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
Then Y is actually Gaussian normal, and the scatter plot of X vs. Y looks like a Ø with a hard line segment in the center...
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Replying to @St_Rev @Meaningness
...but the outer O is a Gaussian blur ring.
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Replying to @St_Rev @Meaningness
So X and Y are perfectly correlated in the center, and uncorrelated outside it.
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I understand the barbell comment now!
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