\frac12\sigma^2u^2 +\int_{\varepsilon<|x|\le1}(e^{iux}-1-iux)\,\nu(dx) +\int_{|x|>1}(e^{iux}-1)\,\nu(dx). $$ The term \(-iux\) in the small-jump integral comes from compensation; there is no such subtraction for the uncompensated large jumps. For real \(y\), Taylor's formula with an integral remainder gives \(|e^{iy}-1-iy|\le y^2/2\). Therefore $$ |e^{iux}-1-iux|\le \frac{u^2x^2}{2}, $$ which is integrable over \(0<|x|\le1\) by the Lévy-measure condition. The small-jump integral consequently converges as \(\varepsilon\downarrow0\). The limiting characteristic function is $$ \mathbb{E}[e^{iuX_t}] =\exp\left(t\left[iu\gamma-\frac12\sigma^2u^2 +\int_{|x|\le1}(e^{iux}-1-iux)\,\nu(dx) +\int_{|x|>1}(e^{iux}-1)\,\nu(dx)\right]\right), $$ which is precisely the Lévy–Khintchine exponent for this truncation. This calculation verifies the drift convention and the compensation in the decomposition. \(\square\)
Worked Decompositions
Worked Example: A Brownian Motion with One Small-Jump Size
Let \(\gamma=\tfrac14\), \(\sigma=\tfrac13\), and \(\nu=2\delta_{1/2}\), where \(\delta_{1/2}\) is unit mass at \(x=\tfrac12\). Write the decomposition and find the mean and variance at time \(t\ge0\).
The jump size satisfies \(|\tfrac12|\le1\), so all jumps are in the compensated small-jump term. The Poisson count \(N_t=N((0,t]\times\{\tfrac12\})\) has rate \(2\). The decomposition is $$ X_t=\frac14t+\frac13B_t+\frac12(N_t-2t) =-\frac34t+\frac13B_t+\frac12N_t. $$ Since \(\mathbb{E}[N_t]=2t\) and \(\operatorname{Var}(N_t)=2t\), while \(B_t\) is independent of \(N_t\), the expectation is \(-\tfrac34t+\tfrac12(2t)=\tfrac14t\). The variance is \(\tfrac19t+\tfrac14(2t)=\tfrac{11}{18}t\).
Check: the compensated form has mean \(\tfrac14t\), because both \(B_t\) and \(N_t-2t\) have mean zero. Its variance is \(\tfrac19t+\tfrac14\cdot2t=\tfrac{2}{18}t+\tfrac{9}{18}t=\tfrac{11}{18}t\), as computed from the expanded form.
Worked Example: Infinitely Many Small Jumps with a Finite Cutoff Variance
Consider the symmetric Lévy measure \(\nu(dx)=\mathbf{1}_{\{0<|x|<1\}}|x|^{-2}\,dx\), with \(\gamma=0\) and \(\sigma=0\). Verify the Lévy-measure condition and compute the variance contributed by jumps with \(\varepsilon<|x|<1\), for \(0<\varepsilon<1\).
The measure has infinite mass near zero: \(\int_{0<|x|<1}|x|^{-2}\,dx=\infty\). But its required square-weighted integral is finite: $$ \int_{0<|x|<1}x^2\,\nu(dx) =\int_{-1}^{1}1\,dx=2. $$ Thus it is a Lévy measure, even though there are infinitely many small jumps in every positive time interval in the Poisson intensity sense. At time \(t\), the cutoff integral has mean zero and variance $$ t\int_{\varepsilon<|x|<1}x^2\,\nu(dx) =t\int_{\varepsilon<|x|<1}1\,dx =2t(1-\varepsilon). $$ The variance of the omitted jumps is \(t\int_{0<|x|\le\varepsilon}x^2\nu(dx)=2t\varepsilon\), which tends to zero.
Check: the total square-weighted integral is the limit \(2t(1-\varepsilon)\to2t\). The omitted variance \(2t\varepsilon\) tends to zero, exactly the Cauchy estimate required for the small-jump limit.
Worked Example: Uncompensated Large Jumps
Let \(\gamma=-1\), \(\sigma=0\), and \(\nu=3\delta_2\). Write \(X_t\) in terms of a Poisson count and determine whether its mean is finite.
Since \(|2|>1\), these jumps are large. The count \(N_t=N((0,t]\times\{2\})\) is Poisson with rate \(3\), and the decomposition is \(X_t=-t+2N_t\). The large-jump sum is finite almost surely on every bounded interval because \(N_t\) is finite almost surely. Here the jump size is constant and the count has finite mean, so \(\mathbb{E}[X_t]=-t+2(3t)=5t\).
Check: the characteristic exponent is \(-iu+3(e^{2iu}-1)\), exactly the drift contribution \(-iu\) plus the uncompensated large-jump contribution. No small-jump compensation term appears.
Common Pitfall: Compensating Every Jump
A tempting but generally invalid rewrite is to compensate the large jumps as well and retain the same drift. For the preceding example, that would replace \(-t+2N_t\) by \(-t+2(N_t-3t)=-7t+2N_t\), changing the mean from \(5t\) to \(-t\). The error is the missing drift adjustment \(t\int_{|x|>1}x\,\nu(dx)=6t\). More generally, changing the truncation convention changes the drift parameter, and that integral may not even be finite for an arbitrary Lévy measure. With the truncation at one, only small jumps are compensated in the displayed decomposition.
Also, independent increments of \(N\) as a random measure do not by themselves guarantee a martingale relative to an arbitrary filtration. If the filtration reveals future jumps at time zero, the conditional mean of a future compensated increment need not be zero. The usual augmented filtration generated by \(B\) and \(N\), with their stated independence, ensures the conditional independence used in the proof.
| Component | Definition in the decomposition | Key condition or property |
|---|---|---|
| Drift | \(\gamma t\) | Depends on the selected truncation convention |
| Brownian part | \(\sigma B_t\) | Continuous Gaussian fluctuations |
| Small jumps | Limit of \(\int x\,\widetilde N\) over \(\varepsilon<|x|\le1\) | \(\int_{|x|\le1}x^2\nu(dx)<\infty\) |
| Large jumps | \(\int_{|x|>1}x\,N(ds,dx)\) | Finite number almost surely on bounded intervals |
Check Your Understanding
For each question, use the truncation \(x\mathbf{1}_{\{|x|\le1\}}\) and assume the filtration and independence conditions stated above.
- Which integral condition on \(\nu\) ensures the compensated small-jump cutoff integrals are Cauchy in \(L^2\)?
- Why does \(\nu(\{|x|>1\})<\infty\) imply that the large-jump sum has finitely many terms almost surely on \([0,T]\)?
- For \(\nu=4\delta_{-1/2}\), identify whether the jump term is compensated or uncompensated and write it using its Poisson count.
- Explain why the conditional-expectation proof for the compensated integral requires a condition on the filtration, not only independent increments of the Poisson random measure.
- How must the drift parameter change if one changes the truncation and compensates jumps that were previously left uncompensated?