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Dose Optimization Under Uncertainty · Tutorial 238 of 497

Worked Example: Selecting a Dose With an Emax Model

A worked hypothetical analysis shows how an Emax fit and its uncertainty can inform dose selection when efficacy must exceed a target.

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What You'll Learn

  • Specify a baseline-adjusted Emax model and identify its assumptions.
  • Interpret nonlinear least-squares estimates and the parameter covariance from a fit.
  • Propagate joint parameter uncertainty to uncertainty in model-mean response.
  • Distinguish uncertainty intervals for mean response from predictive intervals for outcomes.
  • Calculate target-attainment probabilities at candidate doses and apply a stated dose rule.
  • Recognize when the normal and delta-method approximations may be unreliable.

From an Emax Fit to a Dose Decision

A fitted dose-response curve is not itself a dose recommendation. A curve describes the estimated average response at each dose, but the fitted parameters are uncertain, and individual outcomes may vary around that average. When a decision depends on achieving a clinical target, the relevant question is not just whether the estimated mean response exceeds the target. It is also how uncertain that mean is, and how likely an outcome at the chosen dose is to attain the target.

This worked example uses a saturating Emax model to connect those steps. The response is a baseline-adjusted efficacy score, in points, and dose is measured in mg. The numerical data and fit results below are hypothetical. The analysis focuses on efficacy attainment; it does not include a toxicity constraint or a utility penalty for higher doses.

Definition: The baseline-adjusted Emax model is $$ \mu(d)=\frac{E_{\max}d}{ED_{50}+d}, \qquad d\geq 0, $$ where \(\mu(d)\) is the model-mean response at dose \(d\), \(E_{\max}>0\) is the limiting response as dose increases, and \(ED_{50}>0\) is the dose at which the model-mean response is half of \(E_{\max}\). In this tutorial, a target-attainment event is the event that a response at dose \(d\) is at least a specified efficacy threshold.

The baseline-adjusted form sets the response at zero dose to zero. It is appropriate only when the response has been defined relative to baseline in a way that makes this constraint reasonable. If the outcome has a nonzero expected value at zero dose, a baseline parameter \(E_0\) may be needed, giving \(\mu(d)=E_0+E_{\max}d/(ED_{50}+d)\). The examples here use the simpler form.

The model assumes that the average response rises with dose and approaches a plateau, with no downturn at high doses. Its parameters are positive, and the response observations are assumed to be adequately described by this curve plus random error. These are substantive assumptions, not automatic consequences of a good numerical fit. In particular, this model does not describe individual covariate effects, time-varying responses, or toxicity.

Fitting and Checking the Model

For observations \(y_i\) at doses \(d_i\), a common fitting method is nonlinear least squares: choose \(E_{\max}\) and \(ED_{50}\) to minimize the sum of squared residuals between each observation and \(\mu(d_i)\). A likelihood-based fit can instead specify an error distribution, such as independent normal errors with a common variance. In either case, the fitting procedure should respect positive parameter constraints and should account for the actual study design and error structure.

The table gives hypothetical response means from a dose-ranging study. Suppose a nonlinear least-squares fit to the underlying study data, summarized here by these dose-group means, gives rounded estimates \(E_{\max}=100\) points and \(ED_{50}=50\) mg. Suppose a bootstrap calculation from the study data gives the following approximate covariance matrix for the two fitted parameters, in the order \(E_{\max}, ED_{50}\). The covariance matrix is part of the hypothetical fit output; it describes joint uncertainty, not the variability of individual responses.

Dose (mg)Observed mean response (points)
2535
5049
10068
20079
$$ \widehat{\mathrm{Cov}} \begin{pmatrix} E_{\max}\\ ED_{50} \end{pmatrix} = \begin{pmatrix} 64 & 24\\ 24 & 100 \end{pmatrix} $$

The diagonal entries correspond to approximate parameter variances, so the associated standard errors are 8 points for \(E_{\max}\) and 10 mg for \(ED_{50}\). The off-diagonal entry is positive: in the fitted uncertainty distribution, larger values of one parameter tend to occur with larger values of the other. Keeping that dependence matters when calculating uncertainty in a response, because both parameters determine the curve together.

Worked Example: Calculating the Fitted Mean Response

At 100 mg, the fitted mean response is

$$ \widehat{\mu}(100) =\frac{100(100)}{50+100} =\frac{10{,}000}{150} =66.67\ \text{points}. $$

This is the estimated average response under the fitted model, not a guarantee that an individual receiving 100 mg will respond by 66.67 points. The estimate is between the observed means at 50 mg and 200 mg, as expected for an increasing curve. Its proximity to the observed mean of 68 points at 100 mg is a basic fit check, but one dose group alone cannot establish that the Emax shape is adequate.

Useful checks include plotting observations and fitted means together, examining residuals by dose, checking whether residual spread changes with dose, and checking whether the fitted curve is driven by only one or two dose groups. Systematic residual patterns, an apparent decline at high doses, or strong sensitivity to starting values can indicate that the model form or fitting procedure is unsuitable. A plausible-looking curve is not proof that the Emax assumptions hold.

Propagating Parameter Uncertainty

The fitted mean response is a function of two uncertain parameters. A first-order approximation, called the delta method, uses the local slope of that function to translate parameter uncertainty into response uncertainty. Write the parameter vector as \(\theta=(E_{\max},ED_{50})^\mathsf{T}\), and let \(\widehat{\Sigma}\) be its estimated covariance matrix. At a fixed dose \(d\), the gradient is

$$ \nabla_\theta\mu(d)= \begin{pmatrix} \displaystyle\frac{d}{ED_{50}+d}\\[6pt] \displaystyle-\frac{E_{\max}d}{(ED_{50}+d)^2} \end{pmatrix}. $$

The first component describes how much the response changes with \(E_{\max}\); the second describes its change with \(ED_{50}\). The approximate variance of the fitted mean response is the gradient multiplied by the parameter covariance matrix and by the transposed gradient.

$$ \widehat{\mathrm{Var}}\{\widehat{\mu}(d)\} \approx \nabla_\theta\mu(d)^\mathsf{T} \widehat{\Sigma} \nabla_\theta\mu(d). $$

This follows by a first-order Taylor expansion around the fitted parameter estimates: a small parameter deviation \(\Delta\theta\) changes the fitted response by approximately \(\nabla_\theta\mu(d)^\mathsf{T}\Delta\theta\). The variance of that linear combination is the gradient-covariance-gradient expression above. The approximation is most useful when parameter uncertainty is not so large that substantial curvature or parameter-boundary effects dominate.

Worked Example: Uncertainty in the Mean at 100 mg

At \(d=100\) mg and the fitted estimates, the gradient is

$$ \nabla_\theta\mu(100)= \begin{pmatrix} 100/(50+100)\\ -100(100)/(50+100)^2 \end{pmatrix} = \begin{pmatrix} 0.6667\\ -0.4444 \end{pmatrix}. $$

Using the covariance matrix, the approximate variance is

$$ (2/3)^2(64)+(-4/9)^2(100) +2(2/3)(-4/9)(24) =28.4444+19.7531-14.2222 =33.9753\approx33.98\ \text{points}^2. $$

Thus the standard error for the fitted mean is approximately \(\sqrt{33.98}=5.829\) points. A pointwise approximate 90% interval for the model-mean response, using the normal critical value 1.645, is

$$ 66.67\pm1.645(5.829) =66.67\pm9.589, \qquad (57.08,\ 76.26)\ \text{points}. $$

Under the stated large-sample approximation, this is an uncertainty interval for the mean response at 100 mg, reflecting parameter uncertainty. It is not an interval containing 90% of individual responses, and it is not a statement that there is a 90% frequentist probability that a fixed parameter lies in this particular interval.

Mean-Response Uncertainty and Target Attainment

A dose-attainment probability requires an outcome distribution, not just an interval for the model mean. For this example, assume that a response at dose \(d\) is the model mean plus an independent normal residual with standard deviation 5 points. Also use the approximate normal distribution for the uncertain fitted mean obtained from the delta method. Under these assumptions, the approximate variance of a future response combines parameter uncertainty and residual variance:

$$ \widehat{\mathrm{Var}}\{Y(d)\} \approx \widehat{\mathrm{Var}}\{\widehat{\mu}(d)\}+5^2. $$

This calculation does not include between-person heterogeneity beyond the residual term specified here. If the decision concerns a particular population, covariate mix, or individual variability structure, that distribution must be represented in the model rather than silently omitted. A predictive probability is conditional on its model and uncertainty assumptions.

Worked Example: Target Attainment at Three Candidate Doses

Suppose the minimum acceptable response is 60 points. The candidate doses are 50, 100, and 200 mg. For each dose, calculate the delta-method variance of the fitted mean using the same covariance matrix, then add the residual variance of 25 points squared to obtain the approximate predictive standard deviation.

At 50 mg, the fitted mean is \(100(50)/(50+50)=50.00\) points. The gradient is \((0.5,-0.5)^\mathsf{T}\), so the variance of the fitted mean is \(0.25(64)+0.25(100)-12=29.00\), and its standard error is 5.385 points. The predictive standard deviation is \(\sqrt{29+25}=7.348\) points. Therefore

$$ P\{Y(50)\geq60\} \approx 1-\Phi\left(\frac{60-50}{7.348}\right) =1-\Phi(1.361) \approx0.087. $$

At 100 mg, the fitted mean is 66.67 points, the mean-response variance is 33.98 points squared, and the predictive standard deviation is \(\sqrt{33.98+25}=7.680\) points. The attainment probability is

$$ P\{Y(100)\geq60\} \approx 1-\Phi\left(\frac{60-66.67}{7.680}\right) =\Phi(0.868) \approx0.807. $$

At 200 mg, the fitted mean is \(100(200)/(50+200)=80.00\) points. Its gradient is \((0.8,-0.32)^\mathsf{T}\), giving mean-response variance \(0.64(64)+0.1024(100)+2(0.8)(-0.32)(24)=38.91\) points squared. The predictive standard deviation is \(\sqrt{38.91+25}=7.994\) points, and

$$ P\{Y(200)\geq60\} \approx 1-\Phi\left(\frac{60-80}{7.994}\right) =\Phi(2.502) \approx0.994. $$

If the decision rule is to choose the lowest candidate dose with an estimated attainment probability of at least 0.80, 100 mg is selected: its estimated probability is about 0.807, while 50 mg falls well below the threshold. This is a rule applied to the specified candidates and assumptions; it is not a universal dose-selection criterion. The 100 mg estimate is close to the cutoff, so rounding and uncertainty in the probability itself deserve attention.

For comparison, the approximate 90% interval for the mean response at 50 mg is \(50\pm1.645\sqrt{29}\), or 41.09 to 58.91 points; at 200 mg it is \(80\pm1.645\sqrt{38.91}\), or 69.74 to 90.26 points. These are intervals for the dose-specific model mean. The target-attainment calculations instead use predictive standard deviations that also include residual variation. For instance, at 100 mg the approximate 90% predictive interval is \(66.67\pm1.645(7.679)\), or 54.04 to 79.30 points. Confusing these intervals changes the quantity being described.

Why the Joint Uncertainty and Model Checks Matter

The covariance term can materially affect the propagated variance. At 100 mg, its contribution is negative because the gradient with respect to \(ED_{50}\) is negative while the parameter covariance is positive. Omitting that term would give \(28.44+19.75=48.20\) points squared instead of 33.98. Treating the parameters as independent when they are not can therefore misstate uncertainty and, in turn, the estimated probability of meeting the target.

The normal approximation also has limits. The Emax curve is nonlinear, the parameters are constrained to be positive, and a parameter distribution with substantial skewness may not be well summarized by a covariance matrix. The delta method may produce poor intervals when uncertainty is large, the dose is far outside the range supported by observations, or the fitted parameters are near a boundary. In those cases, propagate joint bootstrap or posterior parameter draws through the Emax equation and calculate response and attainment summaries from the resulting draws. If the response residual is non-normal or heteroscedastic, simulate from an appropriate residual model as well.

A common mistake is to select a dose because its estimated mean exceeds the target, then describe that as a high probability of individual target attainment. The correction is to state the event and the distribution being used. A mean above 60 points does not imply that most outcomes exceed 60: at 100 mg the fitted mean is 66.67 points, while the estimated predictive attainment probability is about 0.807 under the assumptions above. Conversely, an uncertainty interval for the mean is not a predictive interval for outcomes.

A further limitation is that the candidate-dose rule optimizes only the efficacy criterion stated here. A clinically appropriate decision may also need to consider a safety limit, dose burden, or consequences of missing the target. Those considerations require explicit decision criteria; they cannot be inferred from the Emax fit alone. The fit provides a model-based summary, while the decision rule states how that summary is used.

1
Check the model and data.
Confirm that the baseline-adjusted saturating response is scientifically plausible, inspect residuals and dose coverage, and verify that the fit is not dominated by a boundary or a single dose group.
2
Propagate joint parameter uncertainty.
Use the full covariance matrix in a delta-method calculation, or use joint parameter draws when nonlinearity or non-normality makes the approximation doubtful.
3
Define the attainment event.
Specify the response threshold and whether the probability concerns the uncertain mean or a future response that also includes residual and population variability.
4
Apply and qualify a decision rule.
Compare candidates using the stated probability threshold or other explicit criterion, and report that the recommendation depends on the model, uncertainty representation, and decision assumptions.

Check Your Understanding

Answer each question, then compare with the concise solutions.

  1. For the Emax model used here, what is the model-mean response at \(ED_{50}\)?
  2. Why does the covariance between \(E_{\max}\) and \(ED_{50}\) enter the variance of the fitted mean response?
  3. At 100 mg, why is the predictive standard deviation larger than the standard error for the fitted mean?
  4. If a dose has a fitted mean above the target, does that by itself establish a high probability of target attainment? Explain.
  5. Name one situation in which a joint-draw approach may be preferable to the delta method.

Solutions. (1) Substituting \(d=ED_{50}\) gives \(\mu(ED_{50})=E_{\max}ED_{50}/(ED_{50}+ED_{50})=E_{\max}/2\). (2) The response depends on both parameters, so its variance includes the cross-product of their sensitivities and covariance; ignoring that term assumes independence. (3) Predictive variability includes the residual variance for individual outcomes in addition to uncertainty in the fitted mean. (4) No. Attainment depends on the outcome distribution and the target, not only on the mean; the response may vary around that mean. (5) Joint draws are useful when parameter uncertainty is skewed, the curve is strongly nonlinear over the uncertainty range, or parameters are near constraints.

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