Rescaling Time in Slope-Based Models

Why time units matter in a slope association

A slope describes change per unit of time. A latent marker slope measured in marker units per day is therefore numerically different from the same physical rate expressed in marker units per year. In a slope-based joint model, this change affects not only the reported slope but also the coefficient multiplying it in the event hazard. Converting one without the other changes the model’s association effect.

The essential check is that the slope contribution to the hazard’s linear predictor stays the same for each subject. This tutorial derives that requirement, then applies it to fixed and random slopes, association coefficients, and hazard ratios. Throughout, the marker is assumed to be measured on the same scale before and after the time conversion.

Definition: Let \(t\) be time in an original unit and let \(u=t/c\) be time in a new unit, where \(c>0\). For a differentiable latent marker trajectory \(m_i(t)\), define its slope in the original unit as \(q_{i,t}=dm_i(t)/dt\). The slope in the new unit is \(q_{i,u}=d\{m_i(cu)\}/du\). Thus \(q_{i,u}=c q_{i,t}\). For example, if \(t\) is measured in years and \(u\) in days, \(c=1/365.25\); if \(t\) is measured in days and \(u\) in years, \(c=365.25\).

Deriving the transformation

Suppose a subject’s latent marker trajectory is \(m_i(t)\), and the slope association enters the hazard through the term \(\alpha_t q_{i,t}\). Here \(\alpha_t\) is the association coefficient when slopes are expressed per original time unit. The chain rule gives the new-unit slope:

q_{i,u} =\frac{d\,m_i(cu)}{du} =\frac{dm_i(t)}{dt}\frac{dt}{du} =cq_{i,t}.

For the same subject and physical time, the slope contribution to the linear predictor must be identical under either unit convention. Therefore \(\alpha_u q_{i,u}=\alpha_t q_{i,t}\). Substituting \(q_{i,u}=cq_{i,t}\) shows that the transformed association coefficient is \(\alpha_u=\alpha_t/c\). This is a re-expression of the same predictor, not a change in the association itself.

\alpha_u q_{i,u} =\frac{\alpha_t}{c}(cq_{i,t}) =\alpha_t q_{i,t}.

If the hazard is written per unit of the event-time scale, its baseline hazard must also be expressed in that scale. For \(u=t/c\), the hazard density transforms as \(\lambda_u(u)=c\lambda_t(cu)\), so the baseline becomes \(\lambda_{0,u}(u)=c\lambda_{0,t}(cu)\). This factor changes the hazard’s units; it does not alter the subject-specific slope contribution or the hazard ratio comparing two subjects at the same physical time.

Key result: Under \(u=t/c\), latent slopes and their association coefficients transform in opposite directions: \(q_{i,u}=cq_{i,t}\) and \(\alpha_u=\alpha_t/c\). Consequently, the product \(\alpha q_i\), and therefore a hazard ratio based on a slope contrast, is invariant to a consistent change of time unit.

Worked conversions

Worked Example: A Slope Association in Days and Years

Suppose time is initially in days and a subject’s latent marker slope is \(-0.008\) marker units per day. The fitted slope association coefficient is \(\alpha_{\text{day}}=0.04\) day per marker unit. Convert both quantities to years, using 365.25 days per year, and verify that the hazard contribution is unchanged.

One year contains 365.25 days, so the slope in marker units per year is

q_{\text{year}} =365.25(-0.008) =-2.922\ \text{marker units per year}.

The association coefficient must be divided by 365.25:

\alpha_{\text{year}} =\frac{0.04}{365.25} =0.000109514\ \text{years per marker unit}.

In day units, the slope contribution is \(0.04(-0.008)=-0.00032\). In year units it is \(0.000109514(-2.922)\approx-0.00032\). Both parameterizations therefore give the same slope contribution, up to rounding.

For a second subject whose slope is \(0.003\) marker units per day greater, the hazard ratio comparing that subject with the first is \(\exp\{0.04(0.003)\}=\exp(0.00012)\approx1.00012\). The slope contrast in years is \(0.003(365.25)=1.09575\) marker units per year, and \(\exp\{0.000109514(1.09575)\}\approx1.00012\) as well. The interpretation is a very small relative hazard increase for this specified slope contrast, not a causal effect.

A linear mixed trajectory makes the random-effect transformation explicit. If \(m_i(t)=\beta_0+b_{0i}+(\beta_{1,t}+b_{1i,t})t\), then under \(u=t/c\) the intercept is unchanged and both the fixed slope and the random slope are multiplied by \(c\). In matrix form, with \(A=\operatorname{diag}(1,c)\), the random-effect covariance matrix transforms as \(D_u=AD_tA^\mathsf{T}\). Thus the random-slope variance is multiplied by \(c^2\), and the intercept–slope covariance is multiplied by \(c\). For \(c>0\), the correlation between random intercept and random slope is unchanged.

Worked Example: Transforming Random-Slope Variability

Suppose a random slope has mean \(0.002\) marker units per day and standard deviation \(0.0015\) marker units per day. The random intercept has standard deviation \(4\) marker units, and the intercept–slope covariance is \(-0.003\) marker-squared per day. Convert these quantities to years.

The slope mean and standard deviation are each multiplied by 365.25:

E(b_{1,\text{year}})=0.002(365.25)=0.7305, \qquad SD(b_{1,\text{year}})=0.0015(365.25)=0.547875.

The random-slope variance in day units is \(0.0015^2=0.00000225\) marker-squared per day-squared. In year units it is

\operatorname{Var}(b_{1,\text{year}}) =0.00000225(365.25)^2 =0.300167\ \text{marker-squared per year-squared}.

The covariance is multiplied by 365.25, giving \(-0.003(365.25)=-1.09575\) marker-squared per year. The intercept standard deviation stays \(4\) marker units. As a check, the original correlation is \(-0.003/(4\times0.0015)=-0.5\); the converted correlation is \(-1.09575/(4\times0.547875)=-0.5\). The distribution has changed numerical units, but not its correlation structure.

A related scale check: a slope standard deviation of \(2\) marker units per year is \(2/365.25=0.005476\) marker units per day, rounded to four significant figures.

Worked Example: Correlation and Rounded Covariance Values

Suppose random intercepts have standard deviation \(5\) marker units, and random slopes in years have standard deviation \(3\) marker units per year. Their correlation is \(-0.08\). Convert the slope distribution to days and verify the correlation, taking care with rounding.

The standard deviation per day is \(3/365.25\approx0.00821355\), and the slope variance per day squared is \((3/365.25)^2\approx0.00006746\). The covariance in year units is \((-0.08)(5)(3)=-1.2\) marker-squared per year. In day units it is

\operatorname{Cov}(b_0,b_{1,\text{day}}) =\frac{-1.2}{365.25} \approx-0.00328542\ \text{marker-squared per day}.

Using the exact covariance, the correlation is \((-1.2/365.25)/\{5(3/365.25)\}=-0.08\), as required. If the covariance and variance are displayed as \(-0.003285\) and \(0.00006746\), then substituting those rounded values gives \(-0.003285/\{5\sqrt{0.00006746}\}\approx-0.07999\). That small discrepancy is due to rounding; it is not a change in the underlying correlation.

Assumptions, diagnostics, and alternatives

These conversions require a consistent, positive, linear change of time units and a trajectory differentiable over the times where its slope is used. The marker’s measurement scale and the physical time points must remain the same. The association must be specified through the slope term \(\alpha q_i\); if a model uses a different transformation of the slope, its full predictor must be re-expressed rather than applying the coefficient rule mechanically. The formula also presumes the same subjects, covariates, and underlying model are being compared.

Useful diagnostics are mostly algebraic and numerical. Check the units beside each quantity, verify that the converted slope equals the original slope times the time conversion factor, and confirm that \(\alpha q_i\) is unchanged for representative subjects. For random effects, check that the slope variance scales by the square of the factor and covariance by the factor itself. If a reported correlation changes materially under a positive unit conversion, a parameter was likely transformed inconsistently, or rounding is too coarse.

The main alternative is to keep the original time unit and report its slopes and association coefficient directly. This avoids conversion but may make values awkward to interpret. Standardizing slopes can aid comparisons across settings, but it changes the coefficient’s scale and requires the standardization factor to be stated; it does not eliminate the need to define time units. Time centering, addressed in the earlier material on time centering in polynomial slope links, changes the reference point, whereas rescaling changes the time unit. These are distinct operations and should not be conflated.

Common mistake and limits

Changing the slope but not its coefficient. Multiplying a per-day slope by 365.25 while retaining the per-day association coefficient multiplies the hazard contribution by 365.25. Divide the coefficient by the same factor.
Using the conversion in the wrong direction. From days to years, a slope is multiplied by 365.25; from years to days, it is divided by 365.25. Write the unit conversion explicitly before calculating.
Rescaling a variance like a standard deviation. A random-slope standard deviation is multiplied by \(c\), but its variance is multiplied by \(c^2\).
Interpreting rounded output as exact. Correlations calculated from rounded variances and covariances can differ slightly from the unrounded correlation. Retain full precision for checks, then round the reported result.

Rescaling time preserves the model’s relative slope-based hazard comparisons only when all related quantities are transformed consistently. It cannot correct a misspecified trajectory, an inappropriate slope association, or unreliable estimation of subject-specific slopes. Unit invariance is a useful consistency check, not evidence that the model assumptions fit the data.

Check Your Understanding

Try each question, then compare with the solutions below.

A slope is \(0.01\) marker units per day. What is its value per year using 365.25 days per year?
If a per-day association coefficient is \(0.12\) day per marker unit, what is the corresponding per-year coefficient?
A random-slope standard deviation is \(0.004\) marker units per day. What is its standard deviation per year, and by what factor does its variance change?
Why can a correlation computed from rounded covariance and variance values differ slightly from the unrounded correlation?

Solutions. (1) Multiply by 365.25: \(0.01(365.25)=3.6525\) marker units per year. (2) Divide by 365.25: \(0.12/365.25\approx0.0003285\) years per marker unit. (3) The standard deviation is \(0.004(365.25)=1.461\) marker units per year; the variance is multiplied by \(365.25^2\). (4) Rounding changes the values in the correlation formula slightly, even though the exact transformed covariance and standard deviations preserve the original correlation.