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Association Structures · Tutorial 179 of 697

Centering and Scaling Current-Value Associations

Learn to re-express a current-value association after centering or scaling a marker without changing the model’s fitted hazard contrasts.

Advanced 10 min read

What You'll Learn

  • Derive the coefficient and intercept transformations for a centered and scaled marker.
  • Interpret the transformed coefficient per chosen marker increment.
  • Verify that hazard ratios between subjects or marker values remain unchanged.
  • Distinguish marker re-expression from a change in the underlying association model.
  • Identify when a change in units, baseline parameterization, or estimation procedure can affect the transformation.

Why Re-express a Current-Value Marker?

A current-value association is often easier to interpret when the marker is centered at a meaningful value or scaled to a familiar increment. For example, an association per one standard deviation may be more interpretable than an association per one unit when one unit is very small. Centering can also make the hazard intercept refer to a subject whose marker is at a clinically relevant value. These changes of reference and units alter numerical coefficients, but—when performed consistently—they do not alter the fitted hazard contrasts.

Let \(m_i(t)\) denote subject \(i\)'s latent marker value at time \(t\), and consider a shared-random-effects model with a current-value term:

$$ h_i(t)=h_0(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha m_i(t)\right\}. $$

Here \(h_i(t)\) is the subject-specific hazard conditional on the latent trajectory and shared random effects, \(h_0(t)\) is the baseline hazard, \(w_i\) is a vector of other covariates, and \(\alpha\) is the current-value association coefficient. The coefficient is on the log-hazard scale per one marker unit. The model links the latent marker value, not a noisy observed measurement at a particular visit, to the hazard.

Definition: For a chosen center \(c\) and positive scale \(s\), define the re-expressed marker by \(m_i^*(t)=(m_i(t)-c)/s\). The center \(c\) is expressed in the original marker units, and \(s\) is the size of one new marker unit in original units. Thus, \(m_i^*(t)=0\) corresponds to an original marker value of \(c\), and a one-unit increase in \(m_i^*(t)\) corresponds to an increase of \(s\) original marker units.

Coefficient and Intercept Transformation

Since \(m_i(t)=c+s m_i^*(t)\), substitution into the hazard model gives

$$ h_i(t)=h_0(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha c+\alpha s\,m_i^*(t)\right\}. $$

The transformed association coefficient is therefore \(\alpha^*=s\alpha\). The constant term \(\alpha c\) can be absorbed into the baseline hazard by defining \(h_0^*(t)=h_0(t)\exp(\alpha c)\). Equivalently, if the log-hazard has a constant intercept \(\gamma_0\), the transformed intercept is \(\gamma_0^*=\gamma_0+\alpha c\). The transformed model is then

$$ h_i(t)=h_0^*(t)\exp\left\{\gamma^\mathsf{T}w_i+\alpha^*m_i^*(t)\right\}. $$

This is an algebraic reparameterization: the transformed marker is a one-to-one affine function of the original marker when \(s>0\), and the parameter changes compensate exactly for that re-expression. The other covariate coefficients remain the same. If the baseline hazard is left in its original form rather than absorbing the constant, the factor \(\exp(\alpha c)\) must remain explicitly in the linear predictor.

Key result: Centering at \(c\) changes the constant part of the hazard predictor but does not change the association coefficient. Scaling by \(s\) changes the coefficient from \(\alpha\) to \(s\alpha\). Together, these changes preserve each fitted hazard and every hazard contrast, provided the constant shift is represented in the baseline hazard or intercept.

Why Hazard Contrasts Are Preserved

At a fixed time, compare two subjects with the same values of \(w_i\), but latent marker values \(m_1(t)\) and \(m_2(t)\). In the original parameterization, their hazard ratio is

$$ \frac{h_1(t)}{h_2(t)} =\exp\left\{\alpha\bigl(m_1(t)-m_2(t)\bigr)\right\}. $$

The shared baseline hazard and the constant centering shift cancel in this comparison. In the transformed parameterization, \(m_1^*(t)-m_2^*(t)=(m_1(t)-m_2(t))/s\), so the contrast is

$$ \exp\left\{\alpha^*\bigl(m_1^*(t)-m_2^*(t)\bigr)\right\} =\exp\left\{s\alpha\frac{m_1(t)-m_2(t)}{s}\right\} =\exp\left\{\alpha\bigl(m_1(t)-m_2(t)\bigr)\right\}. $$

Thus, \(\exp(\alpha)\) is the hazard ratio for a one-unit marker increase in the original scale, while \(\exp(\alpha^*)=\exp(s\alpha)\) is the hazard ratio for an increase of \(s\) original units. These are different increments, not conflicting estimates of the same increment.

Worked Example: Centering and Scaling to Standard-Deviation Units

Suppose a hypothetical fitted model has \(\alpha=0.35\) per mg/L. Re-express the latent marker using \(c=4\) mg/L and \(s=2.5\) mg/L. Calculate the transformed coefficient and compare the hazard ratio for a 2.5 mg/L difference.

The transformed marker is \(m_i^*(t)=(m_i(t)-4)/2.5\). Its association coefficient is

$$ \alpha^*=s\alpha=(2.5)(0.35)=0.875 \quad\text{per transformed marker unit.} $$

The original-scale hazard ratio for a 2.5 mg/L increase is \(\exp\{(0.35)(2.5)\}=\exp(0.875)\approx2.399\). In the transformed scale, the same increase is one unit, giving \(\exp(\alpha^*)=\exp(0.875)\approx2.399\). The two calculations agree.

For a concrete contrast, marker values of 5.5 and 8 mg/L transform to \(m_1^*=0.6\) and \(m_2^*=1.6\). Their difference is one transformed unit, so the hazard ratio is \(\exp\{0.875(1.6-0.6)\}\approx2.399\). In the original scale the difference is 2.5 mg/L, giving the same result. The center changes the reference point, but not this contrast.

Centering Changes the Reference, Not the Slope

With centering alone, \(s=1\). The coefficient remains \(\alpha\), because a one-unit difference in the centered marker is still a one-unit difference in the original marker. The intercept or baseline hazard changes so that its reference now corresponds to the chosen center. This can improve interpretation: the hazard intercept is associated with marker value \(c\), rather than marker value zero. It does not make the association stronger or weaker.

This change of reference is distinct from a baseline-referenced current-value association. In that model, the hazard association involves a difference between the current latent value and a baseline latent value. Merely centering the marker subtracts a fixed constant from every value; it does not subtract each subject’s own baseline value.

Worked Example: Interpreting a Centered Intercept

Consider the hypothetical log-hazard model \(\log h_i(t)=-4+0.20m_i(t)\), with no other covariates and with marker values measured in the stated units. Center the marker at \(c=6\), defining \(m_i^*(t)=m_i(t)-6\). Find the transformed intercept and verify the log hazard at marker values 6 and 9.

Here \(s=1\), so the coefficient stays \(0.20\). The transformed intercept is

$$ \gamma_0^*=-4+\alpha c=-4+(0.20)(6)=-2.8. $$

At \(m_i(t)=6\), the centered value is zero. The original predictor is \(-4+(0.20)(6)=-2.8\); the transformed predictor is \(-2.8+(0.20)(0)=-2.8\). At \(m_i(t)=9\), the centered value is 3. The original predictor is \(-4+(0.20)(9)=-2.2\); the transformed predictor is \(-2.8+(0.20)(3)=-2.2\). The hazard ratio comparing 9 with 6 is \(\exp\{0.20(9-6)\}=\exp(0.6)\approx1.822\) in either parameterization.

The transformed intercept is the log hazard when the marker is 6, while the original intercept is the log hazard when the marker is zero. Neither intercept alone is the association effect.

Changing Units and Reporting Uncertainty

A marker can also be rescaled simply by changing its units. For example, if a new numerical marker value is one hundredth of the old value, then one new unit represents one hundred old units. The coefficient must adjust inversely to the size of the new unit. More generally, if \(x^*=a+bx\), where \(b>0\), the coefficient on \(x^*\) is \(\alpha/b\). This is the same rule as above with \(a=-c/s\) and \(b=1/s\).

Worked Example: Changing Marker Units

Suppose, hypothetically, \(\alpha=0.08\) per mg/dL, and define a new marker \(q=0.01m\), so \(q\) is measured in g/L. Find the coefficient per g/L and the hazard ratio for a 10 mg/dL increase.

Since \(m=100q\), the association term becomes \(0.08m=0.08(100q)=8q\). Therefore the coefficient is \(8\) per g/L. A 10 mg/dL increase is \(0.1\) g/L, so its hazard ratio on the new scale is \(\exp\{8(0.1)\}=\exp(0.8)\approx2.226\). On the original scale it is \(\exp\{0.08(10)\}=\exp(0.8)\approx2.226\). The large-looking coefficient of 8 reflects the smaller numerical unit, not a different fitted association.

For a fixed positive \(s\), uncertainty transforms along with the coefficient: \(SE(\alpha^*)=sSE(\alpha)\), and a confidence interval with endpoints \(L\) and \(U\) for \(\alpha\) becomes \((sL,sU)\) for \(\alpha^*\). A test of \(H_0:\alpha=0\) has the same test statistic and p-value after this positive rescaling, since both estimate and standard error are multiplied by \(s\). If the intercept is also transformed, its uncertainty must account for its covariance with \(\alpha\): for fixed \(c\), \(\gamma_0^*=\gamma_0+c\alpha\), so \(\operatorname{Var}(\gamma_0^*)=\operatorname{Var}(\gamma_0)+c^2\operatorname{Var}(\alpha)+2c\operatorname{Cov}(\gamma_0,\alpha)\).

Conditions, Diagnostics, and Limitations

The exact invariance depends on a consistent affine transformation of the latent marker and its longitudinal model. In particular, the same \(c\) and \(s\) must be used at every time and for every subject; the transformed trajectory must represent \((m_i(t)-c)/s\); and the hazard model must include the corresponding transformed coefficient and intercept or baseline-hazard factor. Under these conditions, each subject’s linear predictor, and therefore fitted hazard, is unchanged.

If \(c\) and \(s\) are calculated from the observed data, the algebraic identities still hold for the resulting values. But when reporting inference, make clear whether they are treated as fixed reference constants or whether their estimation uncertainty is part of the inferential target. Reparameterizing a likelihood does not by itself ensure that every estimation procedure is invariant: constraints, priors, or penalties that are not transformed to match the new parameterization can lead to different estimates.

Useful checks include verifying that the transformed marker equals \((m_i(t)-c)/s\) in the longitudinal model, that the transformed coefficient equals \(s\alpha\), and that the transformed intercept includes \(\alpha c\). For selected marker values, calculate the linear predictor both before and after transformation. A mismatch signals an omitted intercept shift, an inconsistent unit conversion, or a different model rather than a harmless change of scale.

The main alternatives are to retain the original scale, center only, or scale to a convenient increment such as one standard deviation. Retaining the original scale can make units directly interpretable; centering can make the intercept meaningful; scaling can make coefficient magnitudes easier to compare across marker units. None is universally preferable. The choice changes presentation, not the underlying hazard contrasts, when the transformation is handled correctly. It does not correct a misspecified trajectory or establish that the marker causes the event.

Common Mistake

A common error is to multiply the coefficient by \(s\) and then report \(\exp(s\alpha)\) as the hazard ratio for a one-unit increase in the original marker. That interpretation is wrong: the transformed coefficient is per one unit of \(m_i^*\), which equals \(s\) original units. The original one-unit hazard ratio remains \(\exp(\alpha)\); the transformed one-unit hazard ratio is \(\exp(s\alpha)\). Another error is to center the marker but leave the intercept unchanged. That omission shifts every fitted log hazard by \(\alpha c\), so the resulting model is no longer equivalent.

Check Your Understanding

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

  1. A model has \(\alpha=0.12\) per unit. The marker is centered at \(c=5\) and scaled by \(s=4\). What is the transformed coefficient, and what original-scale increase does one transformed unit represent?
  2. Explain why centering alone does not change the current-value coefficient, but generally does change the intercept or baseline hazard.
  3. For \(\alpha=0.25\), compare the hazard ratio for a 2-unit marker increase before and after expressing the marker in units of 2 original units.
  4. State two checks that can help detect an incorrect reparameterization.

Solutions. (1) \(\alpha^*=s\alpha=(4)(0.12)=0.48\) per transformed unit; one transformed unit represents a 4-unit increase in the original marker. (2) Subtracting a constant leaves differences between marker values unchanged, so the coefficient per unit difference is unchanged. But the constant contribution \(\alpha c\) moves into the intercept or baseline hazard. (3) The original hazard ratio is \(\exp\{(0.25)(2)\}=\exp(0.5)\approx1.649\). With \(s=2\), the transformed coefficient is \(0.50\), and the increase is one transformed unit, giving \(\exp(0.50)\approx1.649\). (4) Verify that the transformed trajectory equals \((m_i(t)-c)/s\), and compare original and transformed linear predictors for selected marker values, including the intercept shift \(\alpha c\).

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