# Functions of Characteristic Functions

Suppose that $$\phi(t)$$ is the characteristic function of a random variable. Show that the functions $$\phi^2(t)$$ and $$|\phi(t)|^2$$ are also characteristic functions.

Attempt: So since $$\phi(t)$$ is a characteristic function it satisfies Bochner's theorem but I was told by user Kavi Rama Murthy that this is not helpful.

So Instead we consider our random variable $$X$$, and an identically independently distributed variable $$Y$$, both with the same characteristic function $$\phi(t)$$. The Characteristic function of the random variable $$Z=X+Y$$ is

$$\phi_Z(t)=\mathbf{E}[e^{itZ}]=\mathbf{E}[e^{it(X+Y)}]=\mathbf{E}[e^{itX}e^{itY}]=\mathbf{E}[e^{itX}]\mathbf{E}[e^{itY}]=\phi_X(t)\phi_y(t)=\phi^2(t)$$ So since $$X,Y$$ are i.i.d we have that the characteristic function of the random variable $$Z=X+Y$$ is $$\phi^2(t)$$.

Now I was given a hint, again by Kavi Rama Murthy, to use $$X-Y$$ but I am slightly confused as to how to continue this is my attempt:

Attempt: Consider the random variable $$W=X-Y$$ with $$X,Y$$ i.i.d. then we find the characteristic function of $$Z$$.

$$\phi_Z(t)=\mathbf{E}[e^{itZ}]=\mathbf{E}[e^{it(X-Y)}]=\mathbf{E}[e^{itX}]\mathbf{E}[e^{-itY}]=\phi_X(t)\mathbf{E}[e^{-itY}]$$

If $$X$$ and $$Y$$ are i.i.d. with characteristic function $$\phi$$ a simple calculation show that $$\phi^{2}$$ is the characteristic function of $$X+Y$$ and $$|\phi|^{2}$$ is the characteristic function of $$X-Y$$. Bochner's Theorem is not useful here.
• @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=z\overline {z}$. Nov 28, 2018 at 23:18