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Optimal Sampling for Individual PK Models · Tutorial 35 of 497

Deriving the Individual Fisher Information Matrix

Learn to derive and calculate an individual Fisher information matrix from concentration sensitivities and independent, known residual variances.

Advanced 10 min read

What You'll Learn

  • State the Gaussian residual-error assumptions used to derive the matrix.
  • Arrange time-specific concentration sensitivities in a sensitivity matrix.
  • Derive the information matrix by taking the expected outer product of the score.
  • Weight each sensitivity contribution by its residual variance.
  • Calculate information for homoscedastic and heteroscedastic observations.
  • Diagnose rank deficiency and understand why it limits joint parameter estimation.

From Concentration Sensitivities to Information

For an individual PK model, a sampling schedule specifies the observation times, while the model predicts a concentration at each time as a function of the individual parameter vector. The information contributed by a sample depends on how strongly its predicted concentration changes with those parameters and on how much residual noise obscures that change. Deriving the matrix makes both contributions explicit: sensitivities describe the model response, and residual variances determine the weights.

The matrix is local to a parameter value. In a nonlinear model, sensitivities can change with the parameter values, so a numerical information matrix is calculated at a specified nominal individual parameter vector. This is the conditional, individual-level calculation: it treats that individual’s parameters as the unknown vector of interest and does not integrate over a population distribution.

Definition: Let \(p\) be the number of individual parameters, let \(t_1,\ldots,t_n\) be specified sampling times, and let \(C(t_i;\theta)\) be the model-predicted concentration at time \(t_i\) for parameter vector \(\theta\). The time-specific sensitivity row is \(\mathbf{j}_i(\theta)=\left(\partial C(t_i;\theta)/\partial\theta_1,\ldots,\partial C(t_i;\theta)/\partial\theta_p\right)\). The sensitivity matrix \(J(\theta)\) is the \(n\times p\) matrix whose \(i\)th row is \(\mathbf{j}_i(\theta)\).

Each entry of \(J\) is a local rate of change in predicted concentration with respect to one parameter. Its units are concentration divided by the units of that parameter. The row orientation matters: observations index the rows and parameters index the columns. Under independent errors with known, parameter-independent variances, each row contributes an outer product of its sensitivities, scaled by the inverse of its variance.

Deriving the Matrix Under Independent Gaussian Errors

Assume the concentration observation at time \(t_i\) follows \(Y_i=C(t_i;\theta)+\varepsilon_i\), where the errors are mutually independent and \(\varepsilon_i\sim N(0,\sigma_i^2)\). For this derivation, each \(\sigma_i^2\) is known and does not depend on \(\theta\). Write \(\mathbf{Y}\) for the vector of observations, \(\boldsymbol{\mu}(\theta)\) for the vector of predicted concentrations, and \(R=\operatorname{diag}(\sigma_1^2,\ldots,\sigma_n^2)\) for the residual covariance matrix. The Jacobian of \(\boldsymbol{\mu}\) with respect to \(\theta\) is \(J\).

With these assumptions, the log-likelihood, apart from constants that do not depend on \(\theta\), is the weighted squared residual expression below. Differentiating it gives the score, the vector of first derivatives of the log-likelihood.

$$ \ell(\theta)=-\frac{1}{2}(\mathbf{Y}-\boldsymbol{\mu})^\mathsf{T}R^{-1}(\mathbf{Y}-\boldsymbol{\mu})+\text{constant}, \qquad \mathbf{s}(\theta)=\frac{\partial\ell}{\partial\theta} =J^\mathsf{T}R^{-1}(\mathbf{Y}-\boldsymbol{\mu}). $$

The Fisher information is the expected outer product of the score, with expectation taken under the model at \(\theta\). Since the residual vector has mean zero and covariance \(R\), the expectation simplifies directly. This derivation also shows why the residual variances appear as inverse weights.

$$ \begin{aligned} \mathcal{I}(\theta) &=\operatorname{E}_{\theta}\!\left[\mathbf{s}(\theta)\mathbf{s}(\theta)^\mathsf{T}\right]\\ &=J^\mathsf{T}R^{-1} \operatorname{E}_{\theta}\!\left[(\mathbf{Y}-\boldsymbol{\mu})(\mathbf{Y}-\boldsymbol{\mu})^\mathsf{T}\right] R^{-1}J\\ &=J^\mathsf{T}R^{-1}RR^{-1}J\\ &=J^\mathsf{T}R^{-1}J =\sum_{i=1}^{n}\frac{\mathbf{j}_i^\mathsf{T}\mathbf{j}_i}{\sigma_i^2}. \end{aligned} $$

The last expression is a sum of \(p\times p\) matrices. In the term \(\mathbf{j}_i^\mathsf{T}\mathbf{j}_i\), the sensitivity row is transposed to make an outer product. Thus each observation adds a positive-semidefinite contribution; a more precise observation, with smaller \(\sigma_i^2\), receives greater weight.

The same result is obtained from the expected negative Hessian of the log-likelihood under these assumptions. The observed Hessian at a particular data set need not equal its expectation: in a nonlinear model, it includes residual-dependent terms involving second derivatives of the mean. Fisher information is an expected, model-based quantity, not simply the negative Hessian evaluated at observed data.

Theorem: For independent Gaussian concentration observations with known variances that do not depend on the individual parameters, the individual Fisher information matrix is \(\mathcal{I}(\theta)=J(\theta)^\mathsf{T}R^{-1}J(\theta)\). If all observations have the same residual variance \(\sigma^2\), this reduces to \(\mathcal{I}(\theta)=J^\mathsf{T}J/\sigma^2\).

Worked Calculations

Worked Example: Homoscedastic Errors in an Exponential Concentration Model

Consider the hypothetical model \(C(t;A,k)=A\exp(-kt)\), with \(A=10\ \mathrm{mg/L}\), \(k=0.5\ \mathrm{h}^{-1}\), and observations at \(t_1=1\ \mathrm{h}\) and \(t_2=3\ \mathrm{h}\). Suppose both residual standard deviations are \(0.5\ \mathrm{mg/L}\), independent of each other and of the parameters. The parameter vector is \(\theta=(A,k)\).

The time-specific derivatives are \(\partial C/\partial A=\exp(-kt)\) and \(\partial C/\partial k=-At\exp(-kt)\). At \(t_1=1\), these are \(0.6065\) and \(-6.065\); at \(t_2=3\), they are \(0.2231\) and \(-6.694\). Therefore

$$ J= \begin{pmatrix} 0.6065&-6.065\\ 0.2231&-6.694 \end{pmatrix}, \qquad R^{-1}= \begin{pmatrix} 4&0\\ 0&4 \end{pmatrix}. $$

The information is \(J^\mathsf{T}R^{-1}J=4J^\mathsf{T}J\). Multiplying the sensitivity entries gives

$$ \mathcal{I}(\theta) \approx 4\begin{pmatrix} 0.6065^2+0.2231^2& 0.6065(-6.065)+0.2231(-6.694)\\ 0.6065(-6.065)+0.2231(-6.694)& (-6.065)^2+(-6.694)^2 \end{pmatrix} = \begin{pmatrix} 1.671&-20.69\\ -20.69&326.4 \end{pmatrix}. $$

The diagonal entries describe information about \(A\) and \(k\), respectively, and the off-diagonal entries show that their score components are coupled. The units follow from the parameter units: the \(A,A\) entry has units \((\mathrm{mg/L})^{-2}\), the \(k,k\) entry has units \(\mathrm{h}^2\), and the cross-entry has units \(\mathrm{h}/(\mathrm{mg/L})\). This calculation is conditional on the stated nominal values and error model; it does not establish that the same sensitivities or information apply at other parameter values.

Worked Example: Unequal Residual Variances

Suppose two hypothetical concentration observations have time-specific sensitivity rows, with respect to two parameters, of \((2,1)\) and \((1,3)\), in a chosen set of parameter units. Let the known residual standard deviations be \(1\ \mathrm{mg/L}\) and \(2\ \mathrm{mg/L}\), respectively. The second observation is noisier and must receive less weight.

The weighted sum of outer products gives

$$ \begin{aligned} \mathcal{I} &=\frac{1}{1^2} \begin{pmatrix}2\\1\end{pmatrix} \begin{pmatrix}2&1\end{pmatrix} +\frac{1}{2^2} \begin{pmatrix}1\\3\end{pmatrix} \begin{pmatrix}1&3\end{pmatrix}\\ &= \begin{pmatrix}4&2\\2&1\end{pmatrix} +\frac{1}{4}\begin{pmatrix}1&3\\3&9\end{pmatrix} = \begin{pmatrix}4.25&2.75\\2.75&3.25\end{pmatrix}. \end{aligned} $$

For comparison, treating both variances as if they were \(1\ (\mathrm{mg/L})^2\) would give \(\begin{pmatrix}5&5\\5&10\end{pmatrix}\), incorrectly giving the noisier observation four times the weight it should receive. The relevant weight is inverse variance, not inverse standard deviation.

Worked Example: One Observation Cannot Separate Two Parameters

At a particular hypothetical sampling time, suppose a concentration has sensitivity row \((0.6,-2.4)\) with respect to two parameters, and the residual variance is \(0.16\ (\mathrm{mg/L})^2\). Its contribution is

$$ \mathcal{I} =\frac{1}{0.16} \begin{pmatrix}0.6\\-2.4\end{pmatrix} \begin{pmatrix}0.6&-2.4\end{pmatrix} =6.25\begin{pmatrix}0.36&-1.44\\-1.44&5.76\end{pmatrix} =\begin{pmatrix}2.25&-9\\-9&36\end{pmatrix}. $$

The determinant is \(2.25(36)-(-9)^2=81-81=0\), so the matrix has rank one, not two. One scalar observation supplies sensitivity in only one direction in the two-parameter space. It cannot, by itself, provide independent information in both parameter directions or yield a nonsingular information matrix for their joint estimation.

Assumptions, Diagnostics, and Boundaries

The formula depends on its assumptions. Independent errors make \(R\) diagonal; Gaussian errors with parameter-independent variances give the stated likelihood and derivation. If the model uses correlated errors instead, the general known-covariance expression remains \(J^\mathsf{T}R^{-1}J\), but \(R\) is no longer diagonal and observations cannot be treated as separate inverse-variance contributions. If variances depend on the parameters, the likelihood also contains information from changes in the covariance. For a Gaussian model with differentiable covariance \(R(\theta)\), the expected information has additional covariance-derivative terms; the mean-only formula above is then incomplete.

Useful checks start before interpreting the matrix. Confirm that the sensitivity matrix has \(n\) rows and \(p\) columns, that each variance matches the observation to which it is assigned, and that residual variances are used rather than residual standard deviations in \(R\). Check that the matrix is symmetric and positive semidefinite, as \(J^\mathsf{T}R^{-1}J\) must be when \(R\) is positive definite. A materially negative eigenvalue in a computed matrix usually signals an implementation, rounding, or weighting error.

A singular or nearly singular matrix is a different warning. Singularity means the weighted sensitivity rows do not span all parameter directions; near-singularity means some directions are weakly distinguished. More observations alone do not guarantee full rank: if their sensitivity rows are proportional or nearly proportional, they add little or redundant information about separating parameters. Numerical conditioning and scaling also matter when parameter units differ greatly; a matrix should be assessed with its parameterization in mind.

The most common alternative calculation is to add unweighted sensitivity outer products, \(J^\mathsf{T}J\). That is correct only when all residual variances are equal, apart from a common scaling factor. Another alternative is to use the observed negative Hessian. It may be useful for likelihood-based calculations, but it is data-dependent and is not the same object as expected Fisher information in a nonlinear model. The sensitivity-matrix formula is especially transparent for design calculations because it shows how each proposed sample time and its residual precision contribute.

Key takeaway: Under independent Gaussian errors with known, parameter-independent residual variances, calculate the local individual information as \(J^\mathsf{T}R^{-1}J\). A sample is informative when its concentration sensitivity is large in a parameter direction and its residual variance is small. A nonsingular matrix additionally requires sensitivities that span all parameter directions of interest.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. For sensitivity rows \((1,2)\) and \((3,1)\), with residual variances \(1\) and \(4\), calculate the information matrix.
  2. Why can a schedule with many observations still have a singular information matrix for two parameters?
  3. If every residual standard deviation is doubled while the sensitivities stay fixed, what happens to the information matrix?
  4. What changes in the calculation when residual errors are correlated but their covariance matrix is known and independent of the parameters?

Solutions. For question 1, sum the variance-weighted outer products: \(\begin{pmatrix}1&2\\2&4\end{pmatrix}+\frac14\begin{pmatrix}9&3\\3&1\end{pmatrix}=\begin{pmatrix}3.25&2.75\\2.75&4.25\end{pmatrix}\). For question 2, the rows of the sensitivity matrix may fail to span the two-dimensional parameter space; for example, proportional rows all point in the same parameter direction regardless of their number. For question 3, doubling standard deviations multiplies each variance by four, so \(R^{-1}\), and therefore the information matrix, is divided by four. For question 4, use \(\mathcal{I}=J^\mathsf{T}R^{-1}J\) with the full covariance matrix; the information no longer decomposes into independent per-observation contributions weighted only by their individual variances.

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