For any $\phi$ a characteristic function, is there a characteristic function $\Phi$ such that $\Phi ^2=\phi$?
I know that if $\phi$ is a characteristic function, then $\phi ^2$ is as well. Also that $\sqrt{|\phi|}$ is not always a characteristic function.
But I don’t know how to get to the other way round.
Thanks!!!
In general, not. For example, $\phi(t)=\cos(t)$, the characteristic function for an unbiased $\pm 1$ r.v. cannot be written as $\phi=\Phi^2$ for any characteristic function $\Phi$. (For if there were such a $\Phi$, there would be a probability measure $\mu$ for which $\Phi(t)=\int\exp(itx)\mu(dx)$, and for which $\mu*\mu$ was the uniform distribution on the set $\{-1,1\}$. The support of $\mu$ must be a set $S$ such that $S+S=\{\pm1\}$. Such an $S$ set must have more than $1$ element, or else the cardinality of $S+S$ is too small. But if it has two or more elements, the cardinality of $S+S$ is at least $3$. Therefore there is no such $S$, and hence no such $\mu$, and finally, no such $\Phi$.)
If $\phi$ is the characteristic function of an infinitely divisible distribution, then yes, but this is an exceptionally tiny subset of all characteristic functions.
