The log and logit links — how to read coefficients as biology
A dedicated workbook with worked biological examples for the two most-used link functions in ecology. The goal: when a colleague hands you a glm output, you can say what it means in one sentence.
Why this guide exists
Every wildlife graduate eventually fits a model like glm(count ~ x, family=poisson) or glm(survival ~ x, family=binomial) and stares at the coefficient table. The numbers look small. The standard errors look impressive. But what do they mean?
This guide builds intuition through worked examples. Read it once, do the exercises, and then re-read it after the lab. By the end of Day 2 you should be able to look at a coefficient and immediately picture what the effect would look like if you plotted the fitted curve.
- "A log-link coefficient of $\beta$ means a one-unit change in $x$ multiplies the response by $e^\beta$."
- "A logit-link coefficient of $\beta$ means a one-unit change in $x$ multiplies the odds by $e^\beta$. The probability change depends on where you start."
The log link, conceptually
When you specify family=poisson(link="log") or family=Gamma(link="log"), you are telling R to fit the model
or equivalently
That second form is the key. The mean is a product of two pieces: a baseline ($e^{\beta_0}$) and a multiplicative effect that depends on $x$. As $x$ increases by 1, the mean is multiplied by $e^{\beta_1}$.
How to read a log-link slope
Internalize this table:
| $\beta$ (log scale) | $e^\beta$ | Plain English: per unit increase in $x$… |
|---|---|---|
| −1.00 | 0.37 | response falls to 37% (a 63% drop) |
| −0.69 | 0.50 | response is halved |
| −0.40 | 0.67 | response falls by ~33% |
| −0.10 | 0.90 | response falls by ~10% |
| −0.05 | 0.95 | response falls by ~5% |
| 0 | 1.00 | no effect |
| +0.05 | 1.05 | response rises by ~5% |
| +0.10 | 1.105 | response rises by ~10% |
| +0.20 | 1.22 | response rises by ~22% |
| +0.40 | 1.49 | response rises by ~50% |
| +0.69 | 2.00 | response doubles |
| +1.00 | 2.72 | response increases by a factor of 2.72 |
| +2.00 | 7.39 | response increases ~7.4× |
Multi-unit changes
The multiplicative effect of moving $k$ units of $x$ is $e^{k\beta}$. So if $\beta = 0.05$ per km and you ask about a 10 km increase, the multiplicative effect is $e^{10 \cdot 0.05} = e^{0.5} = 1.65$, a 65% increase.
"Half-life" framing
If your slope is negative, the value of $x$ at which the response is halved (relative to $x = 0$) is
For $\beta = -0.05$ per km, $x_{1/2} = 13.9$ km. That's an intuitive number to report in a paper.
Eight worked biological examples (log link)
Density of salmonberry along an elevation gradient
Negative binomial regression of stem count per plot on elevation (m). $\hat\beta = -0.0030$ per meter.
Read it
Each meter reduces expected count by 0.3% ($e^{-0.003} = 0.997$). Over a 1000 m climb, the expected count falls to $e^{-3} = 0.05$ — 5% of the lowland count. The half-elevation is $0.693/0.003 = 231$ m.
Whale calls per hour vs. ship traffic
Poisson regression of calls/hour on ship-passes-per-day. $\hat\beta = +0.04$ per ship. Effect of a busy 50-ship day vs. a quiet 0-ship day?
Read it
$e^{0.04 \cdot 50} = e^{2} = 7.39$. Wait — that's a 7.4× increase with ships, which is biologically odd if you expected ships to be disturbing. The sign is positive (more ships, more calls) — probably an indicator that whale density and ship traffic both track an underlying productivity gradient. The math is fine; the model is missing a confounder.
Caterpillar abundance vs. fertilizer treatment
Treatment is binary (control vs. fertilizer). Coefficient for fertilizer = $+0.85$.
Read it
$e^{0.85} = 2.34$. Fertilized plots have 2.3× the expected caterpillar count of control plots. If you'd like to report it as "percent increase," that's a 134% increase — but in ecology, "fold change" (×2.3) is cleaner.
Bobcat detections vs. canopy cover and road density
Poisson model with two predictors. Canopy slope $+0.025$ per percent (so 0–100); road density $-0.40$ per (roads/km²).
Read it
A 50-percentage-point canopy increase multiplies expected detections by $e^{0.025 \cdot 50} = e^{1.25} = 3.49$. Adding 1 road/km² multiplies by $e^{-0.40} = 0.67$ (a 33% drop). At a typical urban-fringe site with 3 roads/km², detections are cut to $0.67^3 = 0.30$ of their roadless baseline.
Mortality risk vs. age
Hazard model (a log-link Poisson approximation) for annual mortality. $\hat\beta = +0.08$ per year of age.
Read it
Annual mortality rate rises 8% per year. The mortality doubling time is $0.693/0.08 = 8.7$ years. This is the same as a Gompertz-style demographic model, written GLM-style.
Time-to-detection on cameras (lognormal regression)
$\log(\text{time}) = \beta_0 + \beta_1 \cdot \text{prey availability}$. $\hat\beta_1 = -0.30$ per standardized prey index.
Read it
Each SD of prey availability cuts expected time-to-detection by $1 - e^{-0.30} = 26\%$. Two SDs cut it by $1 - e^{-0.60} = 45\%$ — a $\times 1.82$ rise in the detection rate, not a doubling. Halving the time takes $\ln 2 / 0.30 = 2.31$ SDs of prey. The trap here is arithmetic instinct: effects that add on the log scale multiply on the response scale, so doubling the predictor change does not double the effect.
Multiplicative interaction with a categorical variable
Model: glm(catch ~ temp * gear, family=poisson). Main temp = $-0.05$; gear (trawl vs gillnet ref) = $+0.20$; temp:gearTrawl = $+0.08$.
Read it
For gillnets: temp slope is $-0.05$ → 5% drop per degree.
For trawls: temp slope is $-0.05 + 0.08 = +0.03$ → 3% rise per degree.
Baseline shift for trawls: $e^{0.20} = 1.22$, so at temperature 0 the trawl catches 22% more.
This is two completely different stories on the same axis. The interaction term is doing real biological work.
Offset (effort correction) in a count model
Model: glm(detections ~ habitat + offset(log(hours)), family=poisson). The offset says "the response is rate per hour, not total count." How do you interpret the habitat coefficient?
Read it
Including an offset of $\log(\text{hours})$ is equivalent to modeling the per-hour rate. The habitat coefficient then represents a per-effort multiplicative effect — not confounded with sampling intensity. If habitat coefficient is $+0.5$, the per-hour detection rate is $e^{0.5} = 1.65 \times$ as large in that habitat. Offsets are the single most under-used trick in applied ecology. Use them whenever effort varies across observations.
Try it yourself — log-link exercises
Translate
You fit a Poisson regression of mosquito counts on temperature. The temperature coefficient is $+0.12$ per °C. In one sentence, what does a 5° warming do to expected mosquito counts?
Answer
$e^{0.12 \cdot 5} = e^{0.6} = 1.82$. Five degrees of warming multiplies expected mosquito counts by 1.82 (an 82% increase).
Back-engineer a coefficient
You observe that doubling stream productivity (kJ/m²/yr) is associated with a 50% increase in juvenile salmon density. If productivity enters the model as $\log_2(\text{productivity})$, what log-link slope should you expect?
Answer
$e^\beta = 1.50 \Rightarrow \beta = \ln(1.5) = 0.405$. So you should expect a coefficient near 0.41 on the $\log_2$(productivity) predictor.
Confidence interval back-transformation
A Poisson coefficient is $\hat\beta = 0.40$ with 95% CI $(0.18, 0.62)$ on the log scale. Translate the point estimate and CI to the multiplicative scale.
Answer
$e^{0.40} = 1.49$, with CI $(e^{0.18}, e^{0.62}) = (1.20, 1.86)$. So a one-unit change in $x$ multiplies expected counts by 1.49, with a CI of 1.20× to 1.86×. Never average the endpoints in log space and then exponentiate the average — exponentiate each endpoint and report the asymmetric CI.
The logit link, conceptually
When you specify family=binomial(link="logit") you are fitting
The thing inside the log is the odds — the ratio of the probability of success to the probability of failure. So
As $x$ increases by 1, the odds are multiplied by $e^{\beta_1}$. This is the same multiplicative-effect logic as the log link, just applied to odds instead of means.
Probability ↔ Odds ↔ Log-odds — translation table
| Probability $p$ | Odds $p/(1-p)$ | Log-odds $\eta = \mathrm{logit}(p)$ |
|---|---|---|
| 0.01 | 0.0101 | −4.60 |
| 0.05 | 0.0526 | −2.94 |
| 0.10 | 0.111 | −2.20 |
| 0.25 | 0.333 | −1.10 |
| 0.50 | 1.000 | 0 |
| 0.75 | 3.000 | +1.10 |
| 0.90 | 9.000 | +2.20 |
| 0.95 | 19.000 | +2.94 |
| 0.99 | 99.000 | +4.60 |
R helpers:
plogis(eta) # log-odds → probability
qlogis(p) # probability → log-odds
exp(beta) # coefficient → odds ratioEight worked biological examples (logit link)
Marten site occupancy vs. canopy cover
glm(occupied ~ canopy, family=binomial). Intercept $-2$, canopy slope $+0.04$ per percent (canopy on 0–100 scale).
Read it
At canopy = 0: log-odds = −2 → odds = 0.135 → $p = 0.12$. At canopy = 50: log-odds = 0 → $p = 0.5$. At canopy = 100: log-odds = +2 → $p = 0.88$.
Each percentage-point canopy multiplies odds by $e^{0.04} = 1.041$ (a 4.1% odds increase). A 25-point jump multiplies odds by $e^{1} = 2.72$, almost tripling odds.
Survival of tagged smolts vs. river temperature
$\hat\beta_{\text{temp}} = -0.30$ per °C. Translate.
Read it
$e^{-0.30} = 0.74$. Each degree of warming reduces the odds of survival by 26%. Compounded over a 5° warming, odds are multiplied by $0.74^5 = 0.22$ — a 78% drop in odds.
If the baseline $p$ was 0.80 (odds 4), the new odds is $4 \cdot 0.22 = 0.88$, $p = 0.47$. The probability drops 33 percentage points.
Nest success vs. female age
$\hat\beta_{\text{age}} = +0.15$ per year. Probability at age 5 is 0.5. Estimate $p$ at age 10.
Read it
At age 5, log-odds = 0. At age 10, log-odds = $0 + 5 \cdot 0.15 = 0.75$. Odds = $e^{0.75} = 2.12$. $p = 2.12/3.12 = 0.68$.
So a 5-year increase moves $p$ from 0.5 to 0.68 — an 18 percentage-point increase. (The /4 rule: $0.15 \cdot 5 / 4 \approx 0.19$, close.)
Trap response: dichotomous predictor
A factor predictor "previously trapped" has coefficient $-1.5$.
Read it
$e^{-1.5} = 0.22$. Animals previously trapped have 78% lower odds of being re-captured this period — a strong trap-shy effect. If naive animals are caught at $p = 0.4$ (odds 0.67), trap-experienced are at odds $0.67 \cdot 0.22 = 0.15$, $p = 0.13$. This is exactly the kind of behaviour you'd model with a "trap response" parameter in mark-recapture.
Threshold thinking on the logit scale
You fit a logistic regression of presence on a continuous predictor and find $\hat\beta_0 = -3$, $\hat\beta_1 = +0.06$. At what value of $x$ does the predicted probability cross 0.5?
Read it
$p = 0.5$ when log-odds $= 0$. So $-3 + 0.06 x = 0 \Rightarrow x = 50$. This is the inflection point of the fitted curve — biologists often want this as a "threshold" for habitat suitability.
Marginal effect — the most useful number for talking to managers
For the marten model from B1 with canopy slope $+0.04$, what is the marginal effect of canopy on probability at canopy = 50 (where $p = 0.5$)?
Read it
Marginal effect at $p$: $\dfrac{\partial p}{\partial x} = \beta \cdot p(1-p) = 0.04 \cdot 0.5 \cdot 0.5 = 0.01$ per percentage point of canopy. So at the inflection, each percentage point of canopy raises occupancy probability by 1 percentage point. At canopy = 100 (where $p = 0.88$): marginal effect is $0.04 \cdot 0.88 \cdot 0.12 = 0.0042$ per percentage point — much smaller because the curve is flattening.
Sex × habitat interaction (Titanic-like)
Model: glm(survived ~ sex * class, family=binomial). Main sex (Male) = $-2.5$; main class3 = $-1.2$; sex:class3 = $+0.6$.
Read it
Female 1st-class log-odds = intercept. Male 1st-class log-odds = intercept $- 2.5$.
Female 3rd-class log-odds = intercept $- 1.2$. Male 3rd-class log-odds = intercept $- 2.5 - 1.2 + 0.6 = $ intercept $- 3.1$.
So being male is a $-2.5$ log-odds effect in 1st class but only $-2.5 + 0.6 = -1.9$ in 3rd. The sex disadvantage is smaller in 3rd class — likely because everyone's survival is so low in 3rd class that there's less room for sex differentiation.
Sampling fallacy — base rates again
You fit glm(infected ~ habitat, family=binomial) on a sample where infected animals were oversampled (case-control design). Which coefficients can you still interpret?
Read it
Slopes ($\hat\beta_1, \hat\beta_2, \dots$) are correctly estimated under case-control sampling. The intercept is biased by the sampling design and should not be interpreted as the population-level baseline probability. This is a feature of the logistic likelihood (a famous result from Prentice & Pyke, 1979). It's why epidemiologists love logistic regression for case-control studies.
Try it yourself — logit-link exercises
Convert
If a logistic coefficient is $+0.69$, what does a 1-unit change in $x$ do to the odds?
Answer
$e^{0.69} = 2.00$. The odds double.
Probability from log-odds
Compute $p$ when the log-odds is $-1$. When it's $+2$. When it's $+5$. Use plogis() if needed.
Answer
plogis(c(-1, 2, 5)) = 0.27, 0.88, 0.993.
Find the threshold
A logistic model has intercept $-0.5$ and slope $0.03$ on a covariate. At what covariate value does $p = 0.75$?
Answer
$p = 0.75 \Rightarrow \mathrm{logit}(p) = \log(3) = 1.099$. Solve $-0.5 + 0.03 x = 1.099 \Rightarrow x = 53.3$.
Marginal probability change
At $p = 0.20$, what is the marginal effect on probability of a 1-unit change in $x$, given a logit coefficient $\beta = +0.5$?
Answer
$\beta \cdot p \cdot (1-p) = 0.5 \cdot 0.2 \cdot 0.8 = 0.08$. About an 8 percentage-point increase per unit $x$ at this baseline.
Effect sizes on every scale, side by side
| Link | What's linear | $\beta$ is… | Backtransform | Plain English of a small $\beta$ |
|---|---|---|---|---|
| identity | response | additive change in mean | none | "+1 unit x adds $\beta$ to the response" |
| log | log(response) | multiplicative change in mean | $e^\beta$ | "+1 unit x changes response by ~$100 \beta\%$" |
| logit | log-odds | multiplicative change in odds | $e^\beta$ for OR; $\beta \cdot p(1-p)$ for marginal $p$ | "+1 unit x changes odds by ~$100 \beta\%$; probability change depends on baseline" |
| probit | probit(p) | SDs of a latent Normal | $\Phi$ | "+1 unit x changes the latent score by $\beta$ SDs" |
| cloglog | complementary log-log | multiplicative hazard | $e^\beta$ | survival/hazard interpretation |