Let X_n×p have full column rank p < n and let Ω be a diagonal matrix whose diagonal elements {ω_i}_i=1ⁿ are positive and not all equal. The discussion by Professors Koenker and Portnoy suggested that, for any (nonzero)

, the function

satisfies e_{0, δ} ≥ e_{0.5, δ} and more generally that e_{a, δ} is a decreasing function of a ∈ [0,1] .

We shall show that the second suggestion, and hence the first also, is true for p = 1. Both can fail if p > 1; we illustrate this by an example in which e_{a, δ} is not only nonmonotone but is maximized at a = 0.5.

First let p = 1, so that X = x, a column vector. We calculate that

where

L_Ω is the diagonal matrix with diagonal elements {log ω_i}_i=1ⁿ, and v = Ω^−ax/∥Ω^−ax∥ has unit norm.

Because M, and hence the derivative (1), vanishes when a = 1, our assertion is equivalent to the statement that M > 0 for a ∈ [0,1). Define f (ω) = ω^2−2a and set f_i = f (ω_i). Then

Note that

Furthermore, if any element x_i of x is 0, it can be removed (and the value of n reduced) without affecting the value of

for a diagonal D and hence without affecting the value of e_a,δ. Thus we can assume that all x_i ≠ 0 and hence that all v_i ≠ 0. This together with (3) ensures that (2) is positive (because ω_i ≠ ω_j at least once), thus establishing the monotonicity claim when p = 1.

Counterexample when p > 1. For general p the preceding method results in

where

and V = Ω^−aX(X′Ω^−2aX)^−1/2 satisfies

. The analogue of (2) is that

and the monotonicity claim is equivalent to the positive definiteness of M. However, for p > 1, M often has at least one negative eigenvalue. In such cases the monotonicity properties will depend on the choice of δ. A particular counterexample to both suggestions of Professors Koenker and Portnoy has n = 3, p = 2, and

For these choices the eigenvalues of M are 1.49 and −0.003 when a = 0.5. We choose δ = (1,0.26291)′; then e_{a, δ} is nonmonotone and is maximized at a = 0.5. See Figure 1.

This counterexample can be extended to all n > 3 by adding n − 3 rows of zeros to the bottom of X and continuing the diagonal of Ω arbitrarily.

Plot of ea, δ against a with δ chosen so that the maximum is attained at a = 0.5.