Let $(X_t)_{t \ge 0}$ be a Lévy process on $\mathbb R$ with Lévy measure $\nu$.

Define the jump counting measure $N(t, A) = \lvert\{s \in [0, t] \mathrel: \Delta X_s \in A\}\rvert$ where $\Delta X_s$ denotes the jump height at time $s$.

For a fixed Borel set $A$ such that $0 \notin A$, let $N_t = N(t,A)$.

How can I prove that $E[N_t] = t\nu(A)$?

  • 1
    $\begingroup$ Please don't post content where every single formula is an image. MathOverflow supports MathJax, and that is what you should be using. I have edited accordingly. What is the relationship between $A$ and $B$? $\endgroup$
    – LSpice
    Jan 26 at 18:23
  • $\begingroup$ Sorry, I don't know how to use Mathoverflow. The borel set B is A, I'm confused. $\endgroup$
    – Ginger 17
    Jan 26 at 18:28

1 Answer 1


This follows immediately from (say) the following statements in Schilling - An Introduction to Lévy and Feller Processes:

  • parts b) and c) of Lemma 9.4, stating that $N(\cdot,A)$ is a Poisson process of intensity $\nu(A):=EN(1,A)$, where $A$ is any Borel subset of $\mathbb R^d\setminus\{0\}$

  • Corollary 9.13, stating that the intensity measure $\nu$ coincides with the Lévy measure.

  • $\begingroup$ Thanks. At the moment I have not yet studied the Lévy-Ito decomposition and corollary 9.13 follows from there. Is there no other way to prove it? $\endgroup$
    – Ginger 17
    Jan 26 at 18:57
  • $\begingroup$ There are almost certainly other ways. Even in the linked book, the author says on p. 9: "Nowadays there are at least six possible approaches to constructing processes with (stationary and) independent increments" and then provides details on those approaches. But any complete proof of your desired conclusion would probably be too long to reproduce here as a MathOverflow answer. $\endgroup$ Jan 26 at 19:17
  • $\begingroup$ Ok, thank you very much! $\endgroup$
    – Ginger 17
    Jan 26 at 19:32

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