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Given: $X$, $Y$ iid random variables, $\mathbb E(X) = 0$, $\mathbb E(X^2) = 1$; $X+Y$ and $X-Y$ are independent; $\phi$ is the characteristic function of $X$ and $Y$ and $ \psi: t \rightarrow \dfrac{\phi(t)}{\phi(-t)}$.

What is, in this case, the series expansion of $\phi$ and $\psi$ at $0$?

JohnD
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    Did you mean $X$ and $Y$ are independent? (If so, it's redundant since you already said "i.i.d.".) – Michael Hardy May 10 '13 at 14:22
  • Might you have meant the series expansion of $\psi$ rather than of $\phi$ at $0$? If not, then what does $\psi$ have to do with the question? – Michael Hardy May 10 '13 at 14:23
  • In fact, I meant both, Phi and Psi. And yes, X and Y are Independent. – JohnD May 11 '13 at 19:22
  • Then what does Y have to do with the question? – Did May 11 '13 at 19:24
  • For this question: Nothing. I'd have another one, which I wanted to ask later, but I can do it also right now: Why is this property valid?: if $psi(t) = 1$ for any t in R, then X and Y are Standard Gaussian random variables. – JohnD May 11 '13 at 20:25
  • @Did : As the question is now written, I think it's clear what $Y$ has to do with it. – Michael Hardy May 11 '13 at 23:24
  • @MichaelHardy Indeed. JohnD: Your last comment, written after the radical revision of your question Michael alludes to, makes no sense: NOW, Y is needed, to express the crucial hypothesis that X+Y and X-Y are independent. – Did May 12 '13 at 12:11

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