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Sensitivity to the Association Structure · Tutorial 227 of 697

Time Centering in Slope-Based Association Links

Learn to re-express longitudinal random effects after centering time and determine when a slope association in the event hazard is invariant to that change.

Advanced 11 min read

What You'll Learn

  • Transform fixed and random intercepts when time is recentered
  • Calculate the corresponding random-effects covariance matrix
  • Explain why a linear trajectory’s slope does not depend on its time origin
  • Identify conditions under which a slope-based hazard link is invariant
  • Distinguish a changed parameterization from a changed scientific association
  • Diagnose comparisons that incorrectly hold transformed parameters fixed

Why the Time Origin Matters

A longitudinal trajectory’s intercept is its expected value at time zero. If time zero is changed—for example, from enrollment to a later clinical landmark—the intercept changes its meaning and its numerical value. The slope of a linear trajectory, however, describes change per unit time and does not depend on where time zero is placed. This distinction matters in a joint model when the event hazard is linked to the longitudinal slope: a changed intercept parameterization need not imply a changed hazard association.

The key is to distinguish changing the coordinate used to describe a trajectory from changing the trajectory itself. Centering is only a re-expression of the same model if the fixed effects, random effects, and their covariance are all transformed consistently. A comparison that changes the time origin but leaves some of those parameters fixed may instead compare different models.

Model and Time-Centering Transformation

Consider a linear mixed model for subject \(i\)'s longitudinal measurements \(Y_{ij}\) at physical times \(t_{ij}\):

$$ Y_{ij}=\beta_0+\beta_1t_{ij}+b_{0i}+b_{1i}t_{ij}+\varepsilon_{ij}. $$

Here \(\beta_0\) and \(\beta_1\) are the population intercept and slope, \(b_{0i}\) and \(b_{1i}\) are subject-specific random intercept and slope, and \(\varepsilon_{ij}\) is measurement error. Assume the random-effects vector \(b_i=(b_{0i},b_{1i})^\mathsf{T}\) has mean zero and covariance matrix \(G\), and that the errors have mean zero and are independent of the random effects. These assumptions define the part of the longitudinal model needed for the reparameterization below.

Let \(c\) be a fixed time, measured in the same units as \(t\), and define the centered time \(\tau=t-c\). The new time origin is therefore physical time \(t=c\). The same trajectory can be written

$$ \begin{aligned} m_i(t) &=\beta_0+\beta_1t+b_{0i}+b_{1i}t\\ &=(\beta_0+c\beta_1)+\beta_1\tau +(b_{0i}+cb_{1i})+b_{1i}\tau. \end{aligned} $$
Definition: Under the change of origin \(\tau=t-c\), the centered fixed effects and random effects are \(\beta_0^*=\beta_0+c\beta_1\), \(\beta_1^*=\beta_1\), \(b_{0i}^*=b_{0i}+cb_{1i}\), and \(b_{1i}^*=b_{1i}\). Thus, the centered intercept is the trajectory’s intercept at physical time \(c\), while the slope remains the rate of change per unit time.

The transformation also changes the random-effects covariance. In matrix form, \(b_i^*=A b_i\), where

$$ A= \begin{pmatrix} 1&c\\ 0&1 \end{pmatrix}, \qquad G^*=\operatorname{Var}(b_i^*)=AGA^\mathsf{T}. $$

If \(G\) has entries \(G_{00}=\operatorname{Var}(b_{0i})\), \(G_{01}=\operatorname{Cov}(b_{0i},b_{1i})\), and \(G_{11}=\operatorname{Var}(b_{1i})\), then

$$ G^*= \begin{pmatrix} G_{00}+2cG_{01}+c^2G_{11} & G_{01}+cG_{11}\\ G_{01}+cG_{11} & G_{11} \end{pmatrix}. $$

The random-intercept variance and its covariance with the random slope can therefore change, even though the population slope, random slope, and individual trajectories do not. Since \(A\) is invertible, this is a one-to-one change of coordinates, not a restriction on the possible trajectories.

Why a Linear Slope Link Is Invariant

Suppose the event hazard, conditional on the random effects, is modeled as

$$ \lambda_i(t\mid b_i) =\lambda_0(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha_s m_i'(t)\right\}, \qquad m_i'(t)=\beta_1+b_{1i}. $$

Here \(\lambda_0(t)\) is the baseline hazard, \(w_i\) is a vector of event-model covariates, \(\gamma\) is its coefficient vector, and \(\alpha_s\) is the slope-association parameter. The model assumes that the event hazard depends on the current derivative of the latent linear trajectory through this term. It also assumes that the longitudinal and event processes are conditionally linked through the shared random effects; the same physical time \(t\) is used in both processes.

Because \(\beta_1^*=\beta_1\) and \(b_{1i}^*=b_{1i}\), the derivative under centered time is

$$ \frac{d}{d\tau} \left(\beta_0^*+\beta_1^*\tau+b_{0i}^*+b_{1i}^*\tau\right) =\beta_1^*+b_{1i}^* =\beta_1+b_{1i} =m_i'(t). $$

Thus, at any given physical time, a slope-only association term is unchanged by a shift of the time origin. If the baseline hazard, covariates, and association parameter are also kept on the same physical scale, the conditional hazard is unchanged. This is a parameterization invariance, not evidence that all ways of changing the time scale leave a model unchanged.

Worked Example: Transforming the Intercept and Covariance

Consider a hypothetical biomarker trajectory with \(\beta_0=10\) units and \(\beta_1=0.12\) units per year. Suppose the random-effects covariance, in the order intercept then slope, is

$$ G= \begin{pmatrix} 4&0.30\\ 0.30&0.04 \end{pmatrix}, $$

where the entries have the corresponding squared or product units. Recenter time at \(c=5\) years after the original origin. The fixed effects become \(\beta_0^*=10+5(0.12)=10.6\) units and \(\beta_1^*=0.12\) units per year. The transformed covariance entries are

$$ \begin{aligned} \operatorname{Var}(b_{0i}^*)&=4+2(5)(0.30)+5^2(0.04)=8,\\ \operatorname{Cov}(b_{0i}^*,b_{1i}^*)&=0.30+5(0.04)=0.50,\\ \operatorname{Var}(b_{1i}^*)&=0.04. \end{aligned} $$

So \(G^*=\begin{pmatrix}8&0.50\\0.50&0.04\end{pmatrix}\). The intercept variance has doubled, and the intercept-slope covariance has changed, while the slope variance is unchanged.

For a hypothetical subject with \(b_{0i}=-1\) unit and \(b_{1i}=0.05\) units per year, \(b_{0i}^*=-1+5(0.05)=-0.75\) unit and \(b_{1i}^*=0.05\) units per year. At physical time \(t=8\) years, the original representation gives \(10+0.12(8)-1+0.05(8)=10.36\) units. The centered time is \(\tau=8-5=3\), and the centered representation gives \(10.6+0.12(3)-0.75+0.05(3)=10.36\) units. The same trajectory has been expressed using a different intercept.

Worked Example: Checking a Slope-Based Hazard

Suppose a hypothetical joint model uses a linear biomarker trajectory whose subject-specific slope is \(0.18\) units per year. Let \(\alpha_s=-0.70\) per (unit per year), and suppose the baseline hazard at physical time \(t=6\) years is \(0.030\) per year. For simplicity, take \(\gamma^\mathsf{T}w_i=0\). The slope contribution to the hazard multiplier is

$$ \exp\{\alpha_s m_i'(t)\} =\exp\{(-0.70)(0.18)\} =\exp(-0.126) \approx 0.8816. $$

The conditional hazard is therefore \(0.030(0.8816)=0.02645\) per year. Now center time at \(c=4\) years. The same physical time is \(\tau=6-4=2\) years after the new origin, but the trajectory’s slope remains \(0.18\) units per year. The multiplier is still \(\exp\{(-0.70)(0.18)\}=0.8816\), so the hazard remains \(0.02645\) per year. This conclusion follows because the physical event time and baseline hazard were not changed; only the longitudinal time coordinate was recentered.

The coefficient \(\alpha_s\) is interpreted per unit of biomarker change per year. Recentring time does not change those units. If time were instead converted from years to months, the numerical slope and its association coefficient would need a corresponding scale conversion; that is not merely a change of origin.

When Other Link Parameters Change

A slope-only link is especially simple because the slope coordinate is unchanged by linear centering. A link that directly uses random-effect coordinates must be transformed with them. For example, suppose a contribution to the log hazard is \(\alpha_0 b_{0i}+\alpha_1 b_{1i}\). Since \(b_{0i}=b_{0i}^*-cb_{1i}^*\) and \(b_{1i}=b_{1i}^*\), the identical contribution in centered coordinates is

$$ \alpha_0 b_{0i}+\alpha_1 b_{1i} =\alpha_0 b_{0i}^*+(\alpha_1-c\alpha_0)b_{1i}^*. $$

The numerical coefficient on the centered random slope is therefore not generally the same as the coefficient on the original random slope in this joint level-and-slope expression. The two expressions describe the same association only when the coefficients are transformed consistently.

Worked Example: Re-expressing a Random-Effects Link

For illustration, let a hypothetical log-hazard contribution be \(0.40b_{0i}-2.0b_{1i}\), with \(c=3\) time units. The equivalent centered-coordinate coefficients are \(0.40\) on \(b_{0i}^*\) and \(-2.0-3(0.40)=-3.20\) on \(b_{1i}^*\). Consider a subject with \(b_{0i}=1.2\) and \(b_{1i}=-0.10\). In the original coordinates the contribution is \(0.40(1.2)-2.0(-0.10)=0.68\). The centered intercept is \(b_{0i}^*=1.2+3(-0.10)=0.90\), so the centered expression gives \(0.40(0.90)-3.20(-0.10)=0.68\). The coefficient changed, but the log-hazard contribution did not. The coefficient values alone are not comparable without accounting for their parameterization.

Justification, Diagnostics, and Limitations

The invariance can be seen directly from the algebra: the centered trajectory equals the original trajectory at every physical time, and its derivative with respect to centered time equals the original derivative with respect to physical time. Consequently, the longitudinal fitted values and the slope term in the conditional hazard are identical for each subject. Because the transformation \(b_i^*=A b_i\) is invertible and \(G^*=AGA^\mathsf{T}\), the random-effects distribution is also the same distribution written in new coordinates. A fully reparameterized model therefore represents the same joint model.

In practice, centering can make the intercept scientifically interpretable at a meaningful landmark and may improve numerical behavior when measurements are clustered far from the original zero. It does not guarantee improved estimation: the transformed intercept and slope can remain strongly correlated, and sparse data may still provide weak information about subject-specific slopes.

When two centered and uncentered fits appear to give different slope-link conclusions, check whether the comparison has actually preserved the model. Useful checks include verifying the time conversion \(t=\tau+c\), transforming the fixed and random intercepts, using \(G^*=AGA^\mathsf{T}\), and evaluating both fitted trajectories at the same physical times. Also check that the hazard baseline is indexed by the same physical event time, that \(\alpha_s\) uses the same biomarker and time units, and that the link is based on the derivative rather than a coordinate-specific intercept or slope-like coefficient.

For a polynomial trajectory, the derivative at a given physical time is still invariant if the polynomial is transformed exactly and evaluated at the corresponding centered time. But the coefficient attached to a polynomial basis term need not itself equal the instantaneous slope. A slope link must use the actual derivative intended by the model. More generally, centering does not protect against a misspecified trajectory, an incorrectly transformed covariance, or a change in the time scale.

The main alternative is to retain the original time origin. This leaves the original intercept interpretation intact but may place time zero outside the observed range. Centering at a clinically meaningful landmark changes the intercept interpretation and covariance representation, while preserving the linear trajectory and correctly transformed slope link. Neither choice is universally superior; the essential requirement is to keep the parameterization and interpretation consistent.

Common Pitfall

Common mistake: “After centering, the random intercept variance changed, so the subject trajectories or slope-based hazard association must have changed.”

Correction: The intercept is now the deviation at a different time, so its variance and covariance with slope generally change. Check \(G^*=AGA^\mathsf{T}\) and compare trajectories or derivatives at matched physical times. If the transformed model is used consistently, the linear slope and slope-only hazard contribution are unchanged.

Check Your Understanding

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

  1. A linear trajectory has \(\beta_0=6\) units and \(\beta_1=0.5\) units per year. What is its fixed intercept after centering time at \(c=4\) years?
  2. For the same centering, a subject has \(b_{0i}=1.0\) and \(b_{1i}=-0.2\) units per year. Find \(b_{0i}^*\) and \(b_{1i}^*\).
  3. Does a hazard link \(\alpha_s m_i'(t)\) change just because time is recentered? State the condition needed for your answer.
  4. If a link is \(\alpha_0b_{0i}+\alpha_1b_{1i}\), write its equivalent centered-coordinate expression.

Solutions. 1. \(\beta_0^*=6+4(0.5)=8\) units, the fixed mean at the new origin. 2. \(b_{0i}^*=1.0+4(-0.2)=0.20\) units and \(b_{1i}^*=-0.2\) units per year. 3. No, provided the trajectory is reparameterized consistently and the derivative is evaluated per the same physical time unit; the linear slope is unchanged. 4. Substitute \(b_{0i}=b_{0i}^*-cb_{1i}^*\) and \(b_{1i}=b_{1i}^*\), giving \(\alpha_0b_{0i}^*+(\alpha_1-c\alpha_0)b_{1i}^*\).

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