Problem. I am given an infinitely divisible characteristic function $\varphi(t)$. My task it to prove that $|\varphi(t)|$ is infinitely divisible too.
My attempt. Because $\varphi$ is infinitely divisible then for any $n \in \mathbb{N}$ there exists characteristic functions $\varphi_1(t)= \varphi_2(t) = \ldots \varphi_n(t)$ such that
$$\varphi(t) = \big(\varphi_1(t) \big)^n.$$
I noticed that $|\varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|\varphi(t)|^2 = \varphi(t) \overline\varphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $\overline\varphi(t)$ is infinitely divisible characteristic function because
$$\varphi(t) = \overline\varphi(-t) = \big(\varphi_1(t) \big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?
Best Answer
Let $\varphi_j(t)$ be the characteristic function such that $$ \varphi(t) = [\varphi_j(t)]^j $$ holds. Let $\varphi_2(t) =\phi(t)$. We claim that $\phi$ is infinitely divisible and $\phi_j(t) = \varphi_{2j}(t)$. Note that $$ [\phi(t)]^{2}=\varphi(t)=[\varphi_{2j}(t)]^{2j}, $$ and hence $\frac{\phi(t)}{ (\varphi_{2j}(t))^j}=\omega(t)$ where $\omega(t)\in \{-1,1\}$ for all $t\in\mathbb{R}$ (note that infinitely divisible characteristic function $\varphi$ nowhere vanishes and hence neither does $\varphi_{2j}$.) This gives by continuity of $\omega(t)$ that $\omega \equiv 1$ and $\phi(t) =[\varphi_{2j}(t)]^j$. This shows $\phi$ is infinitely divisible. Now, since $\phi(t)$ is infinitely divisible, so does $\overline{\phi}(t)$ and $\phi(t)\overline{\phi}(t)=|\phi(t)|^2=|\varphi(t)|$.