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I'm working through the CLT proof on Wikipedia trying to get a better intuition, and it made me wonder what an individual distribution looks like after dropping the o(t^2/n) terms from the Taylor series by trying to compute the distribution corresponding to the (1-t^2/(2*n)) characteristic function. However, there seems to be no such distribution.

What is the reason this doesn't exist? Is there an intuition behind it? Is there such a thing as the distribution with the CF closest to (1-t^2/(2*n))?

Gabi
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    What makes you say there's no such distribution? – Glen_b Nov 30 '18 at 16:20
  • @Glen_b Not finding it in any tables and Maple saying that the corresponding integration is undefined :P – Gabi Nov 30 '18 at 16:20
  • Yes, actually, that makes sense; characteristic functions are bounded, but this isn't. – Glen_b Nov 30 '18 at 16:37
  • @Glen_b would it be possible to find the bounded function that is closest in some reasonable/intuitive respect to (1-t^2/(2*n))? – Gabi Nov 30 '18 at 16:39
  • For some choice of 'closest' we possibly could but I'm not sure that tells us anything of value. – Glen_b Nov 30 '18 at 16:40
  • Your question has an interesting and useful answer if you ask it in terms of the cumulant generating function :-). – whuber Nov 30 '18 at 21:09
  • @whuber thanks! I think you're referring to the fact that it translates to "the distribution with zero mean, unit variance, and undefined higher moments". So why is the variance special? is it because it is the moment that is divided by n, with all other higher moments being divided by higher powers of n? – Gabi Dec 01 '18 at 23:12
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    I'm referring to the fact that the Normal is the unique distribution with given first and second cumulants and zero higher cumulants. Why the variance is special is an excellent question. I pondered that a few years ago and tried to explain why at https://stats.stackexchange.com/a/3904/919. – whuber Dec 01 '18 at 23:30

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