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Gaussian GLMs and Weights · Tutorial 344 of 981

Delta-Method Interpretation for Nonidentity Gaussian Links

Use inverse-link derivatives to quantify uncertainty in fitted means, and recognize when a response-scale interval conflicts with the link’s valid domain.

Beginner 10 min read

What You'll Learn

  • Express the fitted mean as an inverse-link transformation of the estimated linear predictor.
  • Calculate a delta-method standard error from the inverse-link derivative and coefficient covariance.
  • Construct symmetric response-scale and transformed linear-predictor confidence intervals.
  • Interpret the standard errors under log and reciprocal Gaussian links.
  • Diagnose when a reciprocal-link confidence interval approaches or crosses the valid mean domain.

From Linear-Predictor Uncertainty to Mean Uncertainty

In a Gaussian regression with a nonidentity link, the fitted mean is obtained by transforming a linear predictor. Coefficient estimates make that predictor uncertain, and the inverse-link function determines how this uncertainty carries through to the mean. The delta method gives a local, first-order approximation to the resulting standard error.

This tutorial focuses on fitted means at a specified covariate vector. It does not describe the uncertainty of a new Gaussian response, which also includes the response’s residual variation. The uncertainty here is about the estimated mean itself.

Definition: Let \(g\) be a link function and \(h=g^{-1}\) its inverse link. At a covariate vector \(x_0\), the linear predictor is \(\eta_0=x_0^\top\beta\), and the fitted mean is \(\mu_0=h(\eta_0)\). If \(\hat\beta\) estimates \(\beta\), then \(\hat\eta_0=x_0^\top\hat\beta\) and \(\hat\mu_0=h(\hat\eta_0)\).

Suppose an estimated covariance matrix for \(\hat\beta\) is \(\widehat V_\beta\). The estimated variance of the linear predictor at \(x_0\) is \(s_{\hat\eta_0}^2=x_0^\top\widehat V_\beta x_0\), so its standard error is \(s_{\hat\eta_0}\). For Gaussian regression with independent observations and precision weights \(w_i\), if \(\operatorname{Var}(Y_i)=\phi/w_i\), a common covariance estimate is \(\widehat V_\beta=\hat\phi(X^\top W X)^{-1}\), where \(W\) has diagonal entries \(w_i\). The dispersion estimate and coefficient covariance must be chosen consistently with the model and inferential procedure.

Delta-method standard error: If \(h\) is differentiable at \(\hat\eta_0\), the first-order estimated standard error of the fitted mean is $$ \widehat{\operatorname{SE}}(\hat\mu_0) = \left|h'(\hat\eta_0)\right|s_{\hat\eta_0} = \left|h'(\hat\eta_0)\right| \sqrt{x_0^\top\widehat V_\beta x_0}. $$

The derivative measures how sensitive the mean is to a small change in the predictor near its fitted value. A small predictor standard error does not by itself guarantee a small mean standard error: the inverse link may be steep at the fitted predictor. Conversely, a shallow part of the inverse link can reduce the propagated uncertainty.

Why the Delta-Method Formula Works

The approximation follows from a first-order Taylor expansion. The result below states the conditions under which this local calculation yields an asymptotic distribution, rather than merely a heuristic.

Theorem: Suppose \(\hat\eta_n\xrightarrow{p}\eta\), \(\sqrt n(\hat\eta_n-\eta)\xrightarrow{d}N(0,v)\) for \(0<v<\infty\), and \(h\) is differentiable at \(\eta\) with finite nonzero derivative. Then $$ \sqrt n\{h(\hat\eta_n)-h(\eta)\} \xrightarrow{d} N\!\left(0,\{h'(\eta)\}^2v\right). $$

Proof. Differentiability at \(\eta\) means that, for \(\delta\) approaching zero, \(h(\eta+\delta)=h(\eta)+h'(\eta)\delta+r(\delta)\), where \(r(\delta)/\delta\to0\). Set \(\delta_n=\hat\eta_n-\eta\). Consistency gives \(\delta_n\xrightarrow{p}0\). The assumed asymptotic normality implies \(\sqrt n\delta_n=O_p(1)\), so $$ \sqrt n r(\delta_n) = \sqrt n\delta_n\,\frac{r(\delta_n)}{\delta_n} \xrightarrow{p}0. $$ (The expression is defined as zero when \(\delta_n=0\).) Therefore, $$ \sqrt n\{h(\hat\eta_n)-h(\eta)\} = h'(\eta)\sqrt n\delta_n+\sqrt n r(\delta_n). $$ The first term converges in distribution to a normal variable with variance \(\{h'(\eta)\}^2v\), and the second converges in probability to zero. Slutsky’s theorem gives the stated limit. \(\square\)

In regression, \(\hat\eta_0=x_0^\top\hat\beta\) is a scalar estimated predictor. Its estimated variance supplies \(s_{\hat\eta_0}^2\), and substituting the fitted predictor into the derivative gives the practical standard-error formula. The approximation is local: it is most informative when uncertainty in the predictor is small enough that the inverse link is nearly linear over the relevant range.

Log-Link Gaussian Means

For the log link, \(g(\mu)=\log(\mu)\) and \(h(\eta)=\exp(\eta)\). The mean must be positive. Since \(h'(\eta)=\exp(\eta)=h(\eta)\), the fitted mean’s estimated standard error is approximately the fitted mean multiplied by the predictor’s standard error.

Log-link formula: For \(\hat\mu_0=\exp(\hat\eta_0)\), $$ \widehat{\operatorname{SE}}(\hat\mu_0) = \hat\mu_0s_{\hat\eta_0}. $$ Thus the delta-method standard error divided by the fitted mean equals \(s_{\hat\eta_0}\).

Worked Example: A Log-Link Mean and Two Confidence Intervals

At a specified covariate vector, suppose \(\hat\eta_0=\log(10)\) and \(s_{\hat\eta_0}=0.20\). Calculate the fitted mean and its approximate 95% confidence intervals using both the delta method on the mean scale and a normal interval on the predictor scale.

The fitted mean is \(\hat\mu_0=\exp\{\log(10)\}=10\). Its delta-method standard error is $$ \widehat{\operatorname{SE}}(\hat\mu_0) = \exp\{\log(10)\}(0.20) = 10(0.20) = 2. $$ Using \(1.96\) as the standard normal critical value, the symmetric mean-scale interval is $$ 10\pm1.96(2) = [6.08,\ 13.92]. $$

The predictor-scale interval is \(\log(10)\pm1.96(0.20)\), or approximately \([1.9106,\ 2.6946]\). Exponentiating its endpoints gives the transformed interval $$ [\exp(1.9106),\ \exp(2.6946)] \approx [6.76,\ 14.80]. $$ Both intervals estimate uncertainty in the fitted mean, but they are not identical. The transformed interval is asymmetric about 10 and stays positive; the delta interval is symmetric about 10. They agree to first order when predictor uncertainty is small.

Reciprocal-Link Gaussian Means

For the reciprocal link, \(g(\mu)=1/\mu\) and \(h(\eta)=1/\eta\). On the positive-mean branch, both \(\mu\) and \(\eta\) must be positive. Since \(h'(\eta)=-1/\eta^2\), the magnitude of the derivative is \(1/\eta^2=\mu^2\). The derivative’s negative sign indicates that the inverse link decreases: increasing the predictor decreases the fitted mean.

Reciprocal-link formula: On the domain \(\hat\eta_0>0\), \(\hat\mu_0=1/\hat\eta_0\), and $$ \widehat{\operatorname{SE}}(\hat\mu_0) = \frac{s_{\hat\eta_0}}{\hat\eta_0^2} = \hat\mu_0^2s_{\hat\eta_0}. $$

Worked Example: A Reciprocal-Link Mean

Suppose \(\hat\eta_0=0.50\) and \(s_{\hat\eta_0}=0.05\). Calculate the fitted mean, its delta-method standard error, and the two 95% intervals.

The fitted mean is \(\hat\mu_0=1/0.50=2\). Because \(h'(0.50)=-1/(0.50)^2=-4\), its delta-method standard error is $$ \widehat{\operatorname{SE}}(\hat\mu_0) = |-4|(0.05) = 0.20. $$ The symmetric mean-scale interval is \(2\pm1.96(0.20)=[1.608,\ 2.392]\), after rounding.

The predictor interval is \(0.50\pm1.96(0.05)=[0.402,\ 0.598]\), which remains inside the positive domain. Because the reciprocal inverse link is decreasing, its endpoints reverse order when transformed: $$ \left[\frac{1}{0.598},\ \frac{1}{0.402}\right] \approx [1.672,\ 2.488]. $$ This transformed interval is asymmetric about 2. The endpoint reversal is essential: transforming the lower predictor endpoint gives the upper mean endpoint.

Constructing and Interpreting Intervals

A common first-order confidence interval for a fitted mean is the symmetric delta-method interval \(\hat\mu_0\pm c\,\widehat{\operatorname{SE}}(\hat\mu_0)\). Here \(c\) is a suitable critical value, such as the standard normal 95% value \(1.96\) for a large-sample approximation. Alternatively, form an interval for the linear predictor, \([\hat\eta_0-cs_{\hat\eta_0},\hat\eta_0+cs_{\hat\eta_0}]\), and transform its endpoints with \(h\), reversing them if \(h\) is decreasing. For small samples, the critical value and coefficient uncertainty should follow the applicable Gaussian regression inference; the delta method itself does not dictate a particular critical value.

Result: If \(h\) is twice continuously differentiable near \(\hat\eta_0\), then as \(s_{\hat\eta_0}\) becomes small, the transformed predictor interval has endpoints equal to the delta-method endpoints to first order: $$ h(\hat\eta_0\pm cs_{\hat\eta_0}) = h(\hat\eta_0)\pm c h'(\hat\eta_0)s_{\hat\eta_0} +O(s_{\hat\eta_0}^2). $$ When \(h\) is decreasing, the transformed interval’s endpoints are reordered to make the lower endpoint come first.

Justification. Taylor’s theorem around \(\hat\eta_0\) gives, for either sign, $$ h(\hat\eta_0\pm cs_{\hat\eta_0}) = h(\hat\eta_0) \pm c h'(\hat\eta_0)s_{\hat\eta_0} +\frac{1}{2}h''(\xi_\pm)c^2s_{\hat\eta_0}^2, $$ where \(\xi_\pm\) lies between \(\hat\eta_0\) and \(\hat\eta_0\pm cs_{\hat\eta_0}\). Continuity of \(h''\) near \(\hat\eta_0\) makes the final term \(O(s_{\hat\eta_0}^2)\). The first-order endpoint displacement has magnitude \(c|h'(\hat\eta_0)|s_{\hat\eta_0}\), which is precisely the delta-method margin. Reordering endpoints when needed produces the stated comparison. \(\square\)

Worked Example: A Reciprocal Interval Near the Domain Boundary

Suppose a reciprocal-link fit has \(\hat\eta_0=0.08\) and \(s_{\hat\eta_0}=0.06\). Calculate the delta-method interval and inspect the predictor interval before transforming it.

The fitted mean is \(1/0.08=12.5\). The derivative is \(-1/(0.08)^2=-156.25\), so the delta-method standard error is \(156.25(0.06)=9.375\). The symmetric 95% interval is $$ 12.5\pm1.96(9.375) = [-5.875,\ 30.875]. $$ Its negative lower endpoint is outside the positive-mean domain.

The predictor interval is \(0.08\pm1.96(0.06)=[-0.0376,\ 0.1976]\), which crosses zero. The reciprocal is undefined at zero, so transforming both endpoints does not produce a finite interval. Intersecting this predictor interval with the positive domain leaves \((0,\ 0.1976]\); its reciprocal image is approximately \([5.061,\ \infty)\). The example shows why the derivative-based standard error can become large near the boundary and why the predictor interval’s domain must be checked before transformation.

Common Mistake and Practical Checks

A frequent mistake is to compute a predictor interval and transform its endpoints without checking whether the whole interval lies in the inverse link’s domain. For a reciprocal link, an interval crossing zero cannot be transformed by taking reciprocals of its two endpoints: zero has no reciprocal, and the inverse link is not continuous across that boundary. Check the domain first. For a log link, a symmetric delta interval can also extend below zero even though every fitted mean under the link is positive.

The two interval constructions answer related but distinct practical needs. The symmetric delta interval is a local approximation on the response scale. The transformed interval reflects the shape of the inverse link and can preserve its domain when the predictor interval lies entirely within it. Neither procedure adds residual variation for a future observation; that would concern prediction rather than uncertainty in the fitted mean.

1
Find predictor uncertainty.
Calculate \(s_{\hat\eta_0}=\sqrt{x_0^\top\widehat V_\beta x_0}\) using the coefficient covariance appropriate to the Gaussian model.
2
Evaluate the inverse link and derivative.
Compute \(\hat\mu_0=h(\hat\eta_0)\) and \(|h'(\hat\eta_0)|\), checking that the fitted predictor is in the valid domain.
3
Report uncertainty with its limitations.
Use \(|h'(\hat\eta_0)|s_{\hat\eta_0}\) for the delta-method standard error. If transforming a predictor interval, check its entire domain and reverse endpoints for a decreasing inverse link.
Key takeaway: The delta method propagates linear-predictor uncertainty through the local slope of the inverse link. For a log link, the approximate standard error is \(\hat\mu_0s_{\hat\eta_0}\); for a positive-branch reciprocal link, it is \(\hat\mu_0^2s_{\hat\eta_0}\). The reciprocal derivative’s steepness near zero makes domain checks especially important.

Check Your Understanding

Use the inverse-link derivative and the fitted predictor’s uncertainty to answer these questions.

  1. For a log link, if \(\hat\mu_0=8\) and \(s_{\hat\eta_0}=0.15\), what is the delta-method standard error of the fitted mean?
  2. For a reciprocal link, why must a predictor interval that crosses zero be checked before transforming it?
  3. Suppose a decreasing inverse link maps a predictor interval \([a,b]\) to the mean scale. In what order should the transformed endpoints be reported?
  4. Explain why a delta-method confidence interval for a fitted mean does not include the Gaussian residual variation of a future response.
  5. What feature of the inverse link makes the reciprocal-link mean standard error especially sensitive to predictor uncertainty near zero?
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