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NEW EDIT TO CLARIFY THE QUESTION

My initial question was about why square difference was used instead of absolute value in the formula of the variance... But I found out that it was already asked and answered here. Very interesting answers! Most of them highlighted the fact it was simpler to use squares and then square root than absolute value. This analysis: Revisiting a 90-year-old debate: the advantages of the mean deviation - Stephen Gorard was also going into this direction.

However, I was intrigued by a comment from @Whuber (which seems to be an important point: "They focus on ease of mathematical calculations (...) A truly fundamental reason that has not been invoked in any answer yet is the unique role played by the variance in the Central Limit Theorem".

Clearly my basic knowledge does not allow me to grasp it correctly... Or probably I just missed the point! There is something that I did not get about the variance and the CLT that could explain why we are using squares and not absolute values. So, I am interested in understanding this point (layman explanations would be ideal)... and if I am just confused, I will delete this question! Thanks for your help.

Mat
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    It is very unclear what you are pointing at. Do you have a link to the particular comment? What is the actual question? What needs to be explained? The role of variance in the central limit theorem? Why isn't the link https://stats.stackexchange.com/questions/3734/what-intuitive-explanation-is-there-for-the-central-limit-theorem doing this already sufficiently? – Sextus Empiricus Oct 02 '21 at 12:17
  • @Sextus The reference is to my comment at https://stats.stackexchange.com/questions/118/why-square-the-difference-instead-of-taking-the-absolute-value-in-standard-devia#comment135385_118. – whuber Oct 02 '21 at 14:16
  • @whuber, I tried to clarify my question... If I am just confused, please let me know, this is also fine :-) – Mat Oct 03 '21 at 14:56
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    @Mat: I think ( whuber: correct me if I'm wrong of course ) what whuber is saying is that, if some other measure of dispersion was used in the denominator ( rather than the sqrt of the squared differences ) when normalizing a random variable's distance from its mean, then the CLT would not hold. You can't put the absolute value of the differences in the denominator and still get the CLT result. – mlofton Oct 03 '21 at 23:43
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    @mlofton That's pretty close to it, understanding that any measure of spread that is *asymptotically* equivalent to the SD will still work in the CLT. – whuber Oct 04 '21 at 11:22
  • Thank you both... I need to digest this information now :) – Mat Oct 04 '21 at 13:30

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