strange bound on correlation for symmetric pdf

Question

I am puzzled by a rather simple fact:

The correlation of a symmetric multivariate pdf seems to be bound from below (increasingly strong with the number of dimensions). That seems unlikely to me.

But I can't find any mistake in my reasoning:

Let $f$ be a symmetric joint pdf for the random variates $X_1, ... X_n$ all in $\mathbb{R}$: $$ \begin{align*} \forall i,j \in {1...n}:& \\ &f(\dots x_{i-1}, x_i, x_{i+1}, \dots x_{j-1}, x_j, x_{j+1}, \dots) = \\ &f(\dots x_{i-1}, x_j, x_{i+1}, \dots x_{j-1}, x_i, x_{j+1}, \dots) \end{align*} $$ Since $f$ is symmetric all the marginal distributions are equal $$ \begin{align*} f_{X_1}(x) &= \int f(x, x_2, \dots) \text{d} x_{2} \dots \text{d} x_{n} = \\ &= \int f(x_2, x, \dots) \text{d} x_{2} \dots \text{d} x_{n} = f_{X_2}(x) \end{align*} $$ and thus all of the following expectation values, variances and covariances are identical. $$ \begin{align*} \forall i,j \in {1...n}, i\ne j:& \\ \epsilon &:= \text{Exp}\left[X_i\right] \\ \nu &:= \text{Var}\left[X_i\right] \\ c &:= \text{Cov}\left[X_i, X_j\right] \\ \rho &:= \rho\left[X_i, X_j\right] = \frac{c}{\nu}\\ \end{align*} $$

The fun begins with $$ \begin{align*} \text{Var}\left[\sum_{i=1}^{n} X_i \right] &= \sum_i \text{Var}\left[X_i\right] + \sum_{i\ne j}\text{Cov}\left[X_i, X_j\right] = \\ &= n \nu + n(n-1)c = n\nu \left(1 + (n-1)\frac{c}{\nu}\right)= n\nu (1 + (n-1)\rho) \end{align*} $$ Since the variance is positive I get $$ \begin{align*} 0<& 1 + (n-1)\rho \\ \Leftrightarrow \rho >& -\frac{1}{(n-1)} \end{align*} $$

This can't be true?!

Edit

If so, wouldn't it mean that the more dimensions I have there is less and less freedom to find symmetric pdfs that are anti-correlated. What's the intuition behind this?

Answer 1

When we have a $n\times n$-correlation matrix of the form $$ \pmatrix{1&\rho&\dots&\dots&\rho\\\rho&\ddots&&&\vdots\\\vdots&&\ddots&&\vdots\\ \vdots&&&\ddots&\rho\\\rho&\dots&\dots&\rho&1} $$ and $n>2$ there is a restriction on how negative the constant $\rho$ can be to guarantee that the matrix is positive semidefinite.

Another proof: Doing row operations we can calculate the determinant of the matrix as \begin{align} &\det\pmatrix{1&\rho&\dots&\rho\\\rho-1&1-\rho&&\\ &\ddots&\ddots&\\&&\rho-1&1-\rho} \end{align} and develop it by the first column. It should not be hard to see that this determinant is $$ (1-\rho)^{n-1}\big(1+(n-1)\rho\big)\,. $$ Since $\rho\in[-1,1]$ the determinant is therefore nonnegative if and only if $$\tag1 \rho\ge-\frac1{n-1}\,. $$ Another remark which simplifies the linked answer a bit: Using the same row operations it is easy to see that the characteristic polynomial of the matrix is $$ p(\lambda)=(1-\lambda-\rho)^{n-1}\Big(1-\lambda+(n-1)\rho\Big) $$ so that the eigenvalues are $$ \lambda_1=1-\rho\,,\quad \lambda_2=1+(n-1)\rho $$ with multiplicities $n-1$ and one, respectively. Since the eigenvalue $1-\rho$ is nonnegative the matrix is positive semidefinite if and only if the inequality (1) holds.