Asymmetric Probability Changes Under Complementary Log-Log

Why a Predictor Shift Does Not Mean a Fixed Risk Change

In a binary regression model, a change in a covariate changes the linear predictor. What that change means on the probability scale depends on the link. Under the complementary log-log link, an equal shift in the predictor multiplies the transformed event risk by a constant; it does not add a constant to the event probability. Consequently, the same shift can produce different probability changes at different starting risks.

This tutorial focuses on that probability-scale asymmetry. We will calculate finite changes at low and high baseline risks, compare upward and downward shifts from the same baseline, and derive how the local probability slope depends on the baseline. The calculations use the inverse link directly; no approximation for rare events is needed.

Definition: For an event probability \(p\) with \(0<p<1\), the complementary log-log link is \(g(p)=\log[-\log(1-p)]\). Its inverse link is \(p=1-\exp\{-\exp(\eta)\}\), where \(\eta=g(p)\) is the linear predictor. A predictor shift of size \(\delta\) changes \(\eta\) to \(\eta+\delta\).

It is useful to write \(H=-\log(1-p)\), a positive transformation of the event probability. Then the link says \(\log(H)=\eta\), so \(H=\exp(\eta)\). A shift of \(\delta\) therefore changes \(H\) to \(\exp(\delta)H\). The resulting probability change depends on the baseline value of \(H\), and hence on the baseline probability.

The Exact Risk Change Under a Predictor Shift

Let \(p_0\) be the baseline probability and \(p_1\) the probability after shifting the predictor by \(\delta\). Since \(\exp(\eta_0)=-\log(1-p_0)\), the inverse link gives an exact expression for \(p_1\):

Result: Under the complementary log-log link, a predictor shift of \(\delta\) transforms a baseline probability \(p_0\) into \(p_1=1-(1-p_0)^{\exp(\delta)}\). This formula holds for every \(0<p_0<1\) and every real \(\delta\).

To verify the expression, set \(H_0=-\log(1-p_0)=\exp(\eta_0)\). After the shift, \(\exp(\eta_1)=\exp(\eta_0+\delta)=\exp(\delta)H_0\). Thus

\begin{aligned} p_1 &=1-\exp\{-\exp(\eta_1)\}\\ &=1-\exp\{-\exp(\delta)H_0\}\\ &=1-\exp\{\exp(\delta)\log(1-p_0)\}\\ &=1-(1-p_0)^{\exp(\delta)}. \end{aligned}

This is an exact finite-shift calculation. When \(\delta>0\), \(\exp(\delta)>1\), so \(p_1>p_0\): the event probability rises. When \(\delta<0\), \(\exp(\delta)<1\), so \(p_1<p_0\): the probability falls. But the size of that rise or fall is not determined by \(\delta\) alone. It also depends on \(p_0\).

Worked Example: A Positive Shift From a Low Baseline Risk

Suppose the baseline event probability is \(p_0=0.05\), and a covariate change raises the linear predictor by \(\delta=0.7\). Calculate the new risk and its change.

The shift multiplier is \(\exp(0.7)\approx2.013752707\). The baseline transformed risk is \(-\log(1-0.05)=-\log(0.95)\approx0.051293294\). After the shift, it is

0.051293294(2.013752707)\approx0.103292010.

Applying the inverse link gives \(p_1=1-\exp(-0.103292010)\approx0.098136\). The probability rises by approximately \(0.098136-0.05=0.048136\), or 4.81 percentage points. The predictor shift is \(0.7\), but the probability change is about \(0.048\), not \(0.7\).

Equal Shifts Can Give Unequal Probability Changes

The exact formula makes the baseline dependence concrete. A shift of \(0.7\) multiplies the transformed risk by about \(2.014\), whether the baseline probability is small or large. But transforming those different baseline values back to probabilities can give quite different changes.

Worked Example: The Same Positive Shift at a Higher Baseline

Now use the same predictor shift, \(\delta=0.7\), with baseline risk \(p_0=0.8\). The baseline transformed risk is \(-\log(0.2)\approx1.609437912\). The shifted value is

1.609437912(2.013752707)\approx3.241010.

Therefore \(p_1=1-\exp(-3.241010)\approx0.960876\). The increase is approximately \(0.960876-0.8=0.160876\), or 16.09 percentage points. This is much larger than the 4.81 percentage-point increase from the baseline risk of \(0.05\), even though the predictor shift is identical.

A very high baseline illustrates why the pattern is not simply “higher baseline means larger increase.” At \(p_0=0.99\), the same shift gives \(-\log(0.01)\exp(0.7)\approx9.2737\), and hence \(p_1\approx1-\exp(-9.2737)\approx0.999906\). The increase is only about \(0.009906\), or 0.99 percentage points. The probability is already close to its upper bound of one.

The response to a predictor shift is therefore not a constant probability difference, and its size does not increase monotonically across all baseline probabilities. It can be small near either boundary of the probability range and larger at intermediate probabilities. A derivative calculation makes this pattern precise.

Result: Write \(q=\exp(\eta)>0\), so \(p=1-\exp(-q)\). The local probability slope with respect to the predictor is \(\frac{dp}{d\eta}=q\exp(-q)\). This slope is maximized at \(q=1\), corresponding to baseline probability \(p=1-\exp(-1)\approx0.6321\).

To prove the result, differentiate the inverse link. Since \(dq/d\eta=q\),

\frac{dp}{d\eta} =\frac{d}{d\eta}\{1-\exp(-q)\} =q\exp(-q).

For \(q>0\), the slope is positive. To find its maximum, differentiate \(q\exp(-q)\) with respect to \(q\):

\frac{d}{dq}\{q\exp(-q)\} =\exp(-q)(1-q).

Because \(\exp(-q)>0\), this derivative is positive for \(q<1\), zero at \(q=1\), and negative for \(q>1\). Thus the slope increases up to \(q=1\) and decreases thereafter. At that point \(p=1-\exp(-1)\), proving the stated maximum. \(\square\)

This result concerns a local slope: the approximate probability change per small unit of predictor change at a particular baseline. It does not say that the finite change for every fixed nonzero \(\delta\) is largest exactly at \(p=1-\exp(-1)\). A finite change evaluates the curve across an interval, not just at its starting point.

Worked Example: Opposite Shifts From the Same Baseline

Take baseline risk \(p_0=0.5\), and compare predictor shifts of \(+0.7\) and \(-0.7\). The baseline transformed risk is \(-\log(0.5)=\log 2\approx0.693147181\).

For the positive shift, \(\exp(0.7)\approx2.013752707\), so the shifted transformed risk is

0.693147181(2.013752707)\approx1.395827012.

Thus \(p_+=1-\exp(-1.395827012)\approx0.752372\), an increase of about \(0.252372\).

For the negative shift, \(\exp(-0.7)\approx0.496585304\), giving

0.693147181(0.496585304)\approx0.3442067.

Thus \(p_-=1-\exp(-0.3442067)\approx0.291218\), a decrease of about \(0.208782\). The two shifts have equal magnitude on the predictor scale, but their probability-scale changes are not equal in magnitude. The link transformation is nonlinear, so equal and opposite predictor shifts need not be symmetric around the baseline probability.

Interpreting a Covariate Shift

If the linear predictor contains a term \(\beta x\), increasing \(x\) by one unit while holding the other covariates fixed changes the predictor by \(\delta=\beta\). The exact risk after that change is therefore \(1-(1-p_0)^{\exp(\beta)}\). For a change of \(c\) units, use \(\delta=c\beta\). These statements describe a conditional comparison at the specified values of the other covariates; they do not assert that every observation has the same baseline risk.

The discrete-time hazard interpretation of the complementary log-log link gives another way to understand the transformed scale: \(-\log(1-p)\) is multiplied by \(\exp(\delta)\). The probability itself is recovered through \(p=1-\exp(-H)\), which is nonlinear and bounded between zero and one. That recovery is why a constant multiplicative change on the transformed scale becomes a baseline-dependent change in risk.

Common Mistakes

Calling \(\delta\) a risk difference. A predictor shift is on the link scale. Compute \(p_1-p_0\) using the inverse link before reporting a probability difference.
Using the odds-ratio interpretation. Under the complementary log-log link, \(\exp(\delta)\) multiplies \(-\log(1-p)\); it is not generally an odds ratio.
Assuming equal and opposite shifts give equal probability changes. The inverse link is nonlinear. As the example at \(p_0=0.5\) shows, the rise and fall differ in magnitude.
Confusing a local slope with a finite change. The derivative describes the response to an infinitesimal predictor change at a point. For a specified nonzero shift, use the exact finite-shift formula.
Assuming probability changes grow continuously with baseline risk. The local slope eventually decreases at high risks, and probabilities near one have limited room to rise.

Key takeaway: A complementary log-log predictor shift \(\delta\) multiplies the transformed risk \(-\log(1-p)\) by \(\exp(\delta)\). The exact probability after the shift is \(1-(1-p_0)^{\exp(\delta)}\), so probability changes depend on the baseline and can be asymmetric for equal positive and negative shifts.

Check Your Understanding

Work through these questions using the exact shift formula and the inverse link.

Starting from baseline risk \(p_0=0.1\), write the exact expression for the risk after a predictor shift of \(0.4\). Is the new risk larger or smaller than \(0.1\)?
Explain why \(\exp(\delta)\) is not, in general, an odds ratio under the complementary log-log link.
For \(p_0=0.5\), compare the probability changes from shifts \(+0.7\) and \(-0.7\). Why are their magnitudes different?
At what value of \(q=\exp(\eta)\) is the local probability slope maximized, and what baseline probability does that correspond to?
In a model with linear predictor containing \(\beta x\), what predictor shift corresponds to increasing \(x\) by three units while holding other covariates fixed?