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I am given a function $$e^{\lambda(\varphi(t) -1)} \tag{1},$$ where $\varphi(t)$ is a characteristic function. I managed to show that $(1)$ is a characteristic function too.
Now I am to show that $(1)$ is an infinitely divisible function. What does it mean?

I know that a distribution is infinitely divisible if it can be expressed as the probability distribution of the sum of an arbitrary number of independent and identically distributed random variables.

Do I have to find the distribution of my characteristic function and then show that it is infinitely divisible?

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    $\begingroup$ Since $\psi:=e^{\lambda(\phi-1)}$ is a characteristic function, we can associate a probability measure, say $\mu$, with $\psi$ through the relation $$\psi(t) = \int e^{it x} \, \mu(dx).$$ You are supposed to show that $\mu$ is infinitely divisible. $\endgroup$
    – saz
    Jan 6, 2019 at 17:37
  • $\begingroup$ @saz thanks for your comment. Why are we considering only $\varphi(t)$? Not the full expression that contains it? $\endgroup$
    – Hendrra
    Jan 6, 2019 at 17:40
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    $\begingroup$ Ah, sorry, my mistake... please see my edited comment. $\endgroup$
    – saz
    Jan 6, 2019 at 17:42
  • $\begingroup$ Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $\mu$ or is there a method of dealing with such things? $\endgroup$
    – Hendrra
    Jan 6, 2019 at 17:59

1 Answer 1

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Hints:

  1. Let $\mu$ be a probability distribution with characteristic function $\psi$. Show that $\mu$ is infinitely divisible if for any $n \in \mathbb{N}$ there exists a characteristic function $\Phi$ such that $$\psi(t) = (\Phi(t))^n, \qquad t \in \mathbb{R}^d.$$
  2. Use Step 1 for $\psi(t) = e^{\lambda (\varphi(t)-1)}$. Try to find a suitable characteristic function $\Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)
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  • $\begingroup$ Thanks! That was easier than I thought. So to sum up you showed that there exist a function $\Phi(t) = e^{\lambda/n(\varphi(t) - 1)}$ which is a characteristic function thus $\mu$ is infinitely divisible and $\psi$ by definition is an infinitely divisible function? $\endgroup$
    – Hendrra
    Jan 6, 2019 at 19:03
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    $\begingroup$ @Hendrra Yes, that's it. $\endgroup$
    – saz
    Jan 6, 2019 at 19:47

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