how to prove that the real part and the modulus of a characteristic function is still a characterisitc function? [closed]

Question

this is a problem from Durret's probability textbook.

Show that if $\varphi$ is a ch.f., then $Re\varphi$ and $|\varphi|^2$ are also ch.f.

I am wondering how to prove this. Actually I'm not even sure how to show something is ch.f. The only idea I have is to use the inversion formula to show that we can get the distribution function of a random variable out of $Re\varphi$, but I couldn't work out the details as it seems pretty complicated.

For the $|\varphi|^2$, since $|\varphi|^2=\varphi\overline{\varphi}$, so I think it suffices to show that $\overline{\varphi}$ is a ch.f. But similar to $Re\varphi$, I am not sure how to show $\overline{\varphi}$ is a ch.f. either.

Answer 1

Although https://math.stackexchange.com/ would be a better place to ask this question, let me answer it.

Since the characteristic function is basically the Fourier transform, let me explain it in terms of the Fourier transform $\varphi=\hat{f}$ and you can translate it to the language of the characteristic functions. I mean the Fourier transform defined by $$ \hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot\xi}\, dx. $$ If $g(x)=\bar{f}(-x)$ is the complex conjugte of the "reflection" of $f$, then moving the conjugate under the sign of the integral and applying change of variables $x\mapsto -x$ yields that $\hat{g}(\xi)=\bar{\hat{f}}(\xi)$. Therefore, $\operatorname{re} \varphi=\operatorname{re} \hat{f}=\widehat{\frac{f+g}{2}}$. Now the fact that $\widehat{f*g}=\hat{f}\hat{g}$ implies that $$ |\varphi|^2=|\hat{f}|^2=\hat{f}\bar{\hat{f}}=\hat{f}\hat{g}=\widehat{f*g}. $$ In probabilistic terms, if independent random variables have distributions $f$ and $g$, then the sum of random variables has the distribution $f*g$.