Rare-Event Approximation For Complementary Log-Log

Why an Exponential Approximation Appears

The complementary log-log link has an inverse link that is asymmetric in the probability scale. When the event probability is small, however, its behavior has a useful simpler form: the probability is close to the exponential of the linear predictor. This approximation can make a coefficient easier to interpret, but only when the probability remains sufficiently small.

Write $\eta$ for a linear predictor and $p$ for the corresponding event probability. The complementary log-log inverse link is $p=1-\exp\{-\exp(\eta)\}$. To examine its rare-event behavior, it helps to distinguish $\exp(\eta)$ from $p$: they are close when $p$ is small, but they are not generally equal.

Definition: Under the complementary log-log inverse link, define $x=\exp(\eta)>0$. Then $p=1-\exp(-x)$, and equivalently $x=-\log(1-p)$. The rare-event approximation is $p\approx x=\exp(\eta)$ when $x$ is small.

The quantity $x=-\log(1-p)$ is sometimes called the cumulative hazard on this scale. The approximation does not say that the probability equals this quantity for every $p$. Instead, it says the difference between $x$ and $p$ becomes small relative to either quantity as $x$ approaches zero.

Deriving and Bounding the Approximation

The identity $p=1-\exp(-x)$ gives an exact starting point. Integrating the derivative of $1-\exp(-x)$ from zero to $x$ yields

p=1-\exp(-x)=\int_0^x \exp(-t)\,dt.

For $t\geq 0$, $\exp(-t)\leq 1$, so the integrand is at most one. Also, $\exp(-t)\geq 1-t$: the function $h(t)=\exp(-t)-1+t$ has $h(0)=0$ and $h'(t)=1-\exp(-t)\geq 0$. Integrating both inequalities gives

x-\frac{x^2}{2}\leq p\leq x.

Theorem: For every $x>0$, $x-x^2/2\leq 1-\exp(-x)\leq x$. In particular, when $0<x<2$, $$ 0\leq \frac{x}{p}-1\leq \frac{x}{2-x}. $$ Moreover, $p/x\to 1$ as $x\to 0$.

Proof. The integral inequalities above establish $x-x^2/2\leq p\leq x$. Since $p=1-\exp(-x)>0$ for $x>0$, division by $p$ is valid. For $0<x<2$, the lower bound on $p$ is positive, and

\frac{x}{p}\leq \frac{x}{x-x^2/2}=\frac{1}{1-x/2}.

Subtracting one gives $x/p-1\leq x/(2-x)$. The upper bound $p\leq x$ gives $x/p-1\geq 0$. Finally, for $0<x<2$, divide the original bounds by $x$ to obtain $1-x/2\leq p/x\leq 1$. Both bounds approach one as $x\to 0$, so the squeeze theorem gives $p/x\to 1$. $\square$

The bound makes the approximation quantitative. It also clarifies which relative error is being discussed: $x/p-1$ measures how much the exponential quantity $x$ exceeds the exact probability $p$, relative to $p$. A bound on this relative error is not the same as an absolute probability difference. The latter satisfies $0\leq x-p\leq x^2/2$.

Worked Example: Checking the Approximation at a Specified Predictor

Suppose $\eta=-3$. Calculate the exact probability, compare it with $\exp(\eta)$, and use the error bound.

First, $x=\exp(-3)\approx 0.0497871$. The exact probability is

p=1-\exp(-0.0497871)\approx 0.0485680.

The exponential approximation is $x\approx 0.0497871$, which exceeds the exact probability by approximately $0.0012191$. Relative to the exact probability, that excess is

\frac{x}{p}-1\approx \frac{0.0497871}{0.0485680}-1\approx 0.0251,

or about $2.51\%$. The theorem's upper bound is $x/(2-x)\approx 0.0255$, which contains the actual relative error. Thus the approximation is quite close here, though it is not exact.

Predictor Changes and Approximate Probability Ratios

Suppose a predictor changes from $\eta_0$ to $\eta_1=\eta_0+\delta$. Let $x_0=\exp(\eta_0)$, and put $r=\exp(\delta)$. Then $x_1=\exp(\eta_1)=r x_0$, while the exact probabilities are $p_0=1-\exp(-x_0)$ and $p_1=1-\exp(-r x_0)$. Consequently, the exact probability ratio is

\frac{p_1}{p_0}=\frac{1-\exp(-r x_0)}{1-\exp(-x_0)}.

When $x_0$ is small and $r$ is fixed, both probabilities are approximately their respective $x$ values, so this ratio is approximately $r=\exp(\delta)$. The following result states that limiting relationship precisely.

Theorem: Fix any finite real number $\delta$, set $r=\exp(\delta)>0$, and let $p(x)=1-\exp(-x)$. Then $$ \lim_{x\to 0^+}\frac{p(rx)}{p(x)}=r. $$

Proof. For $x>0$, write $p(x)=xg(x)$, where $g(x)=p(x)/x$. The preceding theorem established that $g(x)\to 1$ as $x\to 0^+$. Since $r>0$ is fixed, $rx\to 0^+$ as $x\to 0^+$, and therefore $g(rx)\to 1$. Now

\frac{p(rx)}{p(x)} =\frac{rx\,g(rx)}{x\,g(x)} =r\frac{g(rx)}{g(x)} \longrightarrow r\frac{1}{1}=r.

The denominator has a nonzero limit after expressing the ratio in terms of $g$, so the quotient limit is valid. $\square$

In a complementary log-log regression, if a one-unit increase in a predictor changes the linear predictor by $\beta$, the approximate probability ratio for that change is $\exp(\beta)$ when the baseline and changed event probabilities are both small. This is a rare-event approximation, not an exact interpretation of the coefficient. In particular, the coefficient does not make the probability ratio constant over all baseline probabilities.

Worked Example: A Predictor Change When the Baseline Risk Is Small

Suppose the baseline event probability is $p_0=0.02$, and a predictor change is $\delta=\log(2)$. Find the exact changed probability and compare the exact probability ratio with the approximation.

The baseline exponential quantity is obtained from the exact link transformation:

x_0=-\log(1-p_0)=-\log(0.98)\approx 0.0202027.

The predictor change multiplies $x_0$ by $\exp(\delta)=2$, so $x_1=2x_0$. Since $\exp(-x_0)=0.98$, it follows that

p_1=1-\exp(-2x_0)=1-(0.98)^2=0.0396.

The exact probability ratio is $0.0396/0.02=1.98$, while the rare-event approximation is $\exp(\delta)=2$. The approximate ratio exceeds the exact ratio by $0.02$, which is $1.00\%$ of the approximate value (or about $1.01\%$ of the exact value). Here the approximation is close because both probabilities are small.

When the Approximation Becomes Less Reliable

A predictor change that multiplies $x$ by a large factor can move the changed probability out of the rare-event range, even if the baseline probability is small. Then the approximation $p\approx x$ may be reasonable at baseline but poor after the change. For a fixed baseline value $x_0$, the exact probability ratio is $\{1-\exp(-r x_0)\}/\{1-\exp(-x_0)\}$, not simply $r$.

Worked Example: A Larger Predictor Change

Let the baseline event probability be $p_0=0.20$, and consider a predictor change $\delta=\log(1.5)$. Calculate the exact changed probability and compare the ratio with the rare-event approximation.

First,

x_0=-\log(0.80)\approx 0.2231436,\qquad x_1=1.5x_0\approx 0.3347153.

Because $\exp(-x_0)=0.80$, the exact changed probability can be calculated directly:

p_1=1-\exp(-1.5x_0) =1-(0.80)^{1.5} \approx 1-0.7155418 =0.2844582.

Thus $p_1/p_0\approx 0.2844582/0.20=1.4223$, whereas the approximation gives $\exp(\delta)=1.5$. The approximation exceeds the exact ratio by about $0.0777$, or about $5.18\%$ of the approximate value. The changed probability is no longer extremely small, so the difference is more consequential than in the preceding example.

The example also illustrates why an exponential predictor change should not automatically be reported as an exact risk ratio under the complementary log-log link. It multiplies $x=-\log(1-p)$ exactly. The probability ratio is only approximately the same multiplier when the relevant probabilities are small.

Practical Interpretation and Common Pitfall

For reporting, it is useful to state the approximation together with its conditions. For example: “Because the event probabilities are small, a predictor contrast of $\delta$ corresponds approximately to a probability ratio of $\exp(\delta)$.” If the probabilities are not small, report predicted probabilities or calculate the exact ratio from the inverse link instead. The complementary log-log model itself remains defined for all finite $\eta$; it is the shortcut interpretation that has a limited range.

Wrong reasoning: “The complementary log-log inverse link contains $\exp(\eta)$, so a coefficient change of $\delta$ always multiplies the event probability by $\exp(\delta)$.”
Correction: The change multiplies $x=\exp(\eta)$ by $\exp(\delta)$ exactly. Since $p=1-\exp(-x)$, that multiplier is only approximately the probability ratio when the baseline and changed probabilities are small.
Wrong reasoning: “If the baseline event is rare, the approximation must be accurate for any predictor contrast.”
Correction: A large positive contrast can make the changed probability appreciable. Check both probabilities, or compute the exact ratio.

Key takeaway: Under the complementary log-log inverse link, $p=1-\exp\{-\exp(\eta)\}$. For rare events, $p\approx\exp(\eta)$, and a predictor contrast $\delta$ therefore gives an approximate probability ratio $\exp(\delta)$. The exact probability ratio depends on the baseline probability.

Check Your Understanding

Use the inverse link and the rare-event conditions to answer each question.

For $x=0.1$, write the exact probability and the first-order rare-event approximation. Which is larger?
State the exact quantity multiplied by $\exp(\delta)$ when the linear predictor increases by $\delta$.
If $p_0=0.05$ and $\delta=\log(2)$, set up an expression for the exact $p_1$ and the exact ratio $p_1/p_0$.
Why does the limit of the probability ratio as the baseline $x$ approaches zero require the contrast multiplier to be fixed?
Give one reason a small baseline probability alone may not justify using $\exp(\delta)$ as the probability ratio.