Tutorials › Generalized Linear Models › Reading the Log-Mean Scale

Count Effects and Predictions · Tutorial 467 of 981

Reading the Log-Mean Scale

Interpret additive changes in a Poisson model’s log mean and calculate the resulting changes in expected counts while holding other terms fixed.

Intermediate 9 min read

What You'll Learn

  • Distinguish the observed count from its conditional mean in a Poisson model
  • Read the terms of a log-mean model as contributions on an additive scale
  • Calculate how a specified predictor change shifts the linear predictor
  • Convert a log-scale shift into the corresponding change in expected counts
  • Explain why the same log-scale change can produce different count changes at different baselines
  • Recognize the mistake of treating a log-scale coefficient as a count-unit change

From a Count Model to Its Log-Mean Scale

A Poisson model describes a count through its conditional mean. The mean is the expected count at a specified covariate profile; it is not itself the observed count. In a log-mean model, the logarithm of that conditional mean is written as an additive predictor. Reading that predictor correctly is essential: its terms add on the log scale, while the corresponding expected count is obtained by exponentiating the total.

This tutorial focuses on interpreting a change in that total when a predictor or model term changes and all other terms are held fixed. The resulting change in expected counts depends both on the size of the log-scale shift and on the starting profile.

Definition: Let \(Y_i\) be a count response and let \(x_i\) denote the covariates for observation \(i\). A Poisson log-mean model specifies $$ Y_i\mid x_i\sim\operatorname{Poisson}(\mu_i),\qquad \log(\mu_i)=\eta_i, $$ where \(\mu_i=E[Y_i\mid x_i]\) is the conditional mean and \(\eta_i\) is the linear predictor. The log-mean specification requires \(\mu_i>0\), so that \(\log(\mu_i)\) is defined. A general Poisson distribution can also have mean zero; positivity is a restriction of the log-mean specification, not of every Poisson model.

For example, with an intercept and one numeric predictor \(x_i\), the predictor is \(\eta_i=\beta_0+\beta_1x_i\). With several predictors, it can be written \(\eta_i=\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}\). These expressions are additive: each term contributes to the value of \(\eta_i\). The expected count is then \(\mu_i=\exp(\eta_i)\).

The distinction between these scales matters. A coefficient describes a contribution to the log mean, not a fixed number of count units. To find the expected count at a profile, first add the predictor terms at that profile and then exponentiate the result.

Reading Additive Contributions

The intercept is the predictor value when all included covariates are zero, provided that this profile is meaningful for the model. Other terms contribute according to their coefficients and covariate values. For a numeric predictor with coefficient \(\beta_j\), its contribution at value \(x_j\) is \(\beta_jx_j\). To compare two profiles, calculate the predictor at each one; if only one term changes, all the other contributions cancel when finding the difference between the predictors.

Worked Example: A One-Unit Change in a Numeric Predictor

Suppose a fitted model is \(\log(\mu)=0.4+0.3x\). Find the expected counts at \(x=2\) and \(x=3\), and their difference.

At \(x=2\), the log mean is \(0.4+0.3(2)=1.0\). Therefore, $$ \mu(2)=\exp(1.0)\approx 2.718282. $$

At \(x=3\), the log mean is \(0.4+0.3(3)=1.3\), so $$ \mu(3)=\exp(1.3)\approx 3.669297. $$

The expected count changes by \(3.669297-2.718282\approx0.951015\). The predictor increased by \(0.3\), but that does not mean the expected count increased by \(0.3\). The count-scale change is found by evaluating both expected counts and subtracting.

Worked Example: Holding Another Predictor Fixed

Consider the model \(\log(\mu)=-0.2+0.5x-0.4z\). Compare the expected counts at \((x,z)=(2,1)\) and \((3,1)\).

At the first profile, the predictor is \(-0.2+0.5(2)-0.4(1)=0.4\), giving $$ \mu(2,1)=\exp(0.4)\approx1.491825. $$

At the second profile, \(z\) remains fixed at 1, while \(x\) increases by one. The predictor is \(-0.2+0.5(3)-0.4(1)=0.9\), and $$ \mu(3,1)=\exp(0.9)\approx2.459603. $$

The expected count increases by approximately \(2.459603-1.491825=0.967778\). The change in the predictor is \(0.5\), the contribution from the change in \(x\); the term involving \(z\) is the same at both profiles. This is the practical meaning of holding other terms fixed.

Translating a Log-Scale Shift into Counts

The general calculation can be stated directly. Suppose a change to a profile shifts the predictor from \(\eta\) to \(\eta+\delta\), where \(\delta\) is the net change in the additive terms. The starting expected count is \(\exp(\eta)\), and the new expected count is \(\exp(\eta+\delta)\). Their difference is

Theorem: If a change in covariates or model terms shifts the log mean from \(\eta\) to \(\eta+\delta\), with all remaining terms unchanged, the change in conditional expected count is $$ \exp(\eta+\delta)-\exp(\eta) =\exp(\eta)\{\exp(\delta)-1\}. $$

Proof. By the log-mean specification, the starting conditional mean is \(\exp(\eta)\). After the change, the predictor is \(\eta+\delta\), so the new conditional mean is \(\exp(\eta+\delta)\). Subtract the starting mean and use \(\exp(\eta+\delta)=\exp(\eta)\exp(\delta)\): $$ \exp(\eta+\delta)-\exp(\eta) =\exp(\eta)\exp(\delta)-\exp(\eta) =\exp(\eta)\{\exp(\delta)-1\}. $$ This is the difference in the two conditional means, hence the change in expected counts. \(\square\)

This formula makes two features explicit. First, the direction of the change is determined by \(\delta\): since the exponential function is increasing, a positive shift raises the expected count, and a negative shift lowers it. Second, the size of the count-scale change depends on the starting mean \(\exp(\eta)\) as well as on \(\delta\). Thus, the same shift on the log scale need not correspond to the same number of expected counts at different starting profiles.

Worked Example: An Exposure Offset and a Predictor Change

Suppose a model for counts observed over different exposure amounts is \(\log(\mu)=-1+0.6x+\log(t)\), where \(t>0\) is the exposure and its log enters with coefficient fixed at 1. Compare \(x=1\) and \(x=2\) when exposure is held at \(t=4\).

At \(x=1\), the predictor is \(-1+0.6+\log(4)\approx0.986294\). Equivalently, the mean is $$ \mu(1,4)=\exp(-1+0.6+\log(4)) =4\exp(-0.4)\approx2.681280. $$

At \(x=2\), the predictor increases by \(0.6\) to approximately \(1.586294\). The new mean is $$ \mu(2,4)=\exp(-1+1.2+\log(4)) =4\exp(0.2)\approx4.885611. $$

The expected-count difference is \(4.885611-2.681280\approx2.204331\). Exposure is held fixed, so its contribution \(\log(4)\) is present in both predictors and does not change. The coefficient on \(x\) gives the log-scale shift; exponentiating the full predictor gives each expected count.

Why the Starting Mean Matters

For a fixed positive shift \(\delta\), the count difference is \(\exp(\eta)\{\exp(\delta)-1\}\). The factor in braces is fixed, but \(\exp(\eta)\), the starting mean, can vary across profiles. Consequently, a given log-scale shift produces a larger absolute increase when the starting mean is larger.

Result: Fix \(\delta>0\). The increase in expected count caused by shifting the predictor from \(\eta\) to \(\eta+\delta\) is a strictly increasing function of \(\eta\).

Proof. Denote the increase by \(D(\eta)=\exp(\eta+\delta)-\exp(\eta)\). Differentiating with respect to \(\eta\) gives $$ D'(\eta)=\exp(\eta+\delta)-\exp(\eta) =\exp(\eta)\{\exp(\delta)-1\}. $$ For every real \(\eta\), \(\exp(\eta)>0\). Since \(\delta>0\), \(\exp(\delta)-1>0\). Thus \(D'(\eta)>0\) for every \(\eta\), which proves that the increase is strictly larger at a larger starting predictor, and therefore at a larger starting mean. \(\square\)

This result concerns an absolute difference in expected counts. It does not say that the observed count in a particular case must increase by that amount: the response \(Y\) remains random, and an individual observed count can fall even when the model’s conditional mean rises. The comparison is between model-implied conditional means at two covariate profiles.

Common Mistake: Treating a Coefficient as Count Units

A common error is to say, “The coefficient is \(0.5\), so increasing the predictor by one adds \(0.5\) expected counts.” That reasoning treats the coefficient as though it were on the count scale. In a log-mean model, the coefficient contributes to the change in \(\eta=\log(\mu)\), not directly to the change in \(\mu\).

The correction is to compute the predictor at each profile and exponentiate each total. For a one-unit increase in \(x\) with coefficient \(\beta\), the predictor changes by \(\beta\), but the expected-count difference is \(\exp(\eta+\beta)-\exp(\eta)\), where \(\eta\) is the starting predictor. Since this expression includes the starting predictor, a coefficient alone does not determine a fixed count-unit change.

Key takeaway: Add the model terms on the log-mean scale, identify the net predictor shift between profiles, and exponentiate to obtain each conditional mean. Interpret changes in expected counts by comparing those means, not by reading a log-scale coefficient as count units.

Check Your Understanding

Use the log-mean scale carefully in each question.

  1. In \(\log(\mu)=0.2+0.4x\), what is the log mean and expected count at \(x=2\)?
  2. If a profile change shifts the predictor from \(\eta\) to \(\eta-0.3\), does the conditional expected count rise or fall? Explain.
  3. Why is it incorrect to say that a coefficient of \(0.7\) adds \(0.7\) expected counts for a one-unit predictor increase?
  4. Suppose two profiles have the same predictor shift \(\delta>0\), but one has a larger starting mean. Which has the larger absolute increase in expected count, and why?
  5. In a model with an exposure offset \(\log(t)\), what does holding exposure fixed mean when comparing two predictor profiles?
Clinical Biostats