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could you please help me on the following problem?

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 $0<a<1$ 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 $0<a<1/2$.

Do you have some hints or ideas?

Thanks a lot in advance for your help.

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  • $\begingroup$ 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? $\endgroup$ – Kavi Rama Murthy Sep 30 '18 at 11:35
  • $\begingroup$ 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 $\endgroup$ – Piergiacomo Sep 30 '18 at 11:51
  • $\begingroup$ 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. $\endgroup$ – kimchi lover Sep 30 '18 at 13:15
  • $\begingroup$ If for instance $Y$ is an exponential r.v. $\phi_X(u)$ is well defined. That’s actually where I started from. $\endgroup$ – Piergiacomo Sep 30 '18 at 16:04
  • $\begingroup$ Is $b$ an integer? $\endgroup$ – Davide Giraudo Jan 4 at 15:03

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