3个向量是否可能都具有负的成对相关性？

16

给定三个向量，和，和，和以及和之间相关性是否可能都是负的？即有可能吗？ $a$ $b$ $c$ $a$ $b$ $a$ $c$ $b$ $c$

\begin{aligned} corr (a, b) < 0 \\ corr (a, c) < 0 \\ corr (b, c) < 0 \end{aligned}

$\begin{align} \text{corr}(a,b) < 0\\ \text{corr}(a,c) < 0 \\ \text{corr}(b,c) < 0\\ \end{align}$

correlation correlation-matrix

— 安蒂
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3

负相关在几何上意味着居中的矢量相互成钝角。在平面上绘制具有此属性的三个向量的配置应该没有问题。

— ub

它们不能完全负相关（

ρ = - 1

$\rho=-1$ ），但通常可能存在一些负相关，同样由其他相关设置界限。

— karakfa

2

@whuber您的意见似乎与Heikki Pulkkinen的回答相矛盾，后者的回答是说飞机上的矢量不可能实现。如果您支持它，则应将您的评论变成答案。

— RM

2

@RM胡伯和海基之间没有矛盾。这个问题询问数据矩阵

X

$X$ 的

n \times 3

$n\times 3$ 大小。通常，我们会谈论3维的

n

$n$ 数据点，但是这个Q谈论的是

n

$n$ 维的三个“向量” 。Heikki说，如果

，则所有负相关都不会发生

n = 2

$n=2$ （实际上，居中后的两个点始终是完美相关的，因此相关必须为

\pm 1

$\pm 1$ 且不能全部为

- 1

$-1$ ）。Whuber说，

n

$n$ 维度中的3个向量可以有效地位于2维子空间（即

X

$X$ 是排名2），并建议想象一个梅赛德斯标志。

— 变形虫说恢复莫妮卡

1

相关：绑定三个随机变量的相关性。（cc，@amoeba）

— -恢复莫妮卡

19

向量的大小可能为3或更大。例如

\begin{aligned} a & = (- 1, 1, 1) \\ b & = (1, - 9, - 3) \\ c & = (2, 3, - 1) \end{aligned}

$\begin{align} a &= (-1, 1, 1)\\ b &= (1, -9, -3)\\ c &= (2, 3, -1)\\ \end{align}$

相关性是

cor (a, b) = - 0.80... cor (a, c) = - 0.27... cor (b, c) = - 0.34...

$\begin{equation} \text{cor}(a,b) = -0.80...\\ \text{cor}(a,c) = -0.27...\\ \text{cor}(b,c) = -0.34... \end{equation}$

我们可以证明，对于大小为2的向量，这是不可能的：

\begin{aligned} cor (a, b) & < 0 \\ 2 (\sum_{i} a_{i} b_{i}) - (\sum_{i} a_{i}) (\sum_{i} b_{i}) & < 0 \\ 2 (a_{1} b_{1} + a_{2} b_{2}) - (a_{1} + a_{2}) (b_{1} b_{2}) & < 0 \\ 2 (a_{1} b_{1} + a_{2} b_{2}) - (a_{1} + a_{2}) (b_{1} b_{2}) & < 0 \\ 2 (a_{1} b_{1} + a_{2} b_{2}) - a_{1} b_{1} + a_{1} b_{2} + a_{2} b_{1} + a_{2} b_{2} & < 0 \\ a_{1} b_{1} + a_{2} b_{2} - a_{1} b_{2} + a_{2} b_{1} & < 0 \\ a_{1} (b_{1} - b_{2}) + a_{2} (b_{2} - b_{1}) & < 0 \\ (a_{1} - a_{2}) (b_{1} - b_{2}) & < 0 \end{aligned}

$\begin{align} \text{cor}(a,b) &< 0\\[5pt] 2\Big(\sum_i a_i b_i\Big) - \Big(\sum_i a_i\Big)\Big(\sum_i b_i\Big) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - (a_1 + a_2)(b_1 b_2) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - (a_1 + a_2)(b_1 b_2) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - a_1 b_1 + a_1 b_2 + a_2 b_1 + a_2 b_2 &< 0\\[5pt] a_1 b_1 + a_2 b_2 - a_1 b_2 + a_2 b_1 &< 0\\[5pt] a_1 (b_1-b_2) + a_2 (b_2-b_1) &< 0\\[5pt] (a_1-a_2)(b_1-b_2) &< 0 \end{align}$

The formula makes sense: if $a_1$ is larger than $a_2$ , $b_1$ has to be larger than $b_1$ to make the correlation negative.

Similarly for correlations between (a,c) and (b,c) we get

(a_{1} - a_{2}) (c_{1} - c_{2}) < 0 (b_{1} - b_{2}) (c_{1} - c_{2}) < 0

$\begin{equation} (a_1-a_2)(c_1-c_2) < 0\\ (b_1-b_2)(c_1-c_2) < 0\\ \end{equation}$

Clearly, all of these three formulas can not hold in the same time.

— Heikki Pulkkinen
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3

Another example of something unexpected that only happens in dimension three or higher.

— nth

1

With vectors of size

2

$2$ , correlations are usually

\pm 1

$\pm1$ (straight line through two points), and you cannot have three correlations of

- 1

$-1$ with three vectors of any size

— Henry

9

Yes, they can.

Suppose you have a multivariate normal distribution $X\in R^3, X\sim N(0,\Sigma)$ . The only restriction on $\Sigma$ is that it has to be positive semi-definite.

So take the following example $\Sigma = \begin{pmatrix} 1 & -0.2 & -0.2 \\ -0.2 & 1 & -0.2 \\ -0.2 & -0.2 & 1 \end{pmatrix}$

Its eigenvalues are all positive (1.2, 1.2, 0.6), and you can create vectors with negative correlation.

— Kozolovska
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7

let's start with a correlation matrix for 3 variables

$\Sigma = \begin{pmatrix} 1 & p & q \\ p & 1 & r \\ q & r & 1 \end{pmatrix}$

non-negative definiteness creates constraints for pairwise correlations $p,q,r$ which can be written as

p q r \geq \frac{p^{2} + q^{2} + r^{2} - 1}{2}

$pqr \ge \frac{p^2+q^2+r^2-1}2$

For example, if $p=q=-1$ , the values of $r$ is restricted by $2r \ge r^2+1$ , which forces $r=1$ . On the other hand if $p=q=-\frac12$ , $r$ can be within $\frac{2 \pm \sqrt{3}}4$ range.

Answering the interesting follow up question by @amoeba: "what is the lowest possible correlation that all three pairs can simultaneously have?"

Let $p=q=r=x < 0$ , Find the smallest root of $2x^3-3x^2+1$ , which will give you $-\frac12$ . Perhaps not surprising for some.

A stronger argument can be made if one of the correlations, say $r=-1$ . From the same equation $-2pq \ge p^2+q^2$ , we can deduce that $p=-q$ . Therefore if two correlations are $-1$ , third one should be $1$ .

— karakfa
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2

See stats.stackexchange.com/questions/72790/…, inter alia.

— whuber

2

A simple R function to explore this:

f <- function(n,trials = 10000){
  count <- 0
  for(i in 1:trials){
    a <- runif(n)
    b <- runif(n)
    c <- runif(n)
    if(cor(a,b) < 0 & cor(a,c) < 0 & cor(b,c) < 0){
      count <- count + 1
    }
  }
  count/trials
}

As a function of n, f(n) starts at 0, becomes nonzero at n = 3 (with typical values around 0.06), then increases to around 0.11 by n = 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).

— John Coleman
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