Profile-Likelihood Confidence Intervals in QSP Models

From a Profile Curve to Confidence Limits

A profile-likelihood plot is more than a picture of how well a parameter is estimated. Its height at a proposed parameter value can be compared with a likelihood-ratio cutoff to decide whether that value is compatible with the data at a chosen confidence level. The resulting confidence limits can be asymmetric, unlike intervals based on a symmetric standard-error approximation.

This is useful in QSP models because parameter effects can be nonlinear and compensatory: changing one rate may be partly offset by changes elsewhere in the model. A profile curve records the best fit available at each fixed value of the parameter of interest, so the interval reflects that change in fit rather than only the local curvature at the optimum.

Definition: Let $q$ be the scalar parameter of interest, let $\ell(\theta)$ be the log likelihood for the full parameter vector $\theta$, and let $\hat{\theta}$ maximize it, with $\hat{q}$ the corresponding estimate of $q$. The profile log likelihood $\ell_p(q)$ is the maximum log likelihood subject to fixing the parameter of interest at $q$. The profile deviance is $$ D(q)=2\{\ell(\hat{\theta})-\ell_p(q)\}. $$ It is nonnegative and equals zero at a global maximum of the likelihood.

The other model parameters are allowed to take the values that give the best fit for each fixed $q$. Constructing the profile curve is the subject of the earlier material on profile-likelihood construction; here the focus is how to turn that curve into confidence limits. The likelihood-ratio statistic compares the best fit with $q$ fixed to the unrestricted best fit. A larger $D(q)$ means a greater loss of fit from imposing that value.

The Likelihood-Ratio Confidence Set

For a single scalar parameter, under regularity conditions, the likelihood-ratio statistic evaluated at the true value is approximately distributed as a chi-square random variable with one degree of freedom. The main conditions are a correctly specified likelihood, adequate data for the asymptotic approximation, a locally identifiable parameter, a well-behaved likelihood, and a true value in the interior of the parameter space. The observation and residual dependence structure must also be represented appropriately in the likelihood; a time series of observations should not be treated as independent if its errors are dependent.

Let $\chi^2_{1,1-\alpha}$ be the $1-\alpha$ quantile of the chi-square distribution with one degree of freedom. The approximate $100(1-\alpha)\%$ likelihood-ratio confidence set is

C_{1-\alpha}=\{q:D(q)\leq \chi^2_{1,1-\alpha}\}.

For example, the 95% cutoff is $\chi^2_{1,0.95}\approx3.841$. If the profile deviance is below or equal to this cutoff, the corresponding value of $q$ is retained in the confidence set; values above it are excluded. When the set is a single connected region, its lower and upper endpoints are the points where the profile deviance crosses the cutoff. A 95% confidence procedure has approximately 95% coverage across repeated data sets under the stated assumptions. It does not mean that there is a 95% probability that a fixed parameter lies in this particular interval.

Why the cutoff works: Testing a proposed value $q_0$ compares the unrestricted maximum likelihood with the maximum under the restriction $q=q_0$. The likelihood-ratio statistic is $D(q_0)$. Under the regularity conditions above, Wilks' theorem gives $D(q_0)\approx\chi^2_1$ when $q_0$ is the true value. The test retains $q_0$ when $D(q_0)\leq\chi^2_{1,1-\alpha}$. Inverting that test over all possible $q_0$ gives the confidence set shown above.

Worked Confidence Limits

Worked Example: A Nearly Symmetric Profile

Suppose a hypothetical QSP model estimates a drug-elimination rate constant $k$, measured in $\text{day}^{-1}$. The maximum-likelihood estimate is $\hat{k}=4.00\ \text{day}^{-1}$, and the numerical profile around it is described by

D(k)=\frac{(k-4.00)^2}{(0.36)^2}.

This illustrative profile is symmetric and quadratic. For an approximate 95% confidence set, retain values with $D(k)\leq3.841$. Solving for the two cutoff crossings gives

\frac{(k-4.00)^2}{(0.36)^2}\leq3.841 \quad\Longrightarrow\quad |k-4.00|\leq0.36\sqrt{3.841}.

Using $\sqrt{3.841}\approx1.959847$, the half-width is $0.36(1.959847)\approx0.7055449\ \text{day}^{-1}$. Therefore the limits, calculated before rounding, are $4.00-0.7055449=3.2944551$ and $4.00+0.7055449=4.7055449$. To three decimal places, the interval is $[3.294,\ 4.706]\ \text{day}^{-1}$. Its width is approximately $1.411\ \text{day}^{-1}$, calculated from the unrounded endpoints. Under the model and calibration assumptions, values within this range are not rejected by the 5% likelihood-ratio test.

Worked Example: Asymmetric Limits From a Nonlinear Profile

Consider a hypothetical profile for a turnover parameter $q$, measured in $\text{day}^{-1}$, with estimate $\hat{q}=6.00\ \text{day}^{-1}$. Over the range of interest, suppose the fitted profile deviance is

D(q)=\frac{(q-6.00)^2}{\{0.25+0.12(q-6.00)\}^2}.

The denominator is positive around both 95% cutoff crossings. Set $d=q-6.00$ and $c=\sqrt{3.841}\approx1.959847$. At the upper crossing, $d>0$, so $\sqrt{D(q)}=d/(0.25+0.12d)=c$. Rearranging gives

d=\frac{0.25c}{1-0.12c} =\frac{0.489962}{0.764818} \approx0.640625.

At the lower crossing, $d<0$, so $\sqrt{D(q)}=-d/(0.25+0.12d)=c$. Thus

d=-\frac{0.25c}{1+0.12c} =-\frac{0.489962}{1.235182} \approx-0.396672.

The limits are therefore $6.00-0.396672\approx5.603328$ and $6.00+0.640625\approx6.640625\ \text{day}^{-1}$, which round to $5.603$ and $6.641\ \text{day}^{-1}$ to three decimal places. This interval is asymmetric about the estimate: the upper limit is farther away than the lower limit. That asymmetry is part of the information in the profile, not an arithmetic error. Reporting only an estimate plus or minus one symmetric standard error would not reproduce these limits.

What Interval Width Says About Precision

For a single connected interval, its width is the difference between its upper and lower limits, in the units of the parameter. A narrow interval indicates that values far from the estimate cause a substantial loss of fit, so the parameter is relatively precisely estimated by this data set and model. A broad interval indicates that a wider range of values remains compatible with the data at the selected cutoff. This is a statement about practical precision, not about whether the parameter is biologically important or whether the model is scientifically correct.

Worked Example: Comparing Precision From Two Profile Curves

Two hypothetical data sets are analyzed with the same QSP model and the same parameter $r$, a rate constant measured in $\text{day}^{-1}$. Both give $\hat{r}=2.00\ \text{day}^{-1}$. Their profile deviances are, respectively,

D_A(r)=\frac{(r-2.00)^2}{(0.20)^2}, \qquad D_B(r)=\frac{(r-2.00)^2}{(0.60)^2}.

For data set A, the 95% half-width is $0.20\sqrt{3.841}\approx0.391969\ \text{day}^{-1}$, giving $[1.608,\ 2.392]\ \text{day}^{-1}$. Its width is approximately $0.784\ \text{day}^{-1}$. For data set B, the half-width is $0.60\sqrt{3.841}\approx1.175908\ \text{day}^{-1}$, giving $[0.824,\ 3.176]\ \text{day}^{-1}$. Its width is approximately $2.352\ \text{day}^{-1}$.

The point estimates are identical, but data set A produces the narrower profile-likelihood interval and therefore greater practical precision for $r$ under this model. The comparison concerns the profile information actually available in these analyses; it does not by itself identify why the profiles differ.

Computation, Diagnostics, and Alternatives

In practice, a profile is often evaluated on a grid of fixed parameter values, then optimized locally or globally as needed. The cutoff limits can be estimated by interpolation between adjacent grid points, or by solving $D(q)=\chi^2_{1,1-\alpha}$ numerically on each side of the estimate. Because the limits depend on where the curve crosses the cutoff, the grid must be sufficiently fine near those crossings.

Choose the confidence level.
For a 95% interval use $\alpha=0.05$ and the one-degree-of-freedom cutoff $3.841$.

Locate the cutoff crossings.
Find the values of the parameter where the profile deviance equals the cutoff, checking both sides of the maximum when both are present.

Report limits and width in context.
Give the parameter name, units, confidence level, and any meaningful asymmetry or unresolved endpoint.

A sound calculation depends on the profile values being accurate. A numerical optimization that fails at some fixed parameter values can make the profile deviance spuriously high, creating a false cutoff crossing and an interval that is too narrow. Conversely, an inaccurate unrestricted maximum can shift every deviance value. Inspect the curve for abrupt jumps or irregularities, use multiple starting values when optimization is sensitive to initialization, and verify that the reported maximum is at least as good as the fitted constrained values. Refine the grid around apparent crossings and confirm that the limits are stable to tighter optimization tolerances.

A profile need not always produce one neat, closed interval. If the cutoff is not crossed on one side within the explored range, the limit has not been located there; do not report the edge of the computational grid as a confidence limit. A flat profile can imply a very wide or unresolved confidence set. If a profile has separated regions below the cutoff, the likelihood-ratio confidence set is disconnected; replacing it with a single interval may include values that the procedure excludes. Numerical checks should distinguish these shapes from optimization artifacts.

Approach	What determines the limits	Useful feature and caution
Profile-likelihood interval	Cutoff crossings of the likelihood-ratio profile deviance	Can reflect asymmetry and nonlinear loss of fit; depends on a valid profile and suitable likelihood-ratio calibration.
Wald interval	Estimate plus or minus a standard-error multiple, using local curvature	Simple and often quick; can be misleading when the profile is asymmetric, flat, or poorly approximated by a quadratic curve.
Likelihood-ratio test of a specified value	Whether that value's profile deviance exceeds the selected cutoff	Tests one proposed value at a time; inverting the test over values gives the profile-likelihood confidence set.

The chi-square cutoff is an asymptotic calibration, not a universal finite-sample guarantee. Poor identifiability, a small or weakly informative data set, a nonregular likelihood, or a parameter near the edge of its allowed range can undermine the nominal coverage. A profile interval also remains conditional on the chosen model, observation-error assumptions, and data. It does not account automatically for model misspecification or uncertainty about the model structure.

Common Mistake

A frequent error is to treat the profile-likelihood interval as necessarily $\hat{q}$ plus or minus a symmetric standard-error amount, or to interpret its width as a probability that the estimate is correct. The correction is to find the points where $D(q)$ reaches the selected likelihood-ratio cutoff, report any resulting asymmetry, and interpret the interval as a repeated-sampling confidence procedure under its assumptions. Width describes the range of parameter values retained by that procedure; it is not a probability statement about an individual parameter value.

Key takeaway: A profile-likelihood confidence set retains parameter values whose profile deviance does not exceed the chosen likelihood-ratio cutoff. Its limits and width summarize how quickly fit worsens as the parameter moves away from its estimate, subject to the model, numerical optimization, and calibration assumptions.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A profile deviance is $D(q)=(q-10.0)^2/(0.50)^2$. Using the 90% one-parameter cutoff $2.706$, calculate the confidence limits and state the units if $q$ is measured in mg/L.
In repeated samples, what does a 95% likelihood-ratio confidence procedure mean? What does it not mean about the parameter in the interval from one data set?
Two profiles have the same estimate and confidence level, but one has a wider connected interval. What does that indicate about practical precision under the fitted model?
A computed profile crosses the cutoff on the lower side, but on the upper side it remains below the cutoff at the largest value evaluated. Is that largest evaluated value an upper confidence limit?

Solutions. (1) The half-width is $0.50\sqrt{2.706}\approx0.8225\ \text{mg/L}$. The limits are $10.0-0.8225=9.1775$ and $10.0+0.8225=10.8225\ \text{mg/L}$, approximately $[9.178,\ 10.822]\ \text{mg/L}$ to three decimals. (2) Across repeated data sets, the procedure has approximately 95% coverage when its assumptions and calibration hold. It does not assign a 95% probability to the fixed parameter being in the particular interval obtained. (3) The wider interval means a larger range of values remains below the selected likelihood-ratio cutoff, indicating lower practical precision for that parameter in that analysis. (4) No. The upper limit has not been found; the profile may cross beyond the evaluated range or may remain below the cutoff. The grid should be extended and the profile checked rather than reporting its endpoint as a confidence limit.