characteristic function of a convolution of measures

Question

Take the probability measures $\mu,\nu$ on $\mathbb{R}$ and denote $\varphi_{\mu}$ (the same for $\nu$) its characteristic function. Why holds $$\varphi_{\mu *\nu}(t)=\varphi_{\mu}(t)\cdot\varphi_{\nu}(t)$$ where $\mu*\nu$ denotes the convolution of $\mu$ and $\nu$?

Answer 1

I'll answer in the case where the probability densities exist. Let $f(x)$ and $g(x)$ be the given densities and $h(x)$ the result of the convolution.

$h(x)=\int_{-\infty}^{\infty}f(x-u)g(u)du$. Then $\phi_h(t)=\int_{-\infty}^{\infty}h(x)e^{itx}dx=\int_{-\infty}^{\infty}e^{itx}\int_{-\infty}^{\infty}f(x-u)g(u)dudx$

Change order of integration and let $y=x-u$ to get $\phi_h(t)=\int_{-\infty}^{\infty}g(u)\int_{-\infty}^{\infty}e^{it(y+u)}f(y)dydu=\int_{-\infty}^{\infty}g(u)e^{itu}du\int_{-\infty}^{\infty}e^{ity}f(y)dy=\phi_g(t)\phi_f(t)$.

Using Stieljtes integrals, the proof can be used for distribution functions.

Answer 2

Neither do we need to assume that $\mu$ and $\nu$ admit densities with respect to a common reference measure, nor do we need to restrict this result to $\mathbb R$.

Remember that if $(E,\mathcal E)$ is a measurable group and $$\tau:E^2\to E\;,\;\;\;(x,y)\mapsto xy$$ denotes the group operation, then $\mu\ast\nu$ is the pushforward $\tau(\mu\otimes\nu)$ of the product measure $\mu\otimes\nu$ under $\tau$.

Assuming that $E$ is a $\mathbb R$-Banach space (consider as a group with the group operation being the addition) and $\mathcal E=\mathcal B(E)$, we immediately obtain \begin{equation}\begin{split}\varphi_{\mu\ast\nu}(x')&=\int\tau(\mu\otimes\nu)({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\\&=\int\mu({\rm d}x)\int\nu({\rm d}y)e^{{\rm }i\langle x',\:\tau(x,\:y)\rangle}\\&=\int\mu({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\int\nu({\rm d}y)e^{{\rm }i\langle x',\:y\rangle}=\varphi_\mu(x')\varphi_\nu(x')\end{split}\tag1\end{equation} for all $x'\in E'$.