# Strict monotonicity of conditional variances

Let $K \geq 2$ be a positive integer and $C$ be any $K \times K$ non-singular matrix (if necessary, can assume that all $K$ rows of $C$ are needed to span the coordinate row vector $e_1'$). For positive real numbers $q_1, \dots, q_K$, define $$\Sigma\left(q_1, \dots, q_K\right) = CC' + \operatorname{diag} \left(\frac{1}{q_1}, \dots, \frac{1}{q_K}\right)$$ and $$f(q_1, \dots, q_K) = \left[I_K - C'\Sigma^{-1}C\right]_{11}.$$

It is shown in Log-convexity of conditional variances that $f$ is the conditional variance of a normal random variable given $q_i$ i.i.d. observations of some signal $X_i$, and $f$ is a log-convex function.

My question is regarding the monotonicity of $f$. Since $\Sigma$ decreases (in matrix order) in $q_i$, so does $f$. However, there are situations in which the monotonicity is not strict. For instance, when $K=2$, $C = \left(\begin{array}{cc} 1/3 & 2/3 \\ 2/3 & 1/3 \end{array}\right)$ and $q_2 = 3$, the function $f$ is constant regardless of the value of $q_1$.

Conjecture. There always exists some $i \in \{1, \dots, K\}$, such that $f$ is strictly decreasing (equivalently, non-constant) in $q_i$ for all values of other $q_j$.

For $K = 2$, this is correct because if $\partial_1(0,q_2) = \partial_2(q_1,0) = 0$, then by the convexity of $f$ we would deduce $\partial_1(q_1,q_2) = \partial_2(q_1,q_2) = 0$. But it is easy to see $\partial_1(q_1,q_2) + \partial_2(q_1,q_2) > 0$. For general $K$, I would appreciate either a proof or a counterexample.

• Hi, Xiaosheng. What do you mean by "in matrix order"? – Henry.L May 2 '17 at 18:17
• I mean $\Sigma$ decreases by a positive semi-definite matrix – Xiaosheng Mu May 2 '17 at 20:42