Why use an asymmetric link?
A binary outcome has a conditional event probability between zero and one, while a linear predictor can take any real value. A link function connects these scales. The complementary log-log link is one such connection. Its distinctive feature is that it maps probability to predictor asymmetrically: a probability of one half does not correspond to a predictor of zero, and equal changes in the predictor do not produce equal-sized probability changes throughout the range.
This link is useful when an asymmetric response curve is appropriate for the scientific setting. The link itself does not establish that the model is suitable; the response distribution, conditional mean structure, and data collection must still support the analysis.
The link can be used for a Bernoulli response \(Y\), where \(Y=1\) denotes the event and \(p=P(Y=1\mid \text{covariates})\). It can also describe the success probability in a binomial response with a trial count. For the examples below, assume a correctly specified conditional Bernoulli model: the response is coded as zero or one, its conditional mean is \(p\), and the proposed predictor gives \(\eta\). When responses are treated as independent, that independence is also an assumption of the model. A valid probability from the inverse link does not verify these assumptions.
Inverting the link and reading the probabilities
To obtain the inverse, start from \(\eta=\log\{-\log(1-p)\}\). Exponentiating gives \(\exp(\eta)=-\log(1-p)\). Multiplying by negative one and exponentiating again yields the probability formula.
Since \(\exp(\eta)>0\), the quantity \(\exp\{-\exp(\eta)\}\) lies strictly between zero and one. Therefore the fitted event probability \(p\) also lies strictly between zero and one for every finite \(\eta\). The complement probability is \(1-p=\exp\{-\exp(\eta)\}\). Keeping these two probabilities distinct is important: the exponential term is the probability of no event, not the event probability.
Worked Example: Converting Predictors to Event and Complement Probabilities
Suppose a hypothetical binary-outcome model uses the complementary log-log link. Calculate the event probability when \(\eta=-1\), and then when \(\eta=0\).
At \(\eta=-1\), first calculate \(\exp(-1)\approx0.3679\). The complement probability is \(\exp\{-\exp(-1)\}=\exp(-0.3679)\approx0.6922\). Thus the event probability is $$ p=1-\exp\{-\exp(-1)\} \approx 1-0.6922 =0.3078. $$ So the fitted event probability is about 0.3078, and the fitted probability of no event is about 0.6922.
At \(\eta=0\), \(\exp(0)=1\), so $$ p=1-\exp(-1)\approx1-0.3679=0.6321. $$ The predictor of zero therefore corresponds to an event probability of about 0.6321, not 0.5.
The link is strictly increasing. Differentiating its inverse makes the way predictor changes affect probability explicit:
A positive change in \(\eta\) always increases the fitted event probability, but the size of that increase depends on the starting predictor. The derivative is the local change in probability for a one-unit change in \(\eta\); it is not a constant slope. In particular, it tends to zero at both extreme ends of the predictor scale.
Where the asymmetry comes from
The link treats the lower and upper probability boundaries differently. As \(\eta\) becomes very negative, \(\exp(\eta)\) approaches zero. Using the first-order approximation \(1-\exp(-x)\approx x\) for small positive \(x\), with \(x=\exp(\eta)\), gives $$ p=1-\exp\{-\exp(\eta)\}\approx\exp(\eta). $$ Thus the event probability approaches zero approximately exponentially in \(\eta\). As \(\eta\) becomes very positive, the complement probability satisfies $$ 1-p=\exp\{-\exp(\eta)\}, $$ which approaches zero especially rapidly: its negative logarithm is \(\exp(\eta)\), itself increasing exponentially. These two boundary behaviors are not mirror images of one another.
The inverse also shows why the midpoint is not centered at zero. To find the predictor for \(p=0.5\), substitute into the link: $$ g(0.5)=\log\{-\log(0.5)\} =\log(0.6931)\approx-0.3665. $$ So a predictor of about \(-0.3665\) gives an event probability of one half, whereas a predictor of zero gives about 0.6321.
Worked Example: Comparing Link-Scale Distances
Consider hypothetical target probabilities \(p=0.2\), \(p=0.5\), and \(p=0.8\). Convert each to a complementary log-log predictor, then compare the distances on the predictor scale.
For \(p=0.2\), $$ g(0.2)=\log\{-\log(0.8)\} =\log(0.2231)\approx-1.4999. $$ For \(p=0.5\), \(g(0.5)\approx-0.3665\), as calculated above. For \(p=0.8\), $$ g(0.8)=\log\{-\log(0.2)\} =\log(1.6094)\approx0.4759. $$
The predictor distance from probability 0.2 to 0.5 is \(-0.3665-(-1.4999)=1.1334\). The distance from 0.5 to 0.8 is \(0.4759-(-0.3665)=0.8424\). Equal probability increments of 0.3 therefore do not correspond to equal predictor increments. This is one numerical way to see the link’s asymmetry.
The same feature can be seen on the probability scale. For example, increasing \(\eta\) from \(-1\) to \(0\) changes the fitted probability from about 0.3078 to 0.6321, an increase of about 0.3243. Increasing it from \(0\) to \(1\) gives a probability of about 0.9340 at \(\eta=1\), an increase of about 0.3019. These are equal one-unit predictor changes but unequal probability changes. Differences depend on the starting point and should be computed from the inverse link rather than inferred from the coefficient alone.
Worked Example: Checking a Probability Difference
Suppose two hypothetical covariate patterns have complementary log-log predictors \(\eta_0=-1.5\) and \(\eta_1=-0.5\). Find their fitted event probabilities and the difference.
For the first pattern, \(\exp(-1.5)\approx0.2231\), giving $$ p_0=1-\exp\{-\exp(-1.5)\} \approx1-\exp(-0.2231) \approx1-0.8000 =0.2000. $$
For the second pattern, \(\exp(-0.5)\approx0.6065\), so $$ p_1=1-\exp\{-\exp(-0.5)\} \approx1-\exp(-0.6065) \approx1-0.5452 =0.4548. $$ The fitted probability difference is \(p_1-p_0\approx0.4548-0.2000=0.2548\), or about 25.48 percentage points. This is a difference between model-based fitted risks for the two specified patterns; it is not by itself evidence that changing a covariate would cause an individual’s risk to change.
Model assumptions, diagnostics, and alternatives
A complementary log-log model specifies a conditional mean through \(\log\{-\log(1-p)\}=\eta\). In a regression model, \(\eta\) is usually defined by a chosen predictor, such as an intercept plus covariate terms. The resulting probabilities are valid, but validity alone is not evidence that this predictor form or link is appropriate. Check that the response coding matches the intended event, that the conditional response distribution is suitable, and that dependence or clustering has not been ignored. Examine fitted probabilities and residual diagnostics appropriate to the response model, and look for systematic discrepancies across covariate ranges or groups.
The logit and probit links are common alternatives for binary probabilities. Each maps the open probability interval to the real line, but they impose different response curves. The logit link is centered at zero when \(p=0.5\); the complementary log-log link is not, and its mapping is asymmetric. The probit link uses normal quantiles. These distinctions guide model choice, but no link should be selected merely because its inverse produces valid probabilities. Comparing fitted behavior and checking whether the scientific assumptions are plausible remain necessary.
| Feature | Complementary log-log link | Logit link |
|---|---|---|
| Link expression | \(\log\{-\log(1-p)\}\) | \(\log\{p/(1-p)\}\) |
| Predictor corresponding to \(p=0.5\) | About \(-0.3665\) | Zero |
| Response-curve shape | Asymmetric around the probability midpoint | Symmetric around \(p=0.5\) |
A common calculation mistake is to confuse the event probability with its complement. At \(\eta=-1\), \(\exp\{-\exp(-1)\}\approx0.6922\) is \(1-p\). The event probability is \(1-0.6922\approx0.3078\). Another mistake is to assume that \(\eta=0\) means \(p=0.5\), as it does for the logit link. Here it gives \(p=1-\exp(-1)\approx0.6321\). Finally, probability outputs within zero and one do not prove that the link, predictor, or independence assumptions fit the data.
Check Your Understanding
Answer each question, then compare your work with the solutions.
- What is the fitted event probability at \(\eta=-1\), and what is the fitted probability of no event?
- Calculate the complementary log-log predictor for \(p=0.5\). Does it equal zero?
- Calculate the fitted event probability at \(\eta=0\).
- In one sentence, describe what happens to the event probability as \(\eta\) becomes very negative.
- Why does a fitted probability between zero and one not establish that a complementary log-log model is appropriate?
Solutions. (1) The complement is \(\exp\{-\exp(-1)\}=\exp(-0.3679)\approx0.6922\), so the event probability is \(1-0.6922\approx0.3078\). (2) \(g(0.5)=\log\{-\log(0.5)\}=\log(0.6931)\approx-0.3665\), not zero. (3) \(1-\exp\{-\exp(0)\}=1-\exp(-1)\approx0.6321\). (4) As \(\eta\) becomes very negative, the event probability approaches zero and is approximately \(\exp(\eta)\). (5) The inverse link guarantees probabilities in the allowable range, but does not verify the response distribution, predictor specification, independence assumptions, or suitability of the link.