Marker Level Versus Slope at the Same Time

Why Equal Marker Levels Need Not Mean Equal Hazard

At a particular time, two subjects can have the same marker value while their marker trajectories are moving in different directions. One subject’s value may be rising quickly while the other’s is flat or falling. A slope-based association allows the instantaneous event hazard to depend on this rate of change, so equal current marker values do not necessarily imply equal hazards.

This distinction concerns what feature of the latent marker trajectory enters the hazard model. A current-value link uses the marker’s level at time $t$; a slope-based link uses its instantaneous rate of change at time $t$. Here the focus is the slope-based link and the comparison it implies at a shared time. The earlier material on current-value associations provides the contrasting level-based interpretation.

Definition: Let $m_i(t)$ denote subject $i$’s latent marker trajectory, with derivative $m_i'(t)$ at time $t$. A slope-based hazard link has the form $$ h_i(t\mid \mathbf{w}_i)=h_0(t)\exp\{\boldsymbol{\gamma}^{T}\mathbf{w}_i+\alpha_s m_i'(t)\}. $$ Here $h_i(t\mid \mathbf{w}_i)$ is subject $i$’s instantaneous event hazard at time $t$, $h_0(t)$ is a common baseline hazard, $\mathbf{w}_i$ is a vector of covariates, $\boldsymbol{\gamma}$ is its coefficient vector, and $\alpha_s$ is the slope association parameter.

The hazard is conditional on the subject being event-free immediately before $t$, as well as on the covariates and latent trajectory represented in the model. The model assumes a common baseline hazard and an exponential effect of the marker slope. It says that a one-unit increase in slope multiplies the hazard by $\exp(\alpha_s)$, holding the other modeled terms fixed. It does not say that changing a person’s marker slope would cause their hazard to change.

For example, if the marker is measured in marker-units and time in years, then $m_i'(t)$ is measured in marker-units per year. The parameter $\alpha_s$ has reciprocal slope units, so that $\alpha_s m_i'(t)$ is dimensionless. The numerical value of a slope, and therefore the meaning of a one-unit slope difference, depends on the time and marker units used.

Comparing Subjects at the Same Time

Consider two subjects, $A$ and $B$, at the same time $t$. Suppose they have the same covariate values in the hazard model, but their marker slopes differ. Dividing their conditional hazards cancels the common baseline hazard and the common covariate contribution. The resulting comparison is:

\frac{h_A(t)}{h_B(t)} = \exp\{\alpha_s[m_A'(t)-m_B'(t)]\}.

The comparison depends on the difference in slopes, not on whether the subjects’ current marker values are equal. If their slopes are equal too, the slope-based link assigns equal hazards, provided the other modeled terms are equal. If their slopes differ, the hazard ratio is determined by the sign and size of $\alpha_s$ and by the slope difference.

Key result: Under the stated slope-based model, for subjects with equal covariates at the same time, a slope difference of $\Delta=m_A'(t)-m_B'(t)$ gives hazard ratio $\exp(\alpha_s\Delta)$. A ratio above 1 means subject $A$ has the higher conditional instantaneous hazard; a ratio below 1 means subject $B$ does.

This ratio is a comparison of instantaneous hazards at time $t$, not a ratio of event probabilities over a period and not a comparison of survival probabilities. The model’s exponential form makes the ratio multiplicative. Reversing the order of the subjects reverses the slope difference and reciprocates the ratio.

Worked Comparisons

Worked Example: Equal Marker Values, Rising Versus Falling

Suppose two hypothetical subjects have the same latent marker value of 12 marker-units at month 6. At that time, subject $A$’s slope is $0.40$ marker-units per month and subject $B$’s slope is $-0.20$ marker-units per month. Assume their modeled covariates are equal and the fitted slope association parameter is $\alpha_s=0.70$ per (marker-unit per month). Find the hazard ratio for $A$ versus $B$.

The slope difference is $0.40-(-0.20)=0.60$ marker-units per month. Thus,

\frac{h_A(6)}{h_B(6)} =\exp\{0.70(0.60)\} =\exp(0.42) \approx 1.522.

Under this model, subject $A$’s conditional instantaneous hazard at month 6 is about 1.52 times subject $B$’s, or about 52.2% higher. Their equal marker values do not make their hazards equal under a slope-only link, because their rates of change differ. This is a model-based association, not evidence that the rising marker causes the higher hazard.

Worked Example: A Negative Slope Association

In another hypothetical comparison, both subjects have the same current marker value and equal covariates. Subject $A$’s marker slope is $1.2$ marker-units per year and subject $B$’s is $0.4$ marker-units per year. Suppose $\alpha_s=-0.50$ per (marker-unit per year). Calculate the hazard ratio for $A$ versus $B$.

The difference in slopes is $1.2-0.4=0.8$ marker-units per year. The ratio is

\frac{h_A(t)}{h_B(t)} =\exp\{(-0.50)(0.8)\} =\exp(-0.40) \approx 0.6703.

Subject $A$’s conditional instantaneous hazard is about 0.670 times subject $B$’s, or about 33.0% lower at that time under the model. A negative $\alpha_s$ means higher slopes are associated with lower hazards, all else equal. It does not mean that every individual with a rising marker has a low hazard; the comparison depends on the other modeled terms and the slope difference.

Worked Example: From Linear Trajectories to Hazards

Suppose two hypothetical latent trajectories, measured in marker-units over years, are $m_A(t)=8+t$ and $m_B(t)=10$. Compare them at $t=2$ years. Let their covariates be equal, let the baseline hazard at that time be $h_0(2)=0.015$ per year, and let $\alpha_s=0.30$ per (marker-unit per year). First check the levels and slopes, then calculate their hazards.

At $t=2$, $m_A(2)=8+2=10$ and $m_B(2)=10$, so the current values are equal. Differentiating the trajectories gives $m_A'(t)=1$ and $m_B'(t)=0$ marker-units per year. The model therefore gives

h_A(2)=0.015\exp(0.30\times1) =0.015\exp(0.30) \approx 0.02025\text{ per year},

while

h_B(2)=0.015\exp(0.30\times0) =0.015\text{ per year}.

The ratio from these unrounded expressions is $h_A(2)/h_B(2)=\exp(0.30)\approx1.350$. The displayed hazards are rounded, so dividing them may give a slightly different last digit. The model-based ratio is about 1.35: at year 2, subject $A$’s conditional instantaneous hazard is about 35.0% higher. This comparison isolates the slope contribution because the current levels and modeled covariates match.

Why the Slope Comparison Works

The hazard-ratio result follows directly from the assumed model. Write the model for each subject at the same time, with the same covariate vector:

h_A(t)=h_0(t)\exp\{\boldsymbol{\gamma}^{T}\mathbf{w}+\alpha_s m_A'(t)\}, \qquad h_B(t)=h_0(t)\exp\{\boldsymbol{\gamma}^{T}\mathbf{w}+\alpha_s m_B'(t)\}.

Taking the ratio cancels $h_0(t)$. The identical covariate terms also cancel, and the quotient of exponentials is the exponential of the difference:

\frac{h_A(t)}{h_B(t)} = \frac{\exp\{\boldsymbol{\gamma}^{T}\mathbf{w}+\alpha_s m_A'(t)\}} {\exp\{\boldsymbol{\gamma}^{T}\mathbf{w}+\alpha_s m_B'(t)\}} = \exp\{\alpha_s[m_A'(t)-m_B'(t)]\}.

Notice what the derivation does not require: the subjects’ marker values do not enter this slope-only hazard model, so equality of those values is not needed to obtain the ratio. Equal values are useful for the particular contrast in this tutorial because they make clear that a difference in the modeled slopes—not a difference in current levels—accounts for the hazard contrast. If the covariates differ, their contribution does not cancel, and the ratio also includes $\exp\{\boldsymbol{\gamma}^{T}(\mathbf{w}_A-\mathbf{w}_B)\}$.

Assumptions, Diagnostics, and Alternatives

The calculation is meaningful only under the model conditions used to obtain it. In particular, the marker trajectory must have a well-defined slope at the comparison time, the slope must be expressed in the units used to define $\alpha_s$, and the exponential hazard link must be an appropriate working model. The comparison also assumes the same time point, common baseline hazard, and equal modeled covariates. If the model conditions on latent trajectories, the resulting hazards are conditional on those trajectories; uncertainty in estimating them is a separate issue from the algebraic hazard ratio.

A practical diagnostic is to inspect whether the estimated trajectories are sufficiently supported near the times at which slopes are interpreted. Sparse or irregular measurements can make a derivative unstable, especially near the edge of an individual’s observed follow-up. An apparently extreme slope may reflect noisy measurements or trajectory-model misspecification rather than a reliably distinct rate of change. Also check that the sign and magnitude of $\alpha_s$ yield interpretations consistent with the fitted model and the chosen units. A unit conversion changes the numerical slope and parameter scale, even though a correctly expressed product $\alpha_s m_i'(t)$ and its implied hazard ratio remain invariant.

A common mistake is to reason, “The marker values are equal, so the hazards must be equal.” That conclusion is correct only if the hazard model depends on the current value alone (or if all other model terms, including the relevant slope contribution, happen to match). Under the slope-based model, equal levels do not remove the term $\alpha_s m_i'(t)$. The reverse mistake is to infer that different slopes must always give different hazards: if $\alpha_s=0$, or if the slopes are equal, the slope term produces no hazard contrast.

Feature	Slope-based link	Current-value link
Marker quantity in the hazard	Instantaneous slope $m_i'(t)$	Current latent value $m_i(t)$
Equal current values	Do not ensure equal hazards if slopes differ	Give equal marker contributions when the link and other terms match
Interpretation of association parameter	Multiplicative hazard change per one-unit slope increase	Multiplicative hazard change per one-unit current-value increase
Main practical concern	Reliability and units of estimated derivatives	Reliability and scale of estimated current values

These alternatives answer different scientific questions; neither is automatically preferable. A slope link is relevant when the rate of change itself is the intended association. Its limitations include sensitivity to trajectory estimation, dependence on time units, and the restriction that the modeled slope effect is linear on the log-hazard scale. A hazard ratio at a single time also does not summarize the full event-time experience. The choice of link should follow the scientific question and be supported by the data and model assessment, rather than by whether one form gives a more striking association.

Key takeaway: With a slope-based link, equal current marker values can coexist with different conditional instantaneous hazards. For equal covariates at the same time, the hazard ratio is $\exp\{\alpha_s[m_A'(t)-m_B'(t)]\}$; interpret it with the slope’s units and the model’s assumptions in view.

Check Your Understanding

Work each problem, then compare with the solutions.

At the same time, two subjects have equal covariates and slopes of $0.5$ and $0.1$ marker-units per month. If $\alpha_s=0.4$ per (marker-unit per month), what is the hazard ratio for the first subject versus the second?
Two subjects have the same current marker value and equal covariates, but their slopes are also equal. What hazard ratio does the slope-based link assign to them?
Suppose $\alpha_s=-0.2$ per (marker-unit per year), and subject $A$’s slope is 2 marker-units per year higher than subject $B$’s. Which subject has the higher conditional instantaneous hazard, and by what ratio for $A$ versus $B$?
Why is it incorrect to interpret a hazard ratio at time $t$ as the ratio of the subjects’ probabilities of having an event by time $t$?
Name one diagnostic concern specific to using a slope-based link and explain why it matters.

Solutions. (1) The slope difference is $0.5-0.1=0.4$, so the ratio is $\exp\{0.4(0.4)\}=\exp(0.16)\approx1.174$. The first subject’s conditional instantaneous hazard is about 17.4% higher. (2) The slope difference is zero, so the ratio is $\exp(0)=1$. (3) The ratio for $A$ versus $B$ is $\exp\{(-0.2)(2)\}=\exp(-0.4)\approx0.6703$. Thus $A$ has the lower hazard; equivalently, $B$’s hazard is about $1/0.6703\approx1.492$ times $A$’s. (4) A hazard is an instantaneous event rate conditional on being event-free just before that time, whereas an event probability by time $t$ accumulates over time and depends on the hazard history. (5) Derivatives may be unstable when measurements are sparse or noisy, so an extreme estimated slope may not represent a well-supported trajectory rate.

Feature	Slope-based link	Current-value link
Marker quantity in the hazard	Instantaneous slope \(m_i'(t)\)	Current latent value \(m_i(t)\)
Equal current values	Do not ensure equal hazards if slopes differ	Give equal marker contributions when the link and other terms match
Interpretation of association parameter	Multiplicative hazard change per one-unit slope increase	Multiplicative hazard change per one-unit current-value increase
Main practical concern	Reliability and units of estimated derivatives	Reliability and scale of estimated current values