Suppose $\varphi_Y$ is a characteristic function. Let $p \in (0,1),\ q = 1-p$.
Show $\bar{\varphi}(t) = \frac{p}{1-q \varphi_Y(t)}$ is a characteristic function.
My thoughts:
Let X ~ Ge(p).
Then $\varphi_X(t) = \frac{p}{1-q \exp(it)}$
$\varphi_X(\frac{\ln(\varphi_Y(t))}{i}) = \frac{p}{1-q \varphi_Y(t)} = \bar{\varphi}(t)$
$\bar{\varphi}(t) = \varphi_X(\frac{\ln(\varphi_Y(t))}{i}) = E(\exp( i(\frac{\ln(\varphi_Y(t))}{i})X)) = E(\varphi_Y(t)X) = E(X) \varphi_Y(t) = \frac{q\varphi_Y(t)}{p}$
I don't think this can be right, but I also don't know what I did wrong. Can someone point out what I did wrong? Thanks!
First of all logarithms of characteristic functions are not well defined. Logarithms in the complex plane are quite complicated. Even if we assume that $\phi_Y(t)$ is positive for each $t$ your calculations are wrong.
$e^{(\ln (\phi_Y(t)) (X)}=((\phi_Y(t))^{X}$, not $(\phi_Y(t)) X$.