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My textbook states, that a probability measure $P$ is called infinitely divisible, iff

$$\forall n\in \mathbb{N}\ \exists P_n \text{ probability measure: }P=\ast_{i=1}^n P_n$$

A random variable $X$ is called infinitely divisble, iff

$$\forall n\in\mathbb{N}\ \exists X_1,\ldots , X_n \text{ iid random variables: } P^X=P^{X_1+\ldots +X_n}$$

Why are these definitions equivalent? I have problems with the implication

$$\text{$X$ infinitely divisible $\Leftarrow$ $P$ infinitely divisible}$$

I tried following:

Let $n\in\mathbb{N}$. I can find random variables $X$ and $X_1$ such that $P=P^X$ and $P_n=P^{X_1}$, such that $$P^X=P=\ast_{k=1}^n P_n=\ast_{k=1}^n P^{X_1}$$ If I can show there exist $X_2,\ldots, X_n$, such that $X_1,\ldots, X_n$ are iid, it follows

$$P^X=\ast_{k=1}^n P^{X_1}=\ast_{k=1}^n P^{X_k}=P^{X_1+\ldots+ X_n}$$

But how can I prove existence of $X_2,\ldots, X_n$, such that $X_1,\ldots, X_n$ are iid?

user408858
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  • Not an answer, but you are correct to notice that it is not trivial to magically find iid random variables. Kallenberg's Modern Probability has a chapter on this, I think the result is called Skorokhod coupling in it but don't quote me. But it is possible, though not just hand-waving. – I was suspended for talking Sep 30 '19 at 16:50
  • Ok, so I am actually not really looking for a prove how to construct. I am just wondering if there is a theorem which just states that. So what you are saying is: "It is possible to find these variables", right? Because this would already satisfy my needs. – user408858 Sep 30 '19 at 17:00
  • Maybe this post helps as well? https://math.stackexchange.com/questions/2954711/given-a-random-variable-x-1-does-there-always-exsist-an-i-i-d-sequence-x?noredirect=1&lq=1 – user408858 Sep 30 '19 at 17:10
  • Yes that would suffice, Skorokhod coupling says something more specific, I'm sorry. But yes, it is possible, as the post shows. – I was suspended for talking Sep 30 '19 at 18:34

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