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I have the same problem as here: Find characteristic function of ZX+(1-Z)Y with X uniform, Y Poisson and Z Bernoulli. Could you explain last equality? Can I get it without using law of total expectation?

Update

I have some idea, $E $$ e^{it(ZX + (1-Z)Y)} = E $$ (Ze^{itX} + (1-Z)e^{itY}) =$ $ = E $$ (Ze^{itX}) + E((1-Z)e^{itY}) = EZEe^{itX} + E(1-Z)Ee^{itY} = pEe^{itX} + (1-p)Ee^{itY}$

How to prove that $ e^{it(ZX + (1-Z)Y)} = Ze^{itX} + (1-Z)e^{itY} $ ?

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Alexander
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Treat the cases $Z=0$ and $Z=1$ (the only two possible values of $Z$, since $Z$ has Bernoulli distribution).

Davide Giraudo
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    Let $ \psi_1 = e^{it(ZX + (1-Z)Y)} $ and $ \psi_2 = Ze^{itX} + (1-Z)e^{itY} $. Then $ P(\psi_1 \not= \psi_2) = 1 - P(\psi_1 = \psi_2)$ $ = 1 - P(\psi_1 = \psi_2, Z = 0) - P(\psi_1 = \psi_2, Z = 1) $ $ = 1 - P(e^{itY} = e^{itY})P(Z = 0) - P(e^{itX} = e^{itX})P(Z = 1) $ $ = 1 - (1 - p) - p = 0 $. It is right? – Alexander Jan 13 '15 at 13:20