Log Links And Multiplicative Mean Changes

From a Linear Predictor to a Mean Ratio

With an identity link, a coefficient describes an additive change in the mean. A log link gives a different interpretation: it makes a linear change in the predictor correspond to a multiplicative change in the mean. This is useful when a response has a positive mean and effects are naturally described as ratios, such as a mean that is 20% higher or lower.

Let \(\mu(x,z)\) denote the conditional mean of a response at covariate value \(x\) and other covariates \(z\). Consider a model with a log link and a linear predictor that is additive in \(x\):

Here \(q(z)\) collects the terms involving the other covariates, which are held fixed when comparing two values of \(x\). Because the inverse of the log link is the exponential function, the mean is

The model requires a positive conditional mean wherever the log link is used, since \(\log(\mu)\) is defined only for \(\mu>0\). The response itself need not be positive in every observation; the condition here concerns its modeled mean.

Deriving the One-Unit Ratio

Compare two cases with the same \(z\), one at \(x\) and one at \(x+1\). The linear predictor increases by \(\beta_x\). Exponentiating both predictors gives the two means, and their ratio follows by cancellation.

Proof. By the assumed model,

Since \(\mu(x,z)>0\), division by \(\mu(x,z)\) is valid, and gives the stated ratio. \(\square\)

The ratio does not depend on the starting value \(x\) or on the other covariates \(z\), provided the predictor change really is \(\beta_x\). If \(\beta_x=0\), the ratio is 1 and the modeled mean is unchanged. A positive coefficient gives a ratio greater than 1; a negative coefficient gives a ratio between 0 and 1.

The Ratio Is Not an Additive Difference

A ratio compares two means multiplicatively. To obtain their additive difference, multiply the ratio by the starting mean and subtract that starting mean. This yields a difference that generally varies with the starting mean.

Derivation. The ratio result gives \(\mu(x+1,z)=\mu(x,z)\exp(\beta_x)\). Subtract \(\mu(x,z)\) from both sides:

For fixed \(\beta_x\), the factor \(\exp(\beta_x)-1\) is fixed, but the product also contains the starting mean. Thus, unless \(\beta_x=0\), the additive change is not a single constant across different starting means. In contrast, the ratio \(\exp(\beta_x)\) stays the same. \(\square\)

The exact relative change is \(\exp(\beta_x)-1\), so the exact percentage change is \(100\{\exp(\beta_x)-1\}\%\). For a small coefficient, \(\exp(\beta_x)-1\) is approximately \(\beta_x\), making \(100\beta_x\%\) a useful approximation. It is not the exact percentage change in general.

Reading Coefficients Carefully

A log-link coefficient is easiest to interpret by moving through three steps: identify the covariate change, exponentiate the corresponding change in the linear predictor, and state the result as a ratio of means in context. For a one-unit increase in a covariate whose coefficient is \(\beta_x\), that ratio is \(\exp(\beta_x)\). For a change of \(\Delta\) units in the additive model above, it is \(\exp(\beta_x\Delta)\).

A coefficient is tied to the units in which its covariate is recorded. If \(x\) is measured in years, its coefficient describes a one-year increase; if it is measured in months, the coefficient describes a one-month increase. Always state the unit of the comparison. Also distinguish the ratio from its percentage translation: a ratio of 1.2 means the mean is 1.2 times as large, or 20% higher, not 120% higher.

The simple interpretation assumes that increasing \(x\) changes the linear predictor by exactly \(\beta_x\). If the model includes an interaction involving \(x\), the predictor change can depend on the values of the other covariates. In that case, calculate the actual difference between the two linear predictors and exponentiate it; the coefficient on \(x\) alone need not be the ratio for every covariate setting.

Common Mistake

Wrong reasoning: “The coefficient is \(0.25\), so a one-unit increase in \(x\) adds \(0.25\) to the mean.” This treats the coefficient as if the link were the identity link.

Correction: With a log link, a one-unit increase adds \(0.25\) to the log mean. On the mean scale it multiplies the mean by \(\exp(0.25)\), which is about \(1.284\). The corresponding additive change is the starting mean multiplied by \(\exp(0.25)-1\), so it depends on that starting mean.