Moment and Martingale Checks for Jump-Diffusions

Moments as a Check on Jump Models

For a jump model, the drift, Brownian fluctuations, and jumps all contribute to moments, but not in the same way. In particular, the second moment records the square of each jump size. A compensator calculation exposes that contribution, while a characteristic function or an explicit solution can provide an independent check. Exponential transforms require their own integrability check: a formally normalized exponential is not automatically a true martingale.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a standard Brownian motion $B_t$ and an independent Poisson random measure $N(dt,dz)$ with intensity $dt\,\nu(dz)$, where $\nu$ is a finite measure. Write $\widetilde N(dt,dz)=N(dt,dz)-dt\,\nu(dz)$ for its compensated measure. All expectations below are under $\mathbb{P}$, not a risk-neutral measure $\mathbb{Q}$. Brownian integrals are Ito integrals with progressively measurable integrands $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Compensated Poisson integrals are used with predictable integrands $G$ satisfying $\mathbb{E}\int_0^T\int G_s(z)^2\,\nu(dz)\,ds<\infty$; under these conditions they are square-integrable martingales.

A compound Poisson process with rate $\lambda>0$ and independent identically distributed marks $J_1,J_2,\ldots$ is $C_t=\sum_{k=1}^{N_t}J_k$, where $N_t$ has rate $\lambda$. When the marks have a finite second moment, its first two moments are:

\mathbb{E}[C_t]=\lambda t\,\mathbb{E}[J_1], \qquad \operatorname{Var}(C_t)=\lambda t\,\mathbb{E}[J_1^2].

The variance involves the raw second moment of the mark, not just its variance. The reason is that the random number of jumps also fluctuates.

Compound Poisson Moments and Exponential Transforms

Proposition: If $\mathbb{E}[J_1^2]<\infty$, the compound Poisson moments are as displayed above. If $\mathbb{E}[e^{\theta J_1}]<\infty$, then $Z_t=\exp\{\theta C_t-\lambda t(\mathbb{E}[e^{\theta J_1}]-1)\}$ is a true martingale on every bounded time interval. For $\theta$ replaced by $iu$, $u\in\mathbb{R}$, the same formula is a complex-valued martingale, with the expectation always well-defined.

Proof. Conditional on $N_t=n$, the sum $C_t$ has mean $n\mathbb{E}[J_1]$ and variance $n\operatorname{Var}(J_1)$. Since $\mathbb{E}[N_t]=\operatorname{Var}(N_t)=\lambda t$, the conditional-variance formula gives $\operatorname{Var}(C_t)=\mathbb{E}[N_t]\operatorname{Var}(J_1)+\operatorname{Var}(N_t)(\mathbb{E}[J_1])^2=\lambda t\mathbb{E}[J_1^2]$. The conditional-mean formula gives $\mathbb{E}[C_t]=\lambda t\mathbb{E}[J_1]$. For the transform, increments of $C$ after time $s$ are independent of $\mathcal{F}_s$, and $\mathbb{E}[e^{\theta(C_t-C_s)}]=\exp\{\lambda(t-s)(\mathbb{E}[e^{\theta J_1}]-1)\}$. Multiplying by the deterministic normalizing factor therefore yields $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. The assumed exponential moment gives $\mathbb{E}|Z_t|<\infty$, so this conditional-expectation identity proves a true martingale, not merely a local martingale. For $\theta=iu$, $|e^{iu C_t}|=1$, so integrability holds, and the same independent-increment calculation applies. $\square$

Worked Example: Symmetric Compound Poisson Marks

Let the rate be $\lambda=2$, with each mark equal to $1$ or $-1$, each with probability $1/2$. Find the mean, variance, characteristic function, and the normalized real exponential transform.

The mark mean is $\mathbb{E}[J_1]=\tfrac12(1)+\tfrac12(-1)=0$, and its raw second moment is $\mathbb{E}[J_1^2]=\tfrac12(1)+\tfrac12(1)=1$. Thus $\mathbb{E}[C_t]=0$ and $\operatorname{Var}(C_t)=2t$. Also, $\mathbb{E}[e^{iuJ_1}]=\tfrac12e^{iu}+\tfrac12e^{-iu}=\cos u$, so $\mathbb{E}[e^{iuC_t}]=e^{2t(\cos u-1)}$. For every real $\theta$, $\mathbb{E}[e^{\theta J_1}]=\cosh\theta$, which is finite. Hence $e^{\theta C_t-2t(\cosh\theta-1)}$ is a true martingale.

Check: At $u=0$, the characteristic function is $e^{2t(1-1)}=1$. Its first derivative at zero is $0$, consistent with the mean, and the negative second derivative at zero is $2t$, consistent with the variance.

Linear Jump-Diffusion Moments

Consider the finite-activity linear model $dX_t=\alpha X_{t-}\,dt+\sigma X_{t-}\,dB_t+\int h(z)X_{t-}\,N(dt,dz)$, with deterministic $X_0=x$, constants $\alpha,\sigma$, and bounded measurable $h$. The drift coefficient is $\alpha X_{t-}$, the diffusion coefficient is $\sigma X_{t-}$, and a jump with mark $z$ multiplies the state by $1+h(z)$. Since $\nu$ is finite, the jump term is a finite sum on bounded time intervals almost surely.

An explicit solution is $X_t=x\exp\{(\alpha-\tfrac12\sigma^2)t+\sigma B_t\}\prod_{k=1}^{N_t}(1+h(Z_k))$, where $Z_k$ are the jump marks. The empty product is $1$. This formula remains valid when a jump factor is zero; if a factor is negative, the product and the solution can be negative.

Theorem: For every $t\geq0$, $$ \mathbb{E}[X_t]=x e^{k_1t},\qquad \mathbb{E}[X_t^2]=x^2 e^{k_2t}, $$ where $k_1=\alpha+\int h(z)\,\nu(dz)$ and $k_2=2\alpha+\sigma^2+\int(2h(z)+h(z)^2)\,\nu(dz)$. Moreover, $\mathbb{E}[X_t^4]=x^4\exp\{(4\alpha+6\sigma^2+\int((1+h(z))^4-1)\,\nu(dz))t\}$. These formulas include the degenerate case $x=0$.

Proof. First verify the moment bound needed for the expectation calculations. The explicit solution, independence of Brownian motion and the Poisson random measure, the Gaussian moment formula, and the compound Poisson product formula give $$ \mathbb{E}[X_t^4] =x^4 e^{(4\alpha+6\sigma^2)t} \exp\left\{t\int((1+h(z))^4-1)\,\nu(dz)\right\}. $$ Indeed, the Brownian factor has fourth moment $\mathbb{E}[e^{4\sigma B_t-2\sigma^2t}]=e^{6\sigma^2t}$, and the expected product of fourth powers of jump factors is the second exponential factor. Bounded $h$ and finite $\nu$ make this expression finite and bounded for $t$ in any bounded interval. Thus $\mathbb{E}\int_0^T X_s^4\,ds<\infty$.

For $f(y)=y$, the derivatives are $f'(y)=1$ and $f''(y)=0$, so the Brownian Ito correction $\tfrac12\sigma^2X_{s-}^2 f''(X_{s-})\,ds$ is zero. Rewriting the jump term using $\widetilde N$, Ito's formula gives $$ X_t=x+\int_0^t k_1X_{s-}\,ds+\int_0^t\sigma X_{s-}\,dB_s+ \int_0^t\int h(z)X_{s-}\,\widetilde N(ds,dz). $$ The Brownian integrand is square-integrable because $\mathbb{E}\int_0^T X_s^2ds<\infty$, which follows from the fourth-moment bound and Cauchy–Schwarz. The compensated-jump integrand is square-integrable because $h$ is bounded, $\nu$ is finite, and the same second-moment bound holds. Both stochastic integrals are therefore true martingales with zero expectation.

For $f(y)=y^2$, $f'(y)=2y$, $f''(y)=2$, and the Brownian correction is $\tfrac12\sigma^2X_{s-}^2\cdot2\,ds=\sigma^2X_{s-}^2\,ds$. A jump changes $y^2$ by $((1+h)^2-1)y^2=(2h+h^2)y^2$. Consequently the drift after compensation is $k_2X_{s-}^2\,ds$, and $$ X_t^2=x^2+\int_0^t k_2X_{s-}^2\,ds+ \int_0^t 2\sigma X_{s-}^2\,dB_s+ \int_0^t\int(2h(z)+h(z)^2)X_{s-}^2\,\widetilde N(ds,dz). $$ Here it is the fourth-moment bound—not just a second-moment bound—that makes the last two integrands square-integrable: their squared magnitudes are bounded by constants times $X_{s-}^4$, integrable over $[0,T]$ in expectation. Taking expectations is therefore justified. The resulting integral equations are $m_1(t)=x+k_1\int_0^t m_1(s)\,ds$ and $m_2(t)=x^2+k_2\int_0^t m_2(s)\,ds$. Solving these scalar equations gives the stated exponentials, including when $x=0$. $\square$

Worked Example: Exact Moments and a Fourth-Moment Check

Take $x=2$, $\alpha=\tfrac14$, $\sigma=\tfrac12$, constant jump size $h=\tfrac12$, and jump rate $\lambda=2$. Compute the first, second, and fourth moments.

Here $\nu$ has total mass $2$. Thus $k_1=\tfrac14+2\cdot\tfrac12=\tfrac54$. For the second moment, $k_2=2\cdot\tfrac14+(\tfrac12)^2+2(2\cdot\tfrac12+(\tfrac12)^2)=\tfrac12+\tfrac14+2(\tfrac54)=\tfrac{13}{4}$. Therefore $\mathbb{E}[X_t]=2e^{5t/4}$ and $\mathbb{E}[X_t^2]=4e^{13t/4}$.

For the fourth moment, the exponent is $4\cdot\tfrac14+6(\tfrac12)^2+2((1+\tfrac12)^4-1)=1+\tfrac32+2(\tfrac{81}{16}-1)=\tfrac{85}{8}$. Hence $\mathbb{E}[X_t^4]=16e^{85t/8}$.

Check: At $t=0$, these expressions give $2$, $4$, and $16$, respectively, which are $x$, $x^2$, and $x^4$. Also, $\operatorname{Var}(X_t)=4e^{13t/4}-4e^{5t/2}$, which is zero at $t=0$ and nonnegative for $t\geq0$, since $13t/4\geq5t/2$.

Normalized Moments and Martingale Tests

The moment formulas also produce martingales, provided the filtration contains no extra future information: take the usual augmentation of the filtration generated by $B$ and $N$, so future increments are independent of $\mathcal{F}_s$. For the model above, the explicit solution factors into $X_s$ times a future multiplier independent of $\mathcal{F}_s$. That multiplier has first moment $e^{k_1(t-s)}$ and second moment $e^{k_2(t-s)}$. Thus, for $s\leq t$,

\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s e^{k_1(t-s)},\qquad \mathbb{E}[X_t^2\mid\mathcal{F}_s]=X_s^2 e^{k_2(t-s)}.

These conditional expectations are integrable by the moment formulas. It follows that $e^{-k_1t}X_t$ and $e^{-k_2t}X_t^2$ are true martingales. This argument works also for $x=0$, when both processes are identically zero. A local martingale, by contrast, is only a process that becomes a martingale after stopping at a suitable increasing sequence of stopping times; without integrability and the conditional-expectation property, it need not be a true martingale.

Worked Example: Compensator Check for the Second Moment

Use the parameters in the preceding example to compare the correct second-moment drift with a calculation that omits the squared jump contribution.

A jump multiplies $X$ by $3/2$, so it multiplies $X^2$ by $9/4$. Its change in $X^2$ is therefore $(9/4-1)X_{t-}^2=(5/4)X_{t-}^2$. At rate $2$, the compensated drift contribution is $2(5/4)X_{t-}^2=(5/2)X_{t-}^2$. Adding the continuous terms $2\alpha+\sigma^2=\tfrac34$ gives $k_2=\tfrac{13}{4}$, as before.

The wrong calculation keeps only the linear jump term $2h$, producing the jump contribution $2\cdot 2\cdot\tfrac12=2$, rather than $\tfrac52$. It would give the false exponent $\tfrac34+2=\tfrac{11}{4}$, and the false second moment $4e^{11t/4}$. The missing term is the jump square $h^2$, whose contribution is $\lambda h^2=2\cdot\tfrac14=\tfrac12$.

Check: The jump multiplier calculation gives $(1+h)^2-1=9/4-1=5/4$, while the expansion gives $2h+h^2=1+1/4=5/4$. Both yield the same correct contribution.

Common Pitfalls and Scope of the Checks

For an exponential transform, check the moment condition before calling it a martingale. The compound Poisson transform at real $\theta$ requires $\mathbb{E}[e^{\theta J_1}]<\infty$; if this expectation is infinite, the displayed normalization is not a finite expectation identity. For characteristic functions, $\theta=iu$ causes no such problem because the exponential has modulus one.

For jump-diffusion second moments, the Ito drift includes the full jump remainder $f(y+yh)-f(y)-f'(y)yh$ when the jump integral is compensated. For $f(y)=y^2$, that remainder is $h^2y^2$. Finally, square-integrability must be checked for the integrands actually present in the calculation: the second-moment Ito formula has stochastic integrands proportional to $X^2$, so its expectation argument needs a finite fourth moment. The explicit solution supplies that bound here.

Check Your Understanding

Use the stated moment and martingale conditions; when checking a martingale, verify integrability as well as the conditional-expectation identity.

A compound Poisson process has rate $3$ and marks equal to $0$ or $2$, each with probability $1/2$. Find its mean and variance at time $t$.
For a linear jump-diffusion, explain why the second-moment Ito calculation requires control of $\mathbb{E}\int_0^T X_s^4\,ds$.
For constant jump size $h$ and rate $\lambda$, derive the jump contribution to the second-moment exponent from the jump multiplier $1+h$.
State the exponential-moment condition that makes the normalized real transform of a compound Poisson process a true martingale.
Explain why a finite second moment of $X_t$ alone does not establish square-integrability of a Brownian integrand proportional to $X_t^2$.