Back-Transforming Predictions from Log Models

Why Back-Transformation Needs Care

A regression model for a logarithm of the response can be useful when the relationship is more nearly linear on the log scale than on the original scale. But a fitted value on the log scale is not yet a prediction in the response's original units. Exponentiating converts the scale; it does not automatically answer every prediction question. In particular, the exponentiated fitted value and the arithmetic mean response at a given predictor value are generally different.

The distinction matters whenever a prediction is used as an expected number of events, an expected cost, or another arithmetic average. It also matters for intervals: exponentiating their endpoints is a valid scale conversion, but the resulting interval concerns the parameter or outcome represented by the original interval, not a different target.

The Log-Response Model and Direct Conversion

Suppose the response \(Y\) is positive and a model is fitted to its natural logarithm. At predictor value \(x\), write the model as

where \(\eta(x)\) is the fitted systematic part on the log scale and \(\varepsilon\) is the error on that scale. Assume the model is appropriate for the predictor range being used and that the errors have conditional mean zero, \(E(\varepsilon\mid x)=0\). A fitted log-scale value is denoted \(\hat{\eta}(x)\).

The geometric mean is obtained by averaging logarithms and then exponentiating. If \(E(\ln Y\mid x)=\eta(x)\), the conditional geometric mean is \(\exp(\eta(x))\). If the conditional distribution of \(\ln Y\) is normal, its mean and median coincide, so \(\exp(\eta(x))\) is also the conditional median of \(Y\). The arithmetic mean, \(E(Y\mid x)\), averages values on the original scale and is more affected by large responses.

These interpretations depend on the model and its assumptions. In particular, the median interpretation requires the conditional median error on the log scale to be zero; the normal-error model ensures this. The geometric-mean interpretation follows from the conditional mean of \(\ln Y\) being \(\eta(x)\). To interpret a transformed fitted value as an arithmetic mean requires an additional correction.

When the Arithmetic Mean Is the Target

To see why a correction is needed, start from the model and exponentiate both sides: \(Y=\exp(\eta(x))\exp(\varepsilon)\). Taking a conditional expectation gives

The extra factor \(E\{\exp(\varepsilon)\mid x\}\) is generally greater than one when the error has mean zero and is variable. This follows from the convexity of the exponential function: averaging exponentials does not generally equal exponentiating the average. Thus \(\exp(\hat{\eta}(x))\) will tend to be below the arithmetic mean when used as though it were an arithmetic-mean prediction.

A common model-based correction assumes that the conditional log-scale errors are normal with constant variance \(\sigma^2\), so that \(\varepsilon\mid x\sim N(0,\sigma^2)\). For a normal variable, \(E\{\exp(\varepsilon)\mid x\}=\exp(\sigma^2/2)\). The arithmetic-mean prediction is then

where \(\hat{\sigma}^{\,2}\) estimates the conditional variance on the log scale. The correction is not optional if the target is the arithmetic mean and the normal log-error model is being used. Its size grows with the log-scale error variance.

Smearing When Normal Errors Are Not Assumed

The normal correction uses a specific error distribution. An alternative is to estimate the factor \(E\{\exp(\varepsilon)\mid x\}\) from residuals. For fitted log-response residuals \(e_i=\ln(y_i)-\hat{\eta}(x_i)\), a basic smearing estimate is the average of their exponentials:

The resulting arithmetic-mean prediction is \(\exp(\hat{\eta}(x))\hat{S}\). This method does not require normally distributed errors. It does require the observed residuals to represent the conditional log-error distribution relevant to the prediction. In particular, a single pooled factor is most defensible when that distribution is approximately stable across predictor values. If the variability or shape of the errors changes with \(x\), one pooled factor may not estimate the needed conditional factor well.

Back-Transforming Intervals

An interval's endpoints can be exponentiated because the exponential function preserves order. What the transformed interval means depends on what the original interval was constructed to cover.

• Confidence interval for a fitted log-scale location: Exponentiating both endpoints gives an interval for the exponentiated location, such as the conditional geometric mean. Under normal log-scale errors, it is also an interval for the conditional median.
• Prediction interval for an individual log response: If the prediction interval is valid under the log-scale model, exponentiating both endpoints gives a prediction interval for an individual response on the original scale. It remains asymmetric in general, and its coverage is preserved by the one-to-one transformation.
• Interval for an arithmetic mean: A confidence interval for the log-scale location is not automatically an interval for the arithmetic mean. The mean includes a variance or smearing factor, and uncertainty in that factor may also need to be included in the interval procedure.

For example, if a log-scale prediction interval is \([1.40,2.60]\), its response-scale version is \([\exp(1.40),\exp(2.60)]=[4.055,13.464]\) response units. It predicts an individual response, not the conditional average response. An interval for the fitted location and a prediction interval for an individual outcome have different purposes and usually different widths.

Why the Distinction Matters

The target should determine the back-transformation. A geometric mean or median describes a typical location in a multiplicative distribution. An arithmetic mean is appropriate when the question concerns the average total response per individual, such as expected use, cost, or count of a positive outcome. Reporting one as the other can materially understate an average, especially when log-scale variability is substantial.

A common mistake is to say, “The fitted log value is 2, so the predicted average response is \(\exp(2)\).” The reasoning overlooks the factor \(E\{\exp(\varepsilon)\mid x\}\). The correction is to state the target precisely: \(\exp(2)\) is the exponentiated log-scale location, and under normal errors it estimates the median and geometric mean; an arithmetic-mean prediction also accounts for log-scale error variation.