# Proof that the jump measure of a Lévy process is a Poisson random measure

Let $$(X_t)_{t \geq 0}$$ be an $$\mathbb{R}^d$$-valued Lévy process and consider its associated jump measure $$N_t: \Omega \times \mathbb{B}(\mathbb{R}^d \setminus \{0\}) \to \bar{\mathbb{N}}_0$$ given by $$\begin{equation*} N_t(\omega,B):=\#\left\{0 \leq s \leq t \mid \Delta X_s(\omega) \in B\right\} \end{equation*}$$

I am looking for a rigorous proof that, for each $$t \geq 0$$, $$N_t$$ is in fact a Poisson random measure on the measure space $$(\mathbb{R}^d, \mathbb{B}(\mathbb{R}^d \setminus \{0\}), \mu)$$, where $$\mu$$ is the intensity measure $$\mu(B):=t\operatorname{\mathbb{E}}(N_1(B))$$. That is, I would like to prove that $$N_t$$ satisfies the following definition:

$$\mathbf{Definition}$$: Let $$(\Omega, \mathbb{F}, P)$$ be a probability space and $$(\mathcal{X}, \mathbb{E}, \mu)$$ a $$\sigma$$-finite measure space. A Poisson random measure with intensity measure $$\mu$$ is a mapping $$N: \Omega \times \mathbb{E} \rightarrow \mathbb{N}_0$$ satisfying

(i) For every $$\omega \in \Omega,$$ the $$\operatorname{map} B \mapsto N(\omega, B)$$ is a measure on $$(\mathcal{X}, \mathbb{E})$$

(ii) For every $$B \in \mathbb{E}$$, the map $$\omega \mapsto N(\omega, B)$$ is a random variable (i.e. measurable) and $$N(\cdot, B) \sim Pois$$($$\mu(B)$$)

(iii) If $$B_1, \ldots, B_n$$ are disjoint, then $$N(\cdot, B_1), \ldots, N(\cdot, B_n)$$ are mutually independent.

I am aware that Sato, in his book $$\textit{Lévy Processes and Infinitely Divisible Distributions}$$, provides as proof. However, the approach seems quite involved, and I was wondering if a more direct approach is available. In particular, I would like to know if a simple proof of the measurability in condition (ii) is available.

Thank you!

• Check out Theorem 2.3.5 in Applebaum's Lévy Processes and Stochastic Calculus or Proposition 5.15 in Çinlar's Probability and Statistics. The proofs seem less technical. Jun 25 '20 at 16:45

For now this will be only a partial answer. Suppose it has already been proved that the number $$N_t(\cdot,\mathbb R^d)$$ of jumps before time $$t$$ has a Poisson distribution.

Then we can deduce $$\text{(i)},$$ $$\text{(ii)},$$ and $$\text{(iii)}.$$

Proposition $$\text{(i)}$$ just says the jump measure is a measure. (Here I wonder it one should say $$\text{“}$$For almost every $$\omega\in\Omega.\text{''}$$)

Observe that $$\#\{ 0\le s\le t : \Delta X_s\in B\} \mid N_t(\cdot,\mathbb R^d) \sim\operatorname{Binomial}(N_t(\cdot,\mathbb R^d), p(B))$$ where $$p(B) = \dfrac{\mu(B)}{\mu(\mathbb R^d)} \tag 1$$ is the probability that any particular jump is in $$B.$$ Then \begin{align} & \Pr(\#\{0\le s\le t : \Delta_s\in B\} = m) \\[6pt] = {} & \operatorname E(\Pr(\#\{0\le s\le t : \Delta_s\in B\} = m \mid N_t(\cdot,\mathbb R^d)) \\[6pt] = {} & \operatorname E\left( \binom {N_t(\cdot,\mathbb R^d)} m p(B)^m (1-p(B))^{N_t(\cdot, \mathbb R^d)-m} \right) \\[6pt] = {} & \sum_{N=0}^\infty \binom N m p(B)^m (1-p(B))^{N-m} \Pr(N_t(\cdot,\mathbb R^d)=N) \\[6pt] = {} & \sum_{N=0}^\infty \binom N m p(B)^m (1-p(B))^{N-m} \frac{\mu(\mathbb R^d)^N e^{-\mu(\mathbb R^d)}}{N!} \\[6pt] = {} & \frac{\mu(B)^m e^{-\mu(B)}}{m!} \quad(\text{Why? See below.}) \tag 2. \end{align} How do we deduce line $$(2)$$ from what precedes it?

• Note that $$\dbinom Nm=0$$ when $$N so that we can replace the sum by $$\displaystyle \sum_{N=m}^\infty.$$
• Let $$M=N-m,$$ so we have $$\displaystyle \sum_{M=0}^\infty$$ and the exponent $$N-m$$ becomes $$M$$ and the remaining $$N$$s become $$M+m.$$
• Apply the binomial theorem and the power series for the exponential function.

So we conclude that $$N_t(\cdot, B)\sim\operatorname{Poisson}(\mu(B)).$$

Next, how do we know $$N(\cdot, B_1),\ldots, N(\cdot, B_n)$$ are independent?

\begin{align} & \Pr( N(\cdot, B_1)=m_1\ \&\ \cdots\ \&\ N(\cdot, B_n)=m_n) \\[6pt] = {} & \operatorname E(\Pr( N(\cdot, B_1)=m_1\ \&\ \cdots\ \&\ N(\cdot, B_n)=m_n\mid N(\cdot,B_1\cup\cdots\cup B_n) )) \\[6pt] = {} & \operatorname E\left( \binom{N(\cdot,B_1\cup\cdots\cup B_n)}{m_1,\ldots,m_n} p(B_1)^{m_1}\cdots p(B_n)^{m_n} \right) \tag 3 \end{align} where $$\binom N {m_1,\ldots,m_n} = \begin{cases} \dfrac{N!}{m_1!\cdots m_n!} & \text{if } m_1+\cdots+m_n=N, \\[6pt] 0 & \text{otherwise.} \end{cases}$$ Line $$(3)$$ is a sum of infinitely many terms all but one of which are $$0.$$ That one term is \begin{align} & \binom{m_1+\cdots+m_n}{m_1,\ldots,m_n} p(B_1)^{m_1} \cdots p(B_n)^{m_n} \frac{\mu(\mathbb R^d)^{m_1+\cdots+m_n} e^{-\mu(B_1\cup\cdots\cup B_n)} }{(m_1+\cdots+m_n)!} \\[6pt] = {} & \binom{m_1+\cdots+m_n}{m_1,\ldots,m_n} \mu(B_1)^{m_1} \cdots \mu(B_n)^{m_n} \frac{ e^{-\mu(B_1\cup\cdots\cup B_n)} }{(m_1+\cdots+m_n)!} \\[6pt] = {} & \prod_{i=1}^n \frac{\mu(B_i)^{m_i} e^{-\mu(B_i)} }{m_i!}. \end{align}