A Piecewise Medication Dose-Response Model

Why Allow the Dose-Response Slope to Change?

A single straight line assumes that each additional unit of medication dose is associated with the same average change in symptom improvement throughout the dose range. That may be too restrictive. A continuous piecewise-linear model allows the slope to change at one specified dose while keeping the fitted response joined at that dose. It can describe, for example, a steeper increase in average improvement at lower doses followed by a more gradual increase at higher doses.

The change point, called the knot, should be specified before examining the outcome data when the goal is ordinary coefficient inference. A knot chosen after searching many possible doses introduces additional uncertainty that the usual standard errors and tests described here do not account for. The model describes an association between dose and response; by itself, it does not establish that changing the dose causes a change in symptoms.

Definition: Let $x$ be medication dose, let $c$ be a prespecified knot, and let $Y$ be symptom improvement. Define the hinge variable $h=(x-c)_+=\max(0,x-c)$. A continuous piecewise-linear regression model is $$ Y_i=\beta_0+\beta_1x_i+\beta_2(x_i-c)_+ + \varepsilon_i. $$ For doses at or below $c$, the mean response has slope $\beta_1$. Above $c$, its slope is $\beta_1+\beta_2$. Thus, $\beta_2$ is the change in slope after the knot. The fitted mean is continuous at $c$, because the hinge term is zero there.

The usual regression conditions apply: the conditional mean is represented adequately by the stated model; errors have mean zero given the doses; observations are independent; and their errors have constant variance. For the usual small-sample $t$- and $F$-based tests and intervals to be exact, errors are also assumed normally distributed. These conditions concern the errors around the mean model, not the distribution of the doses. In a clinical setting, study design and possible confounding also matter when interpreting the dose-response association.

Interpreting the Model

The coefficient $\beta_1$ is the average change in symptom-improvement score per unit of dose for doses below the knot. The post-knot slope is $\beta_1+\beta_2$, and $\beta_2$ tells how much the slope changes upon passing the knot. If improvement is measured in score points and dose in mg/day, both slopes and their difference have units of score points per mg/day. A negative $\beta_2$ means the slope becomes smaller; it does not necessarily mean that the predicted improvement itself decreases.

At the knot, the predicted mean is $\beta_0+\beta_1c$. There is no sudden jump in the mean at that dose. The intercept $\beta_0$ is the predicted mean response at dose zero; it is substantively meaningful only if zero dose is within a relevant range and the model's interpretation at zero is appropriate.

Worked Example: Reading Slopes on Both Sides of a Knot

Suppose a hypothetical fitted model relates dose $x$, in mg/day, to symptom improvement $Y$, in score points, with a prespecified knot at $c=8$ mg/day: $$ \widehat{Y}=2.5+0.9x-0.5(x-8)_+. $$ Calculate and interpret the predicted means at doses of 6, 8, and 12 mg/day.

At 6 mg/day, the hinge is $(6-8)_+=0$, so $\widehat{Y}=2.5+0.9(6)=7.9$ points. At the knot, the hinge is also zero, giving $\widehat{Y}=2.5+0.9(8)=9.7$ points. At 12 mg/day, the hinge is $12-8=4$, so $\widehat{Y}=2.5+0.9(12)-0.5(4)=11.3$ points.

The slope below 8 mg/day is $0.9$ score points per mg/day. Above 8 mg/day it is $0.9-0.5=0.4$ score points per mg/day. The fitted mean rises on both sides of the knot, but less steeply above it. These are model-based associations, not evidence that increasing the dose would cause the stated improvements.

Fitting and Comparing the Models

For a fixed knot, fitting this model is ordinary least squares regression with three predictors in the model equation: an intercept, dose $x$, and hinge $(x-c)_+$. Its simpler comparison model is the straight-line regression $Y_i=\beta_0+\beta_1x_i+\varepsilon_i$. The straight-line model is nested within the piecewise model: it is obtained by setting $\beta_2=0$. This makes the question of whether the slope changes equivalent to testing $H_0:\beta_2=0$ against $H_a:\beta_2\ne0$.

The knot must be fixed for that test to have its usual interpretation. Under the null, the model has one common slope. Under the alternative, the slope after the knot may differ. A significant test indicates evidence of a slope change at the specified knot under the model assumptions; it does not establish that the knot is the uniquely correct dose or that the change is clinically important.

Theorem: With one prespecified knot, the straight-line model and continuous piecewise-linear model differ by one coefficient, $\beta_2$. Under the regression assumptions, testing $H_0:\beta_2=0$ is equivalent to the one-degree-of-freedom partial $F$-test comparing the two models. The test statistic is $$ F=\frac{(SSE_{\text{linear}}-SSE_{\text{piecewise}})/1}{SSE_{\text{piecewise}}/(n-3)}, $$ where $SSE$ is the sum of squared residuals and $n-3$ is the residual degrees of freedom in the piecewise model. For this single added coefficient, $F=t^2$, where $t$ is the $t$-statistic for $\beta_2$.

The comparison can also be understood through least squares. The hinge variable adds one direction in which the fitted values can vary beyond those available to a straight line. Least squares chooses the coefficient in that direction to minimize the residual sum of squares. Consequently, the piecewise model's in-sample $SSE$ cannot exceed that of the straight-line model. The reduction in $SSE$ alone does not prove that the added term is useful: the $F$-test compares that reduction with the remaining residual variation.

The uncertainty in the slope change is summarized by its standard error and confidence interval. A two-sided test uses $t=\hat{\beta}_2/SE(\hat{\beta}_2)$, with $df=n-3$, and the corresponding confidence interval is $\hat{\beta}_2\pm t^*SE(\hat{\beta}_2)$, where $t^*$ is the relevant critical value. The interval concerns the change in slope, in response units per dose unit. The fitted model's residual standard error, $\sqrt{SSE/(n-3)}$, describes the typical size of residuals in response units under the model.

Worked Example: Fitting and Comparing Models With Hypothetical Data

Consider hypothetical observations of symptom-improvement score $Y$ at doses $x=0,2,\ldots,18$ mg/day. Suppose the observed scores are $6,\ 5.4,\ 10.8,\ 12.2,\ 14.6,\ 17,\ 18.8,\ 17.6,\ 18.4,\ 21.2$, respectively. The knot at $c=10$ mg/day was specified in advance. Assume the observations are independent, the mean is adequately represented by the chosen model, and errors have mean zero and constant variance; normal errors are assumed for the small-sample tests below.

For this data set, the least-squares piecewise fit is $\widehat{Y}=5+1.2x-0.8(x-10)_+$. The estimated slope below 10 mg/day is $1.2$ score points per mg/day. Above 10 mg/day it is $1.2-0.8=0.4$ score points per mg/day, so the estimated change in slope is $-0.8$ score points per mg/day. The residuals from this fit are $1,-2,1,0,0,0,1,-1,-1,1$; their squared values sum to $SSE_{\text{piecewise}}=10$.

With $n=10$ and three fitted coefficients, the piecewise model has $df=10-3=7$, so its residual standard error is $\sqrt{10/7}=1.195$ score points. The straight-line fit for the same observations has slope $284/330=0.8606$ score points per mg/day and $SSE_{\text{linear}}=23.1879$. The partial $F$-statistic is therefore $$ F=\frac{(23.1879-10)/1}{10/7}=9.232. $$ With $df=1$ and $7$, the two-sided $p$-value is about $0.019$. At the $0.05$ significance level, this is evidence against a common slope in favor of a change at the prespecified knot, subject to the assumptions. It is not proof of a biological threshold or a causal dose effect.

The uncertainty for the estimated change can be checked directly. For the hinge variable, its sum of squares after accounting for the intercept and dose is $80-140^2/330=20.606$. Thus $SE(\hat{\beta}_2)=\sqrt{(10/7)/20.606}=0.2633$ score points per mg/day. The $t$-statistic is $-0.8/0.2633=-3.038$, whose square agrees with the partial $F$-statistic up to rounding. Using $t^*=2.365$ for 7 degrees of freedom, the 95% confidence interval for the slope change is $-0.8\pm2.365(0.2633)=[-1.423,-0.177]$ score points per mg/day. In repeated use of this procedure under its assumptions, intervals constructed this way cover the true change in slope 95% of the time; this particular interval is not a probability statement about the fixed parameter.

The linear model's residual standard error is $\sqrt{23.1879/8}=1.702$ score points, larger than the piecewise model's $1.195$. This is a descriptive in-sample comparison. It does not guarantee better predictions for new patients or doses, and the extra coefficient costs degrees of freedom. The estimates also do not account for uncertainty from selecting a knot using these same outcomes.

What the Comparison Can and Cannot Show

The model comparison is useful when a single average slope would obscure a meaningful change in the dose-response pattern. It reports whether the observed improvement in fit is large relative to residual variation, while the coefficient interval conveys uncertainty and direction for the change in slope. Neither result alone establishes clinical importance. For that, the size and units of the slopes, the scale of symptom improvement, the relevant dose range, and clinical context all matter.

Worked Example: Interpreting an Uncertain Slope Change

Suppose a separate hypothetical analysis uses a prespecified knot and estimates the slope change as $\hat{\beta}_2=-0.35$ score points per mg/day, with $SE(\hat{\beta}_2)=0.18$ and $df=16$. The estimated change gives $t=-0.35/0.18=-1.944$. For a two-sided test, the $p$-value is approximately $0.070$.

The 95% critical value is approximately $2.120$, so the interval is $-0.35\pm2.120(0.18)=[-0.732,0.032]$ score points per mg/day. Because the interval includes zero, the data do not provide statistically significant evidence of a slope change at the 0.05 level. This does not demonstrate that the slopes are equal; the interval remains compatible with both a negative change and a small positive change. The appropriate conclusion is about the uncertainty in the specified model, not a claim that the dose has no relationship with improvement.

Common Mistakes and Practical Checks

Reading $\beta_2$ as the post-knot slope. It is the change in slope. The post-knot slope is $\beta_1+\beta_2$.
Assuming the model has a jump at the knot. The hinge is zero at $x=c$, so the fitted mean is continuous. A model with a jump would require a different term.
Calling a smaller training $SSE$ proof of a better model. Adding a coefficient cannot increase the least-squares $SSE$. Use a model comparison and consider uncertainty and prediction performance.
Treating a nonsignificant result as proof of no slope change. A $p$-value above $\alpha$ means the evidence is insufficient to reject the null; inspect the interval to see which changes remain plausible.
Ignoring how the knot was chosen. Standard inference here treats the knot as fixed in advance. Searching the outcomes for a knot and then reporting ordinary coefficient uncertainty can understate uncertainty.
Using the model as a causal claim. A dose-response association does not by itself show what would happen if a patient’s dose were changed.

Key takeaway: With a prespecified knot $c$, the hinge $(x-c)_+$ creates a continuous change in slope. The coefficient on that term is the slope change, while the slope after the knot is the sum of that coefficient and the pre-knot slope. Compare the model with a straight line using the added-term test and interpret its uncertainty in context.

Check Your Understanding

Use the model definition and assumptions above to answer these questions.

For $\widehat{Y}=4+0.7x-0.2(x-6)_+$, what are the slopes below and above the knot, and what does the change in slope mean?
Why is the fitted mean continuous at the knot in a hinge model?
A piecewise model has a lower residual sum of squares than its straight-line counterpart. Why is that alone not enough to conclude that the added term is useful?
What does it mean if a confidence interval for the change in slope includes zero?
Why can ordinary coefficient inference be unreliable if the knot was selected by searching the same outcome data?