Square of the Absolute Value of a Characteristic Function is a Characteristic Function

Question

I am working on the following problem:

Suppose that $\varphi(t)$ is the characteristic function for some random variable.

Show that $|\varphi(t)|^2$ is a characteristic function. For what random variable?

So I know that $\varphi(t)=E\left(e^{itX}\right)$

Also I get that $|\varphi(t)|^2=\varphi(t) \overline{\varphi(t)}=E\left(e^{itX}\right)\overline{E\left(e^{itX}\right)}=E\left(e^{itX}\right)E\left(e^{-itX}\right).$

Here I get stuck trying to continue the calculation.

Any help would be appreciated!

Answer 1

Hint:

If $X$ and $Y$ are independent then: $$\phi_{X+Y}(t)=\phi_X(t)\phi_Y(t)$$

If $Y$ and $-X$ have the same distribution then: $$\phi_Y(t)=\overline{\phi_X(t)}$$

Answer 2

Hint. Let $X'$ be an independent copy of $X$, then $\exp(itX)$ and $\exp(-itX')$ are independent also, but, as $X$ and $X'$ have the same distribution, we have $\overline{\phi(t)} = \mathbf E[\exp(-itX')]$. Now use what you know about the expectation of independent products.

Answer 3

Here is another look at this using Fourier analysis. If $X$ has cdf $F(x) = \mathbb P(X \leqslant x)$, then $-X$ has cdf $G(x) = 1-F(-x)$. Since a probability measure is determined by its cdf, and by the way that the Fourier transform interacts with convolution, $$ \widehat{F\ast G} = \hat F\cdot \hat G = \mathbb E(e^{itX})\cdot \mathbb E(e^{-itX}) = |\varphi(t)|^2. $$ So take a r.v. distributed according to $F\ast G$, say $X - Y$, where $X,Y\sim F$ are independent.