A correlation matrix problem

Question

I have a linear algebraic question about an arbitrary correlation matrix. If

$$ \lambda_{\min} \begin{pmatrix} 1 & \rho_1 & \rho_2 \\ \rho_1 & 1 & \rho_3 \\ \rho_2 & \rho_3 & 1 \end{pmatrix} \ge \alpha $$

does the following hold?

$$ (\rho_1,\rho_3) \begin{pmatrix} 1 & \rho_2 \\ \rho_2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} \rho_1 \\ \rho_3 \end{pmatrix} \le 1-\alpha $$

Thanks so much.

Answer 1

If $\rho_1=\rho_3=0$ it is obvious, so we assume $\rho_1^2+\rho_3^2>0$. First note that $$ A= \begin{pmatrix} 1 & \rho_2 \\ \rho_2 & 1 \end{pmatrix}^{-1}=\frac{1}{\rho_2^2-1}\begin{pmatrix} -1 & \rho_2 \\ \rho_2 & -1 \end{pmatrix}$$ with eigenvalues $\lambda_1=\frac{1}{1-\rho_2}$ and $\lambda_2=\frac{1}{1+\rho_2}$. Since $A$ is symmetric, we know that the Rayleigh quotient of $A$ is upper bounded by $\max\{\lambda_1,\lambda_2\}$, that is $$\frac{\langle Ax,x\rangle}{\langle x,x\rangle} \leq \max\{\lambda_1,\lambda_2\} \qquad \forall x\neq 0.$$ In particular for $z=(\rho_1,\rho_3)^\top$ we obtain the relation $$\langle Az,z\rangle \leq \max\{\lambda_1,\lambda_2\}(\rho_1^2+\rho_3^2)=\max\Big\{\frac{\rho_1^2+\rho_3^2}{1-\rho_2},\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}.$$ So, we need to prove that $$\max\Big\{\frac{\rho_1^2+\rho_3^2}{1-\rho_2},\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}\leq 1-\alpha \iff \min\Big\{1-\frac{\rho_1^2+\rho_3^2}{1-\rho_2},1-\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}\geq \alpha \tag{1} $$

Now, the matrix $$M =\begin{pmatrix} 1 &\rho_1 &\rho_2 \\ \rho_1&1 & \rho_3 \\ \rho_2 & \rho_3 &1 \end{pmatrix}$$ is also symmetric and thus, again by Rayleigh quotient argument we have $$\frac{\langle Mu,u\rangle }{\langle u,u\rangle}\geq \alpha \qquad \forall u\neq 0.\tag{2}$$ The idea is now to prove (1) by plugging in (2) some smart choice of $u$. These choices exist but I could not find nicely elegant ones. Here are some choices obtained with Mathematica: Take $u_{\pm} = (\alpha_{\pm},0,1)$ with $$\alpha_+=\frac{\sqrt{4 \left(\rho_2^2+\rho_2\right)^2-4 \left(\rho_1^2+\rho_3^2\right)^2}-2 \rho_2^2-2 \rho_2}{2 \left(\rho_1^2+\rho_3^2\right)}$$ and $$\alpha_-=\frac{\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4-2 \rho_2^3+\rho_2^2-\rho_3^4}+\rho_2^2-\rho_2}{\rho_1^2+\rho_3^2}$$

Answer 2

Not a proof, but from a numerical experiment (considering many random correlation matrices) I believe that the statement you propose is true. The experiment seems to indicate that a sharp bound is given by $$ \begin{pmatrix} \rho_1 & \rho_3 \end{pmatrix} \begin{pmatrix} 1 & \rho_2 \\ \rho_2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} \rho_1 \\ \rho_3 \end{pmatrix} \le (1-\alpha)^2, $$ which implies your bound since $\lambda_\mathrm{min} \in [0,1]$.

Here is the R code I used for my experiment:

n <- 10000
res <- matrix(0, n, 2)
for (i in 1:n) {
  repeat {
    q <- runif(3, -1, 1)
    A <- matrix(c(1,q[1],q[2],q[1],1,q[3],q[2],q[3],1), 3, 3)
    alpha <- min(eigen(A, symmetric = TRUE, only.values = TRUE)$values)
    if (alpha >= 0) {
      break
    }
  }
  B <- matrix(c(1, q[2], q[2], 1), 2, 2)
  x <- c(q[1], q[3])
  beta <- x %*% solve(B, x)
  res[i,] <- c(alpha, beta)
}
plot(res[,1],res[,2], asp=1, cex=.5)
abline(1, -1)
curve((1-x)^2, add = TRUE)