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I've read that the characteristic function of a probability distribution always exists because it's bounded. However, the characteristic function is still Taylor expanded in terms of the moments of a given probability distribution. Given the the moments don't necessarily exist, why is it that the characteristic function still exists?

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If $X$ is a random variable, then what it means for the characteristic function of $X$ to exist is that the random variable $|e^{itX}|$ must have finite expectation for every $t$. This is automatically true because $|e^{itX}|=1$ for all $t \in \mathbb{R}$, hence $E(|e^{itX}|)=1<\infty$. This is true regardless of whether or not the moments exist. The characteristic function can be Taylor expanded in terms of the moments only if the moments actually exist, but in general they won't.

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    $\begingroup$ Ahh, I see. So the Taylor expansion is not fundamental to the definition of the characteristic function, then? $\endgroup$ Mar 20, 2015 at 6:44
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    $\begingroup$ No, not at all. $\endgroup$
    – shalop
    Mar 20, 2015 at 6:45
  • $\begingroup$ @shalop Why do we take the modulus value of the random variable $e^{itX}$ ? $\endgroup$
    – Esha
    Oct 11, 2020 at 5:01
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    $\begingroup$ By definition, the expectation of a complex-valued random variable is said to exist iff the absolute value of both the real and imaginary parts of the random variable have finite integral, which is equivalent to the modulus of the random variable having finite integral. $\endgroup$
    – shalop
    Oct 11, 2020 at 7:08

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