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 $Z$.

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


1 Answer 1


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.

  • $\begingroup$ "simple" seems to be relative... $\endgroup$ Nov 27, 2018 at 23:44
  • $\begingroup$ I have used your hint to solve the first part but am not sure how to get the negative into an absolute value. $\endgroup$ Nov 28, 2018 at 15:37
  • $\begingroup$ @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 $|z|^{2}=z\overline {z}$. $\endgroup$ Nov 28, 2018 at 23:18

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