Given three vectors $a$, $b$, and $c$, is it possible that correlations between $a$ and $b$, $a$ and $c$, and $b$ and $c$ are all negative? I.e. is this possible?
\begin{align} \text{corr}(a,b) < 0\\ \text{corr}(a,c) < 0 \\ \text{corr}(b,c) < 0\\ \end{align}
It is possible if the size of the vector is 3 or larger. For example
\begin{align} a &= (-1, 1, 1)\\ b &= (1, -9, -3)\\ c &= (2, 3, -1)\\ \end{align}
The correlations are \begin{equation} \text{cor}(a,b) = -0.80...\\ \text{cor}(a,c) = -0.27...\\ \text{cor}(b,c) = -0.34... \end{equation}
We can prove that for vectors of size 2 this is not possible: \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 $a_2$, $b_2$ 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 at 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).
