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Association Structures · Tutorial 187 of 697

Slope Associations on a Log-Marker Scale

Learn to derive, calculate, and interpret a slope association that measures the latent marker’s instantaneous proportional rate of change rather than its absolute rate of change.

Advanced 11 min read

What You'll Learn

  • Derive the log-marker slope as the absolute marker derivative divided by the marker level.
  • Interpret the log-marker derivative as an instantaneous proportional rate of change, with inverse-time units.
  • Calculate hazard ratios for differences in log-marker slopes at the same time.
  • Distinguish equal absolute slopes from equal proportional rates of change.
  • Identify assumptions and diagnostics for using a log-marker slope association.
  • Explain why the association is not automatically causal and is not based on noisy observed slopes.

Why Put the Slope on a Log-Marker Scale?

A marker can change by the same number of units per year in two people while changing by very different proportions of their current levels. An increase of 4 units per year is an 8% instantaneous rate of increase when the marker is 50 units, but a 4% rate when it is 100 units. A slope association based on the derivative of the log latent marker captures this relative, or proportional, rate of change. It is therefore different from an association based on the absolute derivative of the marker itself.

The distinction matters when a percentage change is scientifically more comparable across marker levels than a fixed change in marker units. The derivation is simple, but the scale, units, and interpretation must be kept explicit: the log-marker derivative is measured per unit time, not in marker units per unit time.

Definition and Joint-Model Form

Definition: Let \(m_i(t)>0\) be the differentiable latent marker trajectory for subject \(i\) at time \(t\). The instantaneous log-marker slope is \[ r_i(t)=\frac{d}{dt}\log\{m_i(t)/m_{\mathrm{ref}}\} =\frac{m_i'(t)}{m_i(t)}, \] where \(m_{\mathrm{ref}}>0\) is a fixed reference value and \(m_i'(t)=d m_i(t)/dt\). The quantity \(r_i(t)\) is the instantaneous proportional rate of change of the latent marker, with units of inverse time.

The reference value makes the logarithm’s argument dimensionless. Since the reference is constant, it disappears upon differentiation. The identity follows from the chain rule: the derivative of \(\log\{m_i(t)/m_{\mathrm{ref}}\}\) is \(m_i'(t)/m_i(t)\). This derivative requires a positive marker trajectory and a well-defined derivative at the time of interest.

A slope-linked event hazard can be written as

$$ \lambda_i(t\mid b_i) = \lambda_0(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha_s r_i(t)\right\}. $$

Here \(\lambda_i(t\mid b_i)\) is subject \(i\)’s hazard at time \(t\), conditional on the subject-specific latent trajectory (represented by random effects \(b_i\)) and covariates; \(\lambda_0(t)\) is the baseline hazard; \(w_i\) is a vector of covariates; and \(\gamma\) is its coefficient vector. The parameter \(\alpha_s\) links the instantaneous log-marker slope to the hazard. In this tutorial, the association term is the slope term \(\alpha_s r_i(t)\); no current marker-value term is included.

For \(r_i(t)\) measured in years\(^{-1}\), \(\alpha_s\) has units of years, so their product is dimensionless. At a fixed time and equal covariate values, two subjects’ hazard ratio depends on the difference between their log-marker slopes:

$$ \frac{\lambda_i(t\mid b_i)}{\lambda_j(t\mid b_j)} = \exp\left\{\alpha_s[r_i(t)-r_j(t)]\right\}. $$

This comparison conditions on the same time, so the shared baseline hazard cancels. A positive \(\alpha_s\) means a greater instantaneous proportional marker growth rate is associated with a higher hazard, conditional on the model. A negative \(\alpha_s\) means it is associated with a lower hazard. Neither interpretation by itself establishes a causal effect.

Deriving the Proportional-Rate Interpretation

The link to proportional change follows directly from the derivative definition. Over a short time increment \(\Delta t\), differentiability gives the first-order approximation

$$ \log\{m_i(t+\Delta t)/m_i(t)\} \approx r_i(t)\Delta t. $$

Exponentiating both sides gives \(m_i(t+\Delta t)/m_i(t)\approx \exp\{r_i(t)\Delta t\}\). Thus, for a sufficiently short interval, the relative change is approximately \(r_i(t)\Delta t\), or \(100r_i(t)\Delta t\)% when expressed as a percentage. This is an instantaneous interpretation: if the rate varies over a longer interval, the total proportional change depends on the rate throughout that interval, not just its value at the start.

For comparison, the absolute marker slope is \(m_i'(t)\), with units of marker units per unit time. The two slopes are related by \(m_i'(t)=m_i(t)r_i(t)\). Consequently, the same absolute slope can correspond to different proportional rates at different marker levels, and the same proportional rate can correspond to different absolute slopes.

Worked Example: Calculating the Log-Marker Slope from a Trajectory

Suppose a hypothetical subject has latent marker trajectory

$$ m_i(t)=80\exp(0.09t-0.005t^2), $$

where \(t\) is measured in years and \(m_i(t)\) is in marker units. Find the marker level, absolute slope, and log-marker slope at year 3.

Take natural logarithms and differentiate:

$$ \log\{m_i(t)/80\}=0.09t-0.005t^2, \qquad r_i(t)=0.09-0.010t. $$

At \(t=3\), the instantaneous log-marker slope is \(r_i(3)=0.09-0.010(3)=0.060\) per year. The latent marker level is

$$ m_i(3)=80\exp\{0.09(3)-0.005(3^2)\} =80\exp(0.225) \approx 100.186\text{ units}. $$

By the chain-rule identity, the absolute slope is

$$ m_i'(3)=m_i(3)r_i(3) \approx 100.1858(0.060) \approx 6.011\text{ units per year}. $$

The interpretation is that at year 3 the marker is increasing at about 6.011 units per year, while its instantaneous proportional rate of increase is 0.060 per year, or 6% per year to first order. The latter is not an exact 6% increase over every full year: the trajectory’s rate changes with time.

What the Hazard Ratio Compares

The slope association compares instantaneous log-marker slopes at a specified time. It does not compare marker levels, and it does not generally compare hazards at different times for one subject. For two subjects with the same covariates, a slope difference of \(\Delta r\) gives the same-time hazard ratio \(\exp(\alpha_s\Delta r)\). Because \(\Delta r\) has inverse-time units, the size of the contrast and the units of time used in the model must be reported.

Worked Example: Equal Absolute Slopes, Different Proportional Rates

Consider two hypothetical subjects at time zero. Subject A has \(m_A(0)=50\) units and \(r_A(0)=0.08\) per year. Subject B has \(m_B(0)=100\) units and \(r_B(0)=0.04\) per year. Suppose \(\alpha_s=-1.5\) years, and the subjects have equal covariates.

Their absolute marker slopes are equal:

$$ m_A'(0)=50(0.08)=4\text{ units per year}, \qquad m_B'(0)=100(0.04)=4\text{ units per year}. $$

But their instantaneous proportional rates differ: A’s rate is 0.08 per year and B’s is 0.04 per year. Under the log-marker slope link, the same-time hazard ratio for A versus B is

$$ \frac{\lambda_A(0)}{\lambda_B(0)} = \exp\{-1.5(0.08-0.04)\} = \exp(-0.06) \approx 0.9418. $$

Thus, conditional on the model and equal covariates, A’s hazard is about 0.942 times B’s hazard at that time, or about 5.82% lower. The model distinguishes the subjects despite their equal absolute marker slopes because their proportional rates differ. This result is an association implied by the specified model, not evidence that changing either subject’s marker rate would cause a change in hazard.

Worked Example: Interpreting a Difference in Log-Marker Slope

At a fixed time, suppose two hypothetical subjects have log-marker slopes \(r_A=0.06\) per year and \(r_B=-0.02\) per year. Let \(\alpha_s=2.0\) years, with equal covariates. Calculate and interpret the hazard ratio for A versus B.

The slope difference is \(0.06-(-0.02)=0.08\) per year. Substitution into the hazard-ratio formula gives

$$ \frac{\lambda_A(t)}{\lambda_B(t)} = \exp\{2.0(0.08)\} = \exp(0.16) \approx 1.174. $$

At that time, A’s hazard is about 1.174 times B’s hazard, or about 17.4% higher, conditional on the model and equal covariates. For scale, a slope difference of 0.01 per year corresponds to a hazard ratio of \(\exp\{2.0(0.01)\}=\exp(0.02)\approx1.020\), about 2.02% higher. This describes a one-percentage-point-per-year difference in instantaneous proportional rate, not a one-unit difference in marker level.

Assumptions, Diagnostics, and Limitations

The mathematical definition and hazard interpretation rely on several conditions. The latent marker must be positive and differentiable at the times being evaluated. The longitudinal model must provide a credible trajectory and derivative; the hazard model must correctly specify the baseline hazard, covariate contribution, and log-linear slope association for the intended interpretation. The hazard ratio formula also assumes a comparison at the same time with the same covariates, so that the baseline hazard and covariate terms cancel.

In a joint model, the association is defined using the latent trajectory, not a raw difference between two noisy observed measurements. Taking differences of observed measurements can amplify measurement error and make a purported slope unstable. Using the fitted latent trajectory addresses that distinction, but it does not remove dependence on the longitudinal model’s assumptions or on how well the available observations determine its derivative.

Useful checks include plotting fitted latent trajectories and their derivatives over the observed time range, checking whether the implied marker values remain positive, and examining whether the fitted slopes are plausible rather than being driven by sparse observations or extrapolation. Investigate whether the log-linear hazard contribution is adequate and whether conclusions change under reasonable alternative trajectory specifications. Signs of trouble include implausible rapid changes between observations, highly uncertain derivatives, strong sensitivity to a few observations, or a systematic mismatch between the assumed trajectory and longitudinal data.

The log-marker slope link is not interchangeable with an absolute-slope link. An absolute-slope link asks whether the hazard is associated with change in marker units per time. The log-marker slope link asks whether it is associated with instantaneous proportional change. A current-value link instead uses marker level; it answers a different question. Which scale is useful depends on the scientific meaning of change and the marker’s measurement scale.

A log transformation is not suitable without care when the latent marker can be zero or negative. Adding an arbitrary constant solely to make values positive changes the transformation and its derivative, so it changes the estimand rather than simply fixing notation. Finally, a slope association describes a conditional relationship in the specified joint model. It does not alone show causation, guarantee a constant effect outside the observed time range, or establish that a particular intervention on the marker would change event risk.

Common Mistake

  • Wrong reasoning: “The marker rises by 4 units per year in both people, so their log-marker slopes are equal.”
  • Correction: The log-marker slope is \(m_i'(t)/m_i(t)\), not \(m_i'(t)\). With a marker level of 50 units, a 4-unit-per-year slope gives \(0.08\) per year; with a level of 100 units, it gives \(0.04\) per year.
  • Wrong reasoning: “A slope of 0.06 per year means the marker increases by exactly 6% in every year.”
  • Correction: It is an instantaneous proportional rate. The exact proportional change over an interval depends on how the log-marker slope varies over that interval.
  • Wrong reasoning: “The exponential of the slope term alone is always the hazard ratio across two times.”
  • Correction: The slope term gives a hazard ratio for a slope contrast at the same time when other covariates are equal. Comparing different times for one subject also involves the baseline hazard and any time-varying covariate contributions.
Key takeaway: A log-marker slope is \(m_i'(t)/m_i(t)\), an instantaneous proportional rate with inverse-time units. In a slope-linked hazard, a same-time difference \(\Delta r\) corresponds to a hazard ratio of \(\exp(\alpha_s\Delta r)\), conditional on equal covariates and the model assumptions.

Check Your Understanding

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

  1. A positive latent marker is 40 units at a particular time and has an absolute slope of 2 units per month. What is its instantaneous log-marker slope?
  2. At the same time, subject A has log-marker slope \(0.03\) per month and subject B has slope \(0.01\) per month. If \(\alpha_s=4\) months and covariates are equal, what is the hazard ratio for A versus B?
  3. Why can equal absolute marker slopes correspond to different log-marker slopes?
  4. What is the interpretation of a log-marker slope of \(-0.05\) per year?
  5. Name one condition required to define the log-marker slope and one diagnostic that could reveal an unreliable fitted slope.

Solutions. (1) \(r=m'/m=2/40=0.05\) per month. (2) The slope difference is \(0.02\) per month, so the hazard ratio is \(\exp\{4(0.02)\}=\exp(0.08)\approx1.083\); A’s hazard is about 8.33% higher at that time, conditional on the model. (3) Since \(r=m'/m\), the same absolute slope divided by different marker levels gives different proportional rates. (4) At that instant the marker is decreasing at a proportional rate of 0.05 per year, or 5% per year to first order; it does not imply an exact 5% decrease over every full year. (5) The marker must be positive and differentiable at the time of interest. One diagnostic is to inspect whether the fitted derivative is implausibly large or unstable where longitudinal observations are sparse.

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