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Population Design and the Population Fisher Matrix · Tutorial 99 of 497

Local Dependence on Nominal Population Parameters

You will learn to compare approximate population Fisher matrices at different plausible nominal values while keeping the design fixed, and to interpret changes without treating them as design changes.

Advanced 13 min read

What You'll Learn

  • Define local dependence of an approximate population Fisher matrix on nominal parameter values.
  • Derive the Gaussian fixed-covariance sensitivity form of the information matrix.
  • Calculate how changing a nominal value alters information for a nonlinear mean model.
  • Compare matrix eigenvalues and condition numbers across nominal values.
  • Recognize when information is independent of the nominal values and when a local comparison is unreliable.
  • Distinguish a fixed-design sensitivity analysis from averaging over parameter uncertainty or changing the design.

Why the nominal values matter

An approximate population Fisher matrix is local: its calculation uses parameter values at which model sensitivities and, depending on the approximation, variability terms are evaluated. If the subject design is held fixed but those nominal values change, the resulting matrix can still change. That change describes how the model’s local information behaves at different parameter values; it is not evidence that the design itself changed.

This dependence matters when a design is assessed before the population parameters are well known. A design that appears informative at one plausible parameter value may be less informative at another, particularly when the model is nonlinear. The useful question is therefore not simply whether a matrix is large at one nominal point, but how sensitive the information assessment is to plausible changes in that point.

Definition: Let \(d\) denote a fixed subject design and let \(\theta^*\) denote the nominal population parameter vector used in an approximate information calculation. The local dependence on nominal parameters is the change in \(M_{\mathrm{app}}(d;\theta^*)\) as \(\theta^*\) varies while \(d\), the model, the approximation, and the parameterization are held fixed. Here \(M_{\mathrm{app}}\) denotes the approximate population Fisher information matrix.

A comparison requires more than substituting new numbers. It should make clear which quantities are fixed, which are reevaluated, and which matrix features are being compared. In particular, sensitivity-based approximations reevaluate derivatives at each nominal value. If an approximation also depends on a model-based covariance, that covariance may need to be reevaluated too. Changing the nominal values while inadvertently changing the sampling times, subject allocation, residual-error assumptions, or parameter scaling confounds the comparison.

A local information calculation

A useful justification comes from a Gaussian model with a fixed covariance. Suppose the observation vector \(Y\) has mean \(m(\theta)\) and covariance matrix \(\Sigma\), where \(\Sigma\) is positive definite and does not depend on \(\theta\). Write \(J(\theta)\) for the matrix whose columns are the derivatives of \(m(\theta)\) with respect to the components of \(\theta\). The information from this model at a nominal value \(\theta^*\) is

$$ M(\theta^*)=J(\theta^*)^\mathsf{T}\Sigma^{-1}J(\theta^*). $$

This formula makes the source of local dependence explicit: even with fixed \(\Sigma\), changing \(\theta^*\) can change \(J(\theta^*)\). The result is an exact Fisher information expression under these stated Gaussian assumptions; in a population-design calculation, a sensitivity-based version may instead be an approximation to the population information.

Derivation for a Gaussian mean model

For completeness, the sensitivity expression follows directly from the score. Under the assumptions above, the log likelihood, apart from a constant not depending on \(\theta\), is

$$ \ell(\theta;Y)=-\frac{1}{2}\{Y-m(\theta)\}^\mathsf{T}\Sigma^{-1}\{Y-m(\theta)\}. $$

Differentiating with respect to \(\theta\) gives the score \(U(\theta)=J(\theta)^\mathsf{T}\Sigma^{-1}\{Y-m(\theta)\}\). At the true parameter value, the residual has mean zero and covariance \(\Sigma\). Therefore the expected outer product of the score is

$$ \begin{aligned} E\{U(\theta)U(\theta)^\mathsf{T}\} &=J(\theta)^\mathsf{T}\Sigma^{-1} E[\{Y-m(\theta)\}\{Y-m(\theta)\}^\mathsf{T}] \Sigma^{-1}J(\theta)\\ &=J(\theta)^\mathsf{T}\Sigma^{-1}\Sigma\Sigma^{-1}J(\theta)\\ &=J(\theta)^\mathsf{T}\Sigma^{-1}J(\theta). \end{aligned} $$

The middle equality uses the assumed covariance of the residual, and the final equality uses the symmetry of \(\Sigma^{-1}\). If the covariance depends on the parameters, this derivation is incomplete: derivatives of the covariance contribute to the information. The fixed-covariance formula should not be applied as if those contributions were absent.

Worked Example: Nonlinear Sensitivity at Two Nominal Values

Suppose a hypothetical response measured at time \(t\) has mean \(\mu(t;\kappa)=2-1/(1+\kappa t)\), where \(t\) is in hours and \(\kappa\) is in inverse hours. The response is in response units (RU). Consider 20 independent observations at the fixed time \(t=3\) hours, each with normally distributed error of standard deviation \(0.2\) RU. Compare the scalar information for \(\kappa\) at nominal values \(0.1\) and \(0.2\) per hour. Assume the stated error variance is fixed at both values.

The derivative of the mean with respect to \(\kappa\) is

$$ \frac{\partial\mu(t;\kappa)}{\partial\kappa} =\frac{t}{(1+\kappa t)^2}. $$

For independent normal observations with fixed variance \(\sigma^2\), the information is the number of observations times the squared derivative divided by the variance. Here \(\sigma^2=(0.2)^2=0.04\ \mathrm{RU}^2\). At \(\kappa=0.1\ \mathrm{hour}^{-1}\), the derivative is \(3/(1+0.1\times3)^2=3/1.69=1.77515\ \mathrm{RU}\cdot\mathrm{hour}\). Thus

$$ M(0.1)=\frac{20[3/(1+0.1\times3)^2]^2}{0.04} =1575.58\ \mathrm{hour}^2. $$

At \(\kappa=0.2\ \mathrm{hour}^{-1}\), the derivative is \(3/(1+0.2\times3)^2=3/2.56=1.171875\ \mathrm{RU}\cdot\mathrm{hour}\). Therefore

$$ M(0.2)=\frac{20[3/(1+0.2\times3)^2]^2}{0.04} =686.646\ \mathrm{hour}^2. $$

The design, number of observations, and error variance are unchanged. The local information is about 43.6% as large at the second nominal value: \(686.646/1575.58=0.4358\). This difference arises from the change in the mean sensitivity, not a change in the observations’ planned times. It does not by itself establish which nominal value is correct or how much actual data will be observed.

Comparing matrices, not just entries

For a vector of population parameters, individual matrix entries describe information about particular parameter combinations and their cross-information. A change in an entry is not automatically a change in overall precision: the entries act together. If parameters are on different scales, direct comparisons of eigenvalues or matrix entries can also be dominated by units. The earlier material on scaling parameters before information diagnostics provides the relevant warning: comparisons of matrix summaries require a consistent parameterization and scale.

The following example uses dimensionless parameters and fixed unit variances, so the matrix comparison is directly interpretable in the stated coordinates.

Worked Example: A Two-Parameter Matrix Changes Shape

Consider two independent hypothetical measurements, \(Y_1=a+\epsilon_1\) and \(Y_2=ab+\epsilon_2\), with each error normally distributed with mean zero and variance 1. The fixed design consists of these two measurement types, and \(a\) and \(b\) are dimensionless. The mean vector is \(m(a,b)=(a,ab)^\mathsf{T}\), so its sensitivity matrix is

$$ J(a,b)= \begin{pmatrix} 1&0\\ b&a \end{pmatrix}, \qquad M(a,b)=J^\mathsf{T}J= \begin{pmatrix} 1+b^2&ab\\ ab&a^2 \end{pmatrix}. $$

At nominal values \((a,b)=(1,1)\), the information matrix is \(\begin{pmatrix}2&1\\1&1\end{pmatrix}\). Its trace is 3 and determinant is 1, so its eigenvalues solve \(\lambda^2-3\lambda+1=0\). They are \((3+\sqrt{5})/2=2.618\) and \((3-\sqrt{5})/2=0.3820\). The condition number, the ratio of the largest to smallest eigenvalue, is about 6.854.

At \((a,b)=(1,2)\), without changing the measurement types or variances, the matrix is \(\begin{pmatrix}5&2\\2&1\end{pmatrix}\). Its trace is 6 and determinant is 1, giving eigenvalues \((6+\sqrt{32})/2=5.828\) and \((6-\sqrt{32})/2=0.1716\). The condition number is about 33.97.

Both matrices have the same determinant, but their eigenvalues and condition numbers differ substantially. Thus, even one familiar scalar summary need not tell the whole story. Here the local information becomes more uneven across parameter directions at the second nominal value. This is a statement about the matrices in the specified parameterization; it is not a claim that either individual measurement type has been added or removed.

When the information does not depend on the nominal value

Local dependence is possible, not inevitable. For a model whose mean is linear in its parameters and whose covariance is fixed, the sensitivity matrix is constant. The information matrix is then also constant across nominal parameter values. This provides a useful check: if a calculation for such a model changes when only the nominal parameter values are altered, investigate whether some other input has changed or whether the computation is incorrect.

Worked Example: A Linear Mean with Constant Information

Suppose two independent observations have means \(\mu_1=\theta_0-\theta_1\) and \(\mu_2=\theta_0+\theta_1\), each with fixed normal error variance 1. The measurement locations and variances are fixed. The sensitivity matrix and information matrix are

$$ J= \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}, \qquad M=J^\mathsf{T}J= \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}. $$

No term in \(J\) depends on \(\theta_0\) or \(\theta_1\). Consequently, \(M\) is the same at, for example, \((0,0)\) and \((2,-1)\). The design remains fixed, and, unlike the nonlinear examples, its information does not vary with the nominal values under these model assumptions.

A practical comparison and its failure modes

A disciplined comparison varies nominal values in a controlled way. First fix the design, parameterization, model, and approximation. Then choose a central nominal vector and a set of plausible perturbations, preferably varying one component at a time before examining joint changes. Recompute the whole approximate matrix at each point, including any nominal-dependent sensitivities or covariance terms required by the chosen approximation. Finally, compare matrix features appropriate to the question, while recording units and scale.

A local grid is often more informative than a single pair of points. It can reveal a sharp change, a smooth trend, or a region where the matrix becomes nearly singular. For a matrix intended to represent information, materially negative eigenvalues can signal numerical error, an invalid approximation, or a problem in how the matrix was assembled; a positive semidefinite information matrix should not have negative eigenvalues apart from negligible numerical rounding. A very small eigenvalue indicates weak information in some parameter direction, but does not identify the cause without examining the corresponding direction and model sensitivities.

A local comparison should not be confused with other ways to address uncertain parameter values. Evaluating at several nominal points is a sensitivity analysis. Averaging information over a covariate or parameter distribution is a different operation, and an optimization that protects performance across a range of values poses a different design criterion. Nor does changing the nominal value turn an approximate information matrix into exact marginal information. These approaches answer different questions.

  • Changing more than nominal values. If sample times, subject counts, residual variance, or parameter scaling also change, the observed matrix difference cannot be attributed solely to local parameter dependence.
  • Reading one entry as total precision. Cross-information and parameter scaling matter. Compare the full matrix or suitable summaries, not a single diagonal entry in isolation.
  • Extrapolating too far. A local approximation evaluated at a nominal point need not represent behavior far away, especially when the model is strongly nonlinear.
  • Treating a numerical difference as substantive. Check precision, numerical stability, and units. Small apparent changes may reflect rounding or computation error rather than meaningful model sensitivity.
Key takeaway: With the design fixed, variation in an approximate population Fisher matrix across nominal values measures local sensitivity of the model-based information calculation to those values. Interpret that variation in a consistent parameterization, check the approximation’s assumptions, and do not mistake a local comparison for a design change or an uncertainty-averaged result.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

  1. In a fixed-design comparison, name two inputs besides nominal parameter values that should be held constant.
  2. For a Gaussian mean model with fixed covariance, why can the information change when the nominal parameter vector changes?
  3. In the nonlinear worked example, calculate the ratio of information at \(\kappa=0.2\) to information at \(\kappa=0.1\), and interpret it.
  4. Why does the linear-mean example have the same information at every nominal parameter value?
  5. Does evaluating information at several nominal points equal averaging information over a parameter distribution? Explain.

Solutions. (1) For example, hold the subject design and covariance assumptions fixed; also retain the same model, approximation, and parameterization. (2) The sensitivity matrix \(J(\theta^*)\) can change with \(\theta^*\), and \(M=J^\mathsf{T}\Sigma^{-1}J\). (3) \(686.646/1575.58=0.4358\), so the information at \(0.2\ \mathrm{hour}^{-1}\) is about 43.6% of that at \(0.1\ \mathrm{hour}^{-1}\), under the stated model and fixed design. (4) Its sensitivity matrix is constant in the parameters, and its covariance is fixed, so \(J^\mathsf{T}\Sigma^{-1}J\) is constant. (5) No. Evaluating at several points compares local matrices; averaging requires a specified distribution and an averaging operation, which is a different analysis.

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