Fisher Information Methods for Model-Based Sample Size

Why Use Fisher Information for Sample-Size Planning?

Many clinically relevant endpoints are not parameters estimated directly by a model. They may be a change from baseline predicted at a particular time, a difference between two predicted responses, or a transformation such as a rate ratio. Their precision depends both on the model and on the design: the assessment times, the number of observations, and the variability of those observations. Expected Fisher information provides a way to connect these features to parameter precision and, with the delta method, to precision for derived endpoints.

The resulting sample size is an approximation, not a guarantee. It is based on the assumed model and parameter values, and it is most reliable when the model is identifiable and the estimator has an approximately normal sampling distribution. A plan should therefore be checked against plausible data-generating behavior, especially when the model is strongly nonlinear or the intended sample is small.

Information, Parameter Uncertainty, and the Delta Method

Definition: Let $Y$ denote the observations from a model with parameter vector $\boldsymbol{\theta}$, and let $\ell(\boldsymbol{\theta};Y)$ be its log-likelihood. The expected Fisher information is the matrix $$ \mathcal{I}(\boldsymbol{\theta}) = -E\left[\frac{\partial^2 \ell(\boldsymbol{\theta};Y)} {\partial\boldsymbol{\theta}\,\partial\boldsymbol{\theta}^{\mathsf T}}\right]. $$ Under regularity conditions, it also equals the covariance matrix of the score, the first derivative of the log-likelihood. For $N$ independent participants with the same design, expected information is $N\mathcal{I}_1(\boldsymbol{\theta})$, where $\mathcal{I}_1$ is the information per participant.

When the model is correctly specified, parameters are identifiable, and the estimator is sufficiently close to its large-sample regime, the maximum-likelihood estimator has approximate covariance matrix $[N\mathcal{I}_1(\boldsymbol{\theta})]^{-1}$. This is a local approximation evaluated at assumed parameter values. If observations within a participant are correlated, the participant's full joint likelihood must be used to compute that participant's information; treating repeated observations as independent would generally give the wrong matrix.

Delta method: For a differentiable scalar endpoint $g(\boldsymbol{\theta})$, let $\nabla g$ be its gradient evaluated at the assumed parameter values. Its approximate variance is $$ \operatorname{Var}\{\widehat{g}\} \approx \nabla g^{\mathsf T} [N\mathcal{I}_1(\boldsymbol{\theta})]^{-1}\nabla g =\frac{V_g}{N}, \qquad V_g=\nabla g^{\mathsf T}\mathcal{I}_1(\boldsymbol{\theta})^{-1}\nabla g. $$

For a two-sided confidence interval with nominal confidence level $1-\alpha$, an approximate target half-width $d$ requires $z_{1-\alpha/2}\sqrt{V_g/N}\leq d$. Thus the continuous sample-size requirement is $N\geq z_{1-\alpha/2}^{\,2}V_g/d^2$, rounded up to a whole number. This is a precision calculation; it does not by itself specify power for a hypothesis test. For a test of a scalar contrast with null value zero, approximate standard error $\sqrt{V_g/N}$, two-sided level $\alpha$, and target power $1-\beta$ at an alternative value $\delta$, a corresponding normal-approximation requirement is $N\geq (z_{1-\alpha/2}+z_{1-\beta})^2V_g/\delta^2$.

Worked Example: Precision for a Nonlinear Transformation

Suppose a hypothetical model estimates a dimensionless parameter $\theta$, and the endpoint of interest is $g(\theta)=e^{-\theta}$. At the planning value $\theta=0.6$, assume the expected information per participant is $1/0.09$, so that $\operatorname{Var}(\hat{\theta})\approx0.09/N$. Find the sample size for a 95% confidence interval for $g$ with approximate half-width no greater than 0.05.

The endpoint estimate is $g(0.6)=e^{-0.6}\approx0.5488$. Its derivative is $g'(\theta)=-e^{-\theta}$. The delta method therefore gives $$ \operatorname{Var}(\hat{g}) \approx [g'(0.6)]^2\frac{0.09}{N} =\frac{0.09e^{-1.2}}{N} =\frac{9e^{-1.2}/100}{N}. $$

The per-participant variance coefficient is $V_g=9e^{-1.2}/100\approx0.0271075$. With $z_{0.975}=1.96$, the required sample size is $$ N\geq\frac{1.96^2(0.0271075)}{0.05^2} \approx41.65. $$

Rounding up gives 42 participants. Under the assumed model and its large-sample approximation, this is the number expected to give a 95% interval with an approximate half-width of at most 0.05 on the dimensionless $g$ scale. It does not mean there is a 95% probability that the fixed parameter lies in this particular interval.

Derived Contrasts With Multiple Parameters

For a vector of model parameters, the entire information matrix matters. Inverting only its diagonal entries ignores correlations among parameter estimates and can misstate the precision of a derived endpoint. The gradient of the endpoint determines how uncertainty in each parameter, including their covariance, contributes to the endpoint's variance.

For a longitudinal model, one practical design calculation is to compute each participant's expected information from the planned assessment schedule, sum across participants, invert the resulting matrix, and propagate that covariance through the endpoint gradient. The next example makes this calculation explicit. The design and assumed parameter values are hypothetical.

Worked Example: A Predicted Response at a Later Time

Suppose each hypothetical participant has normally distributed measurements with variance 1 at times $t=0$ and $t=1$, and follows the mean model $m(t;a,b)=ae^{-bt}$. The two measurements within a participant are independent. Plan for $a=2$, $b=0.5$, with half the participants assigned to each time for this simplified design. The endpoint is the model-predicted response at $t=2$, $g(a,b)=ae^{-2b}$. Calculate the approximate sample size for a 95% confidence-interval half-width of 0.25. Here, each participant contributes one measurement at their assigned time.

For a normal observation with known variance 1, the per-observation information is the outer product of the gradient of its mean. At $t=0$, the gradient is $(1,0)^{\mathsf T}$; at $t=1$, it is $(e^{-0.5},-2e^{-0.5})^{\mathsf T}$. Averaging the two contributions gives the per-participant information matrix $$ \mathcal{I}_1= \begin{pmatrix} 0.683940 & -0.367879\\ -0.367879 & 0.735759 \end{pmatrix}. $$ Its determinant is approximately $0.367879$, so it is invertible. Its inverse is approximately $$ \mathcal{I}_1^{-1}= \begin{pmatrix} 2.00000 & 1.00000\\ 1.00000 & 1.85914 \end{pmatrix}. $$

At the planning values, $g=2e^{-1}\approx0.7358$, and the gradient with respect to $(a,b)$ is $\nabla g=(e^{-1},-4e^{-1})^{\mathsf T}$. Substitution into the delta-method expression gives $$ V_g=\nabla g^{\mathsf T}\mathcal{I}_1^{-1}\nabla g \approx3.2137. $$

For a 95% interval and $d=0.25$, $$ N\geq\frac{1.96^2(3.2137)}{0.25^2} \approx197.53. $$

The design therefore requires at least 198 participants under these assumptions. The result is for a single measurement per participant assigned to one of the two times, not for every participant measured at both times. Because the two parameters are correlated, the off-diagonal entries of the inverse information matrix contribute to the endpoint variance.

Assessment-Time Choice Changes Information

Information depends on where measurements are taken, not only on how many participants are enrolled. A measurement time can be informative about one parameter and uninformative about another. Comparing candidate designs using the same assumed model and parameter values helps show how those differences affect the sample size for a specified target.

Worked Example: Comparing Times for an Exponential Rate

In a hypothetical study, the mean response is $m(t)=E_0+Ae^{-kt}$, where $E_0$ and $A=3$ are known, the rate $k=0.4$ is the parameter of interest, and each independent observation has normally distributed error with known standard deviation $\sigma=1.5$. Compare designs that measure one response per participant at $t=1$ or at $t=2$, targeting a standard error for $\hat{k}$ no greater than 0.08.

For one normal observation with known variance $\sigma^2$, information about $k$ is $[\partial m(t)/\partial k]^2/\sigma^2$. Since $\partial m(t)/\partial k=-At e^{-kt}$, the information per participant is $$ \mathcal{I}_1(k;t)=\frac{A^2t^2e^{-2kt}}{\sigma^2}. $$

At $t=1$, this is $9e^{-0.8}/2.25=4e^{-0.8}\approx1.79732$. The standard error approximation is $1/\sqrt{N\mathcal{I}_1}$, so the target requires $$ N\geq\frac{1}{1.79732(0.08^2)} \approx86.93. $$

At $t=2$, the information is $36e^{-1.6}/2.25=16e^{-1.6}\approx3.23034$, giving $$ N\geq\frac{1}{3.23034(0.08^2)} \approx48.37. $$

The respective rounded requirements are 87 and 49 participants. In this particular model, with $E_0$, $A$, and $\sigma$ treated as known, the later time is more informative about $k$ at the specified planning value. This comparison would change if other parameters were estimated, the error structure differed, or the assumed rate differed.

Calculation Workflow and Assumptions

Specify the model and design.
State the response distribution, which parameters are estimated, the assessment schedule, and the dependence structure for observations within each participant.

Choose planning values and the endpoint.
Evaluate information and the endpoint gradient at scientifically plausible parameter values. Write the endpoint as an explicit function of the model parameters.

Calculate information and endpoint variance.
Obtain the expected information per participant for the full observation schedule, invert the matrix, and calculate $V_g=\nabla g^{\mathsf T}\mathcal{I}_1^{-1}\nabla g$.

Apply the target criterion.
Use the confidence-interval half-width formula for a precision objective, or a power formula for a specified test and alternative. Round the resulting sample size up.

Check whether the approximation is credible.
Inspect identifiability, information-matrix conditioning, and parameter plausibility. Consider simulation-based validation if nonlinear behavior or small samples may undermine the normal approximation.

Diagnostics, Alternatives, and Limitations

The information calculation is only useful if the model and planned design can estimate the relevant parameters. A singular information matrix prevents use of the usual inverse-information covariance approximation and indicates a locally uninformative parameter direction under the regular, first-order approximation; it does not by itself establish global nonidentifiability. A nearly singular matrix can produce very large variances, strong parameter correlations, and endpoint estimates that are sensitive to small changes in the assumptions. Numerical checks should confirm that the matrix is symmetric and positive definite and that its inverse is stable.

Other warning signs include a very small predicted response at all scheduled times, a rate parameter that changes little over the observed time range, or a gradient that is nearly zero for the endpoint of interest. These can leave the endpoint weakly informed even if some model parameters are individually estimated with reasonable precision. Also check that the assumed error model and within-participant covariance match the planned data structure. An incorrect independence assumption, for example, can overstate the information contributed by repeated measurements.

Approach	What it provides	Main consideration
Expected Fisher information with the delta method	A fast local approximation to parameter or derived-endpoint precision across candidate designs.	Depends on correct model assumptions, planning values, identifiability, and an adequate large-sample approximation.
Direct analysis of a simpler endpoint	May yield a simpler sample-size calculation when the endpoint itself has a tractable sampling distribution.	May not represent the nonlinear model-based estimand or the planned longitudinal structure.
Simulation under the planned model	Can assess estimator behavior, interval coverage, and power in finite samples under specified scenarios.	Requires explicit data-generating assumptions and more computation; simulation does not resolve uncertainty about which assumptions are plausible.

The delta method linearizes $g$ near the assumed parameter values. If $g$ is sharply curved over the likely range of estimates, its approximation may be poor even when the parameter estimates are approximately normal. Information-based calculations also condition on chosen nuisance-parameter values; they do not automatically account for uncertainty about those values or guarantee robustness to model misspecification. Comparing designs at more than one plausible parameter setting and validating important plans by simulation can reveal material departures from the approximation.

Common Pitfall

A common error is to calculate the variance of a nonlinear endpoint by applying the transformation only to the parameter variance—for example, to report $\operatorname{Var}(\hat{\theta})$ as the variance of $e^{-\hat{\theta}}$. These are different quantities and generally have different units and scales. The correction is to differentiate the endpoint, then propagate the parameter covariance: $\operatorname{Var}(\hat{g})\approx[g'(\theta)]^2\operatorname{Var}(\hat{\theta})$ for a scalar parameter, or use the full gradient and covariance matrix for multiple parameters. A second practical check is arithmetic: retain adequate precision through the calculation, then round the required sample size up only at the end.

Key takeaway: Expected information describes how the planned design estimates model parameters; the delta method translates that uncertainty to a nonlinear endpoint. The resulting sample size is a locally evaluated approximation whose assumptions and numerical stability must be checked.

Check Your Understanding

Use the model-based information and delta-method principles to answer each question. Solutions follow the questions.

If $g(\theta)=\theta^2$, what derivative is used to approximate the variance of $\hat{g}$ at a planning value $\theta_0$?
If $V_g=0.04$, what is the approximate sample size for a 95% interval with half-width $d=0.10$?
Why can the off-diagonal entries of the inverse information matrix matter for a derived endpoint?
What does a singular expected information matrix indicate about the proposed design and model parameters?

Solutions. (1) The derivative is $g'(\theta_0)=2\theta_0$, so the approximate variance is $4\theta_0^2\operatorname{Var}(\hat{\theta})$. (2) $N\geq1.96^2(0.04)/0.10^2=15.3664$, so round up to 16. (3) They represent covariances among parameter estimates; the endpoint gradient can combine those covariances with the variances, changing the endpoint's uncertainty. (4) A singular expected information matrix prevents use of the usual inverse-information covariance approximation and indicates a locally uninformative parameter direction under the regular, first-order approximation; it does not by itself establish global nonidentifiability.