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An approximate D-optimum need not be an exact three-run design

Example

Take candidates $X = {0, 1, 2}$ , regression vector $f (x) = (1, x)^{T}$ , and $p = 2$ . The approximate design with weights $ξ_{0}^{*} = ξ_{2}^{*} = 1 / 2$ and $ξ_{1}^{*} = 0$ has $\det M (ξ^{*}) = 1$ and is D-optimal. For exactly three runs, the largest possible determinant is $8 / 9$ , so no exact three-run allocation attains the approximate optimum.

Facts & Assumptions

Given: The three candidate points, regression vectors, and approximate design in the Example.

[F1]

Approximate designs and linear-model optimality criteria: $M (ξ) = \sum_{x \in X} ξ_{x} f (x) f (x)^{T}$ .

[F2]

Approximate designs and linear-model optimality criteria: an exact $n$ -run design has weights $ξ_{x} = n_{x} / n$ with nonnegative integer counts summing to $n$ .

[F3]

Equivalence criterion for D-optimal approximate designs: a nonsingular approximate design is D-optimal when every candidate leverage $f (x)^{T} M (ξ)^{- 1} f (x)$ is at most $p$ , with equality on its support.

Verification

1.1givenF1F3algebra

The information matrix is $M^{*} = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array})$ , with determinant $1$ and inverse $(\begin{array}{cc} 2 & - 1 \\ - 1 & 1 \end{array})$ . Hence the leverage at $x$ is $u (x) = (1, x) M^{* - 1} (1, x)^{T} = (x - 1)^{2} + 1$ , giving $u (0) = 2$ , $u (1) = 1$ , and $u (2) = 2$ . By [F3], all candidates satisfy $u (x) \leq p = 2$ and equality holds on the support ${0, 2}$ , so $ξ^{*}$ is D-optimal.

1.2givenF1F2algebra

For an exact three-run design let $(n_{0}, n_{1}, n_{2})$ be the nonnegative integer counts with sum $3$ . Substitution into [F1] and the exact-count convention [F2] gives $\det M = (n_{0} n_{1} + 4 n_{0} n_{2} + n_{1} n_{2}) / 9$ .

2.1step 1.2choosealgebra

The ten count triples give determinants as follows: $(3, 0, 0), (0, 3, 0), (0, 0, 3)$ give $0$ ; $(2, 1, 0), (1, 2, 0), (0, 2, 1), (0, 1, 2)$ give $2 / 9$ ; $(2, 0, 1), (1, 0, 2)$ give $8 / 9$ ; and $(1, 1, 1)$ gives $6 / 9$ . Thus the exact maximum is $8 / 9$ , attained by either endpoint allocation with counts $2$ and $1$ .

3.1step 1.1step 2.1algebra∎

Since $8 / 9 < 1 = \det M (ξ^{*})$ , the exact three-run optimum and the approximate optimum differ. This finite calculation concerns only the stated candidate set and run count.

Depends on

Used by

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Sources

Stephen Boyd and Lieven Vandenberghe, Convex Optimization, §7.5 (standard reference, not scraped)