Relationship between $\text{Pois}(\mu)$ and $\text{Pois}(\ell)$. Choose $\mu=\ell\delta_1$? Let $\mu$ denote a finite measure on $(\mathbb{R}, \mathcal{B}(\mathbb{
R}))$. The General Poisson measure w.r.t $\mu$ is defined as
$$
\text{Pois}(\mu)(A) \equiv \exp({-\mu(\mathbb{R})})\sum_{k=0}^\infty \frac{\mu^{*k}(A)}{k!}
$$
for every $A \in \mathcal{B}(\mathbb{
R})$. Here $\mu^{*k}$ denotes the convolution of $\mu$ with itself $k$ times.
The Poisson distribtion with rate $\ell > 0$ is defined as
$$
\text{Pois}(\ell)(A) \equiv \exp({-\ell})\sum_{k \in A} \frac{\ell^k}{k!}
$$
for every $A \in \mathcal{P}(\mathbb{N})$.
My question is how these two are related?
My gut is telling me it is something along the lines of choosing $\mu=\ell\delta_1$. Then
$\exp({-\ell\delta_1(\mathbb{R})})=\exp({-\ell})$ as we want. However it seems to me that even for $k \in A$ we have $\mu^{*k}(A)\neq \ell^k$. Also it would have to be $\mu^{*k}(A) = 0$ for $k \notin A$. How to fix this? Or is the chosen $\mu$ wrong?
 A: Your measure $\mu$ is chosen exactly right. First, $(\ell \delta_1)^{\ast k}=\ell^k\delta_1^{\ast k}$ by the bilinearity of the convolution. Now it is not hard to show by induction that $\delta_1^{\ast k}=\delta_k$, but there is also a probabilistic argument: $\delta_1$ is the distribution of a deterministic random variable that takes the value $1$ almost surely. The $k$-fold convolution $\delta_1^{\ast k}$ is then the distribution of the sum of $k$ independent random variables $X_1,\dots,X_k$, each with distribution $\delta_1$. Of course, since $X_j=1$ a.s. for all $j$, we have $X_1+\dots+X_k=k$ a.s. Thus $X_1+\dots+X_k\sim \delta_k$.
This means
$$
\sum_{k=0}^\infty\frac{\mu^{\ast k}(A)}{k!}=\sum_{k=0}^\infty\frac{\ell^k\delta_k(A)}{k!}=\sum_{k\in A}\frac{\ell^k}{k!}
$$
as desired.
A: *

*Consider the special case $\nu(\mathbb R)=1.$ In this case $\nu$ is a probability measure. Let $X_1,X_2,X_3,\ldots$ be independent random variables with with this distribution. Then the probability distribution of $X_1+\cdots+X_k$ is $\nu^{*k}.$
Now suppose

*

*$K$ is a random variable with a Poisson distribution with expectation $\ell,$ i.e. for $k=0,1,2,3,\ldots$ we have $\Pr(K=k) = e^{-\ell} \ell^k/k!, $ and


*$Y= X_1+\cdots+X_K.$ (This sum is $0$ if $K=0.$)
What then is the probability distribution of $Y\text{ ?}$
Proposition: Let $\mu=\ell\nu.$ Then
$$
\Pr(Y\in A) = \operatorname{Pois}(\mu)(A).
$$
Thus the general Poisson measure with respect to the measure $\mu$ is the compound Poisson distribution of a random variable $Y$ in which the compounding distribution has expected value $\mu(\mathbb R)$ and the compounded distribution is $\mu/\mu(\mathbb R).$
Proof:
\begin{align}
\Pr(Y\in A) & = \operatorname E(\Pr(Y\in A\mid K)) \\[8pt]
& = \sum_{k\,=\,0}^\infty \Pr(Y\in A\mid K=k)\Pr(K=k) \\[8pt]
& = \sum_{k\,=\,0}^\infty \nu^{*k}(A)\cdot \frac{e^{-\ell} \ell^k}{k!} \\[8pt]
& = \operatorname{Pois}(\mu)(A).
\end{align}
