Recent Versus Remote Marker Accumulation

Why the Age of Marker History Matters

A cumulative association lets the event hazard depend on marker history rather than only on the marker’s current value. But “cumulative” does not specify how history is counted. One model can give every past marker value the same weight; another can give a value less influence the farther back in time it occurred. Those choices can imply different exposures, even for trajectories with the same unweighted total.

The distinction is about the weighting of history, not about changing the observed marker trajectory. We write \(m_i(s)\) for subject \(i\)’s latent marker value at time \(s\), and compare two summaries evaluated at event-risk time \(t\). The integrals below use the history from time zero to \(t\); the time origin and exposure window therefore need to be specified consistently.

Definition: The equally weighted cumulative exposure at time \(t\) is \(A_i^{U}(t)=\int_0^t m_i(s)\,ds\). The exponentially recency-weighted cumulative exposure, with decay rate \(\kappa>0\), is \(A_i^{R}(t)=\int_0^t e^{-\kappa(t-s)}m_i(s)\,ds\). In the second expression, the weight depends on the lag \(t-s\): recent history has a weight near 1, while remote history has a smaller weight.

The superscript \(U\) denotes equal weighting, and \(R\) denotes recency weighting. For the exponential kernel, \(\kappa\) controls how rapidly the weight falls as the lag grows. A larger \(\kappa\) discounts remote history more strongly. The kernel is not normalized to integrate to one: these exposures are weighted cumulative totals, not weighted averages.

A common association structure places one of these exposures in a proportional hazards model:

h_i(t\mid m_i)=h_0(t)\exp\{\gamma A_i(t)+\boldsymbol{\eta}^{\mathsf T}\mathbf{x}_i\}.

Here \(h_i(t\mid m_i)\) is the hazard conditional on the marker trajectory and covariates, \(h_0(t)\) is the baseline hazard, \(\mathbf{x}_i\) is a vector of covariates, and \(\boldsymbol{\eta}\) is its coefficient vector. In an equally weighted model, \(A_i(t)=A_i^{U}(t)\); in a recency-weighted model, \(A_i(t)=A_i^{R}(t)\). The association coefficient \(\gamma\) describes the log-hazard change per unit of the chosen exposure, conditional on the other terms in the model.

How the Weights Change a Hazard Contrast

For two marker trajectories \(m_a\) and \(m_b\) compared at the same time \(t\), with the same covariates and the same baseline hazard, the conditional hazard ratio is

\frac{h_a(t)}{h_b(t)} = \exp\{\gamma[A_a(t)-A_b(t)]\}.

Thus the kernel affects the hazard contrast through the exposure difference. Under equal weighting, every unit of time contributes equally to the integral. Under exponential recency weighting, the contribution of a marker difference at time \(s\) is multiplied by \(e^{-\kappa(t-s)}\). If \(\gamma>0\), a larger exposure corresponds to a larger conditional hazard; if \(\gamma<0\), it corresponds to a smaller one. Neither interpretation by itself establishes a causal effect.

The two kernels are connected. As \(\kappa\) approaches zero, the exponential weight approaches one at every lag, so the recency-weighted exposure approaches the equally weighted exposure. To see this, for any bounded trajectory over a finite interval, \(e^{-\kappa(t-s)}\) converges uniformly to 1 as \(\kappa\to0\). Therefore,

\lim_{\kappa\to0} A_i^{R}(t) = \int_0^t \lim_{\kappa\to0} e^{-\kappa(t-s)}m_i(s)\,ds = \int_0^t m_i(s)\,ds = A_i^{U}(t).

This limiting relationship is a useful check on formulas and software. It does not mean that every positive \(\kappa\) produces an equivalent model: positive decay changes the relative influence of different parts of the history. Nor should raw estimates of \(\gamma\) from different kernels be treated as directly comparable without considering the scale of their exposures.

Calculating and Comparing the Exposures

Worked Example: A Constant Marker Trajectory

Suppose a hypothetical subject has latent marker value \(m_i(s)=2\) marker units throughout the four time units from 0 to \(t=4\). Compare the two exposures for \(\kappa=0.5\) per time unit.

For equal weighting,

A_i^{U}(4)=\int_0^4 2\,ds=2(4)=8.

For recency weighting, integrate the exponential kernel:

A_i^{R}(4) = \int_0^4 e^{-0.5(4-s)}2\,ds = 2\left[\frac{1-e^{-0.5(4)}}{0.5}\right] = 4(1-e^{-2}) \approx 3.459.

The recency-weighted total is smaller because marker values earlier than time 4 receive weights below 1. For illustration, if the recency-weighted model has \(\gamma=0.20\) per exposure unit, its conditional hazard ratio for this trajectory versus a zero-marker trajectory is \(\exp(0.20\times3.459)=\exp(0.6918)\approx1.997\). The exposure contrast is \(3.459\), and the ratio is conditional on the same covariates and baseline hazard. This number is an illustration within that model; it is not a direct comparison with a hazard ratio from a separately fitted equally weighted model.

A particularly clear contrast arises when two histories have equal unweighted totals but differ in when the marker elevation occurred. Suppose a marker is elevated by the same amount over two intervals of equal length, one remote and one recent. Equal weighting assigns those elevations the same exposure contribution. Recency weighting does not.

Worked Example: Equal Totals, Different Timing

Consider two hypothetical trajectories evaluated at \(t=4\). Trajectory E is elevated by 2 marker units on \([0,1]\), and trajectory L is elevated by 2 marker units on \([3,4]\); each is zero outside its stated interval. Let \(\kappa=0.5\) per time unit. The intervals are taken to include their endpoints, which do not affect these integrals.

Both equally weighted exposures are the marker elevation multiplied by the interval length:

A_E^{U}(4)=2(1-0)=2, \qquad A_L^{U}(4)=2(4-3)=2.

For an elevation of size 2 on an interval \([a,b]\), the recency-weighted contribution is

2\int_a^b e^{-0.5(4-s)}\,ds = 4\left[e^{-0.5(4-b)}-e^{-0.5(4-a)}\right].

Substituting \([a,b]=[0,1]\) for E gives \(A_E^{R}(4)=4(e^{-1.5}-e^{-2})\approx4(0.22313-0.13534)=0.3512\). For L, substituting \([a,b]=[3,4]\) gives \(A_L^{R}(4)=4(1-e^{-0.5})\approx4(1-0.60653)=1.574\). The recent elevation therefore contributes more than the remote elevation under this kernel, despite equal unweighted totals.

If the recency-weighted model has \(\gamma=0.40\) per exposure unit, the conditional hazard ratio for L versus E is \(\exp\{0.40(1.574-0.3512)\}=\exp(0.4891)\approx1.631\). The model implies a 1.631-fold conditional hazard for L relative to E at time 4, holding the covariates fixed. Under equal weighting, their exposure difference is zero, so the corresponding hazard ratio would be 1 in that model, regardless of its value of \(\gamma\).

A Linear Trajectory and a Full Hazard Contrast

For a trajectory that changes over time, the weighting affects each point along the path. The following calculation shows how to integrate a linear trajectory and then form a contrast against a constant trajectory.

Worked Example: Comparing a Rising Marker With a Constant Marker

Suppose two hypothetical trajectories are observed through \(t=4\): \(m_a(s)=1+0.5s\) and \(m_b(s)=1\). Take \(\kappa=0.25\) per time unit. First, their equally weighted exposures are

A_a^{U}(4)=\int_0^4(1+0.5s)\,ds =4+0.25(4^2)=8, \qquad A_b^{U}(4)=\int_0^4 1\,ds=4.

For recency weighting, set \(u=4-s\), so the rising trajectory becomes \(1+0.5(4-u)=3-0.5u\). Define

I_0=\int_0^4 e^{-0.25u}\,du =\frac{1-e^{-1}}{0.25} \approx2.5285,

and

I_1=\int_0^4 u e^{-0.25u}\,du = \frac{1-e^{-1}(1+1)}{0.25^2} \approx4.2279.

The expression for \(I_1\) follows by integration by parts, using the antiderivative \(-e^{-0.25u}(u/0.25+1/0.25^2)\). Thus

A_a^{R}(4)=3I_0-0.5I_1 =3(2.5285)-0.5(4.2279) \approx5.472.

The constant trajectory has \(A_b^{R}(4)=I_0\approx2.5285\), so the recency-weighted exposure difference is approximately \(2.943\). In contrast, the equally weighted exposure difference is \(8-4=4\). If the recency-weighted model has \(\gamma=0.30\) per exposure unit, its conditional hazard ratio for trajectory \(a\) versus \(b\) is \(\exp(0.30\times2.943)=\exp(0.8829)\approx2.418\). For an equally weighted model with \(\gamma=0.30\) per exposure unit, the corresponding calculation is \(\exp(0.30\times4)=\exp(1.2)\approx3.320\). These are model-specific illustrations: equal numerical values of \(\gamma\) do not guarantee equivalent effect scales across differently weighted exposures.

Choosing and Checking a Weighting Function

The exponential kernel is one explicit way to express a gradual discount of remote history. Its half-life is the lag at which the weight falls to one half. Solving \(e^{-\kappa u}=0.5\) gives \(u=\log(2)/\kappa\). For example, when \(\kappa=0.5\) per time unit, the half-life is about \(1.386\) time units. The half-life can make the decay rate easier to interpret than \(\kappa\) alone.

The kernel should reflect the scientific question and the available follow-up, not simply produce a preferred hazard contrast. A rectangular window, introduced in the earlier material on windowed cumulative marker effects, excludes history before a chosen lag limit. Exponential weighting instead retains all history after time zero, but attenuates remote contributions smoothly. A current-value link uses only \(m_i(t)\), while a cumulative link summarizes a span of past values. These alternatives answer different questions; they are not interchangeable definitions of “the” marker effect.

Several checks help identify problems in the specification or calculation:

Check the lag direction. In \(e^{-\kappa(t-s)}\), the lag is \(t-s\). At \(s=t\), the weight is 1; for \(s<t\), it is less than 1. Reversing the sign would increase remote weights rather than discount them.
Check the limiting case. As \(\kappa\) becomes small, the recency-weighted exposure should approach the equally weighted exposure. A calculation that does not do so may have an incorrect kernel or integration limit.
Check the time origin and risk time. Both exposures above accumulate from time zero to the current risk time \(t\). Changing the origin or using marker history after \(t\) changes the quantity being modeled.
Check numerical integration. With irregular measurement times, a numerical approximation should respect the trajectory representation and the rapidly changing part of the kernel. Compare integration accuracy at finer time steps where feasible.
Check sensitivity to \(\kappa\). A decay rate that is poorly determined or chosen without substantive justification can make conclusions depend strongly on an arbitrary weighting choice. Examine plausible alternatives and report the choice clearly.

Common Pitfall and Limitations

Common mistake: “The two subjects have the same total marker exposure, so any cumulative model must give them the same hazard.” This is true for equal weighting when their unweighted integrals are equal, but it need not be true for a recency-weighted model. The correction is to calculate the exposure under the specified kernel: equal totals do not imply equal weighted totals when the marker elevations occurred at different times.

A second pitfall is to compare estimated association coefficients as if they represented the same contrast. The coefficient multiplies a particular exposure, and changing the kernel changes how much exposure is assigned to a trajectory. Interpret a coefficient together with its kernel and the exposure contrast it multiplies; compare model implications for a clearly specified pair of trajectories when a direct substantive comparison is needed.

Cumulative links also compress a potentially complicated trajectory into a single summary. Different histories can have the same weighted exposure, and the selected kernel may fail to capture a pattern that matters for the event process. A fitted association is conditional on the specified model and its assumptions; it does not by itself demonstrate that changing the marker would change the hazard. These limitations make transparent kernel choices and sensitivity checks important complements to estimation.

Check Your Understanding

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

For \(\kappa>0\), what is the exponential recency weight at the current time, and how does it change for increasingly remote marker values?
A marker is elevated by 3 units for one time unit. Compare its equally weighted contributions if the interval is recent or remote.
For \(\kappa=0.4\), calculate the recency-weighted contribution of a 2-unit elevation on \([2,3]\), evaluated at \(t=4\).
If two trajectories have recency-weighted exposures that differ by 0.8 and the model has \(\gamma=-0.25\), what is the hazard ratio for the trajectory with the larger exposure versus the other?
Why does a small positive \(\kappa\) make the recency-weighted exposure close to the equally weighted exposure?

Solutions. (1) At \(s=t\), the weight is \(e^0=1\); as the lag \(t-s\) increases, \(e^{-\kappa(t-s)}\) decreases. (2) Both contributions are \(3\times1=3\) marker-unit time units under equal weighting. (3) The contribution is \(2\int_2^3 e^{-0.4(4-s)}ds=\frac{2}{0.4}(e^{-0.4}-e^{-0.8})\approx5(0.6703-0.4493)=1.105\). (4) The log hazard ratio is \((-0.25)(0.8)=-0.20\), so the ratio is \(e^{-0.20}\approx0.8187\); the larger exposure is associated with a lower conditional hazard under this negative coefficient. (5) When \(\kappa\) is close to zero, the exponent \(-\kappa(t-s)\) is close to zero across a finite history, so the weights are close to \(e^0=1\), the equal-weight kernel.