Itô's Formula for Finite-Activity Jump Processes

Why a Jump Needs Its Own Correction

For a continuous semimartingale, Itô’s formula adds a second-derivative term because the continuous martingale part has nonzero quadratic variation. A process with jumps requires an additional adjustment: at a jump, the actual change in $f(X)$ is $f(X_t)-f(X_{t-})$, which need not equal the linear approximation $f'(X_{t-})\Delta X_t$. For finitely many jumps, the exact difference can be recorded one jump at a time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. A continuous semimartingale $C$ is a continuous adapted process that can be written as a continuous local martingale plus a continuous finite-variation process. A semimartingale with finitely many jumps on $[0,T]$ can be written $X_t=C_t+\sum_{\tau_j\le t}\Delta X_{\tau_j}$, where $C$ is continuous, the $\tau_j$ are its distinct jump times, and there are almost surely finitely many such times on $[0,T]$. We use $\Delta X_s=X_s-X_{s-}$, where $X_{s-}$ is the left limit.

All integrals against a semimartingale below are semimartingale integrals. Their integrands are predictable; the integrands used here are locally bounded, which makes these integrals well-defined. If a continuous local martingale term is written as a Brownian integral $\int_0^t H_s\,dB_s$, the integral is an Itô integral: $H$ is adapted and progressively measurable, and $H\in\mathcal{L}^2$ when $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The local version requires local square integrability instead. An integral against a local martingale need not be a true martingale without further integrability. We use $[C]_t$ for the quadratic variation of the continuous semimartingale $C$; for $C=B$, $[B]_t=t$.

Definition: For a jump of size $z=\Delta X_s$ from the pre-jump value $x=X_{s-}$, the jump correction for $f$ is $f(x+z)-f(x)-f'(x)z$. It is the exact change in $f$ across the jump minus its first-order approximation.

Itô’s Formula with Finitely Many Jumps

Theorem: Suppose $X$ is a semimartingale with almost surely finitely many jumps on every bounded time interval, and write $X=C+\sum_{\tau_j\le \cdot}\Delta X_{\tau_j}$, with $C$ continuous. For $f\in C^2(\mathbb{R})$, almost surely, for every $t\in[0,T]$, $$ f(X_t)=f(X_0)+\int_0^t f'(X_{s-})\,dX_s +\frac12\int_0^t f''(X_{s-})\,d[C]_s +\sum_{0<s\le t}\bigl(f(X_s)-f(X_{s-})-f'(X_{s-})\Delta X_s\bigr). $$ The sum is over jump times and is finite on $[0,T]$. The integral with respect to $[C]$ is an ordinary Lebesgue–Stieltjes integral; the integral with respect to $X$ is a semimartingale integral.

Proof. Fix a path on which there are finitely many jumps on $[0,T]$, and list them in increasing order as $\tau_1,\ldots,\tau_n$. Between two consecutive jump times, $X$ is a continuous semimartingale: it equals $C$ plus a constant, the sum of the jumps already made. On each such interval, apply the standard Itô formula for a continuous semimartingale to the function $x\mapsto f(x)$ evaluated at $X$. This gives the integral of $f'(X)$ against the continuous part and the correction $\tfrac12\int f''(X)\,d[C]$. At a jump time $\tau_j$, add the exact increment $f(X_{\tau_j})-f(X_{\tau_j-})$. The integral $\int f'(X_{s-})\,dX_s$ includes the jump contribution $f'(X_{\tau_j-})\Delta X_{\tau_j}$. Therefore the part of the actual increment not already included in that integral is precisely $f(X_{\tau_j})-f(X_{\tau_j-})-f'(X_{\tau_j-})\Delta X_{\tau_j}$. Adding the continuous-interval identities and these finitely many jump increments yields the displayed formula. The continuous quadratic variation is that of $C$, since the finite-variation jump sum contributes no continuous quadratic variation. The argument applies on each bounded interval, and localization handles locally bounded derivatives and semimartingale integrals if needed. $\square$

This proof uses the continuous semimartingale version of Itô’s formula, together with finite addition and subtraction at each jump. It does not treat a jump as an infinitesimal increment: the correction is exact even when the jump is large.

Worked Examples

Worked Example: Squaring Brownian Motion Plus a Poisson Process

Let $B$ be Brownian motion and $N$ an independent Poisson counting process of rate $\lambda>0$, on the filtered space above, and set $X_t=B_t+aN_t$ for a constant $a$. Verify the formula for $f(x)=x^2$.

Here $C=B$, so $[C]_t=t$, and every jump of $X$ has size $a$. Since $f'(x)=2x$ and $f''(x)=2$, the formula gives

X_t^2=2\int_0^t X_{s-}\,dX_s+t+a^2N_t.

The jump correction is $(x+a)^2-x^2-2xa=a^2$, which explains the last term. Expanding the semimartingale integral, including both parts of $X_{s-}=B_s+aN_{s-}$, gives $2\int_0^t B_s\,dB_s+2a\int_0^t N_{s-}\,dB_s+2a\sum_{\tau_j\le t}(B_{\tau_j}+aN_{\tau_j-})+t+a^2N_t$. For unit jumps, $N_{\tau_j-}=j-1$.

To check the identity without dropping the continuous part of the product $B_tN_t$, integration by parts gives $B_tN_t=\int_0^t N_{s-}\,dB_s+\sum_{\tau_j\le t}B_{\tau_j}$. Also, the continuous Brownian formula gives $B_t^2=2\int_0^t B_s\,dB_s+t$, and $N_t^2=N_t+2\sum_{j=1}^{N_t}(j-1)$. Substituting these three identities into the expanded expression gives $B_t^2+2aB_tN_t+a^2N_t^2=(B_t+aN_t)^2$, as required.

The Brownian integral with integrand $N_{s-}$ is well-defined in $L^2$ on $[0,T]$: $\mathbb{E}\int_0^T N_{s-}^2\,ds=\int_0^T(\lambda s+(\lambda s)^2)\,ds<\infty$. Thus it is a square-integrable martingale, with conditional mean-zero increments; this fact follows from the Itô integral property and does not assert that every semimartingale integral in the theorem is a true martingale.

Worked Example: An Exponential Across a Single Jump

Take $X_t=t+(\log 2)\mathbf{1}_{\{t\ge1\}}$ on $[0,2]$ and $f(x)=e^x$. This is a finite-variation semimartingale with continuous part $C_t=t$ and one jump of size $\log 2$ at time $1$. There is no continuous quadratic-variation term. Compute the formula at $t=2$.

We have $X_0=0$, $X_{1-}=1$, and $X_2=2+\log 2$, so the left side is $e^{X_2}=2e^2$. The continuous part of $\int f'(X_{s-})\,dX_s$ is $\int_0^1 e^s\,ds+\int_1^2 2e^s\,ds=(e-1)+2(e^2-e)=2e^2-e-1$. Its jump contribution is $e^{X_{1-}}\log 2=e\log 2$. The jump correction is $e^{X_1}-e^{X_{1-}}-e^{X_{1-}}\log 2=2e-e-e\log 2=e(1-\log 2)$. Therefore the right side is $1+(2e^2-e-1)+e\log 2+e(1-\log 2)=2e^2$, matching the left side.

This check separates the linear jump contribution in the integral from the nonlinear correction. Together they give the actual exponential change $e^{X_1}-e^{X_{1-}}=e$.

Worked Example: A Cubic Polynomial and a Finite Jump

Let $X_t=1+t+2\mathbf{1}_{\{t\ge1\}}$ on $[0,2]$ and $f(x)=x^3$. Compute every term at $t=2$. The path starts at $1$, evolves continuously to $2$, jumps to $4$, and then evolves continuously to $5$. Thus $f(X_2)=5^3=125$.

Since $f'(x)=3x^2$ and $f''(x)=6x$, and $X$ has finite variation, $[C]=0$. The continuous part of $\int f'(X_{s-})\,dX_s$ is $\int_0^1 3(1+s)^2\,ds+\int_1^2 3(3+s)^2\,ds=(2^3-1^3)+(5^3-4^3)=7+61=68$. At the jump, the integral contributes $f'(2)\cdot2=3\cdot 2^2\cdot2=24$. The jump correction is $f(4)-f(2)-f'(2)\cdot2=64-8-24=32$. Hence the formula gives $f(X_0)+68+24+32=1+68+24+32=125$, exactly the endpoint value.

As a second check of the jump term, the actual jump in the cubic is $4^3-2^3=56$, and the formula divides it into the linear part $24$ and correction $32$. Their sum is $56$.

What the Formula Says About the Jump Term

The correction is not generally just a second-derivative term. If $f$ has additional smoothness, Taylor’s theorem can describe its size, but the formula itself needs only $f\in C^2$ and uses the exact difference. For a quadratic function, the correction is exactly proportional to the square of the jump. For an exponential, it involves the full exponential of the jump size. A large jump cannot in general be accurately represented by its first-order approximation.

For a continuous path there are no jump times, so the sum vanishes and the formula reduces to continuous-semimartingale Itô calculus. For a finite-variation process with finitely many jumps, the continuous quadratic-variation term vanishes, but the jump correction remains. In the examples, all finite-variation integrals are computed directly; in the Brownian–Poisson example, the Brownian quadratic variation contributes $t$ and the Poisson jumps contribute their own exact corrections.

Takeaway: In the finite-jump formula, the integral $\int f'(X_{s-})\,dX_s$ already contains the linear part $f'(X_{s-})\Delta X_s$ at every jump. Add exactly one correction, $f(X_s)-f(X_{s-})-f'(X_{s-})\Delta X_s$, to recover the full nonlinear change.

Common Mistake: Linearizing a Finite Jump

A tempting but incorrect calculation is to set the change in $e^x$ at a jump of size $z$ equal to $e^x z$. For example, from pre-jump value $x=0$ and jump size $z=\log 2$, this predicts a change of $\log 2$. The actual change is $e^{\log 2}-e^0=2-1=1$. The missing correction is $e^{\log 2}-e^0-e^0\log 2=1-\log 2$; adding it to the linear term gives $\log 2+(1-\log 2)=1$.

Another error is to identify $B_tN_t$ only with the sum of its jump increments. Its jump increments are $B_{\tau_j}$, but the product also has a continuous part. The correct integration-by-parts identity is $B_tN_t=\int_0^tN_{s-}\,dB_s+\sum_{\tau_j\le t}B_{\tau_j}$. Keeping the integral term is essential when checking the square of $B_t+aN_t$.

Check Your Understanding

Use the theorem and keep the integral’s jump contribution separate from the correction.

For $f(x)=x^2$, compute the jump correction at pre-jump value $x$ and jump size $z$.
For a process with continuous part $C$ and one jump of size $z$ at time $\tau$, write the jump correction for $f(x)=e^x$.
In the cubic example, verify that the actual jump in $f(X)$ is the sum of the jump contribution to $\int f'(X_{s-})\,dX_s$ and the correction.
Explain why the quadratic-variation term in the theorem uses $[C]$, not the quadratic variation of the finite jump sum.
State what integrability is needed for $\int_0^t H_s\,dB_s$ to be a square-integrable Itô integral on $[0,T]$.