Why Rate Intervals Need a Transformation
In a Poisson rate model with a log link and an exposure offset, the estimated log rate at a specified covariate profile is a linear combination of fitted coefficients. The rate itself is the exponential of that combination. Because exponentiation is nonlinear, uncertainty in the log rate cannot simply be copied onto the rate scale. The delta method supplies a standard error for the transformed estimate, and the log scale also gives a natural way to construct a confidence interval whose endpoints are positive.
The calculation depends on more than the separate standard errors of the coefficients. If several estimated coefficients contribute to a linear predictor, their covariances contribute too. The exposure unit matters as well: a rate per 1,000 person-years and a rate per person-year have different numerical values, so the estimate and both interval endpoints must use the same unit.
Rate Estimates and Their Delta-Method Standard Errors
Here, \(\mathbf{x}\) includes the intercept coordinate when the model has an intercept. The covariance matrix \(\widehat V\) estimates the covariance of the fitted coefficient vector; it may be the model-based covariance estimate or another justified covariance estimate. The vector \(\mathbf{x}\) is held fixed when calculating the uncertainty for the rate at a specified profile.
To see the formula, regard the rate as a function of the coefficients, \(g(\boldsymbol\beta)=\exp(\mathbf{x}^{\mathsf T}\boldsymbol\beta)\). Its gradient is \[ \nabla g(\boldsymbol\beta)=\exp(\mathbf{x}^{\mathsf T}\boldsymbol\beta)\mathbf{x}. \] The first-order delta-method variance estimate is the gradient, evaluated at \(\hat{\boldsymbol\beta}\), multiplied on both sides of \(\widehat V\). Thus \[ \widehat{\operatorname{Var}}(\hat r) =\exp(2\hat\eta)\mathbf{x}^{\mathsf T}\widehat V\mathbf{x}, \] whose square root is the stated standard error. This is a first-order approximation, not an exact finite-sample variance formula.
Worked Example: A Rate and Its Interval in the Model's Exposure Units
Suppose a fitted model gives a log-rate estimate \(\hat\eta=-2.5\) at the profile of interest, with estimated variance \(0.0616\). The model's rate unit is events per 1,000 person-years. Find the fitted rate and a 95% Wald confidence interval.
The standard error of the estimated log rate is \(\sqrt{0.0616}\), approximately \(0.2481935\). The fitted rate is \[ \hat r=\exp(-2.5)\approx 0.082085 \] events per 1,000 person-years.
First form the interval on the log-rate scale, using \(1.96\) as the standard normal 97.5th percentile: \[ -2.5\pm 1.96\sqrt{0.0616}. \] The margin is approximately \(0.486459\), so the log-scale endpoints are approximately \(-2.986459\) and \(-2.013541\). Exponentiating each endpoint gives \[ \left[\exp\!\left(-2.5-1.96\sqrt{0.0616}\right), \exp\!\left(-2.5+1.96\sqrt{0.0616}\right)\right] \approx [0.05047,\ 0.13352]. \] The fitted rate and interval are all expressed in events per 1,000 person-years.
One person-year is one-thousandth of 1,000 person-years. Therefore, the numerical rate per person-year is the rate per 1,000 person-years divided by 1,000. The point estimate becomes approximately \(0.00008208\) events per person-year, and the interval becomes approximately \([0.00005047,\ 0.00013352]\) events per person-year. Both endpoints were converted using the same factor as the estimate.
Coefficient Covariance Is Part of the Calculation
For a linear predictor containing multiple coefficients, the variance is a quadratic form. In coordinates, if \(\mathbf{x}=(x_1,\ldots,x_p)^{\mathsf T}\), then \[ \widehat{\operatorname{Var}}(\hat\eta) =\mathbf{x}^{\mathsf T}\widehat V\mathbf{x} =\sum_{j=1}^{p}x_j^2\widehat V_{jj} +2\sum_{j<k}x_jx_k\widehat V_{jk}. \] The off-diagonal terms are the covariance contributions. Dropping them is generally not justified; depending on their signs and the profile \(\mathbf{x}\), doing so can either inflate or reduce the estimated uncertainty.
Worked Example: A Binary Predictor and a Negative Covariance
Suppose an intercept-and-group model has fitted coefficient vector \(\hat{\boldsymbol\beta}=(-3,\ 0.4)^{\mathsf T}\) and estimated covariance matrix \[ \widehat V= \begin{pmatrix} 0.04 & -0.01\\ -0.01 & 0.0225 \end{pmatrix}. \] Find a 95% Wald interval for the rate in the group coded \(1\), whose covariate vector is \(\mathbf{x}=(1,\ 1)^{\mathsf T}\).
The estimated log rate is \(\hat\eta=-3+0.4=-2.6\), so the fitted rate is \(\exp(-2.6)\approx 0.07427\) in the model's rate units. Its log-rate variance estimate includes both the coefficient variances and their covariance: \[ \mathbf{x}^{\mathsf T}\widehat V\mathbf{x} =0.04+0.0225+2(-0.01)=0.0425. \] This is positive, and the standard error is \(\sqrt{0.0425}\approx 0.20616\).
The log-scale interval is \[ -2.6\pm1.96\sqrt{0.0425} \approx [-3.00406,\ -2.19594]. \] Exponentiating gives an approximate rate interval of \([0.04959,\ 0.11125]\). Both limits are positive, as required for a rate.
If the covariance were incorrectly omitted, the log-rate variance would be calculated as \(0.04+0.0225=0.0625\), giving a standard error of \(0.25\), rather than \(0.20616\). That interval would not reflect the covariance information in the fitted model.
Wald Intervals on the Log-Rate Scale
To justify this interval, suppose the coefficient estimator is consistent and asymptotically normal: for a positive-definite matrix \(\Sigma\), \[ \sqrt n(\hat{\boldsymbol\beta}-\boldsymbol\beta_0) \xrightarrow{d}N(\mathbf{0},\Sigma), \] and suppose \(n\widehat V\xrightarrow{p}\Sigma\). For fixed \(\mathbf{x}\), the function \(g(\boldsymbol\beta)=\exp(\mathbf{x}^{\mathsf T}\boldsymbol\beta)\) is differentiable, with gradient \(g(\boldsymbol\beta)\mathbf{x}\). The multivariate delta method therefore gives \[ \sqrt n\{g(\hat{\boldsymbol\beta})-g(\boldsymbol\beta_0)\} \xrightarrow{d} N\!\left(0,\ g(\boldsymbol\beta_0)^2\mathbf{x}^{\mathsf T}\Sigma\mathbf{x}\right). \] Because the log-rate estimate is a linear combination of the coefficients, its estimated variance is \(\mathbf{x}^{\mathsf T}\widehat V\mathbf{x}\). Equivalently, the estimated log rate is approximately normal with mean \(\mathbf{x}^{\mathsf T}\boldsymbol\beta_0\) and variance \(\mathbf{x}^{\mathsf T}\widehat V\mathbf{x}\). The resulting Wald interval on that scale has approximate coverage \(1-\alpha\). Exponentiation is strictly increasing, so the transformed interval has the same approximate coverage for the rate. This is a large-sample result; it does not guarantee exact coverage in a small sample. \(\square\)
The interval is often called a log-scale Wald interval. The rate-scale delta-method standard error and this interval are related but not identical procedures: a symmetric interval formed as \(\hat r\pm z\widehat{\operatorname{SE}}(\hat r)\) uses a linear approximation directly on the rate scale, while exponentiating the log-scale endpoints respects the nonlinear transformation and ensures positive limits.
Worked Example: A Continuous Predictor and a Change of Rate Units
Consider a fitted log-rate model with an intercept and a continuous predictor. At the covariate value \(x=2\), the design vector is \(\mathbf{x}=(1,\ 2)^{\mathsf T}\). Let \[ \hat{\boldsymbol\beta}=(-4,\ 0.03)^{\mathsf T}, \qquad \widehat V= \begin{pmatrix} 0.09 & -0.006\\ -0.006 & 0.0016 \end{pmatrix}. \] Suppose the fitted rate unit is events per 10,000 person-days. Calculate the fitted rate and its 95% interval, then convert them to events per person-day.
At \(x=2\), the estimated log rate is \(-4+2(0.03)=-3.94\), so the fitted rate is \(\exp(-3.94)\approx 0.01945\) events per 10,000 person-days. The log-rate variance estimate is \[ \mathbf{x}^{\mathsf T}\widehat V\mathbf{x} =0.09+4(0.0016)+4(-0.006) =0.0724. \] Thus the log-rate standard error is \(\sqrt{0.0724}\approx 0.26907\), and the 95% margin is \(1.96(0.26907)\approx0.52738\).
The log-scale interval is approximately \[ [-3.94-0.52738,\ -3.94+0.52738] =[-4.46738,\ -3.41262]. \] Exponentiating gives approximately \([0.01148,\ 0.03295]\) events per 10,000 person-days. Since one person-day is one ten-thousandth of 10,000 person-days, divide the estimate and both limits by 10,000. The fitted rate is approximately \(0.000001945\) events per person-day, with interval \([0.000001148,\ 0.000003295]\) events per person-day.
Why Changing Units Scales the Whole Interval
Changing the rate's exposure unit changes the numerical scale of the rate, not the underlying uncertainty about the fitted log-rate at a fixed profile. For example, a rate per 1,000 person-years is 1,000 times the rate per person-year. The coefficient covariance calculation for a given profile is unchanged by this conversion; only the rate scale and its interval endpoints are multiplied by the corresponding positive conversion factor.
Indeed, write the original rate estimate as \(\exp(\hat\eta)\). In the new unit it is \(a\exp(\hat\eta)=\exp(\hat\eta+\log a)\). The log-scale interval endpoints are shifted by \(\log a\), becoming \(\hat\eta\pm z\,\widehat{\operatorname{SE}}(\hat\eta)+\log a\). Exponentiating each shifted endpoint gives \(a\exp(\hat\eta\pm z\,\widehat{\operatorname{SE}}(\hat\eta))\), which is exactly \(aL\) and \(aU\). The scale factor is positive, so the order of the endpoints is preserved. \(\square\)
Common Mistakes and Interpretation
- Ignoring coefficient covariance. Using only the diagonal of \(\widehat V\) omits the terms \(2x_jx_k\widehat V_{jk}\). Use the full quadratic form unless the relevant covariances are known to be zero.
- Exponentiating the point estimate but not the interval. A confidence interval for a log rate has endpoints on the log scale. Exponentiate both endpoints to obtain an interval for the rate.
- Converting only the point estimate. A unit conversion multiplies the point estimate and both confidence limits by the same factor. Otherwise, the reported interval and estimate do not refer to the same rate scale.
- Calling a symmetric rate-scale interval the log-scale interval. The interval \(\hat r\pm z\widehat{\operatorname{SE}}(\hat r)\) is a different, first-order approximation. The exponentiated log-scale interval is generally asymmetric around \(\hat r\), and its endpoints remain positive.
- Confusing a rate interval with uncertainty about a future count. This interval quantifies uncertainty in the fitted rate parameter at a specified covariate profile. It is not an interval for the number of events that will occur in a future exposure period.
A useful reporting habit is to state the covariate profile, the rate unit, the point estimate, and the interval together. The covariance matrix determines uncertainty for the chosen profile; the exposure unit determines how the final numerical rate should be read.
Check Your Understanding
Use the full covariance matrix and keep the units attached to every reported rate.
- For a fixed profile \(\mathbf{x}\), write the delta-method standard error of \(\exp(\mathbf{x}^{\mathsf T}\hat{\boldsymbol\beta})\) in terms of \(\widehat V\).
- Why can omitting an off-diagonal covariance make the estimated standard error either too large or too small?
- What operation converts a log-scale Wald interval into a confidence interval for a rate?
- A rate is reported per 100 person-years. Describe how to convert its estimate and interval to rates per person-year.
- In one sentence, distinguish an interval for a fitted rate from an interval for a future count.