Why Combine Optimality Criteria?
A design that performs well for one inferential goal need not perform well for another. D-optimality values overall information through the determinant of the information matrix, whereas A-optimality targets the average parameter variance through the trace of its inverse. A design may therefore improve joint precision across parameters while leaving one parameter comparatively imprecise, or may improve average precision without maximizing generalized information.
A compound criterion makes such competing goals explicit. It does not discover a uniquely correct trade-off: the weights encode the priorities of the design problem. Their interpretation depends on how the component criteria are scaled. This is especially important because a weighted sum of raw criteria such as \(\log\det M\) and \(-\operatorname{tr}(M^{-1})\) can change meaning with units or parameterization. The parameter scaling established in the earlier material on parameter scaling should be fixed before setting up the compound objective.
Benchmark-Relative Component Scores
Let \(M(\xi)\) be the information matrix for design \(\xi\), with \(p\) parameters. The formulas below assume that the design is locally evaluated at specified nominal parameter values, that \(M(\xi)\) is positive definite, and that the local covariance approximation is proportional to \(M(\xi)^{-1}\). Fix the parameterization and a positive-definite reference information matrix \(M_{\mathrm{ref}}\). The reference may represent a baseline design or another agreed benchmark. It must be held fixed when candidate designs are compared.
The logarithms turn ratios into additive scores. For example, doubling the determinant gives a D-score increase of \(\log 2\); halving \(A(\xi)\) gives an A-utility increase of \(\log 2\). These are relative gains, not variances or probabilities. The ratio defining \(d(\xi)\) compares determinants in the same parameterization, and \(a(\xi)\) also requires a consistent parameter scaling because the trace of an inverse information matrix depends on parameter scales.
These weights are not probabilities, shares of sample size, or claims about the truth of either objective. They express a decision rule applied to the specified, normalized component scores. A change of reference or scale can change the practical trade-off represented by the same numerical weight, so report the score definitions and reference along with the weight.
Worked Examples
Worked Example: Deriving the Weight at Which a Preference Changes
Suppose two hypothetical candidate designs, P and Q, have benchmark-relative scores \(d(P)=\log 4\), \(d(Q)=\log 3\), \(a(P)=0\), and \(a(Q)=\log(63/32)\). Determine the weight at which the compound criterion gives them equal scores, and identify the preferred design on either side.
The difference in D scores is \(d(P)-d(Q)=\log(4/3)\), so P has the D advantage. The A-utility difference is \(a(Q)-a(P)=\log(63/32)\), so Q has the A advantage. Equality of compound scores requires
Rearranging gives \(w_D\log(4/3)=(1-w_D)\log(63/32)\), and hence
Thus P is preferred when \(w_D>0.7019\), because its D advantage receives more emphasis; Q is preferred when \(w_D<0.7019\), because Q's A advantage then carries more weight. At the threshold the two scores are equal. The value is calculated from the unrounded logarithms; rounding the separate score differences before solving can alter the last reported digit.
Worked Example: Evaluating Two Candidates at Equal Weights
For two hypothetical designs, let \(d(P)=\log 20\), \(a(P)=-\log(0.525)\), \(d(Q)=\log 15\), and \(a(Q)=-\log(4/15)\). Calculate the scores with \(w_D=0.5\).
For P, calculate the weighted total directly from the unrounded expressions:
For Q:
Therefore Q has the larger compound score at equal weights, by approximately \(2.0149-1.8200=0.1949\) score units. P has the larger D score, but Q's A utility is sufficiently larger to outweigh that disadvantage under this specified rule. The result is conditional on the component definitions, reference, and equal weighting; it does not establish that Q is universally better.
Worked Example: Computing Component Scores From Information Matrices
Consider two hypothetical designs with information matrices \(M_X=\operatorname{diag}(4,1)\) and \(M_Y=\operatorname{diag}(6,0.5)\). Use \(M_{\mathrm{ref}}=I_2\) and equal weights. All matrices are positive definite, so their determinants and inverse traces are defined.
For the reference matrix, \(\det(M_{\mathrm{ref}})=1\) and \(A_{\mathrm{ref}}=\operatorname{tr}(I_2)=2\). For X, \(\det(M_X)=4\) and \(A(X)=1/4+1=1.25\). Thus
and \(U(X)=0.5\log4+0.5\log1.6\approx0.9281\), using the unrounded expressions for the total. For Y, \(\det(M_Y)=6(0.5)=3\), while \(A(Y)=1/6+1/0.5=1/6+2=13/6\). Therefore
and \(U(Y)=0.5\log3+0.5\log(12/13)\approx0.5093\). X is preferred at equal weights. Here X has both the larger determinant and the smaller inverse trace, so it improves both component objectives relative to Y. In contrast to the preceding trade-off example, no balance between these two criteria is needed to rank these candidates.
Why the Weighted Rule Works—and What It Does Not Say
For two designs \(P\) and \(Q\), subtracting their compound scores gives
This identity justifies the threshold calculation: the difference is a linear function of \(w_D\). If P is better on D and Q is better on A, the two contributions have opposite signs, so a crossing may occur within \([0,1]\). If P is at least as good on both components and strictly better on one, then \(U(P)>U(Q)\) for every interior weight \(0<w_D<1\). Such a candidate is not made worse by the stated two objectives. This dominance check can eliminate candidates before more detailed weight analysis.
A weighted sum is not the only way to handle competing goals. One alternative is to report the D and A results separately and let decision-makers choose after seeing the trade-off. Another is a constrained approach: optimize one criterion subject to a minimum acceptable value for the other. A lexicographic rule ranks one objective first and consults the second only when the first ties. These approaches answer different decision questions. The compound score is useful when a transparent numerical trade-off is desired, but its weights should not be selected merely because equal weights seem neutral.
Diagnostics, Failure Modes, and Limitations
Before interpreting a compound score, check that each candidate has a positive-definite information matrix, the same parameterization and nominal model assumptions, and the same fixed reference. A singular matrix makes the determinant zero and the inverse unavailable; the stated finite-score comparison then does not apply. Also verify that the A criterion is computed as the trace of the full inverse matrix, not as the inverse of the trace.
A practical diagnostic is to tabulate both component scores as well as the compound total. If one candidate has a higher total, the component table shows whether that result reflects a genuine trade-off or a calculation or scaling issue. Repeat the ranking over a range of plausible weights. A ranking that changes sharply over a narrow weight interval is sensitive to the chosen preference and should be reported as such, rather than summarized as a robust winner.
The benchmark-relative logarithms provide a common dimensionless scale, but they do not make the weights universal. Different normalizations or references may change the balance implied by a fixed weight. Moreover, a local information matrix and its covariance approximation depend on the assumed model and nominal parameter values; a compound score cannot protect against model misspecification. Finally, a weighted sum reduces several goals to one number and may not reveal all practically relevant trade-offs. The separate component scores and the decision context remain essential.
Common Mistake
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- If a design has \(d=0.4\) and \(a=-0.2\), what is its compound score at \(w_D=0.75\)?
- For two designs, P has \((d,a)=(1,0)\) and Q has \((d,a)=(0,0.5)\). At what weight do their compound scores tie?
- What does \(a(\xi)=\log 2\) mean in terms of \(A(\xi)\) and \(A_{\mathrm{ref}}\)?
- Why is it useful to inspect both component scores after calculating the compound score?
Solutions. 1. \(U=0.75(0.4)+0.25(-0.2)=0.30-0.05=0.25\). 2. Equality requires \(w_D=(1-w_D)0.5\), so \(1.5w_D=0.5\) and \(w_D=1/3\). 3. Since \(\log(A_{\mathrm{ref}}/A)=\log2\), the reference inverse-trace criterion is twice the candidate's: \(A(\xi)=A_{\mathrm{ref}}/2\). 4. The components identify the trade-off or dominance behind the total and help detect an unintended scaling or calculation error.