Why Information Is Local in a Nonlinear Model
In a nonlinear model, the response to a change in a parameter can depend on the parameter values themselves. Fisher information is built from these response sensitivities, so it too depends on the parameter values at which it is evaluated. A design may therefore have different information matrices under different plausible parameter values, even when the observation locations and measurement precision are unchanged.
This tutorial develops that dependence for a two-parameter response model. The goal is to calculate information at explicitly stated nominal parameter values: values treated as the working parameter specification for a local calculation. The result is conditional on those values; it is not automatically a global summary of information across all plausible parameter values.
Model, Assumptions, and the Information Formula
Consider the two-parameter Emax mean response \(m(C;E_{\max},K)=E_{\max}C/(K+C)\), where \(C\) is a nonnegative concentration, \(E_{\max}>0\) is the maximum response, and \(K>0\) is the concentration producing half of \(E_{\max}\). Suppose the observed response at condition \(C_i\) is \(Y_i=m(C_i;\theta)+\varepsilon_i\), with independent errors \(\varepsilon_i\sim N(0,\sigma^2)\). Assume \(\sigma^2\) is known, is the same for all observations, and does not depend on \(\theta\). Also assume the mean is differentiable in the parameters and the nominal values are in the interior of the allowed parameter region.
These assumptions matter. Independence lets the information contributions add. The normal model with constant known variance makes the mean-sensitivity contribution the full information expression used here. If the variance also depends on parameters, its derivatives contribute additional information; the formula below would then be incomplete.
For this model, differentiating the mean gives
The first component describes the change in mean response per unit change in \(E_{\max}\); the second describes the change per unit change in \(K\). The second derivative is negative because increasing \(K\), at fixed positive concentration and \(E_{\max}\), lowers the predicted response. Both components depend on the model settings, and the second also depends directly on \(E_{\max}\).
Why the Local Information Formula Holds
For one observation \(Y_i\), the normal log likelihood, apart from constants not involving \(\theta\), is \(-[Y_i-m(C_i;\theta)]^2/(2\sigma^2)\). Differentiating with respect to \(\theta\) gives the score contribution \([Y_i-m(C_i;\theta)]g(C_i;\theta)/\sigma^2\). The expected outer product of this score contribution is
because the residual has mean zero and variance \(\sigma^2\). Independence makes cross-products between different observations have expectation zero, so summing the individual contributions gives the information matrix. Substituting \(\theta_0\) into every sensitivity gives the local matrix \(I(\theta_0)\). This calculation also shows why one must state the nominal values before reporting a numerical matrix.
Worked Example: Information at One Concentration
Suppose one hypothetical observation is planned at \(C=2\) concentration units, the nominal values are \(E_{\max}=100\) response units and \(K=2\) concentration units, and the known residual standard deviation is \(\sigma=5\) response units. Calculate the local information contribution.
First evaluate the sensitivity vector:
The observation contributes \(gg^\top/\sigma^2\), and \(\sigma^2=25\):
The negative off-diagonal entry follows from the opposite signs of the two sensitivities at this concentration. The matrix entries have units determined by their parameter pair: the first entry is in inverse response-units squared, the off-diagonal entries in inverse response-unit times inverse concentration-unit, and the last entry in inverse concentration-units squared.
Changing the Nominal Values Changes the Answer
Keep the same observation condition and measurement variance, but change the nominal \(K\) from 2 to 4 concentration units while retaining \(E_{\max}=100\) response units. The sensitivity becomes
Worked Example: Recalculating at a Different Nominal Value
At \(C=2\), \(E_{\max}=100\), and \(K=4\), calculate the local information contribution and compare it with the preceding example.
The new sensitivity vector is
Using the same variance, \(\sigma^2=25\):
This differs from the earlier matrix even though the observation condition, error variance, and \(E_{\max}\) are unchanged. At the new \(K\), the model is less sensitive to both parameters at \(C=2\), so the corresponding information contributions are smaller. It would be incorrect to treat the first matrix as the information at every plausible \(K\).
Combining Contributions at Specified Nominal Values
For several independent observations, evaluate each sensitivity at the same nominal parameter vector, form its outer product, divide by the variance, and add the resulting matrices. The following example makes the calculation explicit. It uses one hypothetical observation at each of two concentrations; no population-level or study claim is intended.
Worked Example: A Two-Condition Local Information Matrix
Suppose observations are planned at \(C_1=1\) and \(C_2=4\) concentration units. Let \(E_{\max}=100\) response units, \(K=2\) concentration units, and \(\sigma=5\) response units. Calculate the local information matrix for these two observations.
At \(C_1=1\), the sensitivity is \(g_1=(1/3,-100/9)^\top\approx(0.3333,-11.1111)^\top\). At \(C_2=4\), it is \(g_2=(2/3,-400/36)^\top=(2/3,-100/9)^\top\approx(0.6667,-11.1111)^\top\). The sum of the outer products is
Divide every entry by \(\sigma^2=25\):
The lower-right sum is \(10000/81+10000/81=20000/81\approx246.9136\), not a sum based on either sensitivity squared only once or on a different parameter value. Dividing it by 25 gives approximately \(9.8765\). This direct check is useful because a single arithmetic error in an outer product can substantially alter a reported information entry.
Interpreting and Checking a Local Calculation
The matrix \(I(\theta_0)\) describes expected information under the stated model when the true parameter value is \(\theta_0\), or serves as a local approximation when \(\theta_0\) is a working value. It is not an observed-data quantity: the realized responses do not enter this expected-information calculation. It also does not establish that the nominal values are correct. It answers a conditional question: given these values, model, variance, and observation conditions, what information does the specified design provide locally?
Several checks help catch calculation or modeling failures:
- Check the gradient analytically and numerically. A small finite-difference change in each parameter should produce a mean change close to the corresponding derivative times that parameter change.
- Check units. Each matrix entry must have reciprocal units corresponding to the two parameters indexing its row and column.
- Check the variance factor. With the stated model, multiplying the residual variance by a constant divides every information entry by that constant.
- Check nominal-value sensitivity. Recalculate at other scientifically plausible values. Large changes indicate that a single local calculation is a fragile summary.
- Check the assumptions. Dependence between observations, unequal variances, or a variance that depends on parameters changes the calculation. Do not use the constant-variance formula without justification.
A common mistake is to compute derivatives at the nominal value but describe the resulting matrix as if it applied throughout the parameter space. The correction is to report the value explicitly, for example, “\(I(E_{\max}=100,K=2)\), under known \(\sigma=5\),” and explain that another nominal point may yield another matrix.
Local Information and Other Summaries
Local expected information differs from observed information, which is calculated from the realized data and the likelihood curvature at the fitted or specified parameter values. The two may be close in some settings, but they are not interchangeable. It also differs from a summary that averages performance across a range or distribution of parameter values: a single nominal calculation does not perform that averaging.
When uncertainty about the parameter values is material, a practical check is to repeat the local calculation at several plausible nominal values and report how the matrices change. This is a sensitivity analysis, not a replacement for stating the assumed model. Its purpose is to reveal whether conclusions depend strongly on one chosen parameter point. Local information remains useful because it gives a precise, interpretable calculation at a specified point; its limitation is precisely that locality.
Do not infer from one entry alone that the entire design is informative or uninformative. The matrix summarizes joint local sensitivity, and its entries must be read together with the model, parameter units, nominal values, and variance assumptions. Assessing structural problems in the matrix or translating it into particular design criteria requires additional tools beyond this calculation.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- For the Emax model, write the sensitivity vector at a general concentration \(C\).
- At \(C=2\), \(E_{\max}=100\), \(K=2\), and \(\sigma=5\), what is the information contribution for one observation?
- Why does changing \(K\) change the local information matrix even if the design concentration and variance stay fixed?
- If the variance were multiplied by four while the sensitivities remained the same, how would the information matrix change under the stated model?
- Name two assumptions that would require reconsidering the formula used in this tutorial.
Solutions. (1) \(g(C;E_{\max},K)=(C/(K+C),-E_{\max}C/(K+C)^2)^\top\). (2) The contribution is \(\begin{pmatrix}0.01000&-0.2500\\-0.2500&6.250\end{pmatrix}\), obtained from \(gg^\top/25\). (3) Both derivative components involve \(K\), so substituting a different value changes the sensitivity vector and its outer product. (4) Every entry is divided by four because information is proportional to \(1/\sigma^2\). (5) Examples include dependent observations, unequal variances under a formula that assumes a common variance, or variance depending on the parameters.