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Question

Is it possible for $X$, $Y$ and $Z$ to have the same distribution and satisfy $X=U(Y+Z)$ where $U$ is uniform on $[0,1]$ and $Y$, $Z$ are independent of $U$ and of one another?

The above question is from Grimmett and Stirzaker.

My attempt

We translate the condition into characteristic functions. Let $\phi(t)=Ee^{it X}$ be the characteristic function of $X$. Then $$ \phi(t)=Ee^{itUY}Ee^{itUZ}=(Ee^{itUX})^2=\left[\int_0^1 \int e^{itux}\,dF(x)\, du\right]^2=\left[\int_0^1 \phi(tu)\, du\right]^2 $$ using the independence and equality of distribution assumptions. We can write the above equation as $$ \phi (t)=\frac{1}{t^2}\left[\int_0^t \phi(y)\, dy\right]^2 $$ but I am not sure where to proceed from here. I guess we have to solve a differential equation. Put $\Phi(t)=\int_0^t \phi(y)\, dy$. Then we have that $$ \Phi'(t)=\frac{1}{t^2}\Phi(t)^2 $$ but I am unable to solve this differential equation.

Any help is appreciated.

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  • $\begingroup$ Incredibly enough, that ODE can be solved setting "$u = \Phi(t)$" so that "$\dfrac{du}{u^2} = \dfrac{dt}{t^2}.$" $\endgroup$ – Will M. Dec 9 '18 at 17:36
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Rearranging your differential equation gives $d\Phi/\Phi^2=dt/t^2$ so $1/\Phi=1/t+C$, i.e. $\Phi=\frac{t}{1+Ct}$. Hence $\phi=\frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|\phi|\le 1$ would fail for some $t\in\Bbb R$). This implies $X,\,Y,\,Z$ are identically $0$, which works.

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