Why the Error Scale Matters
A regression model describes not only how its systematic part changes with predictors, but also how observations vary around that part. For a positive outcome such as concentration, cost, or time, a log-outcome model describes variation in a particular way: equal errors on the log scale correspond to equal ratios on the original scale. The resulting errors are multiplicative, not additive.
This distinction is important when an outcome’s deviations are naturally proportional. An error of 5 units has a different relative size for a reference level of 20 than for a reference level of 200. By contrast, multiplying either reference level by the same factor gives the same proportional change. A log-outcome model encodes the latter pattern.
The Log-Outcome Model
Let \(Y_i\) be a strictly positive outcome for observation \(i\), and let \(x_i\) denote its predictor or predictors. A log-outcome regression models the natural logarithm of the outcome as a linear function of the predictors plus an error:
With several predictors, the linear part can contain additional terms, but the error interpretation is unchanged. Write that linear part as \(\eta_i\). Then the model is \(\log(Y_i)=\eta_i+\varepsilon_i\). The error \(\varepsilon_i\) is an observation’s deviation from the modeled value on the log scale; it is not a deviation measured in the original units of \(Y_i\).
The standard regression conditions apply to the scale on which the model is fitted. A common set of assumptions is that \(E(\varepsilon_i\mid x_i)=0\), that \(\operatorname{Var}(\varepsilon_i\mid x_i)=\sigma^2\) is constant, and that observations’ errors are independent conditional on their predictors. Normal errors, \(\varepsilon_i\sim N(0,\sigma^2)\), are an additional condition when exact small-sample \(t\)-based inference is required. These are assumptions about errors in the log outcome, not about additive errors in \(Y_i\).
The logarithm requires \(Y_i>0\) for every observation included in this model. An observed zero or negative outcome cannot be entered directly as \(\log(Y_i)\). Also, the model form and error conditions need scientific and data-based justification; choosing a logarithm does not make them true automatically.
From Addition on One Scale to Multiplication on Another
The change in error structure follows directly by exponentiating the model. This algebra is the central result for interpreting a log-outcome regression.
\(\log(Y_i)=\eta_i+\varepsilon_i\).
\(Y_i=\exp(\eta_i+\varepsilon_i)\).
\(Y_i=\exp(\eta_i)\exp(\varepsilon_i)\).
Define \(B_i=\exp(\eta_i)\), the positive reference level determined by the model’s linear part, and \(U_i=\exp(\varepsilon_i)\), the multiplicative error factor. The original-scale equation is therefore \(Y_i=B_iU_i\). Dividing by the reference level gives a direct interpretation:
Thus the error factor tells how many times the observed outcome is the reference level. If \(\varepsilon_i=0\), then \(U_i=1\), so \(Y_i=B_i\). If \(\varepsilon_i>0\), then \(U_i>1\), and the outcome is above the reference level; if \(\varepsilon_i<0\), then \(0<U_i<1\), and it is below. In either case, the outcome remains positive.
The proportional difference from the reference level is \(\exp(\varepsilon_i)-1\), not exactly \(\varepsilon_i\). For small errors, \(\exp(\varepsilon_i)-1\) is approximately \(\varepsilon_i\), which can make the log error a useful approximation to a fractional change. For larger errors, use the exponential formula rather than treating the log error as an exact percentage.
Worked Example: A Positive Log Error Means a Multiplicative Increase
Suppose a hypothetical concentration model has linear part \(\eta=3.20+0.178x\). At \(x=4\), the model’s reference level is calculated from \(\eta\), and the observation has log-scale error \(\varepsilon=\log(1.10)\). Find the original-scale outcome implied by these values.
Check the setup. This example stipulates a positive outcome and a log-scale error. The predictor value is used in the stated model; the example illustrates the transformation rather than claiming that regression assumptions have been established from data.
First calculate the linear part:
The reference level is approximately \(\exp(3.912)=50.0\) concentration units. Since \(\varepsilon=\log(1.10)\), the multiplicative factor is
Therefore the outcome is \(1.10\) times the reference level:
The observation is 5.0 concentration units above the reference level, or 10% higher. The error itself, \(\log(1.10)\approx 0.09531\), is expressed on the log scale; its exact proportional interpretation comes from exponentiating it.
Equal Log Errors Give Equal Ratios, Not Equal Differences
If two observations have the same log-scale error, each has the same multiplicative factor relative to its own reference level. Their absolute differences can still be very different because the reference levels differ. Conversely, a fixed additive difference on the outcome scale does not correspond to a fixed log-scale error.
Worked Example: The Same Log Error at Two Reference Levels
Consider two hypothetical observations with reference levels of 20 and 200 units. Each has log-scale error \(\varepsilon=0.10\). Find the factor and the two outcomes.
Check the setup. Both reference levels and the resulting outcomes are positive, as required. The example stipulates the same log error for both observations; it does not infer that this error structure fits any particular data set.
The shared factor is
At the reference level of 20 units:
The absolute increase is about \(22.103-20=2.103\) units. At the reference level of 200 units:
The absolute increase is about \(221.034-200=21.034\) units. In both cases the outcome is approximately 10.517% above its reference level, since \(100(\exp(0.10)-1)\approx10.517\%\). The absolute changes differ by a factor of ten because the reference levels differ by a factor of ten.
Worked Example: A Fixed Additive Error Is Not a Fixed Log Error
Suppose two hypothetical reference levels are 20 and 200 units, and an analyst considers an additive increase of 5 units at each level. Calculate the corresponding log errors and compare them.
Check the setup. Each starting value and each value after adding 5 units is positive, so both logarithms are defined. This is a comparison of error scales, not a claim that an additive-error model is appropriate.
For the reference level of 20 units, the resulting value is \(25\). Its log error, measured relative to the reference level, is
For the reference level of 200 units, the resulting value is \(205\), giving
Although each increase is 5 units, the first is a 25% increase and the second is a 2.5% increase. Their log errors are therefore quite different. A common additive error on the original scale and a common additive error on the log scale describe different patterns of variation.
What the Assumptions Imply—and What They Do Not
If the errors have constant variance on the log scale, observations have a common error variance in their logged outcomes under the model. Equivalently, the model describes variability through the distribution of the multiplicative factors \(U_i=\exp(\varepsilon_i)\). When errors have the same distribution across predictor values, those factors have the same distribution across predictor values as well. This does not mean that deviations in the original units have the same size at every reference level.
The condition \(E(\varepsilon_i\mid x_i)=0\) concerns the average error after taking logarithms. It should not be silently replaced by a condition about the average of \(U_i\) on the original scale. The fitted linear expression describes the modeled log outcome; the error assumptions must be understood on that same scale. Translating a fitted log-scale model into particular original-scale summary predictions requires additional care.
In practice, the errors are not observed directly because the true coefficients are unknown. A fitted regression produces residuals, which are estimates of the errors. Plots and summaries of those residuals can help assess whether the log-scale model’s patterns are plausible. For example, residual spread that grows with the fitted log outcome may be evidence against constant error variance on the modeled scale. Such checks do not prove the assumptions; they help identify important departures.
Common Mistake: Treating a Log Error as an Outcome-Unit Error
A frequent mistake is to say that an error of \(0.10\) means the observed outcome is 0.10 units above the model’s reference level, or exactly 10% above it. Neither interpretation is generally correct. The error is in log units, and its exact factor is \(\exp(0.10)\approx1.10517\). The exact increase is therefore about 10.517% of the reference level, while the absolute increase depends on the reference level.
Another mistake is to describe the model as having additive errors on the original scale just because the regression equation has an additive error term. That addition occurs in \(\log(Y_i)\). Exponentiation turns it into multiplication on the original scale. Always identify the scale of the equation before interpreting its error.
Check Your Understanding
Use the scale of the model carefully when answering each question.
- If a log-scale error is zero, what is the multiplicative error factor, and how does the outcome compare with its reference level?
- An error of \(0.20\) occurs in a log-outcome model. Write the exact expression for the proportional difference from the reference level. Is \(0.20\) itself the exact proportional difference?
- Two observations have the same log error but reference levels of 30 and 300 units. What is the same about their error interpretation, and what may differ?
- Why must every outcome used in a log-outcome model be strictly positive?
- When a regression is fitted to \(\log(Y)\), on which scale should the constant-variance assumption for the errors be assessed?