Quadratic Variation of Jump Processes

Why Jumps Add Their Own Variation

For a continuous Brownian-driven process, quadratic variation records the accumulated variance generated by its diffusion. A jump process contributes in a different way: a single jump of size \(z\) contributes \(z^2\), no matter how finely the time interval containing that jump is subdivided. Thus, for a process with both components, quadratic variation has a continuous part and a sum over actual jumps. Covariation has the corresponding cross-product of simultaneous jumps.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). A Brownian integrand \(H_s\) below is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Its Ito integral is \(\int_0^t H_s\,dB_s\). We use the standard continuous Ito quadratic-variation result: this integral has quadratic variation \(\int_0^t H_s^2\,ds\). The result is the continuous part of the calculation; here we derive how finitely many jumps extend it.

Consider processes of the form \(X_t=X_0+A_t+\int_0^t H_s\,dB_s+J_t\), where \(A\) is continuous and has finite variation on every bounded time interval, and \(J\) has finitely many jumps on each such interval and is constant between jumps. The initial value \(X_0\) does not affect quadratic variation. The finite-activity assumption means precisely that the number of jumps on every bounded interval is finite almost surely. At a jump time, \(\Delta X_s=\Delta J_s\), because the other terms are continuous.

The Quadratic-Variation Formula

Proof. First consider the continuous part \(C^X_t=A_t+\int_0^tH_s\,dB_s\). Its quadratic variation is \(\int_0^tH_s^2\,ds\). Indeed, the continuous finite-variation part has zero quadratic variation, and its cross sums with the continuous Ito integral vanish by Cauchy–Schwarz: the sum of squared increments of the finite-variation part tends to zero, while the quadratic-variation sums of the Ito integral are bounded in probability. This uses the standard continuous Ito quadratic-variation result stated above.

Now take any sequence of partitions whose mesh tends to zero. Since there are finitely many jumps on \([0,t]\), sufficiently fine partitions put distinct jump times in distinct partition intervals. In an interval containing a jump at \(s\), the increment of \(J\) is \(\Delta J_s\); intervals with no jump have zero increment of \(J\). Consequently the sum of squared increments of \(J\) is eventually \(\sum_{0<s\le t}(\Delta J_s)^2\). In the cross sum between \(C^X\) and \(J\), only intervals containing jumps can contribute. There are finitely many such intervals, and the increment of the continuous path \(C^X\) in each tends to zero as the mesh tends to zero. Hence that cross sum tends to zero. Expanding the square of each increment of \(C^X+J\) gives the asserted formula for \([X]\).

For covariation, apply the quadratic-variation formula to \(X+Y\) and \(X-Y\), then use polarization. The continuous terms satisfy \(\tfrac14((H+K)^2-(H-K)^2)=HK\); the jump terms satisfy \(\tfrac14((\Delta X+\Delta Y)^2-(\Delta X-\Delta Y)^2)=\Delta X\,\Delta Y\). This proves the covariation formula, including when one or both jump sizes are zero. \(\square\)

The shorthand multiplication rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) describe the continuous contributions. They do not replace the jump term: at a jump, the contribution to quadratic variation is the square of the full jump size. In particular, \([B]_t=t\), whereas a finite-activity pure-jump process \(J\) has \([J]_t=\sum_{s\le t}(\Delta J_s)^2\).

Covariation and Shared Jumps

Covariation records alignment as well as size. Continuous Brownian components driven by the same \(B_t\) contribute the product of their integrands integrated over time. Jumps contribute only when both processes jump at the same time, and the contribution is the product of their jump sizes. A jump in only one process contributes nothing to their covariation.

Compound Poisson Expectations

For a compound Poisson process \(J_t=\sum_{k=1}^{N_t}Y_k\), assume \(N_t\) has rate \(\lambda>0\), the jump sizes \(Y_k\) are independent and identically distributed, and they are independent of \(N_t\). If \(\mathbb{E}[Y_1^2]<\infty\), then the pathwise jump contribution is \([J]_t=\sum_{k=1}^{N_t}Y_k^2\). Conditioning on \(N_t=n\), the expectation of this sum is \(n\mathbb{E}[Y_1^2]\). Averaging over the Poisson count, whose mean is \(\lambda t\), gives \(\mathbb{E}[J]_t=\lambda t\,\mathbb{E}[Y_1^2]\). For \(X_t=\int_0^tH_s\,dB_s+J_t\), if also \(\mathbb{E}\int_0^tH_s^2\,ds<\infty\), the quadratic-variation formula and expectation yield \(\mathbb{E}[X]_t=\mathbb{E}\int_0^tH_s^2\,ds+\lambda t\,\mathbb{E}[Y_1^2]\).

Common Pitfall: Replacing Realized Jumps by Their Mean

A common error is to replace a Poisson counting process by its mean rate when computing quadratic variation. For a unit-jump Poisson process \(N_t\), the correct formula is \([N]_t=N_t\): every realized jump contributes \(1^2=1\). Writing \([N]_t=\lambda t\) confuses the random realized count with its expectation. For example, with rate \(1\), at time \(2\) on a path with \(N_2=3\), the correct value is \([N]_2=3\), while the incorrect substitution gives \(\lambda t=1\cdot2=2\). The latter is \(\mathbb{E}[N_2]\), not the pathwise quadratic variation.

Likewise, the ordinary continuous multiplication table cannot be applied to a jump increment by pretending its square vanishes. For a unit jump, \((\Delta N)^2=1\), not \(0\). Jump sizes must be squared first and then summed over the actual jump times. The expectation formula is useful for averages, but it does not replace the pathwise formula.