Учитывая три вектора , и , возможно ли, чтобы корреляции между и , и , а также и были отрицательными? Т.е. возможно ли это?
correlation
correlation-matrix
Антти А
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Ответы:
Это возможно, если размер вектора равен 3 или больше. Например
Корреляции следующие:
Мы можем доказать, что для векторов размера 2 это невозможно:
Формула имеет смысл: если больше, чем a 2 , b 1 должно быть больше, чем b 1, чтобы сделать корреляцию отрицательной.a1 a2 b1 b1
Аналогично для соотношений между (a, c) и (b, c) получаем
Clearly, all of these three formulas can not hold in the same time.
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Yes, they can.
Suppose you have a multivariate normal distributionX∈R3,X∼N(0,Σ) .
The only restriction on Σ is that it has to be positive semi-definite.
So take the following exampleΣ=⎛⎝⎜1−0.2−0.2−0.21−0.2−0.2−0.21⎞⎠⎟
Its eigenvalues are all positive (1.2, 1.2, 0.6), and you can create vectors with negative correlation.
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let's start with a correlation matrix for 3 variables
non-negative definiteness creates constraints for pairwise correlationsp,q,r which can be written as
For example, ifp=q=−1 , the values of r is restricted by 2r≥r2+1 , which forces r=1 . On the other hand if p=q=−12 , r can be within 2±3√4 range.
Answering the interesting follow up question by @amoeba: "what is the lowest possible correlation that all three pairs can simultaneously have?"
Letp=q=r=x<0 , Find the smallest root of 2x3−3x2+1 , which will give you −12 . Perhaps not surprising for some.
A stronger argument can be made if one of the correlations, sayr=−1 . From the same equation −2pq≥p2+q2 , we can deduce that p=−q . Therefore if two correlations are −1 , third one should be 1 .
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A simple R function to explore this:
As a function of
n
,f(n)
starts at 0, becomes nonzero atn = 3
(with typical values around 0.06), then increases to around 0.11 byn = 15
, after which it seems to stabilize:So, not only is it possible to have all three correlations negative, it doesn't seem to be terribly uncommon (at least for uniform distributions).
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