# Question involving Characteristic Functions and the Existence of a Distribution

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.

• Incredibly enough, that ODE can be solved setting "$u = \Phi(t)$" so that "$\dfrac{du}{u^2} = \dfrac{dt}{t^2}.$" – Will M. Dec 9 '18 at 17:36

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.