From a Linear Equation to a Jump Multiplier
A linear diffusion changes continuously between jumps, but a multiplicative jump changes the current state by a factor at the instant it occurs. Solving the equation therefore requires tracking both effects: the Brownian exponential between jumps and the multiplier applied at each jump. Compensation must also be handled carefully, because replacing a Poisson process by its compensated version changes the drift.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) and an independent Poisson process \(N_t\) of rate \(\lambda\ge0\). Write \(\widetilde N_t=N_t-\lambda t\). The Brownian integrals below are Ito integrals: for a progressively measurable integrand \(H\), \(\int_0^T H_s\,dB_s\) is defined when \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The compensated Poisson integral \(\int_0^T H_s\,d\widetilde N_s\) is defined for predictable \(H\) with \(\mathbb{E}\int_0^T H_s^2\lambda\,ds<\infty\). Raw Poisson integrals here are finite jump sums on bounded time intervals.
Consider constants \(a,b,c\in\mathbb{R}\), initial value \(x_0\), and the compensated linear equation \(dX_t=aX_{t-}\,dt+bX_{t-}\,dB_t+cX_{t-}\,d\widetilde N_t\), with \(X_0=x_0\). The drift is \(aX_{t-}\) in this compensated form. Since \(d\widetilde N_t=dN_t-\lambda\,dt\), the same equation in raw Poisson form is \(dX_t=(a-\lambda c)X_{t-}\,dt+bX_{t-}\,dB_t+cX_{t-}\,dN_t\). At a jump, the last term produces \(\Delta X_t=cX_{t-}\), hence the multiplier \(1+c\).
The Explicit Solution
Proof. Between Poisson jumps the raw-form equation is \(dX_t=(a-\lambda c)X_t\,dt+bX_t\,dB_t\). Put \(\kappa=a-\lambda c\) and define \(Y_t=\exp((\kappa-b^2/2)t+bB_t)\) on an interval with no jump. Ito's formula for \(f(t,x)=\exp((\kappa-b^2/2)t+bx)\) gives \(f_t=(\kappa-b^2/2)f\), \(f_x=bf\), and \(f_{xx}=b^2f\). Including the Ito correction explicitly, $$ dY_t=f_t\,dt+f_x\,dB_t+\frac12 f_{xx}\,dt =\left(\kappa-\frac{b^2}{2}\right)Y_t\,dt+bY_t\,dB_t+\frac12b^2Y_t\,dt =\kappa Y_t\,dt+bY_t\,dB_t. $$ Thus the continuous factor has exactly the required between-jump dynamics. At a jump time of \(N\), multiplying by \((1+c)^{N_t}\) changes the state from its left limit by the factor \(1+c\), so \(\Delta X_t=cX_{t-}\). These continuous dynamics and jump updates give the raw-form equation, and therefore also its compensated form. The formula starts at \(x_0\). For uniqueness, \(N\) has finitely many jumps on each bounded interval almost surely. Between successive jumps, the linear Brownian equation has a unique solution: multiplying any solution by the reciprocal of its positive Brownian exponential and applying Ito's product rule gives a constant process. At each jump, the next value is forced to be \((1+c)\) times the preceding value. Induction over the finitely many jumps proves uniqueness, including \(c=-1\), when the process is set to zero at the first jump and remains zero thereafter. \(\square\)
The formula can also be read through the Doléans-Dade stochastic exponential. For a semimartingale \(M\), \(\mathcal{E}(M)\) denotes the solution of \(dZ_t=Z_{t-}\,dM_t\), \(Z_0=1\). Here take \(M_t=bB_t+c\widetilde N_t\). Its continuous Brownian part contributes \(\exp(bB_t-\tfrac12b^2t)\); its compensated jumps contribute \((1+c)^{N_t}\exp(-\lambda ct)\). Consequently $$ X_t=x_0e^{at}\mathcal{E}(bB+c\widetilde N)_t =x_0\exp\!\left(\left(a-\lambda c-\frac{b^2}{2}\right)t+bB_t\right)(1+c)^{N_t}. $$ This representation does not by itself assert that \(X\) is a martingale. A martingale claim requires separate integrability and conditional-expectation checks.
For constant coefficients the displayed process has finite second moments on bounded intervals: the Brownian exponential has finite moments, and a Poisson random variable has finite exponential moments. In particular, \(\mathbb{E}\int_0^T X_s^2\,ds<\infty\), so the Brownian integrand \(bX_{s-}\) and compensated Poisson integrand \(cX_{s-}\) meet the stated square-integrability requirements. The formulas below are therefore ordinary square-integrable integrals on each finite horizon.
Worked Example: A Raw Poisson Equation
Solve \(dX_t=\tfrac14X_{t-}\,dt+\tfrac12X_{t-}\,dB_t+2X_{t-}\,dN_t\), where \(N\) has rate \(\tfrac32\) and \(X_0=2\). Then evaluate the explicit expression at a time \(t=1\) for a path with \(B_1=0\) and \(N_1=1\).
The raw drift is \(a-\lambda c=\tfrac14-\tfrac32\cdot2=\tfrac14-3=-\tfrac{11}{4}\). The Ito exponent has the additional correction \(-\tfrac12b^2=-\tfrac12(\tfrac12)^2=-\tfrac18\), so its time coefficient is \(-\tfrac{11}{4}-\tfrac18=-\tfrac{22}{8}-\tfrac18=-\tfrac{23}{8}\). The jump multiplier is \(1+c=3\). Therefore $$ X_t=2\exp\!\left(-\frac{23}{8}t+\frac12B_t\right)3^{N_t}. $$
For the specified path values, substitution gives \(X_1=2e^{-23/8}e^0 3^1=6e^{-23/8}\). To check the dynamics, between jumps Ito's formula gives drift \(-\tfrac{11}{4}X_t\,dt\) and diffusion \(\tfrac12X_t\,dB_t\), exactly the raw equation's continuous terms. At a jump, the factor \(3^{N_t}\) increases by \(3\), so \(X_t=3X_{t-}\) and \(\Delta X_t=2X_{t-}\), exactly the coefficient of \(dN_t\).
Worked Example: Reading the Compensated Drift
Solve \(dX_t=X_{t-}\,dt+X_{t-}\,d\widetilde N_t\), where \(N\) has rate \(2\) and \(X_0=1\). Check the result on an interval with no jumps and at a jump.
Here \(a=1\), \(b=0\), \(c=1\), and \(\lambda=2\). The raw-form drift is \(a-\lambda c=1-2=-1\), and each jump multiplies the state by \(1+c=2\). The solution is $$ X_t=e^{-t}2^{N_t}. $$ Between jumps \(N_t\) is constant, so ordinary differentiation gives \(dX_t=-X_t\,dt\). In raw form, the compensated equation is \(dX_t=X_{t-}\,dt+X_{t-}(dN_t-2dt)=-X_{t-}\,dt+X_{t-}\,dN_t\), which has precisely that between-jump drift. At a jump, \(2^{N_t}\) doubles, as required. For example, at \(t=1\) with \(N_1=2\), direct substitution gives \(X_1=e^{-1}2^2=4/e\); the jump multiplier applied twice to the between-jump factor \(e^{-1}\) gives the same value \(4/e\).
Worked Example: A Jump That Sends the State to Zero
Take \(dX_t=-X_{t-}\,d\widetilde N_t\), with rate \(\lambda=1\) and \(X_0=5\). Find the solution and check what happens at the first jump.
The coefficients are \(a=0\), \(b=0\), and \(c=-1\). Thus \(a-\lambda c=0-1(-1)=1\), while \(1+c=0\). The explicit formula, with \(0^0=1\), is $$ X_t=5e^t0^{N_t}. $$ Before the first jump \(N_t=0\), so \(X_t=5e^t\). At the first jump the multiplier is zero, and \(X_t=0\). It remains zero at later times. Independently, expanding the compensated equation gives \(dX_t=X_{t-}\,dt-X_{t-}\,dN_t\). Between jumps this yields \(dX_t=X_t\,dt\), hence \(5e^t\) before the first jump; at a jump it yields \(\Delta X_t=-X_{t-}\), hence the post-jump value is zero. This verifies both parts of the formula.
Compensation, Sign, and Positivity
The distinction between the two equation forms is easy to lose because both use the same jump coefficient. The coefficient \(c\) determines the relative jump \(cX_{t-}\), but the drift depends on whether the Poisson process is compensated.
| Equation form | Drift between jumps | Jump multiplier |
|---|---|---|
| \(dX=aX_-dt+bX_-dB+cX_-dN\) | \(aX_-\) | \(1+c\) |
| \(dX=aX_-dt+bX_-dB+cX_-d\widetilde N\) | \((a-\lambda c)X_-\) | \(1+c\) |
If \(x_0>0\) and \(c>-1\), every jump multiplier is positive, so the explicit solution remains positive. If \(c=-1\), the first jump sends the process to zero, where it stays. If \(c<-1\), the multiplier is negative, and each jump reverses the sign; positivity is not preserved. These statements follow directly from the solution and do not require taking a logarithm of \(1+c\).
Common Pitfall: Keeping the Wrong Drift
A frequent error is to use the compensated-equation parameter \(a\) as the between-jump drift. For the second example, this would produce the incorrect candidate \(e^t2^{N_t}\). On a no-jump interval its derivative is \(e^t2^{N_t}=X_t\), so it has drift \(+X_t\,dt\). But the given equation expands to \(dX_t=-X_{t-}\,dt+X_{t-}\,dN_t\), whose drift between jumps is \(-X_t\,dt\). The correct candidate \(e^{-t}2^{N_t}\) has derivative \(-e^{-t}2^{N_t}=-X_t\) between jumps and doubles at each jump. The compensation correction \(-\lambda c\) is essential.
The other common error is to use the continuous Ito exponential alone and omit the jump factor. The factor \(\exp((a-\lambda c-\tfrac12b^2)t+bB_t)\) describes evolution between jumps; it does not produce \(\Delta X_t=cX_{t-}\). That jump condition is supplied by \((1+c)^{N_t}\). The jump multiplier is \(1+c\), not \(c\), because \(cX_{t-}\) is the increment, not the post-jump state.
Explicit solutions are useful because they separate continuous fluctuations, deterministic drift adjustment, and jump-by-jump state changes. They also provide candidates that can be checked directly with Ito's formula and the jump update. Whether such a process is a true martingale, and which moments it has, are separate questions; the formula alone should not be used to assume either property.
Check Your Understanding
For each question, distinguish the compensated drift from the jump multiplier before using the solution formula.
- Write the raw Poisson form of \(dX_t=\alpha X_{t-}\,dt+\beta X_{t-}\,dB_t+\gamma X_{t-}\,d\widetilde N_t\), and state the drift between jumps.
- For a compensated equation with \(a=0\), \(b=0\), \(c=\tfrac12\), and rate \(\lambda=4\), find the time coefficient in the exponential and the multiplier at each jump.
- Explain why \(c=-1\) requires no logarithm in the explicit formula, and describe the state after the first jump.
- For the raw equation \(dX_t=3X_{t-}\,dt+X_{t-}\,dB_t-X_{t-}\,dN_t\), identify the jump multiplier and say whether a positive initial value remains positive after a jump.
- Check a proposed explicit solution by computing both its between-jump Ito differential and its jump increment. What must those two computations match?