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Let $H$ be a $\mathbb R$-Hilbert space, $\mu$ be a finite measure on $\mathcal B(H)$ with $\mu(\{0\})=0$ and $(L_t)_{t\ge0}$ be a $H$-valued càdlàg Lévy process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $$L_t=\sum_{i=1}^{N_t}Y_i\;\;\;\text{for all }t\ge0\tag1$$ for some $H$-valued independent identically distributed process $(Y_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$ with $Y_1\sim\lambda^{-1}\mu$ for some $\lambda>0$ and some càdlàg Poisson process $(N_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$.

Let $t\ge0$ and $B\in\mathcal B(H\setminus\{0\})$. I would like to show that $$\left|\left\{s\in[0,t]:\Delta L_s\in B\right\}\right|=\sum_{i=1}^{N_t}1_B(Y_i),\tag2$$ where $$\Delta L_s:=L_s-L_{s-}=L_s-\lim_{r\to s-}L_r\;\;\;\text{for }s\ge0.$$ How can we do that?

Remark: Since the measure is not involved in $(2)$ it might be unimportant for the claim, but we may note that $$Z_n:=\sum_{i=1}^nY_i\;\;\;\text{for }n\in\mathbb N$$ is a time-homogenous Markov chain and hence $$L_t=Z_{N_t}\;\;\;\text{for all }t\ge0$$ is a time-homogeneous Markov process.

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$\newcommand{\De}{\Delta}$ This is not really a probability problem, since the equality \begin{equation}\label{1}\tag{1} l_t:=|\{s\in[0,t]\colon\De L_s\in B\}|=\sum_{i=1}^{N_t}1_B(Y_i)=:r_t \end{equation} holds almost everywhere on $\Omega$, for "almost" any joint distribution of the involved random variables -- provided only that $0\notin B$ and for all real $t\ge0$ \begin{equation*} N_t=\sum_{i=1}^\infty 1(\tau_i\le t), \tag{1.5} \end{equation*} where $0=\tau_0<\tau_1<\cdots$, so that for all natural $i$ and all real $t\ge0$ \begin{equation}\label{2}\tag{2} i\le N_t\iff\tau_i\le t. \end{equation} (Of course, here $\tau_1<\tau_2<\cdots$ are the times of the jumps of the Poisson process $N_\cdot$.)

Indeed, then for any $j\in\{0,1,\dots\}$ and any real $t\ge0$ \begin{equation*} L_t=\sum_{i=1}^{N_t}Y_i=\sum_{i=1}^j Y_i\quad\text{if}\quad \tau_j\le t<\tau_{j+1}. \end{equation*} Hence, for each real $s\ge0$, we have $\De L_s=Y_j$ if $s=\tau_j$ for some $j\in\{1,2,\dots\}$, and $\De L_s=0$ if $s\ne\tau_j$ for any $j\in\{1,2,\dots\}$. It follows that \begin{equation*} l_t=\sum_{j=1}^\infty 1(Y_j\in B,\tau_j\le t). \end{equation*}

On the other hand, for all real $t\ge0$ \begin{equation*} r_t=\sum_{i=1}^{N_t}1_B(Y_i) =\sum_{i=1}^\infty 1(Y_i\in B,i\le N_t)=\sum_{i=1}^\infty 1(Y_i\in B,\tau_i\le t), \end{equation*} by (\ref{2}).

Thus, $l_t=r_t$ for all real $t\ge0$, so that (\ref{1}) does hold.

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  • $\begingroup$ You didn't specify your sequence $(\tau_k)_{k\in\mathbb N_0}$, but I guess $\tau_0:=0$ and $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t>0\right\}.$$ Since we know that there is a $\operatorname P$-null set $N$ such that $N(\omega)$ is nondecreasing and $$\Delta N_t(\omega)\in\{0,1\}\;\;\;\text{for all }t\ge0$$ for all $\omega\in\Omega\setminus N$, we can conclude that $$N_{\tau_k}(\omega)=k\;\;\;\text{for all }k\in\mathbb N_0\text{ and }\omega\in\Omega\setminus N.$$ This should yield your eq. $(2)$. $\endgroup$
    – 0xbadf00d
    Commented Oct 23, 2020 at 4:44
  • $\begingroup$ @0xbadf00d : What you said in this comment is of course correct, but I don't understand the point of the comment. Is it to detail how (2) is obtained? In fact, (2) (as well as your description of the $\tau_i$'s) follow immediately from (1.5) and the condition $0=\tau_0<\tau_1<\cdots$. Anyway, other than this, are you satisfied with this answer? $\endgroup$ Commented Oct 23, 2020 at 13:05
  • $\begingroup$ @0xbadf00d : To have a closure here, are you satisfied with my answer and the previous comment? $\endgroup$ Commented Jan 8, 2021 at 17:08
  • $\begingroup$ I'll come back to this question and accept your answer soon. $\endgroup$
    – 0xbadf00d
    Commented Jan 9, 2021 at 11:46

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