Does real characteristic function imply even density

Question

Does real characteristic function imply even density?

So, I was wondering over this. I know that when a characteristic function is real, it is even. Also, I know this result when the characteristic function is absolutely integrable, by the inversion formula. But can we generalize it?

1) Does real characteristic function imply that density exists and it is even?

2) How about the case when we know that density exists? Can we conclude about evenness?

Please prove or provide a counterexample.

Answer 1

A characteristic function is real iff the distribution is symmetric. So a random variable taking values $\pm 1$ with probability $\frac 1 2$ each has real characteristic function but it does not have density. If the characteristic function of a random variable $X$ is real and there is density $f$ then $f(-x)=f(x)$ almost everywhere because $f(-x)$ is the density of $-X$ and $-X$ has same distribution as $X$.

Answer 2

If $X$ is a real-valued random variable, $f$ its characteristic function, then $x \longmapsto f(-x)$ is the characteristic function of $-X$.

So $f$ real iff $f$ even iff $X$ and $-X$ have the same characteristic function iff $X$ and $-X$ have the same law.

Therefore, obviously, if $X$ has a density, $-X$ has the same, so the density is even.

However, it is possible for a random variable to be symmetrical even though it doesn’t have a density: consider $X=1$ with probability $1/2$ and $X=-1$ with probability $-1/2$.

Answer 3

If $X$ has characteristic function $\phi(t)$, then $\phi^\ast(t)=\phi(-t)$ is the characteristic function of $-X$. So the following are equivalent:

• $\forall t\in\Bbb R\left(\phi(t)\in\Bbb R\right)$;
• $\forall t\left(\phi^\ast(t)=\phi(t)\right)$;
• $\forall t\left(\phi(-t)=\phi(t)\right)$;
• $X$ has the same distribution as $-X$.

Insofar as the distribution of $X$ can be described with something like a pdf or pmf, possibly with a Dirac comb, evenness thereof is equivalent to the last of these.