Delta-Method Intervals for Expected Counts

From Linear-Predictor Uncertainty to Mean Uncertainty

In a Poisson log-link model, the mean at a covariate profile is obtained by exponentiating its linear predictor. Uncertainty is often first calculated for that predictor, because it is a linear combination of estimated coefficients. To report uncertainty for the expected count itself, we need to carry that uncertainty through the exponential transformation.

The delta method gives a first-order approximation to the variance after a smooth transformation. Applied to the exponential function, it says that uncertainty in the linear predictor is multiplied, approximately, by the fitted mean. This leads to a simple standard error and a direct-scale Wald interval. The approximation is useful, but it does not guarantee that the interval respects the positive range of a Poisson mean or has accurate coverage in a given sample.

If the model includes a known offset, it is added to \(\eta_0\) before exponentiation. The derivative of the predictor with respect to \(\beta\) remains \(x_0\), so the variance calculation below is unchanged. In applications, \(\widehat V\) is typically obtained from the fitted model’s coefficient covariance matrix; it is the covariance matrix of the coefficient estimates, not a matrix of coefficient standard errors placed on the diagonal.

The Delta-Method Standard Error

For a differentiable function \(g\), the first-order delta method approximates the variance of \(g(\hat\theta)\) by the squared derivative of \(g\) at the true parameter, multiplied by the variance of \(\hat\theta\). Here \(g(\eta)=\exp(\eta)\), whose derivative is also \(\exp(\eta)\). Thus, near the fitted predictor, a change in \(\eta_0\) of size \(d\) corresponds to a change in the mean of approximately \(\mu_0d\).

The variance result follows directly by a first-order expansion. It is important that the predictor variance includes the covariances among coefficient estimates: for a design vector with several nonzero entries, those covariances contribute to \(x_0^\mathsf{T}\widehat Vx_0\).

Why the Approximation Works

The justification is asymptotic. The result below states the conditions that connect uncertainty in the estimated coefficients to uncertainty in the fitted mean. It assumes the usual regular large-sample behavior of the coefficient estimator; it does not assert that a normal approximation is exact for a finite sample.

For a regular, identifiable Poisson log-link model with a fixed-dimensional parameter and a nonsingular information matrix, asymptotic normality and a consistent model-based covariance estimate are standard large-sample conditions. The theorem also states the needed conditions directly, so it can be used when a different valid covariance estimator is employed.

Proof. Write \(\hat\eta_0=x_0^\mathsf{T}\hat\beta\), so \(\hat\eta_0-\eta_0=x_0^\mathsf{T}(\hat\beta-\beta)\). The assumed convergence and the linear transformation rule for normal limits give $$ \sqrt n(\hat\eta_0-\eta_0)\xrightarrow{d} N\bigl(0,x_0^\mathsf{T}\Sigma x_0\bigr). $$ In particular, \(\hat\eta_0-\eta_0=O_p(n^{-1/2})\), so it converges in probability to zero. Taylor’s theorem for the exponential function gives, for some value \(\xi_n\) between \(\hat\eta_0\) and \(\eta_0\), $$ \hat\mu_0-\mu_0 =\exp(\hat\eta_0)-\exp(\eta_0) =\exp(\xi_n)(\hat\eta_0-\eta_0). $$ Because \(\hat\eta_0\xrightarrow{p}\eta_0\), also \(\xi_n\xrightarrow{p}\eta_0\), and hence \(\exp(\xi_n)/\mu_0\xrightarrow{p}1\). Slutsky’s theorem now yields $$ \sqrt n\,\frac{\hat\mu_0-\mu_0}{\mu_0} \xrightarrow{d}N\bigl(0,x_0^\mathsf{T}\Sigma x_0\bigr). $$ Meanwhile, \(n\widehat V\xrightarrow{p}\Sigma\) implies \(n x_0^\mathsf{T}\widehat Vx_0\xrightarrow{p}x_0^\mathsf{T}\Sigma x_0\). Dividing by the resulting consistent estimate of the limiting standard deviation proves the displayed standardized convergence. The limiting standard normal distribution assigns probability \(1-\alpha\) to \([-z_{1-\alpha/2},z_{1-\alpha/2}]\); therefore the corresponding Wald interval has asymptotic coverage \(1-\alpha\). \(\square\)

When a Direct-Scale Interval Can Be Poor

The direct-scale interval uses a symmetric normal approximation around \(\hat\mu_0\). But a Poisson mean must be positive, and exponentiation creates a skewed transformation: equal-sized changes in the predictor do not translate into equal-sized changes in the mean. When predictor uncertainty is appreciable, a symmetric interval on the mean scale can extend below zero or poorly represent that asymmetry. These are warnings about the approximation, not automatic tests that tell us exactly how inaccurate coverage will be.

Common Mistakes and Practical Checks

• Dropping the square. The approximate variance is \(\hat\mu_0^2\hat v_0\), while the standard error is \(\hat\mu_0\sqrt{\hat v_0}\). Confusing these gives the wrong scale.
• Ignoring coefficient covariances. Compute \(x_0^\mathsf{T}\widehat Vx_0\) using the full covariance matrix. Keeping only diagonal entries is generally incorrect.
• Calling the interval exact. Its standard-normal justification is asymptotic and depends on the estimator and covariance estimate behaving as specified. The delta method is a local approximation.
• Treating a negative endpoint as a negative fitted mean. The fitted mean is an exponential and is positive. The endpoint instead warns that the direct-scale interval does not respect the parameter’s range.
• Confusing mean uncertainty with count variability. This interval concerns the model’s expected count at a profile. A future observed count also varies around that mean; that additional variation is not included here.

In practice, check the predictor variance, the resulting standard error, and the interval endpoints before interpreting the result. A negative endpoint or a very wide interval deserves particular caution. Even if both endpoints are positive, the direct-scale interval remains an approximation whose accuracy cannot be established from endpoint positivity alone.