Levy Exponents of Compound Poisson Processes

From Poisson Counts to an Exponent

In a compound Poisson process, a Poisson count determines how many jumps occur, and independent marks determine the jump sizes. Its characteristic function has a particularly useful form: an exponential whose exponent is linear in elapsed time. This exponent records the rate and distribution of the jumps together, while the measure describing jump sizes makes the representation precise.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a Poisson process of rate $\lambda$, where $\lambda\ge0$, and let $(Y_k)_{k\ge1}$ be independent and identically distributed real-valued marks, independent of $N_t$. Write $F$ for the distribution of a mark. The compound Poisson process is $X_t=\sum_{k=1}^{N_t}Y_k$, with an empty sum equal to zero. Its increments over intervals of length $h$ have the same distribution as $X_h$.

Definition: For a real-valued random variable $Z$, its characteristic function is $\phi_Z(u)=\mathbb{E}[e^{iuZ}]$, $u\in\mathbb{R}$. For a compound Poisson process of rate $\lambda$, its Lévy exponent in the untruncated convention is $\psi(u)=\lambda(\mathbb{E}[e^{iuY_1}]-1)$. Thus an increment of length $h$ has characteristic function $\exp(h\psi(u))$.

The subtraction of $1$ is essential: it accounts for the possibility of no arrivals. The exponent is zero at $u=0$, as every characteristic function must equal one there. No moment assumptions on the marks are needed for this characteristic-function formula, since $|e^{iuY_1}|=1$.

Definition: The finite jump measure of the compound Poisson process is the measure $\nu$ on $\mathbb{R}\setminus\{0\}$ given by $\nu(B)=\lambda\mathbb{P}(Y_1\in B)$ for measurable $B\subseteq\mathbb{R}\setminus\{0\}$. Its total mass is $\nu(\mathbb{R}\setminus\{0\})=\lambda\mathbb{P}(Y_1\ne0)$. In this convention, $\psi(u)=\int_{\mathbb{R}\setminus\{0\}}(e^{iux}-1)\nu(dx)$. Marks equal to zero make no change to $X_t$, so they are omitted from the jump measure. If $\lambda=0$, the process is identically zero and $\nu$ is the zero measure.

Some descriptions instead use the finite measure $\lambda F$ on all of $\mathbb{R}$, including a possible atom at zero. Integrating $e^{iux}-1$ against that measure gives exactly the same exponent, because the integrand vanishes at zero. The jump-measure convention above removes that irrelevant atom. Since $\nu$ is finite, it satisfies the usual Lévy-measure integrability condition $\int (1\wedge x^2)\nu(dx)<\infty$: the integrand is at most $1$, so the integral is at most $\nu(\mathbb{R}\setminus\{0\})$.

Deriving the Characteristic Function

Theorem: Suppose $\lambda\ge0$, $N_t$ is a Poisson process of rate $\lambda$, and the marks are independent and identically distributed with law $F$, independently of $N_t$. For $h\ge0$, the increment over an interval of length $h$ satisfies $$ \mathbb{E}[e^{iu(X_{s+h}-X_s)}] =\exp\!\left(h\lambda\left(\int_{\mathbb{R}}e^{iux}F(dx)-1\right)\right) =\exp\!\left(h\int_{\mathbb{R}\setminus\{0\}}(e^{iux}-1)\nu(dx)\right). $$ Here $\nu(B)=\lambda F(B)$ for measurable $B\subseteq\mathbb{R}\setminus\{0\}$. The exponent per unit time is $\psi(u)=\int_{\mathbb{R}\setminus\{0\}}(e^{iux}-1)\nu(dx)$.

Proof. By stationarity of the Poisson construction and independence of its marks, the increment $X_{s+h}-X_s$ has the same distribution as a sum of $N_h$ independent marks. Let $\phi_Y(u)=\mathbb{E}[e^{iuY_1}]$. Conditional on $N_h=m$, independence gives $$ \mathbb{E}[e^{iu(X_{s+h}-X_s)}\mid N_h=m]=\phi_Y(u)^m. $$ A Poisson random variable of mean $\lambda h$ has probability $e^{-\lambda h}(\lambda h)^m/m!$ of equaling $m$. Therefore $$ \mathbb{E}[e^{iu(X_{s+h}-X_s)}] =\sum_{m=0}^{\infty}e^{-\lambda h}\frac{(\lambda h)^m}{m!}\phi_Y(u)^m =e^{-\lambda h}\exp(\lambda h\phi_Y(u)) =\exp\bigl(\lambda h(\phi_Y(u)-1)\bigr). $$ The series is the exponential series; it converges since $|\phi_Y(u)|\le1$. Also, $\phi_Y(u)-1=\int_{\mathbb{R}}(e^{iux}-1)F(dx)$, and the integrand is zero at $x=0$. Multiplication by $\lambda$ therefore gives the integral against $\nu$ on $\mathbb{R}\setminus\{0\}$. If $h=0$ or $\lambda=0$, both sides equal $1$, so the formula includes those cases. $\square$

The derivation also identifies the roles of the ingredients: the mark law supplies $\phi_Y(u)$, while the Poisson count turns the average over possible numbers of marks into an exponential. The time factor $h$ appears linearly, so the same exponent describes increments of every length.

Worked Example: A Fixed Jump Size

Suppose the arrival rate is $\lambda=2$ and each mark is $Y_k=3$ with probability one. Find the jump measure, the exponent, and the characteristic function of an increment of length $h=\tfrac12$.

Every mark is nonzero, so $\nu=2\delta_3$, where $\delta_3$ places unit mass at $3$. Its total mass is $2$. The exponent per unit time is $$ \psi(u)=\int(e^{iux}-1)\nu(dx) =2(e^{3iu}-1). $$ For an increment of length $\tfrac12$, the characteristic function is $$ \exp\!\left(\tfrac12\psi(u)\right) =\exp(e^{3iu}-1). $$

Check: At $u=0$, $e^{3iu}=1$, so the exponent for the increment is $0$ and its characteristic function is $e^0=1$. The jump measure has mass $2$, which equals the arrival rate because every arrival produces a nonzero jump.

Infinite Divisibility from the Exponent

Definition: A probability distribution is infinitely divisible if, for every positive integer $n$, it is the distribution of a sum of $n$ independent, identically distributed random variables. The summand distribution may depend on $n$.

Theorem: Every increment of a compound Poisson process is infinitely divisible. More precisely, for any $h\ge0$ and positive integer $n$, the increment law over length $h$ is the law of a sum of $n$ independent compound Poisson variables, each with arrival rate $\lambda h/n$ and mark law $F$.

Proof. First take $h>0$ and $\lambda>0$. Construct $n$ independent compound Poisson variables, each with rate $\lambda h/n$ and mark law $F$. The characteristic function of each variable is, by the theorem just proved, $$ \exp\!\left(\frac{\lambda h}{n}(\phi_Y(u)-1)\right). $$ Independence means that the characteristic function of their sum is the product of these $n$ characteristic functions. That product is $$ \left[\exp\!\left(\frac{\lambda h}{n}(\phi_Y(u)-1)\right)\right]^n =\exp\!\left(\lambda h(\phi_Y(u)-1)\right), $$ which is the characteristic function of the original increment. By the standard uniqueness theorem for characteristic functions, the sum and the increment have the same distribution. If $h=0$ or $\lambda=0$, the increment is zero with probability one; it is the sum of $n$ independent variables that are all zero. Thus the definition holds in every case. $\square$

This proof identifies a specific $n$th convolution root: it is itself compound Poisson, with the same mark law and $1/n$ of the original interval's arrival rate. In exponent form, dividing the exponent by $n$ gives the characteristic function of one summand. The finite jump measure of that summand is $(h/n)\nu$, while the sum of the $n$ measures is $h\nu$.

Worked Example: Two Possible Jump Sizes

Let arrivals have rate $3$, and let each mark equal $-1$ with probability $\tfrac13$ or $2$ with probability $\tfrac23$. Find the jump measure and exponent, then give the two identical summand laws whose sum has the same distribution as an increment of length $h=1$.

The mark probabilities sum to $\tfrac13+\tfrac23=1$. The jump measure assigns mass $3\cdot\tfrac13=1$ to $-1$ and mass $3\cdot\tfrac23=2$ to $2$; thus $\nu=\delta_{-1}+2\delta_2$, with total mass $1+2=3$. The exponent is $$ \psi(u)=(e^{-iu}-1)+2(e^{2iu}-1) =e^{-iu}+2e^{2iu}-3. $$ For an interval of length $1$, the characteristic function is $e^{\psi(u)}$.

For $n=2$, take two independent compound Poisson variables, each with rate $3/2$ and the same mark probabilities. Each has exponent $\psi(u)/2$, so their sum has characteristic function $$ e^{\psi(u)/2}e^{\psi(u)/2}=e^{\psi(u)}. $$ By uniqueness of characteristic functions, their sum has the required increment law.

Check: At $u=0$, $\psi(0)=1+2-3=0$, so the characteristic function equals $1$. Each summand has jump-measure masses $(3/2)(1/3)=1/2$ at $-1$ and $(3/2)(2/3)=1$ at $2$; doubling these masses recovers $1$ and $2$, the original jump measure.

A Continuous Jump-Size Distribution

Worked Example: Exponential Marks

Let the arrival rate be $\lambda=3$, and let each positive mark have the exponential density $2e^{-2x}$ for $x>0$. Find the jump measure and exponent. Also evaluate the exponent at $u=1$.

The jump measure has density $3\cdot2e^{-2x}=6e^{-2x}$ on $x>0$, so $\nu(dx)=6e^{-2x}\,dx$ there and is zero for $x\le0$. Its total mass is $$ \int_0^\infty 6e^{-2x}\,dx =6\left[-\frac12e^{-2x}\right]_{0}^{\infty} =6\left(0+\frac12\right)=3. $$ The mark characteristic function follows by integrating: $$ \mathbb{E}[e^{iuY_1}] =\int_0^\infty 2e^{-2x}e^{iux}\,dx =2\int_0^\infty e^{-(2-iu)x}\,dx =\frac{2}{2-iu}. $$ The integral converges because the real part of $2-iu$ is $2>0$. Hence $$ \psi(u)=3\left(\frac{2}{2-iu}-1\right) =\frac{3iu}{2-iu}. $$ At $u=1$, multiply numerator and denominator by $2+i$: $$ \psi(1)=\frac{3i}{2-i} =\frac{3i(2+i)}{(2-i)(2+i)} =\frac{-3+6i}{5}. $$

Check: The mass calculation gives $3$, matching the arrival rate because exponential marks are positive and therefore nonzero with probability one. Also $\psi(0)=0$ from $3iu/(2-iu)$, and the direct expression $3(2/(2-iu)-1)$ gives $3(1-1)=0$ at $u=0$.

Interpretation and a Common Error

The exponent is useful because independent increments multiply characteristic functions, while their exponents add. It also separates time from jump structure: an interval of length $h$ uses the measure $h\nu$, so dividing an interval into equal pieces divides the measure and exponent by the same factor. The jump measure records rates of nonzero jumps by size; it is not itself a probability distribution unless normalized by its mass.

A common mistake is to write $\lambda\mathbb{E}[e^{iuY_1}]$ as the exponent. For the fixed-jump example with $\lambda=2$ and $Y_1=3$, that incorrect expression gives $2e^{3iu}$. At $u=0$ it equals $2$, so exponentiating gives $e^2$, not the required characteristic-function value $1$. The correction is to subtract the no-jump contribution: $\psi(u)=2(e^{3iu}-1)$, which is zero at $u=0$. The subtraction is not optional; it is what makes the Poisson mixture a normalized probability law.

Check Your Understanding

Use the definitions and formulas above. In each calculation, include a check at $u=0$ or a check of the jump-measure mass.

If the arrival rate is $4$ and every mark equals $-2$, write the jump measure and the exponent.
A mark equals $0$ with probability $\tfrac14$ and $1$ with probability $\tfrac34$, at arrival rate $2$. What is the total mass of the jump measure, and why is it not $2$?
For an interval of length $h$, what arrival rate and mark law give one of the $n$ identical compound Poisson summands in the infinite-divisibility representation?
For a jump measure $\nu=2\delta_1+\delta_3$, compute $\psi(u)$ and verify $\psi(0)=0$.
Explain why replacing $\lambda(\mathbb{E}[e^{iuY_1}]-1)$ by $\lambda\mathbb{E}[e^{iuY_1}]$ cannot give a characteristic exponent.