# Show that $\psi(t)=e^{\lambda (\varphi(t)-1)}$ is infinitely divisble for any characteristic function $\varphi$

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?

• 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.
– saz
Jan 6, 2019 at 17:37
• @saz thanks for your comment. Why are we considering only $\varphi(t)$? Not the full expression that contains it? Jan 6, 2019 at 17:40
• Ah, sorry, my mistake... please see my edited comment.
– saz
Jan 6, 2019 at 17:42
• 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? Jan 6, 2019 at 17:59

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 $$e^x = (e^{x/n})^n$$.)
• 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? Jan 6, 2019 at 19:03