Log-Scale Intervals for Positive Parameters

Why Work on the Log Scale?

A parameter such as a rate, variance, or mean is often known to be positive. A conventional Wald interval adds and subtracts a margin on the parameter’s original scale, so its lower endpoint can fall below zero. That is not merely inconvenient: a negative endpoint cannot represent a possible value of a positive parameter.

When an estimator is positive, taking its logarithm offers another route. The logarithm maps positive values to the whole real line, where a normal approximation can be used to form a familiar symmetric interval. Exponentiating the two endpoints returns an interval on the original scale. That interval has positive endpoints and is generally asymmetric around the estimate.

The earlier material on delta-method intervals provides the main approximation behind this construction: a smooth transformation of an asymptotically normal estimator is often asymptotically normal as well. Here the transformation is \(g(\theta)=\log\theta\), and the focus is the resulting interval and its interpretation.

Construction and Conditions

Fix a confidence level \(1-\alpha\), where \(0<\alpha<1\), and let \(z_{1-\alpha/2}\) be the \(1-\alpha/2\) quantile of the standard normal distribution. Suppose an estimated standard error \(\widehat{\operatorname{se}}_{\log}\) is available for \(\log\hat{\theta}\). The ordinary two-sided interval on the log scale is

Exponentiating its lower and upper endpoints gives the interval

The estimated standard error is on the log scale. If the standard error is instead first derived for \(\hat{\theta}\), the delta method often gives the approximation \(\widehat{\operatorname{se}}_{\log}\approx\widehat{\operatorname{se}}(\hat{\theta})/\hat{\theta}\). This follows because the derivative of \(\log\theta\) is \(1/\theta\), evaluated at the estimate for a plug-in standard error. A model may also provide a direct standard error for the log estimator.

For the approximation to justify a confidence interval, the log estimator must have an appropriate large-sample normal approximation, and the estimated standard error must consistently estimate its scale. The following result states one convenient version of these requirements.

Proof. On the event \(\hat{\theta}>0\), the logarithm is strictly increasing. Taking logarithms of the two endpoint inequalities shows that the coverage event is equivalent to

Since \(\widehat{\operatorname{se}}_{\log}>0\) with probability tending to one under the stated consistency condition, divide by it on that event. Equivalently, the resulting event is

By the assumed convergence in distribution and Slutsky’s theorem, the fraction converges in distribution to \(N(0,1)\). The standard normal distribution assigns probability \(1-\alpha\) to the interval \([-z_{1-\alpha/2},z_{1-\alpha/2}]\), and has probability zero at either endpoint. Therefore the probability of the displayed event converges to \(1-\alpha\). The event \(\hat{\theta}\leq0\) has probability tending to zero, so it does not change the limit. \(\square\)

What the Endpoints Mean

Write \(c=z_{1-\alpha/2}\widehat{\operatorname{se}}_{\log}\), so \(c\geq0\). The endpoints are \(\hat{\theta}e^{-c}\) and \(\hat{\theta}e^c\). Both are positive whenever \(\hat{\theta}>0\). Their geometric mean is exactly the estimate:

Thus the estimate is the center on the log scale, or equivalently the geometric center on the original scale. The endpoints are not generally equally far from \(\hat{\theta}\) in ordinary units. When \(c>0\), the upper distance is \(\hat{\theta}(e^c-1)\), while the lower distance is \(\hat{\theta}(1-e^{-c})\). Their difference is

The inequality follows because \(e^c+e^{-c}>2\) for \(c>0\). Hence the interval is strictly asymmetric around \(\hat{\theta}\) when \(c>0\). When \(c=0\), the endpoints coincide with the estimate, so there is no strict asymmetry. The multiplicative interpretation is direct: the upper endpoint is \(\hat{\theta}\) multiplied by \(e^c\), and the lower endpoint is \(\hat{\theta}\) multiplied by \(e^{-c}\).

Practical Checks and a Common Mistake

A log-scale interval is useful when positive values span a range where relative differences are more meaningful than absolute differences. For example, multiplying a rate estimate by 1.2 represents the same proportional change whether the original rate is small or large. The interval’s endpoints reflect that multiplicative scale.

Before reporting one, check that the estimated parameter is positive for the observed data, that the standard error corresponds to the log scale, and that the normal approximation is credible for the sample size and model. If a chosen estimator can be zero in finite samples, the logarithm is not defined at those samples; this construction cannot be computed there without changing the estimator or method.

• Common mistake: adding and subtracting the log-scale margin from the original estimate. That produces \(\hat{\theta}\pm c\), which is not the interval derived from normal inference for \(\log\theta\). The correction is to exponentiate the log endpoints, yielding \(\hat{\theta}e^{-c}\) and \(\hat{\theta}e^c\).
• Common mistake: using the standard error of \(\hat{\theta}\) as if it were the standard error of \(\log\hat{\theta}\). The delta-method plug-in approximation is \(\widehat{\operatorname{se}}(\hat{\theta})/\hat{\theta}\), not simply \(\widehat{\operatorname{se}}(\hat{\theta})\).
• Common mistake: claiming strict asymmetry whenever the interval is exponentiated. The margin satisfies \(c\geq0\). The interval is strictly asymmetric around the estimate when \(c>0\); if \(c=0\), both endpoints equal the estimate.