Characteristic Function and Generalized Binomial Theorem

I have the characteristic functions of two random variables $$X$$ and $$Y$$, denoted $$\phi_X(u)$$ and $$\phi_Y(u)$$, such that: $$\begin{equation*} \phi_X(u) = \left(a + (1-a)\phi_Y(u) \right)^b \end{equation*}$$ with $$0 and $$b>0$$. I know the density of $$Y$$ and would like somehow to know the one of $$X$$.

If $$1/2\le a<1$$, I can write: $$\begin{equation*} \phi_X(u) = a\left(1 + \frac{(1-a)}{a}\phi_Y(u) \right)^b \end{equation*}$$ and I can apply the generalized binomial theorem because $$\frac{(1-a)}{a}<1$$ and $$|\phi_Y(u)|<\le 1$$. Because I know the law of $$Y$$ I can get a sort of infinite mixture of densities that is fine for me.

However, I do not know how to proceed in case $$0.

Do you have some hints or ideas?

• Characteristic functions are, in general, complex valued and there is a problem in defining $z^{b}$ for a complex number $z$. Are yo assuming that $\phi_Y$ is positive? – Kavi Rama Murthy Sep 30 '18 at 11:35
• No I am not restricted to this assumption. I know for sure that both characteristic functions are well defined for $b>0$ and $0<a<1$. thanks – Piergiacomo Sep 30 '18 at 11:51
• Why do you know that $\phi_Y$ is a characteristic function? There are, for instance, very few characteristic functions $\phi_Z$ such that for each $b>0$, the function $\phi_Z^b$ is a characteristic function. Your $a+(1-x)\phi_X$ must be one of them, if I understand your question right. – kimchi lover Sep 30 '18 at 13:15
• If for instance $Y$ is an exponential r.v. $\phi_X(u)$ is well defined. That’s actually where I started from. – Piergiacomo Sep 30 '18 at 16:04
• Is $b$ an integer? – Davide Giraudo Jan 4 at 15:03