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

Question

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?

Answer 1

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 $e^x = (e^{x/n})^n$.)