yes!
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I’m sorry, I don’t know any stats/pt, so this is probably dumb but my intuition is that this
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Replying to @Meaningness @karlrohe and
only would apply to a middle slice like in
@The_Lagrangian ‘s plot. If IQ were sum of additive features2 replies 0 retweets 0 likes -
Replying to @Meaningness @karlrohe and
I would think asymptotic independence would look like a barbell.
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can you say another 140 while I google “asymptotic independence”?
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Replying to @Meaningness
@karlrohem Well, to posit an extreme example: suppose X is a Gaussian normal, and Y is defined to be = X if X \in [-1, 1]...
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Replying to @St_Rev @Meaningness
...and if X not \in [-1,1], Y is randomly sampled from a Gaussian normal until it lands outside [-1,1].
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Replying to @St_Rev @Meaningness
for tail dependence, it needs to be the opposite. X=Y for extreme values, i.e. outside [-1,1].
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for tail independence, you don't need to do anything special to the joint gaussian. it already is tail independent.
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Replying to @karlrohe @Meaningness
I was trying to describe asymptotic independence, as that is the term you brought into the discussion.
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And yes independence -> asymptotic independence but that's useless for describing what the latter is.
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Replying to @St_Rev @Meaningness
Correlated Bivariate gaussians are tail independent.
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. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.