Jump Calculus Pitfalls and Diagnostic Problems

Three Checks for Jump Calculus

Jump calculations can look formally similar to Brownian calculations while obeying different bookkeeping rules. At a jump time, the value just before the jump and the value just after it are distinct. A stochastic integral against a jump process uses the predictable, pre-jump integrand. A nonlinear function changes by its full increment across a jump, not merely by its derivative times the jump size. And a raw jump sum differs from a compensated sum by a compensator integral. Keeping these three distinctions explicit prevents many errors.

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 process $N_t$ of rate $\lambda>0$, whose increments are independent of the past. The Brownian integral $\int_0^t H_s\,dB_s$ is the Itô integral for progressively measurable $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For the Poisson process, $H$ is predictable (so its value at a jump is determined by information immediately before the jump). When $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, the compensated integral is defined by $\int_0^t H_s(dN_s-\lambda ds)=\int_0^t H_s\,dN_s-\lambda\int_0^t H_s\,ds$. In the finite-activity examples below, raw sums are also used when their terms are integrable.

Definition: For a process $X$ with left limits, $X_{t-}$ denotes its value immediately before $t$, and $\Delta X_t=X_t-X_{t-}$ is its jump. The raw integral $\int_0^t H_s\,dN_s$ sums $H_s$ over jump times of $N$; because $H$ is predictable, each term is evaluated just before the jump. The compensated integral subtracts the rate-weighted time integral $\lambda\int_0^t H_s\,ds$.

The Pre-Jump Rule and the Nonlinear Correction

Consider a finite-activity jump-diffusion $X_t=X_0+\int_0^t b_s\,ds+\int_0^t \sigma_s\,dB_s+\int_0^t \gamma_s\,dN_s$, where the coefficients used in the integrals are predictable. Assume the Brownian integrand is in $\mathcal{L}^2$ on each finite time interval, and the finite-activity jump terms are well-defined and integrable. Between the finitely many jumps on a bounded interval, $X$ evolves continuously. The exact jump in $X$ at a jump time $s$ of $N$ is $\Delta X_s=\gamma_s$.

Theorem: If $f$ is twice continuously differentiable and the displayed terms are integrable, then $$ f(X_t)=f(X_0)+\int_0^t f'(X_{s-})b_s\,ds+\int_0^t f'(X_{s-})\sigma_s\,dB_s+\frac12\int_0^t f''(X_{s-})\sigma_s^2\,ds+\int_0^t\bigl(f(X_{s-}+\gamma_s)-f(X_{s-})\bigr)\,dN_s. $$ At a jump, the last integrand is the full function increment. Equivalently, its jump contribution can be written as $f'(X_{s-})\gamma_s$ plus the nonlinear correction $f(X_{s-}+\gamma_s)-f(X_{s-})-f'(X_{s-})\gamma_s$.

Proof. On any interval containing no jump of $N$, $X$ has continuous dynamics $dX_s=b_s\,ds+\sigma_s\,dB_s$. The continuous Itô formula, with function $f$, derivatives $f'$ and $f''$, and correction $\tfrac12 f''(X_s)\sigma_s^2\,ds$, gives the first three integral terms over that interval. At a jump time $s$, the continuous part does not jump, so $X_s=X_{s-}+\gamma_s$. The change in $f(X)$ at that time is exactly $f(X_{s-}+\gamma_s)-f(X_{s-})$. Add the continuous Itô identities over the finitely many intervals between jumps and add these exact jump increments. The resulting identity is the displayed formula. Adding and subtracting $f'(X_{s-})\gamma_s$ gives the equivalent correction form. $\square$

This formula uses the jump multiplication rules $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ for the continuous part, but it does not replace a jump by an infinitesimal Taylor approximation. At a jump, the nonlinear remainder is generally nonzero.

Worked Example: Squaring a Process with Unit Jumps

Let $X_t=1+N_t$ with $N_0=0$, and take $f(x)=x^2$. Compute the exact jump contribution to $X_t^2$, and compare it with the derivative-only calculation.

Here $f'(x)=2x$, $f''(x)=2$, and each jump has size $\gamma=1$. Immediately before the $k$-th jump, $X_{s-}=k$, so the exact change in $f$ is $(k+1)^2-k^2=2k+1$. The derivative-only term gives $f'(k)\gamma=2k$; the nonlinear correction is $(k+1)^2-k^2-2k=1$.

After $n$ jumps, summing the exact increments gives $1+\sum_{k=1}^n(2k+1)=1+n(n+1)+n=(n+1)^2$. The derivative-only sum instead gives $1+\sum_{k=1}^n2k=1+n(n+1)$, which is short by $n$.

Check: for $n=2$, the path ends at $X=3$, so $X^2=9$. The exact increments are $3$ and $5$, giving $1+3+5=9$; the derivative-only increments are $2$ and $4$, giving $1+2+4=7$. The missing amount is the two unit corrections.

Raw Sums and Compensated Sums

Proposition: For a predictable $H$ for which the integrals exist, $$ \int_0^t H_s\,dN_s = \int_0^t H_s(dN_s-\lambda ds)+\lambda\int_0^t H_s\,ds. $$ If $H$ is square-integrable, the compensated integral is a true martingale on $[0,T]$: it is integrable and its conditional expectation at a later time, given the current filtration, equals its current value.

Proof. The identity follows directly from the definition of the compensated integral. For the martingale assertion, first take a predictable step process $H_s=\sum_j H_j\mathbf{1}_{(t_j,t_{j+1}]}(s)$, where each $H_j$ is bounded and $\mathcal{F}_{t_j}$-measurable. On a step interval, the compensated increment is $H_j(N_v-N_u-\lambda(v-u))$. The Poisson increment is independent of $\mathcal{F}_u$ and has mean $\lambda(v-u)$. Thus its conditional mean is zero. Adding the step increments proves the martingale conditional-expectation identity. The compensated Poisson isometry gives a second moment $\mathbb{E}\int_0^T H_s^2\lambda\,ds$; approximating a square-integrable predictable integrand by predictable step processes therefore extends the martingale identity in $L^2$. In particular, the integral is integrable, and the identity holds for all $0\le u\le v\le T$. $\square$

The raw integral is not generally a martingale: its conditional mean increment includes $\lambda\mathbb{E}[\int_u^v H_s\,ds\mid\mathcal{F}_u]$. The compensator removes this predictable drift. Do not replace a raw sum with a compensated integral without also retaining the compensator term.

Worked Example: A Raw Sum Is Not a Compensated Sum

Let $N$ have rate $\lambda=2$, let $H_s=1$, and consider $T=1$. Suppose the observed path has three jumps by time $1$. Compute the raw, compensator, and compensated values.

The raw integral is the number of jumps: $\int_0^1 1\,dN_s=N_1=3$. The compensator is $\lambda\int_0^1 1\,ds=2\cdot1=2$. Therefore the compensated value on this path is $3-2=1$.

Check: the compensated process is $N_t-2t$. Its expectation at $t=1$ is $\mathbb{E}[N_1]-2=2-2=0$, as required by the martingale property. The particular observed value $1$ is not an expectation; it is one path’s realized compensated increment.

Diagnostic Problems: Track the Value at the Jump

Worked Example: Detecting a Post-Jump Integrand Error

Suppose a path of a rate-$\lambda$ Poisson process has two jumps before $T$, and let $H_s=N_{s-}$. Compute the raw integral using the predictable value, then compare it with the incorrect post-jump sum.

Before the first jump, $N_{s-}=0$; before the second, $N_{s-}=1$. Thus $\int_0^T N_{s-}\,dN_s=0+1=1$. If one incorrectly uses the value after each jump, the two terms are $1$ and $2$, giving $3$. The discrepancy is $3-1=2$, one extra unit at each jump.

In general, at a unit jump, $N_s=N_{s-}+1$. Summing the post-jump terms instead of the predictable terms adds $1$ per jump, so the incorrect sum is $\int_0^T N_{s-}\,dN_s+N_T$. Check: for exactly two jumps this gives $1+2=3$, matching the direct post-jump calculation. The error is not a different convention for the same integral; it is a different random sum.

A reliable diagnostic is to write down the value immediately before each jump before evaluating any jump integral. For a predictable integrand $H$, the jump contribution to $\int H\,dN$ at time $s$ is $H_s\Delta N_s$, with $H_s$ determined by the information just before $s$. In a process written with left limits, this is often displayed as $H_{s-}$.

Freeze the pre-jump state.
Record $X_{s-}$, $H_s$, and the jump size before updating the state.

Update the state.
Use $X_s=X_{s-}+\Delta X_s$ to evaluate any nonlinear function exactly.

Separate raw and compensated parts.
Write the raw sum as the compensated integral plus its compensator; do not discard either term.

Common Pitfalls and a Compact Audit

Wrong: For $X=1+N$, write the jump contribution to $d(X^2)$ as $2X_{s-}\,dN_s$, because $f'(x)=2x$. This omits the nonlinear correction. Correction: the exact increment is $(X_{s-}+1)^2-X_{s-}^2=2X_{s-}+1$. The missing $1$ occurs at every jump.

Wrong: Treat $\int H\,dN$ and $\int H(dN-\lambda ds)$ as interchangeable. Correction: the former equals the latter plus $\lambda\int H\,ds$. The compensated integral is a martingale under the stated square-integrability condition; the raw integral generally has a nonzero conditional drift.

Wrong: Evaluate an integrand at the state after the jump it is multiplying. Correction: stochastic jump integrals use predictable pre-jump values. To test a proposed identity, evaluate it on a path with one jump: if the integrand changes because of that same jump, the proposed post-jump sum is not the predictable integral.

Object	At a jump time $s$	Diagnostic question
Predictable jump integral	Uses the pre-jump integrand	Was the integrand fixed before the jump?
Nonlinear function change	$f(X_s)-f(X_{s-})$	Did the calculation retain the full increment?
Compensated integral	Raw sum minus compensator	Was the compensator added back when converting to a raw sum?

Check Your Understanding

In each problem, state the pre-jump value or compensator explicitly before simplifying.

If $X_t=2+N_t$ and $f(x)=x^2$, what is the exact change in $f(X)$ at a unit jump from a pre-jump state $x$? What is the nonlinear correction beyond $f'(x)$?
On a path with exactly three jumps, what is $\int_0^T N_{s-}\,dN_s$? Compare it with the sum formed using post-jump values.
For a rate-$4$ Poisson process on $[0,2]$, compute the compensator of $\int_0^2 1\,dN_s$. If the observed count is $5$, what is the compensated value?
Explain why square-integrability of a predictable $H$ is relevant to the true-martingale assertion for $\int H(dN-\lambda ds)$.
Write the identity converting $\int_0^T H_s\,dN_s$ into a compensated integral and a time integral, and identify which term is the compensator.