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Jump Processes and Levy Processes · Tutorial 458 of 600

Generators of Jump-Diffusion Processes

Learn how drift, diffusion, and jump compensation each contribute to a jump-diffusion generator and to its Dynkin identity.

Advanced 12 min read

What You'll Learn

  • Derive the generator for a jump-diffusion with finite-activity jumps.
  • Distinguish uncompensated jump sums from compensated jump integrals.
  • Explain why compensation produces a first-derivative correction in the generator.
  • State conditions that make the Dynkin identity valid.
  • Compute generators on test functions for finite and compensated jump models.

From the SDE to the Generator

A generator records the instantaneous expected change of a test function of a process. For a diffusion, it contains a drift term and a second-derivative term. When jumps are present, the change at a jump is generally not well described by a derivative: a jump can move the process a finite distance, so the generator must account for the full difference between the function before and after the jump. Whether the jump integral is compensated determines how that difference is written.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) and a Poisson random measure \(N(dt,dz)\) on a mark space \(E\), with compensator \(dt\,\nu(dz)\). Write \(\widetilde N(dt,dz)=N(dt,dz)-dt\,\nu(dz)\). Assume \(B_t\) and \(N\) have the usual independent-increment structure relative to the filtration. Brownian integrals below are Itô integrals with progressively measurable integrands \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Compensated Poisson integrals use predictable integrands \(G\) satisfying \(\mathbb{E}\int_0^T\int_E G_s(z)^2\,\nu(dz)\,ds<\infty\). For finite-activity, uncompensated integrals, the predictable integrands are assumed to have finite expected absolute jump sum.

Definition: For a time-homogeneous Markov process \(X_t\) with transition semigroup \(P_t f(x)=\mathbb{E}[f(X_t)\mid X_0=x]\), its generator on functions for which the limit exists is \(\mathcal{A}f(x)=\lim_{h\downarrow0}(P_hf(x)-f(x))/h\). The corresponding extended generator is the drift operator \(\mathcal{L}\) in Itô’s formula: \(f(X_t)-f(X_0)-\int_0^t\mathcal{L}f(X_s)\,ds\) is a local martingale. Under the integrability conditions stated below, it is a true martingale and gives Dynkin’s identity.

Finite-Activity Jumps

Consider a solution with initial state \(X_0=x\) of \(dX_t=b(X_{t-})\,dt+\sigma(X_{t-})\,dB_t+\int_E\gamma(X_{t-},z)\,N(dt,dz)\). Here \(b\) is the drift coefficient, \(\sigma\) is the diffusion coefficient, and \(\gamma(y,z)\) is the jump size from state \(y\) at mark \(z\). Suppose \(\nu(E)<\infty\), and assume the coefficients and jump sizes are regular enough for this SDE and the jump version of Itô’s formula. For a twice continuously differentiable test function \(f\), Itô’s formula has drift terms \(f'b+\tfrac12 f''\sigma^2\), together with the actual function change \(f(X_{s-}+\gamma)-f(X_{s-})\) at each jump. The latter is not approximated by its linear part.

Theorem: For the finite-activity SDE above, the generator on a twice continuously differentiable test function is \(\mathcal{L}f(y)=b(y)f'(y)+\tfrac12\sigma(y)^2f''(y)+\int_E[f(y+\gamma(y,z))-f(y)]\,\nu(dz)\). If \(f\), \(f'\), and \(f''\) are bounded, the coefficients are bounded, and the displayed terms are integrable, then \(\mathbb{E}[f(X_t)\mid\mathcal{F}_s]=f(X_s)+\mathbb{E}[\int_s^t\mathcal{L}f(X_{u-})\,du\mid\mathcal{F}_s]\) for \(0\le s\le t\). This is the Dynkin identity.

Proof. Apply Itô’s formula for jump-diffusions, with its derivatives displayed: the continuous drift contribution is \(f'(X_{u-})b(X_{u-})\,du\), and the continuous quadratic-variation contribution is \(\tfrac12 f''(X_{u-})\sigma(X_{u-})^2\,du\), since \(dB_u\,dB_u=du\), \(dB_u\,du=0\), and \(du\,du=0\). The jump contribution is \(\int_E[f(X_{u-}+\gamma(X_{u-},z))-f(X_{u-})]N(du,dz)\). Thus the remaining terms are the Brownian integral with integrand \(f'(X_{u-})\sigma(X_{u-})\) and the compensated Poisson integral with integrand \(f(X_{u-}+\gamma(X_{u-},z))-f(X_{u-})\), after replacing \(N\) by \(\widetilde N+du\,\nu(dz)\). Under the stated boundedness and integrability assumptions, the Brownian integrand is in \(\mathcal{L}^2\), and the compensated jump integrand is square-integrable; both integrals are martingales. Taking conditional expectations therefore removes their increments and gives the Dynkin identity. Dividing the identity from \(0\) to \(h\) by \(h\), and using continuity in probability of \(X\) at zero together with continuity and local boundedness of \(\mathcal{L}f\), gives \(\lim_{h\downarrow0}(\mathbb{E}_x f(X_h)-f(x))/h=\mathcal{L}f(x)\). This identifies the generator wherever these conditions hold. \(\square\)

Worked Example: Constant-Size Poisson Jumps

Let \(dX_t=dt+2\,dB_t+2\,dN_t\), where \(N_t\) is a Poisson counting process of rate \(3\), independent of \(B_t\). Find the generator for \(f(x)=x^2\), and evaluate it at \(x=1\).

The derivatives are \(f'(x)=2x\) and \(f''(x)=2\). The drift contribution is \(1\cdot2x=2x\). The diffusion contribution is \(\tfrac12\cdot 2^2\cdot2=4\). A jump takes \(x\) to \(x+2\), so its change in \(f\) is \((x+2)^2-x^2=4x+4\). Multiplication by the jump rate gives \(3(4x+4)=12x+12\). Consequently \(\mathcal{L}f(x)=14x+16\), and \(\mathcal{L}f(1)=14+16=30\).

Check: the same value follows by adding the separate contributions at \(x=1\): drift \(2\), diffusion \(4\), and jumps \(3[(1+2)^2-1^2]=3(9-1)=24\). Their sum is \(2+4+24=30\).

Why Compensation Changes the Formula

Now consider an SDE with a compensated jump integral: \(dX_t=b(X_{t-})\,dt+\sigma(X_{t-})\,dB_t+\int_E\gamma(X_{t-},z)\,\widetilde N(dt,dz)\). Assume the last integral is defined in the square-integrable sense; in particular, require \(\mathbb{E}\int_0^T\int_E\gamma(X_{s-},z)^2\,\nu(dz)\,ds<\infty\). For the generator formula below, suppose also that \(\int_E\gamma(y,z)^2\,\nu(dz)<\infty\) for the states considered and that \(f\) has bounded first and second derivatives. Taylor’s formula then bounds the compensated remainder by a constant times \(\gamma^2\), making the jump part of the generator integrable.

Theorem: Under these conditions, the compensated-jump generator is \(\mathcal{L}f(y)=b(y)f'(y)+\tfrac12\sigma(y)^2f''(y)+\int_E[f(y+\gamma(y,z))-f(y)-f'(y)\gamma(y,z)]\,\nu(dz)\). If the Brownian and compensated Poisson terms in Itô’s formula are square-integrable and \(\mathcal{L}f(X_s)\) is integrable on bounded time intervals, the Dynkin identity holds with this \(\mathcal{L}\).

Proof. In Itô’s formula, the continuous terms are \(f'(X_{u-})b(X_{u-})\,du\) and \(\tfrac12f''(X_{u-})\sigma(X_{u-})^2\,du\); these use the explicit derivatives \(f'\) and \(f''\) and the Brownian correction \(dB_u\,dB_u=du\). The jump change is represented by integrating \(f(X_{u-}+\gamma)-f(X_{u-})\) against \(N\). Replacing \(N\) by \(\widetilde N+du\,\nu(dz)\) gives a compensated integral and the drift term \(\int_E[f(X_{u-}+\gamma)-f(X_{u-})]\,\nu(dz)\,du\). However, the SDE itself already has the continuous drift \(b\,du\) and the jump martingale \(\int\gamma\,\widetilde N\). The linear jump part in the Itô decomposition is therefore carried by the compensated integral as \(\int f'(X_{u-})\gamma(X_{u-},z)\,\widetilde N(du,dz)\). Subtracting this linear part from the compensator drift leaves precisely \(\int_E[f(y+\gamma)-f(y)-f'(y)\gamma]\,\nu(dz)\). The Brownian martingale integrand is \(f'(X_{u-})\sigma(X_{u-})\); the compensated Poisson martingale integrand is \(f(X_{u-}+\gamma)-f(X_{u-})\). The stated square-integrability assumptions make both true martingales, so taking conditional expectations gives Dynkin’s identity. The remainder is integrable because Taylor’s formula gives \(|f(y+\gamma)-f(y)-f'(y)\gamma|\le\tfrac12\|f''\|_\infty\gamma^2\). \(\square\)

The subtraction of \(f'(y)\gamma\) is essential: it reflects that the jump integral has already been centered by subtracting its compensator. It is not an arbitrary correction. For small jumps, the difference is of second order in the jump size, which is why a square-integrability condition on \(\gamma\) can suffice even when \(\int_E|\gamma|\,\nu(dz)\) is infinite.

Worked Example: A Compensated Jump Generator

Consider \(dX_t=dt+dB_t+\int_E \tfrac12 X_{t-}\,\widetilde N(dt,dz)\), where \(\nu(E)=2\) and \(\gamma(x,z)=x/2\) for every mark \(z\). Compute \(\mathcal{L}f\) for \(f(x)=x^2\), then evaluate it at \(x=2\).

Again \(f'(x)=2x\), \(f''(x)=2\). The drift term is \(2x\), and the diffusion term is \(\tfrac12\cdot1^2\cdot2=1\). For each mark, the jump correction is \((x+x/2)^2-x^2-(2x)(x/2)=\tfrac94x^2-x^2-x^2=\tfrac14x^2\). Integrating over the mark measure multiplies this by \(\nu(E)=2\), giving \(\tfrac12x^2\). Thus \(\mathcal{L}f(x)=2x+1+\tfrac12x^2\), and \(\mathcal{L}f(2)=4+1+\tfrac12\cdot4=7\).

Check: at \(x=2\), a jump has size \(1\), so the correction per mark is \(3^2-2^2-(4)(1)=9-4-4=1\). The total jump contribution is \(2\cdot1=2\); adding the drift contribution \(4\) and diffusion contribution \(1\) again gives \(7\).

Dynkin’s Identity and Its Scope

The Dynkin identity converts a generator calculation into an expectation statement. In unconditional form, when \(X_0=x\), it reads \(\mathbb{E}_x[f(X_t)]-f(x)=\mathbb{E}_x[\int_0^t\mathcal{L}f(X_{s-})\,ds]\). In conditional form it applies between times \(s\) and \(t\). The identity is valid only when the local martingale from Itô’s formula is a true martingale, for example when its Brownian and compensated Poisson integrands are square-integrable. Without that condition, taking expectations of a local martingale need not preserve its value, so the identity cannot be asserted merely from the formal generator expression.

For an infinite-activity compensated measure, the jump integral in the generator should be understood as the combined remainder \(f(y+\gamma)-f(y)-f'(y)\gamma\), not as three separate integrals unless each is individually finite. If the measure has a finite first moment, algebraic rearrangement may sometimes express the same operator using a different drift coefficient. The drift convention in the SDE must be kept fixed when writing the generator: changing from compensated to uncompensated jumps changes the drift as well.

Worked Example: Checking the Compensation Correction

Let \(dX_t=\int_E 1\,\widetilde N(dt,dz)\), where \(\nu(E)=2\), and take \(f(x)=x^2\). Find the generator at \(x=3\), and compare it with the result from the equivalent uncompensated description.

For the compensated equation, \(b=0\), \(\sigma=0\), and \(\gamma=1\). The correction is \((x+1)^2-x^2-2x\cdot1=1\), so \(\mathcal{L}f(x)=2\cdot1=2\), and \(\mathcal{L}f(3)=2\).

Since \(\nu(E)=2\) and the jump size is \(1\), the compensated equation can also be written \(dX_t=-2\,dt+\int_E1\,N(dt,dz)\). Its uncompensated generator has drift \(-2\) and jump rate \(2\). At \(x=3\), its drift contribution is \((-2)(2\cdot3)=-12\), and its jump contribution is \(2[(3+1)^2-3^2]=2(16-9)=14\). The total is \(-12+14=2\), agreeing with the compensated formula.

Check: the two drift conventions differ by \(-\int_E\gamma\,\nu(dz)=-2\), while the operators agree after that drift change. Omitting the derivative correction in the compensated formula would give a jump term \(2[(3+1)^2-3^2]=14\), which is not the generator of the compensated SDE.

Common Pitfall: Linearizing a Finite Jump

A common error is to replace the finite-activity jump change \(f(x+\gamma)-f(x)\) by \(f'(x)\gamma\). For \(f(x)=x^2\), \(x=1\), and jump size \(\gamma=2\), the exact change is \((1+2)^2-1^2=8\), whereas the linear approximation is \(f'(1)\cdot2=2\cdot2=4\). At jump rate \(3\), the correct jump contribution is \(3\cdot8=24\), not \(3\cdot4=12\). The linear term appears in the compensated formula only as part of the subtraction that centers the jump measure; it does not replace the full finite jump change.

Thus the generator’s jump term is determined both by the size of the function change and by the representation of the SDE. The Dynkin identity then supplies a direct interpretation: under its integrability conditions, the expected change in \(f(X_t)\) is the time integral of the expected generator value.

Check Your Understanding

Use the formulas and conditions above to answer each question.

  1. For an uncompensated jump of size \(\gamma(x,z)\), why does the generator use the full increment \(f(x+\gamma)-f(x)\)?
  2. For \(f(x)=x^2\), compute the compensated correction when \(\gamma(x,z)=cx\), and express it in terms of \(c\) and \(x\).
  3. State one sufficient integrability condition that makes a Brownian integral in Itô’s formula a true martingale.
  4. Explain how the drift changes when a finite-mean compensated jump integral is rewritten as an uncompensated integral.
  5. Why does a formal local-martingale decomposition alone not justify taking expectations in Dynkin’s identity?