Itô's Formula for Jump-Diffusions

Why Jumps Change Itô’s Formula

For a continuous diffusion, Itô’s formula accounts for the accumulation of small Brownian increments through a second-derivative correction. A jump is different: it can move the process a finite distance instantaneously. Its contribution to $f(X_t)$ is therefore the exact difference $f(X_{t-}+\Delta X_t)-f(X_{t-})$, not just the linear approximation $f'(X_{t-})\Delta X_t$. A jump-diffusion formula must retain both contributions: the Brownian second-order term and the full nonlinear jump increment.

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 random measure $N(dt,dz)$ with finite compensator $\nu(dz)\,dt$, where $\nu(dz)=\lambda F(dz)$, $\lambda<\infty$, and $F$ is a probability measure on the mark space. Write $\widetilde N(dt,dz)=N(dt,dz)-\nu(dz)\,dt$. The finite intensity ensures that there are almost surely finitely many jumps on each bounded time interval.

Definition: The uncompensated jump-diffusion convention is $dX_t=a_t\,dt+\sigma_t\,dB_t+\int \gamma_t(z)\,N(dt,dz)$. The compensated convention is $dX_t=b_t\,dt+\sigma_t\,dB_t+\int \gamma_t(z)\,\widetilde N(dt,dz)$. Here $\gamma_t(z)$ is the jump size at mark $z$. The displayed drift is the drift in that convention; in particular, the two drift coefficients need not agree.

Assume throughout the formula below that $a,b,\sigma$ are progressively measurable and bounded, $\gamma_t(z)$ is predictable and bounded, and the same jump-size function is used in the two conventions being compared. Take $f\in C^2(\mathbb{R})$ with $f'$ and $f''$ bounded. These conditions ensure the finite-horizon integrability needed here. In particular, the Brownian integrand $f'(X_{t-})\sigma_t$ is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T |f'(X_{t-})\sigma_t|^2dt<\infty$. The compensated jump integrands below are predictable and square-integrable against $\nu(dz)\,dt$; the uncompensated jump sums have finite expected absolute variation. Brownian integrals are Itô integrals.

The Formula in Both Conventions

Theorem: For the uncompensated equation, the Itô formula is $$ \begin{aligned} f(X_t)=f(X_0) &+\int_0^t f'(X_{s-})a_s\,ds +\int_0^t f'(X_{s-})\sigma_s\,dB_s +\frac12\int_0^t f''(X_{s-})\sigma_s^2\,ds\\ &+\int_0^t\int \bigl[f(X_{s-}+\gamma_s(z))-f(X_{s-})\bigr]\,N(ds,dz). \end{aligned} $$ For the compensated equation, it is $$ \begin{aligned} f(X_t)=f(X_0) &+\int_0^t f'(X_{s-})b_s\,ds +\int_0^t f'(X_{s-})\sigma_s\,dB_s +\frac12\int_0^t f''(X_{s-})\sigma_s^2\,ds\\ &+\int_0^t\int \bigl[f(X_{s-}+\gamma_s(z))-f(X_{s-})\bigr]\,\widetilde N(ds,dz)\\ &+\int_0^t\int \bigl[f(X_{s-}+\gamma_s(z))-f(X_{s-})-f'(X_{s-})\gamma_s(z)\bigr]\,\nu(dz)\,ds. \end{aligned} $$

Proof. First consider the uncompensated equation. Between jumps, $X$ follows the continuous equation $dX_s=a_s\,ds+\sigma_s\,dB_s$. Applying the continuous Itô formula on each interval between successive jumps gives the drift term $f'(X_{s-})a_s\,ds$, the Itô integral with integrand $f'(X_{s-})\sigma_s$, and the correction $\tfrac12 f''(X_{s-})\sigma_s^2\,ds$. At a jump with mark $z$, the state changes from $X_{s-}$ to $X_{s-}+\gamma_s(z)$, so the change in $f(X)$ is exactly $f(X_{s-}+\gamma_s(z))-f(X_{s-})$. Adding these changes over the finitely many jumps gives the integral against $N$ in the first formula. The assumed integrability makes each term well-defined.

For the compensated equation, the identity $\widetilde N=N-\nu(dz)\,ds$ rewrites its continuous drift between jumps as $b_s-\int\gamma_s(z)\nu(dz)$. Apply the same interval-by-interval argument. The jump sum is the integral of the full increment against $N$, while the continuous drift contributes \[ f'(X_{s-})\left(b_s-\int\gamma_s(z)\nu(dz)\right)ds. \] Now use $N=\widetilde N+\nu(dz)\,ds$ in the jump sum. The compensator part of that sum adds $\int [f(X_{s-}+\gamma_s(z))-f(X_{s-})]\nu(dz)\,ds$. Combining it with the drift term leaves $f'(X_{s-})b_s\,ds$ and the correction $\int[f(X_{s-}+\gamma_s(z))-f(X_{s-})-f'(X_{s-})\gamma_s(z)]\nu(dz)\,ds$, as stated. $\square$

The Brownian correction comes from $dB_t\,dB_t=dt$; the jump correction comes from using the exact finite increment. In the usual differential shorthand, $dB_t\,dt=0$ and $dt\,dt=0$. Unlike the Brownian second-order term, the jump remainder need not be negligible: its size depends on the full jump size.

Worked Exponential Calculations

Worked Example: An Uncompensated Poisson Jump

Let $N_t$ be a rate-one Poisson process, let $c=\log 2$, and set $X_t=t+B_t+cN_t$. Apply Itô’s formula to $f(x)=e^x$, then check the expectation independently.

Here $f'(x)=e^x$ and $f''(x)=e^x$. Between jumps, the drift is $1$ and the diffusion coefficient is $1$; a jump changes $X$ by $c$. The continuous drift and second-derivative contributions are $e^{X_{t-}}dt$ and $\tfrac12 e^{X_{t-}}dt$. The exact jump increment is $e^{X_{t-}+c}-e^{X_{t-}}=e^{X_{t-}}(2-1)=e^{X_{t-}}$, since $e^c=2$. Therefore $$ de^{X_t}=\frac32e^{X_{t-}}dt+e^{X_{t-}}dB_t+e^{X_{t-}}dN_t. $$

For the check, the explicit expression is $e^{X_t}=e^{t+B_t}2^{N_t}$. Independence of $B$ and $N$, together with $\mathbb{E}[e^{B_t}]=e^{t/2}$ and $\mathbb{E}[2^{N_t}]=\exp(t(2-1))=e^t$, gives $\mathbb{E}[e^{X_t}]=e^{5t/2}$. The formula gives the same mean: the Brownian integral has mean zero by square-integrability, and the Poisson jump sum has expectation $\int_0^t\mathbb{E}[e^{X_{s-}}]ds$ by the compensator identity. Thus the mean $m(t)$ satisfies $m(t)=1+\tfrac52\int_0^t m(s)ds$, whose solution is $m(t)=e^{5t/2}$. The integrals are square-integrable on bounded time intervals because the explicit expression has finite second moments there.

Worked Example: A Compensated Jump and Its Drift Correction

Let $N_t$ again have rate one, let $c=\log 2$, and define $X_t=B_t+c(N_t-t)$. Apply the compensated formula to $f(x)=e^x$.

The compensated SDE has drift $b=0$, diffusion coefficient $\sigma=1$, and jump size $c$. The derivatives are $f'(x)=e^x$ and $f''(x)=e^x$, so the Brownian correction is $\tfrac12e^{X_{t-}}dt$. At a jump, the increment is $e^{X_{t-}+c}-e^{X_{t-}}=e^{X_{t-}}$. The compensator remainder is \[ e^{X_{t-}+c}-e^{X_{t-}}-c e^{X_{t-}} =e^{X_{t-}}(1-c). \] Consequently $$ de^{X_t}=\left(\frac12+1-c\right)e^{X_{t-}}dt +e^{X_{t-}}dB_t+e^{X_{t-}}\widetilde N(dt). $$

Independently, $e^{X_t}=e^{B_t}2^{N_t}e^{-ct}$, so \[ \mathbb{E}[e^{X_t}] =e^{t/2}\exp(t(2-1))e^{-ct} =\exp\left(\left(\frac32-c\right)t\right). \] The drift coefficient in the formula is $\tfrac12+1-c=\tfrac32-c$, exactly the exponent in this expectation. The Brownian and compensated Poisson integrals have mean zero: each integrand is square-integrable on bounded intervals, as follows from the finite second moments of $e^{B_t}2^{N_t}e^{-ct}$. This check also displays why compensation changes the drift correction.

Worked Example: Two Jump Sizes in an Exponential Test

Let $N^+_t,N^-_t$ be independent rate-one Poisson processes, independent of $B_t$, and take $c=\log 2$. Consider \[ dX_t=dt+2\,dB_t+c(dN^+_t-dt)-c(dN^-_t-dt). \] The compensated jump drift terms cancel, so $X_t=t+2B_t+cN^+_t-cN^-_t$. Apply the formula to $e^{X_t}$.

Again $f'=f''=e^x$. The displayed drift contributes $e^{X_{t-}}dt$, and the Brownian correction is $\tfrac12(2^2)e^{X_{t-}}dt=2e^{X_{t-}}dt$. For a positive jump, the exact increment is $e^{X_{t-}}(2-1)=e^{X_{t-}}$; its compensator remainder is $e^{X_{t-}}(1-c)$. For a negative jump, the exact increment is $e^{X_{t-}}(\tfrac12-1)=-\tfrac12e^{X_{t-}}$; its remainder is $e^{X_{t-}}(\tfrac12-1+c)=e^{X_{t-}}(c-\tfrac12)$. Adding the two remainders gives $\tfrac12 e^{X_{t-}}$, since $(1-c)+(c-\tfrac12)=\tfrac12$. Thus the total drift coefficient is $1+2+\tfrac12=\tfrac72$, and $$ de^{X_t}=\frac72e^{X_{t-}}dt+2e^{X_{t-}}dB_t +e^{X_{t-}}\widetilde N^+(dt)-\frac12e^{X_{t-}}\widetilde N^-(dt). $$

For an independent check, the explicit expression gives \[ \mathbb{E}[e^{X_t}] =e^t e^{(2^2)t/2}\exp(t(2-1))\exp\bigl(t(\tfrac12-1)\bigr) =e^{(1+2+1-1/2)t}=e^{7t/2}. \] Here the four factors are respectively the deterministic drift, Brownian exponential moment, positive-jump exponential moment, and negative-jump exponential moment. The resulting exponent agrees with the formula. On bounded time intervals the exponential integrands have finite second moments, so both stochastic integrals are true square-integrable martingales and have mean zero.

Why the Convention Matters

Compensation is not merely a notational choice. Since $\int\gamma\,\widetilde N=\int\gamma\,N-\int\gamma\,\nu\,dt$, keeping the same displayed drift while changing from $N$ to $\widetilde N$ changes the actual process. The jump correction must change with it. In the compensated formula, the remainder $f(x+\gamma)-f(x)-f'(x)\gamma$ is what remains after the linear part of the jump is accounted for by the compensated integral.

Common mistake: using the uncompensated correction with a compensated SDE. In the second example, omitting the $-f'\gamma$ term would give drift coefficient $\tfrac12+(2-1)=\tfrac32$. That is wrong: the correct coefficient is $\tfrac32-\log 2$, as the direct expectation calculation confirms. The missing amount $\log 2$ is precisely the linear jump contribution subtracted when $N$ is replaced by $\widetilde N$.

Another tempting shortcut is to replace every jump increment by $f'(X_{t-})\Delta X_t$. That is only the first-order part of a Taylor expansion, not the actual change of $f(X)$. For exponential functions, the difference is visible in $e^{x+\gamma}-e^x$, which generally is not $e^x\gamma$. The exact increment is essential even when the process also has a Brownian component.

Check Your Understanding

For each question, keep track of whether the jump integral is compensated or uncompensated.

For an uncompensated jump of size $c$, write the exact increment of $f(X)$ at the jump and explain why it is not generally $f'(X_{-})c$.
For $f(x)=e^{ux}$, calculate $f'(x)$, $f''(x)$, and the compensated jump remainder for a jump of size $\gamma$.
If $dX_t=b\,dt+\sigma\,dB_t+c\,\widetilde N(dt)$ with rate $\lambda$, identify the compensator remainder in the formula for $e^{X_t}$.
Explain why the Brownian Itô integral and the compensated Poisson integral in the examples have mean zero, and state the integrability condition used.
In the two-mark example, recompute the positive- and negative-jump remainders and verify their sum is $\tfrac12e^{X_{t-}}$.

Itô's Formula for Jump-Diffusions

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