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Association Structures · Tutorial 175 of 697

Subject-Specific Interpretation of Current-Value Links

Translate a current-value association coefficient into time-specific, subject-specific hazard comparisons using latent marker values.

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What You'll Learn

  • Define a subject-specific hazard ratio conditional on latent marker trajectories.
  • Derive the hazard ratio for two subjects at the same time.
  • Interpret the association coefficient in marker units and on the hazard-ratio scale.
  • Calculate how a hazard ratio changes as two linear trajectories diverge or cross.
  • Separate the marker contribution from other covariate differences.
  • Identify common interpretation errors involving noisy measurements, timing, and uncertainty.

Why the Same Association Can Give Different Subject-Specific Comparisons

In a current-value joint model, the hazard is linked to a subject’s latent marker value at the time under consideration. The association coefficient is shared across subjects, but the marker values need not be: different latent trajectories can therefore imply different hazard comparisons at different times. Interpreting the coefficient for a particular pair of subjects requires specifying the time and the latent values being compared.

The key comparison is between subjects’ hazards at the same time, conditional on their latent trajectories and covariates. It describes a ratio of instantaneous event rates, not a ratio of event probabilities. It is also not generally the hazard ratio obtained after averaging over the population distribution of random effects.

Model and Subject-Specific Hazard Ratio

Let \(Y_{ij}\) denote the marker measurement for subject \(i\) at measurement time \(t_{ij}\). A standard longitudinal model separates the latent marker trajectory from measurement error:

$$ Y_{ij}=m_i(t_{ij})+\epsilon_{ij}, \qquad m_i(t)=\mathbf{x}_i(t)^\top\boldsymbol{\beta} +\mathbf{z}_i(t)^\top\mathbf{b}_i. $$

Here \(m_i(t)\) is subject \(i\)’s latent marker value, \(\boldsymbol{\beta}\) contains population trajectory parameters, and \(\mathbf{b}_i\) contains that subject’s random effects. The observed measurement can differ from \(m_i(t)\) because of \(\epsilon_{ij}\). In the current-value hazard, the linked quantity is \(m_i(t)\), not an individual noisy measurement \(Y_{ij}\).

For the linear current-value association considered here, write the hazard conditional on the subject’s random effects as

$$ h_i(t\mid\mathbf{b}_i) = h_0(t)\exp\left\{ \boldsymbol{\gamma}^{\top}\mathbf{w}_i+\alpha m_i(t) \right\}. $$

The baseline hazard is \(h_0(t)\), \(\mathbf{w}_i\) is a vector of baseline covariates, \(\boldsymbol{\gamma}\) is its coefficient vector, and \(\alpha\) is the current-value association coefficient. This interpretation assumes the specified conditional hazard model is appropriate: the current latent marker enters linearly on the log-hazard scale, the same baseline hazard applies to the subjects being compared, and the comparison conditions on their latent effects. It also relies on the joint model’s assumptions for the longitudinal measurements, event process, and censoring.

Definition: At time \(t\), the subject-specific hazard ratio for subjects \(i\) and \(j\), conditional on their latent trajectories and covariates, is \[ \frac{h_i(t\mid\mathbf{b}_i)}{h_j(t\mid\mathbf{b}_j)} = \exp\left\{ \boldsymbol{\gamma}^{\top}(\mathbf{w}_i-\mathbf{w}_j) +\alpha\bigl[m_i(t)-m_j(t)\bigr] \right\}. \] If the subjects have the same covariates, this reduces to \[ \exp\left\{\alpha\bigl[m_i(t)-m_j(t)\bigr]\right\}. \]

The marker-only comparison follows by dividing the two conditional hazards at the same \(t\): the common factor \(h_0(t)\) cancels, as do the covariate terms when \(\mathbf{w}_i=\mathbf{w}_j\). Thus \(\alpha\) is a log-hazard change per one-unit increase in latent marker value, holding time and other covariates fixed. A difference of \(d\) marker units corresponds to a hazard ratio of \(\exp(\alpha d)\). Reversing the order of the subjects reverses the sign of \(d\) and takes the reciprocal of the ratio.

Worked Example: A Two-Unit Difference at One Time

Worked Example: A Two-Unit Difference at One Time

Suppose the fitted current-value coefficient is \(\alpha=0.35\) per marker unit. At week 6, two subjects with the same modeled covariates have latent marker values \(m_A(6)=8\) and \(m_B(6)=6\). Find and interpret the conditional hazard ratio for A versus B at week 6.

The latent marker difference is \(8-6=2\) marker units. Since the covariates match, the covariate contrast is zero:

$$ \frac{h_A(6\mid\mathbf{b}_A)}{h_B(6\mid\mathbf{b}_B)} = \exp\{0.35(8-6)\} = \exp(0.70) = 2.014. $$

At week 6, A’s conditional instantaneous event hazard is about 2.01 times B’s, given these latent marker values and the shared covariate values. Equivalently, the model implies about a 101% higher instantaneous hazard for A at that time. This is not a claim that A is twice as likely to experience the event over follow-up.

Trajectory Differences Make the Ratio Time-Dependent

A fixed \(\alpha\) does not require a fixed subject-specific hazard ratio over time. If the subjects’ latent marker difference changes, the exponent \(\alpha[m_i(t)-m_j(t)]\) changes with it. For example, a random-intercept-and-slope trajectory can be written as \(m_i(t)=\beta_0+\beta_1t+b_{0i}+b_{1i}t\). For subjects with the same fixed trajectory design and covariates, their difference is

$$ m_i(t)-m_j(t) = (b_{0i}-b_{0j})+(b_{1i}-b_{1j})t, \qquad HR_{ij}(t) = \exp\left\{\alpha\left[(b_{0i}-b_{0j})+(b_{1i}-b_{1j})t\right]\right\}. $$

The subject-specific ratio can therefore increase, decrease, or pass through 1 as trajectories change relative position. A ratio of 1 at a particular time means the two subjects have equal conditional hazards at that instant under this model; it does not mean their hazards were equal at other times or that their future event probabilities are identical.

Worked Example: Trajectories That Cross

Consider two hypothetical latent marker trajectories, measured in marker units, with \(t\) in weeks: \(m_A(t)=12+0.4t\) and \(m_B(t)=10+0.7t\). Suppose \(\alpha=-0.20\) per marker unit and the covariates are the same. Calculate the hazard ratio for A versus B at weeks 2 and 10, and identify when the trajectories cross.

At week 2, \(m_A(2)=12+0.4(2)=12.8\) and \(m_B(2)=10+0.7(2)=11.4\). The difference is \(1.4\), so

$$ HR_{AB}(2)=\exp\{-0.20(12.8-11.4)\} =\exp(-0.28) =0.7558. $$

At week 2, A’s conditional hazard is about 0.756 times B’s, or about 24.4% lower. At week 10, \(m_A(10)=16\) and \(m_B(10)=17\), so

$$ HR_{AB}(10)=\exp\{-0.20(16-17)\} =\exp(0.20) =1.221. $$

At week 10, A’s conditional hazard is about 1.22 times B’s. To find the crossing time, solve \(12+0.4t=10+0.7t\), giving \(2=0.3t\) and \(t=6.667\) weeks. At that time the marker difference is zero, so the marker-only hazard ratio is \(\exp\{(-0.20)(0)\}=1\). The direction of the subject comparison changes because the trajectories cross; the association coefficient itself has not changed.

Separating Marker Differences From Covariate Differences

When subjects differ in baseline covariates as well as latent marker value, the full conditional hazard ratio includes both contributions. Reporting only \(\exp\{\alpha[m_i(t)-m_j(t)]\}\) answers a marker-only comparison with the covariate contrast removed; it is not the full comparison of those two subjects unless their covariates match. The distinction is important when describing a particular pair rather than isolating the marker association.

Worked Example: A Covariate Contrast Changes the Full Ratio

Suppose a binary treatment indicator is included as a covariate, with coefficient \(\gamma=0.40\) for treated versus untreated, and \(\alpha=0.25\) per marker unit. At time \(t\), subject A is treated and subject B is untreated. Their latent marker difference is \(m_A(t)-m_B(t)=-1.2\), and all other covariates match. Calculate both the marker-only and full conditional hazard ratios for A versus B.

The marker-only contribution is

$$ \exp\{0.25(-1.2)\} = \exp(-0.30) = 0.7408. $$

The treatment covariate contrast contributes \(0.40(1-0)=0.40\) to the log-hazard contrast. Therefore the full comparison is

$$ \exp\{0.40+0.25(-1.2)\} = \exp(0.10) = 1.105. $$

Although A’s lower latent marker value contributes a marker-only ratio below 1, the full conditional hazard ratio is about 1.105 because the treatment covariate contrast offsets that contribution. This is a model-based conditional association, not by itself a causal effect of treatment.

Justification and Practical Interpretation

The interpretation can be verified directly from the hazard specification. At a fixed time \(t\), substitute the model for each subject’s hazard and divide:

$$ \frac{h_0(t)\exp\{\boldsymbol{\gamma}^{\top}\mathbf{w}_i+\alpha m_i(t)\}} {h_0(t)\exp\{\boldsymbol{\gamma}^{\top}\mathbf{w}_j+\alpha m_j(t)\}} = \exp\left\{ \boldsymbol{\gamma}^{\top}(\mathbf{w}_i-\mathbf{w}_j) +\alpha[m_i(t)-m_j(t)] \right\}. $$

This calculation explains both the usefulness and the boundaries of the interpretation. The common baseline hazard cancels for a same-time comparison. It does not cancel when comparing hazards at different times, because \(h_0(t_1)\) and \(h_0(t_2)\) need not be equal. A hazard ratio at one time is therefore defined by the two subjects’ hazards at that same time, not by comparing one subject at week 2 with another at week 10.

For a fixed positive marker difference \(d\), the marker-only ratio is \(\exp(\alpha d)\). If \(\alpha>0\), a higher latent marker corresponds to a higher conditional hazard; if \(\alpha<0\), it corresponds to a lower one. The size of the ratio depends on both \(\alpha\) and the chosen difference \(d\). In particular, \(\exp(\alpha)\) is the ratio for a one-unit marker difference, not automatically the ratio for every pair of subjects.

Comparisons, Diagnostics, and Limitations

ComparisonWhat it answersKey qualification
Subject-specific, conditionalHow do the hazards compare for specified latent trajectories and covariates at time \(t\)?Requires the subjects’ latent values at that time and conditions on their random effects.
Marker-only contrastWhat ratio is associated with the latent marker difference?Holds other modeled covariates equal; it is not the full pairwise ratio if covariates differ.
Population-averaged comparisonHow do hazards compare after accounting for variation across the random-effects distribution?Generally cannot be obtained by simply substituting an average marker value into the conditional hazard formula.

The subject-specific interpretation is most useful when the scientific question concerns conditional differences between latent profiles, or when presenting fitted comparisons at times of substantive interest. The main alternatives answer different questions: a population-averaged analysis targets a marginal comparison, while a marker-only contrast deliberately holds other covariates equal. Neither should be presented as interchangeable with the full conditional comparison.

Several checks help prevent a plausible-looking but incorrect interpretation:

  • Check the time. Evaluate both latent values at the same stated time, using the time unit in the fitted trajectory. Do not compare marker values measured at different times as though they formed a same-time contrast.
  • Check the contrast direction. For A versus B, use \(m_A(t)-m_B(t)\) and \(\mathbf{w}_A-\mathbf{w}_B\). Reversing the order must give the reciprocal hazard ratio.
  • Check the scale and units. If \(\alpha\) is per marker unit, a \(d\)-unit difference uses \(\alpha d\). Changing the marker scale changes the numerical coefficient and requires consistent units in the calculation.
  • Use the latent value. Substituting a single noisy measurement for \(m_i(t)\) ignores the model’s separation of trajectory and measurement error. The earlier material on current-value association with noisy measurements develops that distinction.
  • Account for uncertainty. Estimated trajectory values and \(\alpha\) are uncertain. A plug-in ratio is an estimate; formal uncertainty for the ratio should reflect uncertainty in both the association parameter and the latent trajectory contrast.

The model’s linear current-value form is also a limitation: it imposes a constant change in log hazard per marker unit over the modeled range. Poor trajectory fit, sparse measurements near the time of interest, or extrapolation beyond observed follow-up can make a subject-specific ratio unreliable. A calculated ratio does not repair model misspecification, and conditional hazard ratios alone do not supply cumulative event probabilities or causal conclusions.

Key takeaway: For two subjects compared at the same time, the current-value contribution to their conditional hazard ratio is \(\exp\{\alpha[m_i(t)-m_j(t)]\}\). State the time, the order of the comparison, the marker units, and whether other covariates are held equal.

Check Your Understanding

For each question, use the conditional current-value interpretation and show which contrast is being calculated.

  1. Suppose \(\alpha=0.18\) per mmol/L, and two subjects with equal covariates have latent marker values 7.1 and 5.6 mmol/L at the same time. Find the hazard ratio for the subject at 7.1 versus the subject at 5.6, and interpret it.
  2. Let \(m_A(t)=4+0.3t\) and \(m_B(t)=5+0.1t\), with time measured in weeks and \(\alpha=-0.40\) per marker unit. Find the marker-only hazard ratio for A versus B at weeks 5 and 10.
  3. Suppose the marker contribution to the log-hazard contrast for A versus B is \(0.20(3)\), while the covariate contribution is \(-0.50\). Find the full conditional hazard ratio and distinguish it from the marker-only ratio.
  4. At one time, a subject’s observed marker measurement is 9 units, while the fitted latent value is 8 units. Which value belongs in the current-value hazard comparison, and why?

Solutions. (1) The difference is \(7.1-5.6=1.5\) mmol/L, so \(HR=\exp\{0.18(1.5)\}=\exp(0.27)=1.310\). The higher-marker subject has an estimated conditional hazard about 1.31 times as large at that time, with covariates equal. (2) At week 5, \(m_A=5.5\) and \(m_B=5.5\), so \(HR=\exp\{-0.40(0)\}=1\). At week 10, \(m_A=7\) and \(m_B=6\), so \(HR=\exp\{-0.40(1)\}=0.6703\). (3) The marker-only ratio is \(\exp(0.60)=1.822\). The full log contrast is \(-0.50+0.60=0.10\), giving \(HR=\exp(0.10)=1.105\). (4) Use the fitted latent value of 8 units in the model-based comparison. The observed value of 9 includes measurement error and is not the latent current value specified in the hazard model.

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