Deriving Link Effects Along a Linear Biomarker Trajectory

Why derive the link along the trajectory?

A current-value association and a slope association answer different questions about a patient’s latent biomarker history. The first links hazard to the biomarker level at the event time; the second links hazard to its instantaneous rate of change. With a linear trajectory, these contributions can be written out explicitly. Doing so makes clear which part varies with time, how patient profiles compare, and what units the association coefficients must have.

The calculations below concern the association component of a joint model, conditional on the latent trajectory and with other covariates held equal. They do not by themselves describe how the longitudinal or event submodels are estimated. The focus is the algebra that translates a fitted linear trajectory into a hazard multiplier.

Model and assumptions

Definition: For patient $i$, let the latent biomarker trajectory be linear in time: $$ m_i(t)=\beta_0+b_{0i}+(\beta_1+b_{1i})t. $$ Here $\beta_0$ and $\beta_1$ are the population intercept and slope, and $b_{0i}$ and $b_{1i}$ are patient-specific random intercept and slope. The instantaneous slope is $m_i'(t)=\beta_1+b_{1i}$. A current-value-and-slope association model has hazard $$ h_i(t)=h_0(t)\exp\{\boldsymbol{\gamma}^{\mathsf T}\mathbf{w}_i+\alpha_v m_i(t)+\alpha_s m_i'(t)\}, $$ where $h_0(t)$ is the baseline hazard, $\mathbf{w}_i$ is a vector of covariates, and $\boldsymbol{\gamma}$ is its coefficient vector. The association coefficients are $\alpha_v$ for current value and $\alpha_s$ for current slope.

This form assumes that the hazard at time $t$ depends on the latent trajectory through the specified value and slope at $t$, that the trajectory is linear over the time range being interpreted, and that $\alpha_v$ and $\alpha_s$ are constant over time. The random effects determine patient-specific trajectories; the hazard expression is conditional on those latent effects. For profile comparisons below, patients are compared at the same time and have equal covariates unless stated otherwise, so the baseline hazard and covariate terms cancel.

Units provide an important check. If time is measured in years and the biomarker in units $U$, then $m_i(t)$ has units $U$, while $m_i'(t)$ has units $U/\text{year}$. The exponent must be dimensionless. Consequently, $\alpha_v$ has units $1/U$, and $\alpha_s$ has units $\text{year}/U$, the reciprocal of the slope units. In particular, $\alpha_s$ is not measured in $1/(U\cdot\text{year})$.

Substitution into the log hazard

The linear form makes the slope constant for each patient. Substituting both the level and slope into the log hazard gives

\begin{aligned} \log h_i(t) &=\log h_0(t)+\boldsymbol{\gamma}^{\mathsf T}\mathbf{w}_i +\alpha_v\{\beta_0+b_{0i}+(\beta_1+b_{1i})t\} +\alpha_s(\beta_1+b_{1i})\\ &=\log h_0(t)+\boldsymbol{\gamma}^{\mathsf T}\mathbf{w}_i +\alpha_v(\beta_0+b_{0i}) +(\alpha_v t+\alpha_s)(\beta_1+b_{1i}). \end{aligned}

Thus, the current-value term contributes an intercept-like part, $\alpha_v(\beta_0+b_{0i})$, and a part that changes linearly with time, $\alpha_v(\beta_1+b_{1i})t$. The slope term contributes $\alpha_s(\beta_1+b_{1i})$, which is constant over time for that patient’s linear trajectory. That constant can still distinguish patients with different slopes. It does not mean the entire hazard is constant: $h_0(t)$ may vary, and the current-value component may also change.

Same-time hazard ratios between profiles

At a common time $t$, compare patients $i$ and $j$ with the same covariates. Dividing their hazards cancels $h_0(t)$; taking the log of the ratio leaves the difference in their association components. The resulting contrast is

\log\!\left\{\frac{h_i(t)}{h_j(t)}\right\} =\alpha_v\{m_i(t)-m_j(t)\} +\alpha_s\{m_i'(t)-m_j'(t)\}, \qquad \frac{h_i(t)}{h_j(t)} =\exp\!\left[ \alpha_v\Delta m(t)+\alpha_s\Delta m'(t) \right],

where $\Delta m(t)=m_i(t)-m_j(t)$ and $\Delta m'(t)=m_i'(t)-m_j'(t)$. This is a conditional, same-time hazard ratio for the specified profiles, with other covariates equal. A value above one means the first profile has a higher instantaneous hazard at that time under the model; it is not a causal conclusion.

Derivation: Write $h_i(t)=h_0(t)\exp\{\boldsymbol{\gamma}^{\mathsf T}\mathbf{w}_i+\alpha_vm_i(t)+\alpha_sm_i'(t)\}$ and the corresponding expression for $h_j(t)$. With $\mathbf{w}_i=\mathbf{w}_j$, division cancels both $h_0(t)$ and the common covariate multiplier. The exponential difference rule, $\exp(A)/\exp(B)=\exp(A-B)$, then gives the stated log contrast and hazard ratio.

Worked Example: Comparing Two Profiles That Differ in Level and Slope

Suppose the biomarker is measured in $U$ and time in years. For a hypothetical model, let $\alpha_v=-0.30$ per $U$ and $\alpha_s=0.80$ years per $U$. At year 2, patient A has latent trajectory $m_A(t)=8+1.2t$, while patient B has $m_B(t)=6+0.4t$. Assume their covariates are equal.

At year 2, A’s value is $8+1.2(2)=10.4\,U$, and B’s is $6+0.4(2)=6.8\,U$. Their slopes are $1.2\,U/\text{year}$ and $0.4\,U/\text{year}$, respectively. Thus $\Delta m=10.4-6.8=3.6\,U$, and $\Delta m'=1.2-0.4=0.8\,U/\text{year}$.

The log hazard ratio is $$ (-0.30\ \text{per }U)(3.6\,U) +(0.80\ \text{years per }U)(0.8\,U/\text{year}) =-1.08+0.64=-0.44. $$ The units cancel in both products. Therefore,

\frac{h_A(2)}{h_B(2)}=\exp(-0.44)=0.6440.

Under this model and at year 2, A’s instantaneous hazard is about 0.644 times B’s, or about 35.6% lower, conditional on the stated profiles and equal covariates. The negative current-value contribution is partly offset by the positive slope contribution; neither contribution alone is the final comparison.

Isolating the slope contribution

Profiles with the same current value but different slopes are useful for seeing what the slope term adds. At a fixed time, their current-value contributions cancel, leaving a hazard ratio determined by $\alpha_s$ and the slope difference. Conversely, if their slopes match, the slope contributions cancel and the level difference determines the contrast.

Worked Example: Equal Current Values, Different Slopes

Suppose $\alpha_v=0.25$ per $U$ and $\alpha_s=-0.50$ years per $U$. At year 3, hypothetical patient C has trajectory $m_C(t)=7+t$, and patient D has trajectory $m_D(t)=10$. Their covariates are equal.

At year 3, $m_C(3)=7+3=10\,U$, while $m_D(3)=10\,U$. Their value difference is $0\,U$. Their slopes are $1\,U/\text{year}$ and $0\,U/\text{year}$, so the slope difference is $1\,U/\text{year}$. Hence $$ \log\!\left\{\frac{h_C(3)}{h_D(3)}\right\} =(0.25\ \text{per }U)(0\,U) +(-0.50\ \text{years per }U)(1\,U/\text{year}) =-0.50. $$ Therefore $h_C(3)/h_D(3)=\exp(-0.50)=0.6065$. Given the fitted negative slope association, C’s steeper upward slope corresponds to a lower conditional hazard than D’s at that time, despite their equal current values. This is a model-based association, not evidence that increasing the biomarker reduces risk.

Within-patient change across time

A different question compares one patient’s association multiplier at two times. For the same patient, the random effects—and therefore the slope—are fixed under a linear trajectory. The slope contribution $\alpha_s m_i'(t)$ is consequently unchanged between the two times. The change in the log association multiplier is

\{\alpha_vm_i(t_1)+\alpha_sm_i'(t_1)\} -\{\alpha_vm_i(t_0)+\alpha_sm_i'(t_0)\} =\alpha_v\{m_i(t_1)-m_i(t_0)\}.

This is a change in the association component, not generally the ratio of the total hazards at $t_1$ and $t_0$. The total hazard ratio also includes $h_0(t_1)/h_0(t_0)$; if covariates change with time, their contribution must also be included.

Worked Example: Following One Linear Trajectory Over Time

For a hypothetical patient, let $m_i(t)=5+0.75t$, with time in years and biomarker units $U$. Suppose $\alpha_v=0.40$ per $U$ and $\alpha_s=1.20$ years per $U$. Compare years 1 and 4.

The values are $m_i(1)=5.75\,U$ and $m_i(4)=8.00\,U$; the slope is $0.75\,U/\text{year}$ at both times. The association component changes by $$ 0.40(8.00-5.75)=0.40(2.25)=0.90. $$ Equivalently, the slope contribution is $1.20(0.75)=0.90$ at each time and cancels in the difference. Thus the association multiplier at year 4 divided by that at year 1 is $\exp(0.90)=2.460$. The total hazard ratio across these times is not necessarily 2.460: it also depends on the baseline hazard ratio $h_0(4)/h_0(1)$.

Checks, alternatives, and limitations

A useful calculation check is to write the units beside every factor before multiplying. The product $\alpha_vm_i(t)$ is dimensionless because $(1/U)U=1$; $\alpha_sm_i'(t)$ is dimensionless because $(\text{year}/U)(U/\text{year})=1$. A result with units remaining in the exponent signals a unit or coefficient-label error. Also check the comparison being made: a profile contrast at the same time can cancel the baseline hazard, but a within-patient comparison across times generally cannot.

The derivation depends on the linear trajectory and the chosen link. With a nonlinear latent trajectory, its derivative may vary over time, so the slope contribution need not be constant. Sparse or noisy repeated measurements can also make individual slopes weakly determined; then slope-linked hazard contrasts may be sensitive to trajectory assumptions. Assess whether a linear trajectory is adequate over the relevant time range and whether the data support the patient-specific slopes being interpreted.

Association specification	Same-time profile contrast when covariates match	What is not represented
Current value only	$\exp\{\alpha_v\Delta m(t)\}$	A distinct association with slope after conditioning on current value
Slope only	$\exp\{\alpha_s\Delta m'(t)\}$	A distinct association with current level
Current value and slope	$\exp\{\alpha_v\Delta m(t)+\alpha_s\Delta m'(t)\}$	Any trajectory feature not included in the specified link

These alternatives impose different scientific association structures; comparing their interpretations is not the same as establishing which is correct. A combined link can also be difficult to interpret when level and slope are strongly related in the observed trajectories. The coefficients are conditional on the other included component and on the model specification. A linear trajectory and constant association coefficients may be poor approximations if trajectories bend or the association changes over time. The derivation describes what the stated model implies; it does not guarantee that the model is adequate.

Common mistake

A frequent error is to call $\alpha_s$ “per biomarker unit per year.” Read literally, that means $1/(U\cdot\text{year})$. Multiplying by a slope in $U/\text{year}$ would then leave units $1/\text{year}^2$ in the exponent, which is invalid. The correct units are $\text{year}/U$, the reciprocal of $U/\text{year}$. Another error is to claim that the slope contribution makes the hazard constant over time. It is constant only for a patient’s slope term under a linear trajectory; the baseline hazard and current-value contribution can still vary.

Key takeaway: Under a linear latent trajectory, the same-time log hazard ratio between profiles is $\alpha_v$ times their value difference plus $\alpha_s$ times their slope difference. The slope term is constant over time within a patient, but it can change comparisons between patients. Interpret each contrast with its time, units, covariates, and conditioning assumptions stated.

Check Your Understanding

Use the stated linear model and assume equal covariates unless a question says otherwise. Brief solutions follow the questions.

Biomarker units are $U$, time is years, $\alpha_v=-0.2$ per $U$, and $\alpha_s=0.5$ years per $U$. At a common time, profile A exceeds profile B by $2\,U$ in value and $1\,U/\text{year}$ in slope. Find the hazard ratio $h_A/h_B$.
For one patient with constant linear slope, what happens to the slope contribution when comparing association multipliers at two times?
A proposed slope coefficient is reported in units $1/(U\cdot\text{year})$, while the slope is measured in $U/\text{year}$. Explain the unit problem and give the correct units.
At the same time, two profiles have equal values and slopes differing by $-0.4\,U/\text{year}$. If $\alpha_s=0.75$ years per $U$, calculate their hazard ratio in the order profile 1 divided by profile 2.

Solutions. 1. The log ratio is $(-0.2)(2)+(0.5)(1)=-0.4+0.5=0.10$, so $h_A/h_B=\exp(0.10)=1.105$. 2. It cancels: the patient’s slope and $\alpha_s$ are unchanged, so $\alpha_sm_i'(t_1)-\alpha_sm_i'(t_0)=0$. 3. The proposed units leave $1/\text{year}^2$ after multiplication by $U/\text{year}$, rather than a dimensionless exponent. The correct units are $\text{year}/U$. 4. The value contribution is zero; the log ratio is $0.75(-0.4)=-0.30$, so the hazard ratio is $\exp(-0.30)=0.7408$.

Association specification	Same-time profile contrast when covariates match	What is not represented
Current value only	\(\exp\{\alpha_v\Delta m(t)\}\)	A distinct association with slope after conditioning on current value
Slope only	\(\exp\{\alpha_s\Delta m'(t)\}\)	A distinct association with current level
Current value and slope	\(\exp\{\alpha_v\Delta m(t)+\alpha_s\Delta m'(t)\}\)	Any trajectory feature not included in the specified link