From a Marker Rate to a Hazard Ratio
A slope-based association asks whether the rate at which a longitudinal marker is changing is related to the event hazard. The coefficient is not itself a hazard ratio: it is a multiplier on the marker’s rate of change in the log hazard. To interpret it, specify two rates, take their difference, multiply by the coefficient, and exponentiate.
Let \(m_i(t)\) denote the instantaneous slope of subject \(i\)'s latent marker trajectory at time \(t\). If the marker is measured in points and time in years, \(m_i(t)\) is measured in points per year. A slope-linked hazard model can be written as
Here \(h_i(t)\) is the hazard for subject \(i\) at time \(t\), \(h_0(t)\) is the baseline hazard, \(\mathbf{w}_i\) is a vector of other covariates, \(\boldsymbol{\gamma}\) is their coefficient vector, and \(\alpha_s\) is the slope-association coefficient. The slope is the latent trajectory’s rate of change, not the change in a single noisy observed measurement.
The formula follows by dividing the two hazards. Their common baseline hazard cancels, as do the other covariate terms when those terms are held equal. The hazard ratio is therefore determined by the slope difference, not by either slope in isolation. A positive coefficient means a larger slope is associated with a larger hazard, conditional on the model’s other terms; it does not by itself imply that a marker increase causes an event.
Assumptions and Interpretation
This interpretation relies on the stated model form and several conditions. The event hazard is modeled as proportional to the exponential of the linear predictor; the slope enters linearly with a coefficient \(\alpha_s\) that does not vary with time in this model. The latent trajectory must have a defined slope at the time of comparison, and the slope’s units must be specified. The hazard ratio compares subjects at the same time and with the other covariates held fixed. If a model includes additional association terms, their contributions must also be held fixed for this slope-only comparison.
The coefficient is most interpretable through a contrast. For a slope difference \(\Delta m=m_A(t)-m_B(t)\), the log hazard ratio is \(\alpha_s\Delta m\). Thus, \(\alpha_s\) describes the change in log hazard per one-unit increase in slope, where that unit is one marker-unit per one time-unit. A slope difference of zero gives a hazard ratio of one, regardless of the coefficient value.
A confidence interval for \(\alpha_s\) can be converted to an interval for a specified hazard ratio by multiplying both coefficient limits by the slope contrast and exponentiating. If the contrast is negative, multiplying reverses the order of the limits, so the transformed lower and upper limits must be reordered. The interval quantifies uncertainty under the fitted model; it is not a probability statement about a fixed parameter.
Worked Example: Comparing an Increasing and a Declining Marker
Suppose a hypothetical model uses marker units of points and time in years. Its estimated slope-association coefficient is \(\hat{\alpha}_s=0.18\) year per point, with standard error \(SE=0.07\) year per point. At a given time, compare a trajectory increasing at \(1.0\) point per year with one declining at \(0.5\) point per year. Assume the other covariates are equal.
The slope contrast is \(1.0-(-0.5)=1.5\) points per year. The units cancel in the product with the coefficient: \((0.18\ \text{year per point})(1.5\ \text{points per year})=0.27\). Therefore the estimated hazard ratio is
Under this model, at the specified time and conditional on the other terms, the increasing trajectory has an estimated hazard 1.310 times that of the declining trajectory, or about 31.0% higher. This is a comparison of hazards, not a comparison of event probabilities.
For an approximate 95% Wald interval for the coefficient, calculate \(0.18\pm1.96(0.07)=0.18\pm0.1372\), giving limits \(0.0428\) and \(0.3172\) year per point. Because the slope contrast is positive, the corresponding log-hazard-ratio limits are \(0.0428(1.5)=0.0642\) and \(0.3172(1.5)=0.4758\). Exponentiating gives an approximate 95% interval of \([1.066,1.609]\) for the hazard ratio. The interval is compatible with a modest increase through a substantially larger increase; it does not establish that the true hazard ratio has a 95% probability of lying in these limits.
This calculation assumes that the coefficient’s sampling distribution is adequately approximated by a normal distribution for the Wald interval and that the model’s slope-link form is appropriate. It uses the same fixed contrast throughout; it does not account for uncertainty in an estimated slope contrast.
Why the Coefficient Changes with Time Units
The numerical value of \(\alpha_s\) depends on the units used to express the slope. Changing years to months changes the numerical slope even though the physical trajectory is unchanged. The coefficient must change in the opposite direction so that their product, and hence the model’s linear predictor, remains unchanged.
For a marker measured per year, let \(m_{\mathrm{year}}\) be its numerical slope in marker-units per year. The same rate expressed per month is \(m_{\mathrm{month}}=m_{\mathrm{year}}/12\), because a year contains 12 months. Invariance of the slope contribution requires \(\alpha_{s,\mathrm{year}}m_{\mathrm{year}}=\alpha_{s,\mathrm{month}}m_{\mathrm{month}}\). Substitution gives the conversion rule:
The units confirm the direction: the year-based coefficient has units years per marker-unit, while the month-based coefficient has units months per marker-unit. The value becomes 12 times as large when expressed in months. More generally, if one time-unit contains \(c\) of the new time-units, then the numerical slope per new unit is the old numerical slope divided by \(c\), and the coefficient is multiplied by \(c\).
Worked Example: Converting a Year-Based Coefficient to Months
Suppose a hypothetical model with marker units of mmol/L and time in years has \(\hat{\alpha}_{s,\mathrm{year}}=-0.24\) year per (mmol/L). Convert it to a model expressing time in months, then compare a rate of \(1.5\) mmol/L per year with a rate of zero.
The converted coefficient is \(\hat{\alpha}_{s,\mathrm{month}}=12(-0.24)=-2.88\) months per (mmol/L). The rate \(1.5\) mmol/L per year is \(1.5/12=0.125\) mmol/L per month. Relative to a zero slope, the year-based calculation gives
Using months gives the same result: \(\exp\{(-2.88)(0.125-0)\}=\exp(-0.36)=0.6977\). Thus, under this hypothetical model, the specified increasing rate is associated with an estimated hazard about 0.698 times the hazard at zero slope, holding other terms fixed. The negative coefficient means that larger slopes have smaller modeled hazards; it does not establish a causal protective effect.
The unit check is essential: \((-0.24\ \text{year per (mmol/L)})(1.5\ \text{mmol/L per year})\) is dimensionless. The converted product, \((-2.88\ \text{months per (mmol/L)})(0.125\ \text{mmol/L per month})\), is also dimensionless.
Worked Example: Converting a Coefficient to Weeks
A hypothetical model has a slope coefficient of \(0.15\) year per marker-unit when time is measured in years. Express the coefficient in weeks per marker-unit, and calculate the hazard ratio for a slope of \(2\) marker-units per year versus zero. Use 52 weeks per year.
One year is 52 weeks, so the coefficient in weeks is \(52(0.15)=7.80\) weeks per marker-unit. The slope of \(2\) marker-units per year is \(2/52=0.03846\) marker-units per week. The year-based log hazard ratio is \(0.15(2)=0.30\), giving \(\exp(0.30)=1.350\). With weeks, it is \(7.80(0.03846)=0.3000\), again giving \(\exp(0.3000)=1.350\).
The numerical coefficient is larger in weeks because the numerical rate per week is smaller than the rate per year. The hazard ratio does not change: it describes the same physical contrast under two unit systems.
Diagnostics, Alternatives, and Limitations
A basic diagnostic is to compute a hazard ratio using two equivalent time-unit representations. If the results differ beyond rounding, check whether the slope and coefficient were both converted consistently. Also inspect whether the proposed slope contrasts are plausible over the time range in which subjects are observed. A coefficient can produce extreme hazard ratios for implausibly large contrasts even when the model’s fitted behavior over the observed range is moderate.
The slope link imposes a particular interpretation: subjects with the same current marker value but different rates of change can have different hazards through the slope term. This differs from the current-value association, which relates the hazard to the latent marker level. Which association is scientifically useful depends on whether the question concerns the marker’s present level or its rate of change. A slope model can also be sensitive to how well the longitudinal data identify individual trajectory rates; noisy or sparse measurements may make those rates uncertain.
The model’s linear slope contribution is a limitation. It implies that equal increments in slope have equal effects on log hazard, and it may be inadequate if the association changes with the rate or over time. Further, the estimated association is conditional on the model specification and measured covariates; it is not automatically a causal effect. Diagnostics should therefore consider the trajectory fit, the event model’s fit, uncertainty in the slope association, and whether the chosen slope contrasts are substantively meaningful.
- Unit mismatch: A slope in marker-units per year must be multiplied by a coefficient in years per marker-unit. A coefficient in marker-units per year has the wrong units.
- Reversing the conversion: When changing years to months, multiply the coefficient by 12 and divide the numerical slope by 12. Doing both in the same direction changes the model contribution.
- Interpreting the coefficient alone: The coefficient is not a hazard ratio. A hazard ratio requires a specified slope contrast and exponentiation.
- Overstating the result: A modeled association between slope and hazard does not demonstrate that changing the marker’s rate would cause a change in event risk.
Check Your Understanding
Try each problem, then compare your reasoning with the brief solutions below.
- A coefficient is \(0.10\) year per point. Calculate the hazard ratio for a slope of \(3\) points per year versus a slope of \(1\) point per year.
- A coefficient is \(-0.20\) year per (mmol/L). What is its value in months per (mmol/L)?
- With the month-based coefficient from question 2, calculate the hazard ratio for \(0.5\) mmol/L per year versus zero slope.
- Explain why the coefficient’s units are the reciprocal of the slope’s units.
- A colleague converts a year-based coefficient to months by dividing it by 12 but also divides the slope by 12. What is wrong with this procedure?
Solutions. (1) The contrast is \(3-1=2\) points per year, so \(\mathrm{HR}=\exp\{0.10(2)\}=\exp(0.20)=1.221\). (2) Multiply by 12: \(-0.20(12)=-2.40\) months per (mmol/L). (3) The rate is \(0.5/12=0.04167\) mmol/L per month, so \(\mathrm{HR}=\exp\{(-2.40)(0.04167)\}=\exp(-0.10)=0.9048\). (4) The product of the coefficient and slope is a log hazard ratio, which is dimensionless; therefore their units must cancel. (5) The coefficient should be multiplied by 12 when changing years to months, while the numerical slope is divided by 12. Dividing both makes the product too small by a factor of 144 rather than preserving it.