Interpreting a Three-Part Association Link

Reading a Three-Part Association Link

A marker trajectory can carry several kinds of information about an event hazard. Its current value describes the marker level at the event time; its slope describes how quickly that level is changing; and its cumulative exposure summarizes the marker history up to that time. A three-part association link puts these summaries into one hazard model. To interpret it, calculate the three contributions separately before combining them.

The calculations below compare hazards at the same time for individuals with the same other covariates but different marker trajectories. They describe the association specified by the model, conditional on its components; they do not, by themselves, establish that changing a marker would cause a change in event risk.

Definition: Let $m_i(t)$ be individual $i$'s latent marker trajectory, defined on the time interval from the chosen origin $0$ to time $t$. Its current value is $m_i(t)$, its slope is $m_i'(t)$, and its cumulative exposure is $C_i(t)=\int_0^t m_i(u)\,du$. A three-part association link enters these quantities into the conditional hazard as $$ h_i(t\mid m_i(\cdot),\mathbf{x}_i) = h_0(t)\exp\left\{\boldsymbol{\gamma}^{\mathsf T}\mathbf{x}_i+ \alpha_v m_i(t)+\alpha_s m_i'(t)+\alpha_c C_i(t)\right\}. $$ Here $h_0(t)$ is the baseline hazard, $\mathbf{x}_i$ is a vector of other covariates, and $\boldsymbol{\gamma}$, $\alpha_v$, $\alpha_s$, and $\alpha_c$ are coefficients. The model assumes a positive baseline hazard and a well-defined, differentiable trajectory for the slope and an integrable trajectory for cumulative exposure. The association coefficients are constant over time in this specification.

The exponent has a separate contribution from each component. For an individual at time $t$, these are $\alpha_v m_i(t)$, $\alpha_s m_i'(t)$, and $\alpha_c C_i(t)$. Each contribution is on the log-hazard scale. Exponentiating their sum gives the multiplicative association with the hazard, apart from the baseline hazard and other covariates.

Units provide a useful check. If the marker is measured in marker units and time in years, then $m_i(t)$ has marker units, $m_i'(t)$ has marker units per year, and $C_i(t)$ has marker-unit-years. Therefore $\alpha_v$ has units per marker unit, $\alpha_s$ has units years per marker unit (the reciprocal of marker units per year), and $\alpha_c$ has units per marker-unit-year. Each product in the exponent is dimensionless, as required.

Comparing Two Trajectories at a Fixed Time

Let trajectories A and B be evaluated at the same time $t$, with the same covariates $\mathbf{x}$. Dividing their hazards cancels the common baseline hazard and covariate contribution. Taking the logarithm shows exactly how the three component differences combine.

\log\left\{\frac{h_A(t)}{h_B(t)}\right\} = \alpha_v\{m_A(t)-m_B(t)\} +\alpha_s\{m_A'(t)-m_B'(t)\} +\alpha_c\{C_A(t)-C_B(t)\}.

Thus, the hazard ratio is the exponential of the sum on the right. A positive total implies a higher model-based hazard for A than B at that time; a negative total implies a lower one. A component’s contribution can be positive or negative depending on both its coefficient and the direction of the trajectory difference. The comparison is conditional: it holds the other covariates and the model’s association coefficients fixed.

Worked Example: Same Current Value, Different Slopes and Histories

Suppose the marker is measured in units and time in years. Use $\alpha_v=0.30$ per marker unit, $\alpha_s=0.40$ years per marker unit, and $\alpha_c=-0.10$ per marker-unit-year. At $t=2$ years, compare the hypothetical trajectories $m_A(t)=5+0.5t$ and $m_B(t)=4+t$.

For A, the current value is $m_A(2)=5+0.5(2)=6$, the slope is $m_A'(2)=0.5$, and the cumulative exposure is $$ C_A(2)=\int_0^2(5+0.5u)\,du =5(2)+0.5\frac{2^2}{2}=11. $$ For B, $m_B(2)=4+2=6$, $m_B'(2)=1$, and $$ C_B(2)=\int_0^2(4+u)\,du =4(2)+\frac{2^2}{2}=10. $$

The current-value difference is $6-6=0$, so its log-hazard contribution is $0.30(0)=0$. The slope difference is $0.5-1=-0.5$, contributing $0.40(-0.5)=-0.20$. The cumulative-exposure difference is $11-10=1$, contributing $-0.10(1)=-0.10$. Therefore, $$ \log\left\{\frac{h_A(2)}{h_B(2)}\right\}=0-0.20-0.10=-0.30, \qquad \frac{h_A(2)}{h_B(2)}=\exp(-0.30)=0.7408. $$

Under this model, at two years and for equal covariates, A's hazard is about 0.741 times B's hazard, or about 25.9% lower. The equal current values do not make the hazards equal: the slope and cumulative-exposure terms account for the difference.

Worked Calculations for Other Trajectory Shapes

The same procedure applies when current values differ or when the trajectories are curved. The necessary summaries are evaluated at the comparison time; the cumulative term is an integral over the full trajectory from the specified origin, not just the marker value observed at that time.

Worked Example: A Higher Current Value but Lower Accumulation

Keep the coefficients from the previous example. Compare the hypothetical trajectory $m_A(t)=5+0.5t$ with $m_C(t)=4+0.5t$ at $t=2$ years. Both have slope $0.5$ marker units per year. Their current values are $m_A(2)=6$ and $m_C(2)=5$. Their cumulative exposures are $C_A(2)=11$, as calculated above, and $$ C_C(2)=\int_0^2(4+0.5u)\,du =4(2)+0.5\frac{2^2}{2}=9. $$

For A minus C, the current-value difference is $1$, giving contribution $0.30(1)=0.30$. The slope difference is $0$, giving contribution $0.40(0)=0$. The cumulative difference is $11-9=2$, giving contribution $-0.10(2)=-0.20$. The total log-hazard difference is $0.30+0-0.20=0.10$, so the hazard ratio is $$ \frac{h_A(2)}{h_C(2)}=\exp(0.10)=1.105. $$

At two years, A's hazard is about 1.105 times C's, or about 10.5% higher, conditional on equal covariates. The positive current-value contribution is partly offset by the negative cumulative-exposure contribution. This is a model-based comparison, not a causal effect.

Worked Example: A Curved Trajectory

Use the same coefficients and compare $m_D(t)=4+0.3t+0.1t^2$ with $m_E(t)=4+0.5t$ at $t=2$ years. For D, the current value is $4+0.3(2)+0.1(2^2)=5$. Its slope is $m_D'(t)=0.3+0.2t$, so $m_D'(2)=0.7$. Its cumulative exposure is $$ C_D(2)=\int_0^2(4+0.3u+0.1u^2)\,du =8+0.3\frac{2^2}{2}+0.1\frac{2^3}{3} =8.8667. $$ For E, the current value is $5$, the slope is $0.5$, and $C_E(2)=9$.

The current-value difference is $5-5=0$, the slope difference is $0.7-0.5=0.2$, and the cumulative difference is $8.8667-9=-0.1333$. Their contributions are $0.30(0)=0$, $0.40(0.2)=0.0800$, and $-0.10(-0.1333)=0.01333$. Adding them gives $0.09333$, and therefore $$ \frac{h_D(2)}{h_E(2)}=\exp(0.09333)=1.098. $$

The model implies a hazard about 1.098 times as high for D as for E at two years, conditional on equal covariates. The units check is important here: multiplying the slope difference $0.2$ marker units per year by $0.40$ years per marker unit gives the dimensionless contribution $0.0800$.

Interpretation, Checks, and Limitations

Each coefficient describes an association conditional on the other two trajectory summaries and the covariates. For example, $\exp(\alpha_v)$ is the hazard ratio for a one-marker-unit higher current value when slope, cumulative exposure, and covariates are held fixed. Likewise, $\exp(\alpha_s\Delta s)$ is the slope component’s hazard ratio for a slope difference $\Delta s$, holding the other summaries fixed. Such conditional contrasts need not correspond to changes that are easy to make in a real trajectory: current value, slope, and accumulated exposure may be strongly related.

Before interpreting a calculation, check that both trajectories are evaluated at the same time and use the same cumulative-exposure origin. Verify that the derivative uses the model’s time units and that the integral covers the intended interval. A cumulative total changes if the origin or time scale changes; this is why its definition is part of the model, not a harmless computational detail.

In an applied joint model, $m_i(t)$ is typically a latent trajectory estimated from longitudinal data rather than a curve known without error. Sparse or noisy measurements can make estimated slopes unstable, while the integral can be sensitive to uncertainty in the trajectory over the whole interval. A numerical calculation using fitted trajectories and coefficients illustrates the link, but it does not account for their estimation uncertainty unless that uncertainty is incorporated into the inferential procedure.

A three-part link differs from a current-value-only, slope-only, or cumulative-only link by including all three summaries at once. This allows the model to represent several forms of association, but it also makes each coefficient conditional on the other terms. A weighted or windowed cumulative summary changes what part of the marker history contributes; it is not interchangeable with the unweighted integral defined here. Model choice and assessment require attention to the scientific question, trajectory fit, uncertainty, and whether the resulting conditional terms are interpretable.

Common pitfall: It is incorrect to calculate the hazard ratio using only the current-value difference because the two marker values at the event time are equal. The model also includes slope and cumulative exposure, so the correct log-hazard ratio is the sum of all three coefficient-weighted differences. Another frequent error is assigning $\alpha_s$ units of “per marker unit per year.” Since the slope itself has units marker units per year, its coefficient must have reciprocal units: years per marker unit.

Summarize each trajectory.
At the chosen event time, calculate its current value, derivative, and integral from the specified origin.

Form component differences.
Subtract trajectory B's summary from trajectory A's summary for all three components.

Apply the coefficients and combine.
Multiply each difference by its coefficient, add the contributions, and exponentiate to obtain the conditional hazard ratio.

Check Your Understanding

For each question, use the three-part link and state the interpretation in context. Solutions follow the questions.

At $t=1$ year, a hypothetical trajectory is $m(t)=3+2t$. With $\alpha_v=0.10$ per marker unit, $\alpha_s=0.20$ years per marker unit, and $\alpha_c=-0.05$ per marker-unit-year, calculate all three contributions and the hazard multiplier relative to a zero-component reference.
Two trajectories have equal current values and equal cumulative exposures at a comparison time, but A's slope is 0.4 marker units per year higher than B's. If $\alpha_s=-0.25$ years per marker unit, what is the hazard ratio for A versus B from the slope term?
Why must the two trajectories in a hazard-ratio calculation be evaluated at the same time and use the same cumulative-exposure origin?
For a marker measured in marker units and time measured in years, what units must $\alpha_v$, $\alpha_s$, and $\alpha_c$ have?

Solutions. (1) The current value is $m(1)=3+2=5$, the slope is $2$, and $C(1)=\int_0^1(3+2u)\,du=3+1=4$. The contributions are $0.10(5)=0.50$, $0.20(2)=0.40$, and $-0.05(4)=-0.20$, totaling $0.70$. The hazard multiplier is $\exp(0.70)=2.014$ relative to the stated zero-component reference. (2) The log-hazard difference is $(-0.25)(0.4)=-0.10$, so the slope-term hazard ratio is $\exp(-0.10)=0.9048$; A's hazard is about 9.52% lower from this component. (3) The baseline hazard may vary with time, so unequal times do not isolate the association terms; different origins also define different cumulative exposures. (4) The units are per marker unit, years per marker unit, and per marker-unit-year, respectively, making each coefficient-weighted component dimensionless.

Interpreting a Three-Part Association Link

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