¿Es posible que 3 vectores tengan todas las correlaciones negativas por pares?

Dados los tres vectores $a$ , $b$ y $c$ , ¿es posible que las correlaciones entre $a$ y $b$ , $a$ y $c$ , y $b$ y $c$ sean todas negativas? Es decir, ¿es esto posible?

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

Es posible si el tamaño del vector es 3 o mayor. Por ejemplo

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

Las correlaciones son

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

Podemos demostrar que para vectores de tamaño 2 esto no es posible:

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

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

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

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

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.

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

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

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 $1$.

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).