# A correlation matrix problem

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.

• Not that it looks like a terribly hard question, of course, but it would be still polite to at least tell whether the answer is "Yes" or "No" when voting to close, if you find it totally obvious... Jun 28, 2017 at 23:09
• @Koltchinskii it is polite to accept an answer if you are satisfied with it. Jul 3, 2017 at 17:31

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 $$\alpha_-=\frac{\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4-2 \rho_2^3+\rho_2^2-\rho_3^4}+\rho_2^2-\rho_2}{\rho_1^2+\rho_3^2}$$

• Apparently, $u_{\pm}=(\alpha_{\pm},0,-(\rho_3^2+\rho_1^2))$ with $$\alpha_{\pm}=\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4\pm 2 \rho_2^3+\rho_2^2-\rho_3^4}\pm \rho_2^2+\rho_2$$ also works well.
– Surb
Jun 29, 2017 at 12:45
• Did you find $u$ by setting $u=(a,b,1)$ and solve the two quadratic equations of $(a,b)$ where $\frac{\langle Mu,u\rangle }{\langle u,u\rangle}=1-\frac{\rho_1^2+\rho_3^2}{1-\rho_2},1-\frac{\rho_1^2+\rho_3^2}{1+\rho_2}$?
– Hans
Sep 23, 2018 at 22:13

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)