Combining Current Value and Cumulative Exposure

Why Combine Current Value and Accumulated History?

A marker can be associated with event risk in two different ways. Its value at the present time may indicate a subject’s immediate state, while the amount of marker experienced over the preceding follow-up may represent a longer-term burden. A model containing only one of these features cannot represent both associations at once. Combining them makes it possible to ask whether current marker level and prior cumulative exposure each contribute to the hazard, conditional on the other.

The marker in a joint model is the subject-specific latent trajectory, not an error-prone measurement taken at an appointment. Write \(m_i(t)\) for subject \(i\)'s latent marker value at time \(t\), as specified by the longitudinal submodel. Define cumulative exposure from the chosen time origin \(0\) to time \(t\) as \(A_i(t)\). The integral ends at \(t\), so it uses the trajectory history available before the event time, not marker values after the event.

Definition: A current-value-plus-cumulative-exposure association link is a hazard model in which the log hazard depends on both \(m_i(t)\), the latent marker value at time \(t\), and \(A_i(t)=\int_0^t m_i(s)\,ds\), the area under that latent trajectory from the time origin through \(t\).

Model and Interpretation

One standard specification, conditional on the shared random effects \(b_i\) and baseline covariates \(w_i\), is

h_i(t\mid b_i,w_i) = h_0(t)\exp\left\{\gamma^{\mathsf T}w_i+\alpha m_i(t)+\eta A_i(t)\right\}, \qquad A_i(t)=\int_0^t m_i(s)\,ds.

Here \(h_0(t)\) is the baseline hazard, \(\gamma\) is the vector of covariate coefficients, and \(\alpha\) and \(\eta\) are association parameters. The longitudinal model and its random effects determine \(m_i(t)\). Thus the survival model links the event process to the latent trajectory rather than treating a noisy observed marker as if it were the true value.

The model assumes that, conditional on the shared random effects and included covariates, the hazard has the displayed proportional-hazards form with a linear contribution from the two association features. In particular, \(\alpha\) and \(\eta\) are constant over follow-up in this specification. The trajectory must be defined over the interval being integrated, and the time origin and measurement units must be fixed and used consistently. A joint-model analysis also relies on adequate longitudinal and event submodels; its interpretation depends on their assumptions, including the specified random-effects structure and the treatment of censoring.

The coefficients have different units. If \(m_i(t)\) is measured in marker units and time in months, then \(\alpha\) is a log-hazard coefficient per marker unit, while \(\eta\) is a log-hazard coefficient per marker-unit-month. At a fixed time \(t\), a contrast between two subjects or two specified trajectories is

\log(\mathrm{HR}(t)) = \alpha\{m_1(t)-m_0(t)\} + \eta\{A_1(t)-A_0(t)\},

when their baseline covariates are held equal. Exponentiating gives the hazard ratio. The \(\alpha\) term is the current-value contribution, conditional on the cumulative-exposure difference; the \(\eta\) term is the cumulative contribution, conditional on the current-value difference. These are conditional model interpretations, not claims that changing a marker causes a change in hazard.

Worked Example: Combining Two Contributions to a Hazard Ratio

Suppose a hypothetical fitted model has \(\alpha=0.20\) per mg/L and \(\eta=-0.05\) per (mg/L-month). At month 2, compare two hypothetical subjects with equal covariates and trajectories \(m_1(s)=4+0.5s\) and \(m_0(s)=3+0.5s\), where the marker is measured in mg/L and time in months.

At month 2, their current marker values are \(m_1(2)=4+0.5(2)=5\) mg/L and \(m_0(2)=3+0.5(2)=4\) mg/L. The current-value difference is therefore \(1\) mg/L. Their cumulative exposures are

A_1(2)=\int_0^2(4+0.5s)\,ds =4(2)+0.5\frac{2^2}{2}=9\ \text{mg/L-month}, \qquad A_0(2)=\int_0^2(3+0.5s)\,ds =3(2)+0.5\frac{2^2}{2}=7\ \text{mg/L-month}.

The cumulative-exposure difference is \(2\) mg/L-month. The log hazard ratio is

0.20(1)+(-0.05)(2)=0.10, \qquad \mathrm{HR}(2)=\exp(0.10)=1.105.

The current-value component alone contributes \(\exp\{0.20(1)\}=1.221\), while the cumulative component contributes \(\exp\{-0.05(2)\}=0.9048\). Their product is approximately \(1.105\). Under this model, the first subject's hazard at month 2 is estimated to be about 1.105 times the second subject's, conditional on the stated covariates and model structure. The opposing contributions partly offset; neither component should be interpreted as the total contrast by itself.

Separating Current and Cumulative Contrasts

The contrast equation gives a precise way to separate components. To isolate the current-value contribution, compare trajectories with equal cumulative exposure at the time of interest but different current values. To isolate the cumulative contribution, compare trajectories with equal current values but different accumulated exposure. Such comparisons are algebraic contrasts under the fitted model. Whether the data contain enough trajectories like these to estimate the two coefficients precisely is a separate question.

Worked Example: Holding Cumulative Exposure Fixed

Use the hypothetical coefficients \(\alpha=0.20\) per mg/L and \(\eta=-0.05\) per (mg/L-month). At month 2, compare \(m_A(s)=4\) with \(m_B(s)=3+s\), with time in months and marker in mg/L.

For trajectory \(A\), the current value is \(4\) mg/L and the cumulative exposure is \(4(2)=8\) mg/L-month. For trajectory \(B\), the current value is \(3+2=5\) mg/L, and

A_B(2)=\int_0^2(3+s)\,ds=3(2)+\frac{2^2}{2}=8\ \text{mg/L-month}.

The cumulative exposures match, so their difference is zero; the current-value difference, \(m_B(2)-m_A(2)\), is \(1\) mg/L. Thus

\log(\mathrm{HR}_{B:A}(2))=0.20(1)+(-0.05)(0)=0.20, \qquad \mathrm{HR}_{B:A}(2)=\exp(0.20)=1.221.

This contrast isolates the current-value term: trajectory \(B\) has an estimated hazard 1.221 times that for trajectory \(A\) at month 2 under the model. The trajectories have different histories, but their accumulated exposure through month 2 is identical.

Worked Example: Holding the Current Value Fixed

Again compare hypothetical trajectories at month 2, now using \(m_A(s)=4\) and \(m_C(s)=3+s/2\). Both have current value \(4\) mg/L at that time. The cumulative exposures are

A_A(2)=4(2)=8\ \text{mg/L-month}, \qquad A_C(2)=\int_0^2(3+s/2)\,ds=3(2)+\frac{2^2}{4}=7\ \text{mg/L-month}.

The current-value difference \(m_C(2)-m_A(2)\) is zero, and the cumulative-exposure difference \(A_C(2)-A_A(2)\) is \(-1\) mg/L-month. Therefore,

\log(\mathrm{HR}_{C:A}(2)) = 0.20(0)+(-0.05)(-1)=0.05, \qquad \mathrm{HR}_{C:A}(2)=\exp(0.05)=1.051.

The model estimates a hazard about 1.051 times as high for trajectory \(C\) at month 2, despite the equal current values, because its cumulative exposure is lower and \(\eta\) is negative. The example also shows why the sign of a cumulative coefficient must be interpreted together with the direction of the exposure contrast.

Linear Trajectories and a Short Derivation

For a linear latent trajectory, both components can be calculated directly. Suppose \(m_i(s)=a_i+b_i s\), where \(a_i\) is the marker level at the chosen time origin and \(b_i\) is its rate of change. Integrating gives

A_i(t) = \int_0^t(a_i+b_i s)\,ds = a_i t+\frac{b_i t^2}{2}.

Substitution into the log hazard shows how the intercept and slope contribute to each association feature:

\alpha m_i(t)+\eta A_i(t) = \alpha(a_i+b_i t)+\eta\left(a_i t+\frac{b_i t^2}{2}\right).

This expression follows by integrating the trajectory and then inserting the resulting current value and area into the hazard link. It also illustrates a key distinction: the current term depends on \(a_i+b_i t\), while the history term depends on both \(a_i t\) and \(b_i t^2/2\). A slope difference can therefore affect the hazard through both the current value at \(t\) and the accumulated path to \(t\); that is not an additional slope association parameter.

Worked Example: A Contrast Along Linear Trajectories

Consider two hypothetical linear marker trajectories, measured in units per month: \(m_1(s)=3+0.4s\) and \(m_0(s)=2.5+0.2s\). Let \(\alpha=0.15\) per marker unit and \(\eta=0.02\) per marker-unit-month. Compare their hazards at month 3, holding covariates equal.

The current values are \(m_1(3)=3+0.4(3)=4.2\) and \(m_0(3)=2.5+0.2(3)=3.1\), so the current difference is \(1.1\) marker units. The cumulative exposures are

A_1(3)=3(3)+\frac{0.4(3^2)}{2}=10.8, \qquad A_0(3)=2.5(3)+\frac{0.2(3^2)}{2}=8.4 \quad \text{marker-unit-months}.

The cumulative difference is \(2.4\) marker-unit-months. The combined log hazard ratio and hazard ratio are

0.15(1.1)+0.02(2.4)=0.213, \qquad \mathrm{HR}_{1:0}(3)=\exp(0.213)=1.237.

The current-value contribution is \(0.165\) on the log-hazard scale, and the cumulative contribution is \(0.048\). Thus the model estimates a hazard 1.237 times as high for trajectory 1 at month 3, conditional on equal covariates. This is a trajectory contrast, not a causal effect of setting either marker path.

Assumptions, Diagnostics, and Alternatives

The combined link is useful only if its two features are sufficiently supported by the data and the chosen trajectory model. Current value and cumulative exposure are often strongly correlated: subjects with persistently high marker values may have both high \(m_i(t)\) and large \(A_i(t)\). If the data provide little independent variation in these features, many combinations of \(\alpha\) and \(\eta\) can produce similar fitted hazards. The coefficients may then have large standard errors or strong posterior dependence, even when predictions of overall risk appear stable.

Check the estimated uncertainty and dependence of the association parameters, and examine whether the observed fitted trajectories provide variation that distinguishes current levels from accumulated histories. Verify the time origin, integration interval, marker scale, and time units: changing units changes coefficient magnitudes, and changing the time origin changes the cumulative feature. Numerical integration should be accurate over the full interval; for a model-defined linear trajectory, the analytic integral above provides a direct check. Also assess whether the proportional-hazards form and the longitudinal trajectory adequately describe the data. A poor trajectory model can distort both the current value and its integral.

The main simpler alternatives use only the current value or only cumulative exposure. A current-value-only link is easier to interpret when the scientific question concerns contemporaneous status, but it omits accumulated history. A cumulative-only link represents burden over time, but cannot distinguish two subjects with the same cumulative exposure and different current values. The combined link allows both terms but requires enough information to estimate them separately. It does not establish that either association is causal, and it cannot by itself correct misspecification of the longitudinal process, event model, or censoring assumptions.

Common mistake: A reader may say, “The current marker difference is one unit, so the hazard ratio is \(\exp(\alpha)\),” even when the subjects also have different cumulative exposures. That calculation drops part of the specified model. The correct contrast is \(\exp\{\alpha\Delta m(t)+\eta\Delta A(t)\}\); \(\exp\{\alpha\Delta m(t)\}\) alone is appropriate only when the cumulative-exposure difference is zero or when explicitly reporting that component separately.

Key takeaway: The combined link assigns separate terms to the latent marker now and its accumulated history. Calculate both differences at the event time, apply their respective coefficients, add on the log-hazard scale, and exponentiate only after summing.

Check Your Understanding

Answer each question, then compare your work with the solutions.

If \(m_i(s)=2+0.5s\), what is \(A_i(4)\)?
With \(\alpha=0.10\) and \(\eta=0.03\), what is the hazard ratio when the current-value difference is \(2\) units and the cumulative-exposure difference is \(-1\) unit-month?
For a contrast with equal current values but cumulative-exposure difference \(3\) unit-months, which coefficient determines the log hazard ratio?
Why can the current-value and cumulative-exposure coefficients be hard to estimate separately?

Solutions. (1) \(A_i(4)=\int_0^4(2+0.5s)\,ds=2(4)+0.5(4^2/2)=8+4=12\) unit-months. (2) The log hazard ratio is \(0.10(2)+0.03(-1)=0.17\), so the hazard ratio is \(\exp(0.17)=1.185\). (3) The current-value difference is zero, so the log hazard ratio is \(\eta(3)\); here it is \(0.03(3)=0.09\), and the hazard ratio is \(\exp(0.09)=1.094\). (4) Subjects with high current values often also have high accumulated exposure, so the two features may vary together and provide limited information for distinguishing their separate contributions.