Interpreting Log-Scale Fits on the Original Response Scale

Why the Scale of a Prediction Matters

A regression with a logged response models changes in $\log Y$, not directly in $Y$. Exponentiating a fitted value returns a prediction on the original response scale, but it does not automatically return the arithmetic conditional mean. Confusing these quantities can affect forecasts, comparisons between groups, and estimates of expected amounts in hypothetical insurance settings.

The key distinction is between the fitted center of the log response and the arithmetic average of the original response. The latter depends on the distribution of the log-scale errors. We will make that dependence explicit, interpret slopes as multiplicative effects, and work through both normal-error and empirical retransformation methods. This topic is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: A log-response regression has a positive response $Y$ and specifies $\log Y=\eta(\mathbf{x})+\varepsilon$, where $\eta(\mathbf{x})$ is the linear predictor, often $\mathbf{x}^{T}\boldsymbol{\beta}$, and $\varepsilon$ is the error on the log scale. The fitted log response is $\hat{\eta}(\mathbf{x})$. Its exponentiation, $\exp(\hat{\eta}(\mathbf{x}))$, is a fitted value on the original response scale, whose interpretation depends on the conditional error distribution.

Three Response-Scale Quantities

Assume $Y>0$, and write the model as $\log Y=\eta(\mathbf{x})+\varepsilon$. A common regression assumption is $E[\varepsilon\mid\mathbf{x}]=0$, so $\eta(\mathbf{x})=E[\log Y\mid\mathbf{x}]$. That assumption concerns the log response. It does not imply that $\exp(\eta(\mathbf{x}))=E[Y\mid\mathbf{x}]$.

If the conditional median of $\varepsilon$ is zero, then $\exp(\eta(\mathbf{x}))$ is the conditional median of $Y$. In particular, this holds when $\varepsilon\mid\mathbf{x}\sim N(0,\sigma^2)$. The same exponentiated value is also the conditional geometric mean when the conditional mean of $\log Y$ is $\eta(\mathbf{x})$. The arithmetic conditional mean instead satisfies

E[Y\mid\mathbf{x}] =\exp(\eta(\mathbf{x}))E[\exp(\varepsilon)\mid\mathbf{x}].

The factor $E[\exp(\varepsilon)\mid\mathbf{x}]$ is the retransformation factor. It is at least 1 when $E[\varepsilon\mid\mathbf{x}]=0$ and the required expectation exists: since the exponential function is convex, Jensen's inequality gives $E[\exp(\varepsilon)\mid\mathbf{x}]\geq \exp(E[\varepsilon\mid\mathbf{x}])=1$. Thus, under these conditions, simply exponentiating the fitted log response generally understates the arithmetic conditional mean.

Normal Errors and the Mean Adjustment

Suppose, more specifically, that the conditional errors are normal with mean zero and variance $\sigma^2(\mathbf{x})$: $\varepsilon\mid\mathbf{x}\sim N(0,\sigma^2(\mathbf{x}))$. Then $Y\mid\mathbf{x}$ is lognormal, and its arithmetic conditional mean is

E[Y\mid\mathbf{x}] =\exp\!\left(\eta(\mathbf{x})+\frac{\sigma^2(\mathbf{x})}{2}\right).

This correction follows from the normal density, not from a general property of every regression error. To see why it holds, let $\sigma^2>0$ and use the conditional normal density:

\begin{aligned} E[\exp(\varepsilon)] &=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty} \exp(e)\exp\!\left(-\frac{e^2}{2\sigma^2}\right)\,de\\ &=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty} \exp\!\left(-\frac{(e-\sigma^2)^2}{2\sigma^2} +\frac{\sigma^2}{2}\right)\,de\\ &=\exp\!\left(\frac{\sigma^2}{2}\right). \end{aligned}

The second equality completes the square, using $e-e^2/(2\sigma^2)=-(e-\sigma^2)^2/(2\sigma^2)+\sigma^2/2$. The remaining integral is the integral of a normal density centered at $\sigma^2$, so it equals 1 after its normalizing constant is included. Multiplying this expectation by $\exp(\eta(\mathbf{x}))$ gives the stated conditional mean.

If the variance changes with the predictors, use the variance at each predictor value. A single correction factor is appropriate only when the conditional variance is constant, or when a separate argument justifies a common retransformation factor.

Worked Example: Constant Variance and Two Conditional Means

A hypothetical model for a positive claim amount $Y$, measured in dollars, is $\log Y=4.2+0.1x+\varepsilon$, with $\varepsilon\mid x\sim N(0,0.09)$. Find the fitted conditional medians and arithmetic conditional means at $x=4$ and $x=7$, and compare the means.

Method and assumption. The conditional error is normal with variance $0.09$ at both predictor values. Therefore the conditional median is $\exp(\eta)$, while the arithmetic conditional mean is $\exp(\eta+0.09/2)$. The mean adjustment in the exponent is $0.045$.

At $x=4$, the fitted log response is $4.2+0.1(4)=4.6$. The conditional median is $\exp(4.6)\approx 99.4843$, or $99.48 to the nearest cent. The arithmetic conditional mean is $\exp(4.6+0.045)=\exp(4.645)\approx 104.0634$, or $104.06.

At $x=7$, the fitted log response is $4.2+0.1(7)=4.9$. The conditional median is $\exp(4.9)\approx 134.2898$, or $134.29. The arithmetic conditional mean is $\exp(4.9+0.045)=\exp(4.945)\approx 140.4709$, or $140.47.

The ratio of arithmetic conditional means is $\exp(4.945)/\exp(4.645)=\exp(0.3)\approx 1.3499$. Thus the mean at $x=7$ is about 1.3499 times the mean at $x=4$. The medians have the same ratio because the variance, and hence the correction factor, is constant. Check: both the fitted log response and the variance correction increase the mean calculation from the corresponding exponentiated fitted log response.

Common mistake avoided: reporting $\exp(4.6)$ and $\exp(4.9)$ as arithmetic means. Those values are conditional medians under this model; the variance adjustment is needed for arithmetic means.

Interpreting Slopes on the Original Scale

In a multiple regression, suppose $\eta(\mathbf{x})=\beta_0+\beta_1x_1+\cdots+\beta_px_p$. Holding the other predictors fixed, increasing $x_j$ by $\Delta$ changes the fitted log response by $\beta_j\Delta$. Therefore the ratio of exponentiated fitted values is $\exp(\beta_j\Delta)$. If the conditional error has median zero, this is also the ratio of conditional medians.

The corresponding percentage change is $100\{\exp(\beta_j\Delta)-1\}\%$, not $100\beta_j\Delta\%$ exactly. The latter is a small-change approximation when $\beta_j\Delta$ is close to zero. For arithmetic conditional means, the ratio is also $\exp(\beta_j\Delta)$ if the retransformation factor is the same at the two predictor settings. With normal errors, this condition holds when the conditional variance is unchanged. If the variance changes, the arithmetic-mean ratio includes the change in the variance adjustment as well.

Worked Example: A Slope With Predictor-Dependent Variance

A hypothetical model for a positive payment amount $Y$, in dollars, is $\log Y=3.2+0.15x+\varepsilon$, where $\varepsilon\mid x\sim N(0,0.04+0.01x)$. Compare the conditional medians and arithmetic conditional means at $x=2$ and $x=6$.

Method and assumption. Use $\exp(\eta)$ for each conditional median and $\exp(\eta+\sigma^2(x)/2)$ for each arithmetic conditional mean. Since the conditional variance changes with $x$, do not use one common mean adjustment.

At $x=2$, $\eta=3.2+0.15(2)=3.5$, and the conditional variance is $0.04+0.01(2)=0.06$. The conditional median is $\exp(3.5)\approx 33.1155$, or $33.12. The arithmetic conditional mean is $\exp(3.5+0.06/2)=\exp(3.53)\approx 34.1240$, or $34.12.

At $x=6$, $\eta=3.2+0.15(6)=4.1$, and the conditional variance is $0.04+0.01(6)=0.10$. The conditional median is $\exp(4.1)\approx 60.3403$, or $60.34. The arithmetic conditional mean is $\exp(4.1+0.10/2)=\exp(4.15)\approx 63.4340$, or $63.43.

The median ratio is $\exp(4.1-3.5)=\exp(0.6)\approx 1.8221$. The mean ratio is $\exp(4.15-3.53)=\exp(0.62)\approx 1.8589$. The extra difference comes from the larger conditional variance at $x=6$, which increases its retransformation adjustment. Check: the mean ratio exceeds the median ratio, consistent with the higher variance adjustment at $x=6$.

Common mistake avoided: applying the variance $0.06$ at both values, or interpreting the coefficient $0.15$ as an exact 15% increase in the response. The variance is predictor-dependent, and a one-unit increase corresponds to a multiplicative factor of $\exp(0.15)$, not exactly 1.15.

When Normality Is Not Assumed

The normal-error formula is not a universal retransformation rule. For a general conditional error distribution, the required quantity remains $E[\exp(\varepsilon)\mid\mathbf{x}]$. One practical estimate is an empirical smearing factor: average the exponentiated log-scale residuals from the fitted model, then multiply the exponentiated fitted log response by that average. This approach does not require normally distributed errors. For it to represent the conditional mean at different predictor values, the residual distribution—and thus the retransformation factor—must be sufficiently stable across those values. If it varies with the predictors, one overall factor may not be suitable.

Worked Example: Empirical Retransformation With Residuals

A hypothetical log-response regression has fitted log response $\hat{\eta}=4.0$ for a portfolio segment. Three illustrative log-scale residuals used to estimate a common smearing factor are $-0.2$, $0$, and $0.2$. Estimate the arithmetic response-scale mean using these residuals.

Method and assumption. Use the empirical average of the exponentiated residuals as the retransformation factor. This example assumes that these residuals are representative of the conditional error distribution for the segment.

The exponentiated residuals are $\exp(-0.2)\approx0.818731$, $\exp(0)=1$, and $\exp(0.2)\approx1.221403$. Their average is \[ (0.818731+1+1.221403)/3 =3.040134/3 \approx1.013378. \] The exponentiated fitted log response is $\exp(4.0)\approx54.5982$. The estimated arithmetic mean is therefore $54.5982(1.013378)\approx55.3286$, or $55.33 to the nearest cent.

Check: the estimated mean exceeds $\exp(4.0)$, as expected here because the smearing factor is greater than 1. The calculation uses the residual-based factor rather than assuming the normal-error correction.

Common mistake avoided: treating the three residuals as though their average were the smearing factor. The factor is the average of their exponentials, not the exponential of their average.

Common Exam Traps and a Practical Sequence

A log-response fit is easy to misread because several quantities can look like a “prediction.” Before reporting a number, identify the requested target: a fitted log response, an exponentiated fitted value, a conditional median, or an arithmetic conditional mean. State the error assumption needed for that interpretation.

Using exponentiation alone for an arithmetic mean. Under normal log-scale errors, include $\sigma^2(\mathbf{x})/2$ in the exponent.
Calling every exponentiated fitted value a median. It is a conditional median when the conditional error median is zero; the usual centered normal model satisfies this condition.
Using the wrong variance. If variance depends on predictors, calculate the correction separately at each predictor setting.
Reading a slope as an additive dollar change. A log-scale coefficient describes a multiplicative effect after exponentiation.
Assuming a common smearing factor without justification. A single empirical factor relies on the retransformation behavior being sufficiently stable across the predictor settings of interest.

Write the fitted log response.
Substitute the specified predictor values into $\hat{\eta}(\mathbf{x})$, keeping units and held-fixed predictors clear.

Identify the requested response-scale target.
Exponentiating the fitted log response gives a response-scale fitted value; interpret it as a median only under the appropriate conditional error assumption.

Apply the retransformation factor for a mean.
Use the normal variance adjustment only under conditional normality, or use a justified estimate of $E[\exp(\varepsilon)\mid\mathbf{x}]$.

Check comparisons and units.
Compare response-scale ratios for slopes, account for changes in the retransformation factor, and round only after the calculation is complete.

Key takeaway: Exponentiating a fitted log response does not generally give the arithmetic conditional mean. Under centered normal log-scale errors, the mean is $\exp(\eta+\sigma^2/2)$; more generally, it is $\exp(\eta)E[\exp(\varepsilon)\mid\mathbf{x}]$. A log-scale slope becomes a multiplicative response-scale effect, subject to any change in the retransformation factor.

Check Your Understanding

For each problem, distinguish the response-scale target and apply only the assumptions stated.

A hypothetical model is $\log Y=2.4+0.2x+\varepsilon$, with $\varepsilon\mid x\sim N(0,0.25)$. At $x=3$, find the conditional median and arithmetic conditional mean, rounded to the nearest cent. (Answer: median $\exp(3)\approx\$20.09$; mean $\exp(3.125)\approx\$22.76$.)
A model is $\log Y=4.2-0.05t+\varepsilon$, with centered normal errors of constant variance. What is the ratio of conditional medians at $t=8$ to $t=2$, and what is the ratio of arithmetic conditional means? (Answer: both ratios are $\exp(-0.3)\approx0.7408$.)
A fitted log response is $3.2$, and three residuals for estimating a common smearing factor are $-0.1,0,0.1$. Estimate the arithmetic mean using their exponentials and round to the nearest cent. (Answer: smearing factor $\approx1.003336$; estimated mean $\exp(3.2)(1.003336)\approx\$24.61$.)
Suppose $\log Y=1+0.3x+\varepsilon$, with $\varepsilon\mid x\sim N(0,0.02+0.02x)$. Find the ratio of arithmetic conditional means at $x=3$ to $x=1$. (Answer: $\exp[(1.94)-(1.32)]=\exp(0.62)\approx1.8589$.)
A log-response regression has a coefficient of $0.07$ for predictor $z$. Holding other predictors fixed, express the fitted median percentage change for an increase of 5 units in $z$, assuming the conditional error median is zero. (Answer: $100\{\exp(0.35)-1\}\%\approx41.91\%$ increase.)

What You'll Learn