If $L_t=\sum_{i=1}^{N_t}Y_i$ is a compound Poisson process, then $\left|\left\{s\in[0,t]:\Delta L_s\in B\right\}\right|=\sum_{i=1}^{N_t}1_B(Y_i)$

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.

$$\newcommand{\De}{\Delta}$$ This is not really a probability problem, since the equality $$$$\label{1}\tag{1} l_t:=|\{s\in[0,t]\colon\De L_s\in B\}|=\sum_{i=1}^{N_t}1_B(Y_i)=:r_t$$$$ holds almost everywhere on $$\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$$ $$$$\label{2}\tag{2} i\le N_t\iff\tau_i\le t.$$$$ (Of course, here $$\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$$, so that (\ref{1}) does hold.
• You didn't specify your sequence $(\tau_k)_{k\in\mathbb N_0}$, but I guess $\tau_0:=0$ and $$\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. $(2)$. Oct 23, 2020 at 4:44
• @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? Oct 23, 2020 at 13:05