Log-Log Links For Opposite-Sided Asymmetry

Why Use the Log-Log Link?

Binary-response links need not change probabilities symmetrically around the middle of the probability scale. The complementary log-log link gives a particularly rapid approach to probability one at one end of its predictor scale. The log-log link reverses which end has that rapid approach: it approaches zero especially quickly on one side, while its approach to one is comparatively gradual.

This difference matters when a binary response has an asymmetric pattern across covariate values. The link determines how a linear predictor translates into probabilities; it does not make a coefficient a constant probability difference. We will define the log-log convention used here, derive its inverse and slope, and compare its limiting behavior with complementary log-log regression.

Definition and Inverse Link

Let $Y$ be a Bernoulli response with conditional mean $\mu=P(Y=1\mid x)$, where $0<\mu<1$. In a binary generalized linear model, write the linear predictor as $\eta=x^\mathsf{T}\beta$. The log-log link used in this tutorial is

g(\mu)=-\log[-\log(\mu)].

The minus sign outside the logarithm is part of the convention. It makes the link increase with $\mu$, so a positive linear-predictor change corresponds to a higher fitted probability.

Definition: The log-log link is $g(\mu)=-\log[-\log(\mu)]$, for $0<\mu<1$. Its inverse-link function is $$ g^{-1}(\eta)=\mu=\exp[-\exp(-\eta)],\qquad -\infty<\eta<\infty. $$

To obtain the inverse, set $\eta=-\log[-\log(\mu)]$. Then $-\eta=\log[-\log(\mu)]$, so $\exp(-\eta)=-\log(\mu)$. Exponentiating once more gives $\mu=\exp[-\exp(-\eta)]$. This lies strictly between zero and one for every finite $\eta$.

Worked Example: Recovering Predictors From Probabilities

Find the log-log linear predictor corresponding to fitted probabilities $0.2$ and $0.8$.

For a probability $p$, the predictor is $\eta=-\log[-\log(p)]$. At $p=0.2$,

-\log(0.2)=1.6094379\ldots,\qquad \eta=-\log(1.6094379\ldots)\approx -0.4759.

Substituting the unrounded value verifies the answer: $\exp(-\eta)=-\log(0.2)$, and therefore $\exp[-\exp(-\eta)]=\exp[\log(0.2)]=0.2$.

At $p=0.8$, first calculate $-\log(0.8)=0.2231436\ldots$, then use that unrounded inner value:

\eta=-\log[-\log(0.8)] =-\log(0.2231436\ldots)\approx 1.4999.

The two probabilities are equally far from $0.5$, but their predictors are not opposites: $-0.4759$ and $1.4999$. This is one indication that the link treats the two sides of the probability scale differently.

Derivative and Coefficient Direction

The inverse link determines how a change in the predictor changes the fitted probability. Differentiating $\mu(\eta)=\exp[-\exp(-\eta)]$ by the chain rule gives

\frac{d\mu}{d\eta} =\exp[-\exp(-\eta)]\exp(-\eta) =\mu\exp(-\eta)>0.

Thus, for a model $\eta=x^\mathsf{T}\beta$, increasing a covariate $x_j$ by one unit while holding the other covariates fixed changes the predictor by $\beta_j$. A positive $\beta_j$ increases the fitted probability, and a negative $\beta_j$ decreases it. The size of the probability change depends on the starting predictor, because $d\mu/d\eta$ varies with $\eta$.

Result: For a differentiable predictor $\eta(x)$, a log-log model satisfies $$ \frac{d\mu}{dx} =\mu\exp(-\eta)\frac{d\eta}{dx}. $$ In particular, if $\eta=\beta_0+\beta_1x$, then $d\mu/dx=\beta_1\mu\exp(-\eta)$. The sign of the slope is the sign of $\beta_1$, but its magnitude depends on $x$.

The coefficient also has an exact interpretation on a transformed scale. Since the link equation implies $\log[-\log(\mu)]=-\eta$, changing $x_j$ by $d$, with other covariates fixed, multiplies $-\log(\mu)$ by $\exp(-\beta_jd)$. This statement concerns a transformed probability quantity, not the probability itself.

Worked Example: A Fitted Probability Change and Its Local Slope

Suppose $\eta=-0.5+0.8x$, where $x$ is a binary exposure. Calculate the fitted probabilities at $x=0$ and $x=1$, their difference, and the probability slope at $x=0$ when $x$ is instead treated as continuous.

At $x=0$, $\eta=-0.5$, so

\hat\mu(0)=\exp[-\exp(0.5)] =\exp(-1.6487213\ldots) \approx 0.1923.

At $x=1$, $\eta=0.3$, giving

\hat\mu(1)=\exp[-\exp(-0.3)] =\exp(-0.7408182\ldots) \approx 0.4767.

The fitted probability difference for this particular change from $x=0$ to $x=1$ is

0.476724\ldots-0.192295\ldots =0.284429\ldots\approx 0.2844.

For a continuous $x$, the derivative is $0.8\mu\exp(-\eta)$. At $x=0$, using the unrounded fitted probability and predictor,

\left.\frac{d\mu}{dx}\right|_{x=0} =0.8(0.192295\ldots)\exp(0.5) =0.253636\ldots\approx 0.2536.

The derivative is a local rate of change, whereas $0.2844$ is the finite fitted difference across a full unit change. They are different summaries and need not be equal.

Why the Asymmetry Is Opposite to Complementary Log-Log

For comparison, the complementary log-log link is $g_{\mathrm{cloglog}}(\mu)=\log[-\log(1-\mu)]$, with inverse $1-\exp[-\exp(\eta)]$. The log-log inverse has the identity

g_{\mathrm{loglog}}^{-1}(\eta) =1-g_{\mathrm{cloglog}}^{-1}(-\eta).

This identity expresses the opposition precisely: the log-log probability at predictor $\eta$ is one minus the complementary log-log probability at predictor $-\eta$. The two links therefore put their especially rapid tail behavior on opposite ends of the predictor scale.

For the log-log link, as $\eta\to-\infty$, $e^{-\eta}\to\infty$ and $\mu=\exp[-e^{-\eta}]\to0$. This is faster than exponential decay in $-\eta$: writing $\eta=-a$, for any $k>0$,

\frac{\exp(-e^a)}{\exp(-ka)} =\exp(-e^a+ka)\longrightarrow 0 \qquad (a\to\infty),

because $e^a$ eventually grows faster than $ka$. At the other end, as $\eta\to\infty$, let $t=e^{-\eta}\to0$. The expansion $e^{-t}=1-t+O(t^2)$ yields

1-\mu=1-\exp(-e^{-\eta}) =e^{-\eta}+O(e^{-2\eta}).

So the log-log probability approaches one with an approximately exponential tail. Complementary log-log has the reverse pattern: its probability is approximately $e^\eta$ as $\eta\to-\infty$, while its remaining distance from one is $\exp(-e^\eta)$ as $\eta\to\infty$.

Worked Example: Comparing Both Links at the Same Predictors

Compare log-log and complementary log-log fitted probabilities at $\eta=-1$ and $\eta=1$. Use the inverse formulas directly.

At $\eta=-1$, the log-log probability is $\exp[-\exp(1)]\approx0.0660$. The complementary log-log probability is $1-\exp[-\exp(-1)]\approx0.3078$.

At $\eta=1$, the log-log probability is $\exp[-\exp(-1)]\approx0.6922$. The complementary log-log probability is $1-\exp[-\exp(1)]\approx0.9340$.

The log-log link is much closer to zero at the negative predictor and approaches one more gradually at the positive predictor. Complementary log-log shows the opposite-sided pattern. These comparisons hold the numerical predictor fixed; they do not assert that the links produce identical fits when separately estimated from data.

Limiting Probabilities and Interpretation

For every finite linear predictor, the log-log inverse gives a probability strictly between zero and one. Its limiting probabilities are

\lim_{\eta\to-\infty}\exp[-\exp(-\eta)]=0, \qquad \lim_{\eta\to\infty}\exp[-\exp(-\eta)]=1.

These limits help explain which covariate patterns can be represented smoothly. A very negative predictor drives the probability extremely close to zero, while a very positive predictor brings it toward one at a comparatively slower rate. A finite coefficient does not produce a probability of exactly zero or one.

The coefficient’s transformed-scale interpretation can be checked directly. For a one-unit change in $x_j$, $\eta$ changes by $\beta_j$, and

\frac{-\log(\mu_{\mathrm{new}})} {-\log(\mu_{\mathrm{old}})} =\exp(-\beta_j).

This is not a risk ratio $\mu_{\mathrm{new}}/\mu_{\mathrm{old}}$, nor is $\beta_j$ a fixed risk difference. To report an effect on the probability scale, calculate fitted probabilities at specified covariate values or use the derivative at a specified value. This is the same general distinction between link-scale coefficients and probability-scale effects that matters in other binary regression models.

Common Mistake

Reversing the log-log convention. Writing $\log[-\log(\mu)]$ without the leading minus sign gives a decreasing link and a different coefficient-direction interpretation. Under the convention here, $g(\mu)=-\log[-\log(\mu)]$ is increasing.
Calling a coefficient a constant risk difference. The coefficient changes $\eta$; the probability effect depends on the current $\eta$. Compute the inverse-link values or the derivative for the covariate values of interest.
Assuming the same asymmetry as complementary log-log. The rapid approach to an endpoint occurs on the opposite predictor tail. Check the limits of the inverse link rather than inferring the tail from the name.
Rounding an intermediate value too early. Nested exponentials can magnify small rounding discrepancies. Retain full precision through the calculation and round only the final result.

Key takeaway: With $g(\mu)=-\log[-\log(\mu)]$, the inverse is $\exp[-\exp(-\eta)]$, it increases with the predictor, and its unusually rapid approach to zero occurs at the negative-predictor end. Complementary log-log places the rapid tail on the opposite side.

Check Your Understanding

Use the definitions and results above to answer the following questions.

Derive the inverse of $g(\mu)=-\log[-\log(\mu)]$ by solving for $\mu$.
For a model with $\eta=\beta_0+\beta_1x$, what is the sign of $d\mu/dx$ when $\beta_1<0$? Explain using the inverse-link derivative.
Why is a one-unit log-log coefficient not itself a risk ratio for the probability?
Which predictor tail makes the log-log fitted probability approach zero faster than exponentially, and what happens at the other tail?
State the identity relating the log-log inverse at $\eta$ to the complementary log-log inverse at a transformed predictor.