Show that $\int_{0}^{1} \phi(yt)\, dy$ is a characteristic function, where $\phi(t)$ is itself a characteristic function. I tried:
\begin{align}\int_{0}^{1} \phi(yt)\, dy &= \int_{0}^{1} E(e^{iytX})\, dy\\
&= \int_{0}^{1} E(\cos ytX + i\sin ytX)\, dy\\
&= \int_{0}^{1} E(\cos ytX)\, dy + i\int_{0}^{1} E(\sin ytX)\, dy\\
&= \text{??}
\end{align}
No idea where to go from here.
 A: This is just a matter of translating between expectations and integrals, which are basically the same thing.
In probabilistic terms it's quite simple - just introduce an independent $\mathrm{Uniform}(0,1)$ variable $Y$, and note that $E(e^{itXY})=\int_0^1 E(e^{ityX})dy.$
In measure-theoretic terms, a function $\psi$ is a characteristic function iff it can be expressed as
$$\psi(t)=\int_{\Omega} e^{itZ(\omega)} d\omega\tag{*}$$ for some measure space $\Omega$ of measure $1$ and some measurable function $Z:\Omega\to\mathbb R.$ You are given some measurable $X:\Omega_1\to\mathbb R$ and you need to construct $\Omega$ and $Z$ such that
$$\int_{0}^1 \int_{\Omega_1} e^{ityX(\omega_1)} d\omega dy=\int_{\Omega}e^{it Z(\omega)} d\omega$$
In these terms the answer is to take $\Omega=\Omega_1\times [0,1]$ and $Z(\omega_1,y)=X(\omega_1)y.$
A: Use Bochner's theorem which states a function is a characteristic function if and only if it is 


*

*positive semi-definite

*The value at $0$ is $1$


So let be $$F(t) = \int_{0}^{1} \phi(yt)\, dy$$ then F is positive semi-definite because $\phi$ is and obviously $$F(0) = \int_{0}^{1} \phi(y\cdot0)\, dy = \int_{0}^{1} 1\, dy = 1$$
Hence $F$ is a characteristic function.
