Why the Link Changes the Curvature
Newton–Raphson uses the curvature of the log-likelihood at the observed data, while Fisher scoring uses its expectation under the model. These matrices need not generally agree. In a canonical-link generalized linear model with fixed dispersion, however, the regression-coefficient Hessian has no data-dependent residual term. The observed and expected information therefore coincide at every coefficient value where the model is defined—not only at the maximum-likelihood estimate.
This is a statement about the curvature for the regression coefficients. It does not say that the response data do not matter: the data still determine the score and hence the update direction. Nor does it assert the same equality for dispersion parameters or for every noncanonical link.
Here \(x_i\) is the fixed predictor vector for observation \(i\), and \(\beta\) is the coefficient vector. The form above allows different fixed positive scale factors \(a_i(\phi)\); for example, known prior weights can be incorporated into them. We assume the model is regular enough that its stated likelihood derivatives exist. The exponential-family mean and variance identities are \(E_\beta[Y_i]=b'(\theta_i)\) and \(\operatorname{Var}_\beta(Y_i)=a_i(\phi)b''(\theta_i)\).
Deriving the Score and Hessian
Under the canonical link, \(\theta_i=x_i^\mathsf{T}\beta\). Since \(x_i\) does not depend on \(\beta\), differentiating the contribution \(\ell_i\) gives
Differentiating once more gives the Hessian contribution:
The response \(Y_i\) appears in the score but not in this Hessian. That cancellation is the key fact. Summing over independent observations yields the score vector \(U(\beta)=\nabla_\beta\ell(\beta)\) and the information matrix
Proof. For each observation, the second derivative derived above is \(-b''(\theta_i)x_i x_i^\mathsf{T}/a_i(\phi)\), which is a function of the fixed design, \(\beta\), and fixed dispersion only. Thus \(J(\beta)\) is exactly the displayed sum for every possible response vector. Taking its expectation under the model leaves it unchanged: \(I(\beta)=E_\beta[J(\beta)]=J(\beta)\). This proves the claim. \(\square\)
The same result can be written using the design matrix \(X\). Let \(W(\beta)\) be diagonal with entries \(b''(x_i^\mathsf{T}\beta)/a_i(\phi)\). Then \(J(\beta)=I(\beta)=X^\mathsf{T}W(\beta)X\). This matrix may be singular if the design does not identify all coefficients; equality still holds, although an update requiring its inverse would then need further attention.
Three Worked Examples
Worked Example: Bernoulli Logistic Regression
Take three independent Bernoulli responses with predictor vectors \(x_1=(1,0)^\mathsf{T}\), \(x_2=(1,1)^\mathsf{T}\), and \(x_3=(1,2)^\mathsf{T}\). Use the logit link, so \(\theta_i=x_i^\mathsf{T}\beta\), and evaluate at \(\beta=(0,\log 2)^\mathsf{T}\). Suppose the observed responses are \(Y=(1,0,1)\).
For a Bernoulli response, \(b(\theta)=\log(1+e^\theta)\), \(b'(\theta)=p\), and \(b''(\theta)=p(1-p)\), where \(p=e^\theta/(1+e^\theta)\). The three linear predictors are \(0,\log 2,\log 4\), giving fitted probabilities \(1/2,2/3,4/5\). Thus the information weights are \(1/4,2/9,4/25\).
The score is \(\sum_i x_i(Y_i-p_i)\). Its intercept component is \(1/2-2/3+1/5=1/30\), and its slope component is \(0(1/2)+1(-2/3)+2(1/5)=-4/15\). Therefore \(U(\beta)=(1/30,-4/15)^\mathsf{T}\).
The observed information is the weighted sum \(\sum_i p_i(1-p_i)x_ix_i^\mathsf{T}\):
The same matrix is the expected information: the Hessian has these weights regardless of whether the observed response is 0 or 1. In particular, the observed outcomes affect the score above, not the information matrix at this \(\beta\).
Worked Example: Poisson Regression With a Log Link
Consider two independent Poisson counts with \(x_1=1\), \(x_2=2\), log link, and coefficient \(\beta=\log 2\). The canonical parameter is \(\theta_i=\log\mu_i=x_i\beta\). At this value the means are \(\mu_1=2\) and \(\mu_2=4\). Suppose the observed counts are \(Y_1=1\) and \(Y_2=3\).
For the Poisson family, \(b(\theta)=e^\theta\), so \(b'(\theta)=b''(\theta)=\mu\), and \(a(\phi)=1\). The score is \(\sum_i x_i(Y_i-\mu_i)\), which here equals \(1(1-2)+2(3-4)=-1-2=-3\).
The observed information is \(\sum_i \mu_i x_i^2=2(1)^2+4(2)^2=2+16=18\). Because the Hessian does not contain \(Y_i\), its expectation is also \(18\). The equality holds at this coefficient value even though the observed score is nonzero.
Worked Example: Gaussian Regression With an Identity Link
Let \(Y_1,Y_2\) be independent normal observations with known variance \(\sigma^2=2\), predictors \(x_1=1,x_2=2\), and identity link. The normal canonical parameter is \(\theta_i=\mu_i\), so the identity link is canonical. Evaluate at \(\beta=1.5\), and take observed values \(Y_1=2,Y_2=4\).
In canonical exponential-family form, \(a(\phi)=\sigma^2=2\) and \(b(\theta)=\theta^2/2\). Hence \(b''(\theta)=1\). The fitted means are \(\mu_1=1.5\) and \(\mu_2=3\). The score is \(\sum_i x_i(Y_i-\mu_i)/2=[1(0.5)+2(1)]/2=1.25\).
The observed information is \(\sum_i x_i^2/2=(1+4)/2=2.5\). It has no dependence on the observed values, so the expected information is also \(2.5\). This is the familiar fixed-variance least-squares curvature expressed as the same observed–expected information equality.
When Newton and Fisher Scoring Give the Same Step
For a nonsingular observed information matrix, a Newton–Raphson coefficient update is \(\beta_{\mathrm{new}}=\beta+J(\beta)^{-1}U(\beta)\). Fisher scoring replaces \(J(\beta)\) by \(I(\beta)\). Under the theorem’s assumptions those matrices are equal, so the two updates agree exactly, provided the inverse exists.
Proof. The Newton increment is \(\Delta_N=J(\beta)^{-1}U(\beta)\), and the Fisher-scoring increment is \(\Delta_F=I(\beta)^{-1}U(\beta)\). By the theorem, \(I(\beta)=J(\beta)\). Since this common matrix is nonsingular, the two inverses are equal, and therefore \(\Delta_N=\Delta_F\). \(\square\)
This conclusion concerns the proposed full coefficient increment. It does not guarantee that every full step increases the likelihood, that the likelihood has a finite maximizer, or that the algorithm converges from every starting value. Those are separate questions about optimization and the likelihood surface.
Why the Canonical Link Matters
The cancellation can be seen by comparing a general link with the canonical one. Write the canonical parameter as a twice-differentiable function \(h(\eta_i)\) of the linear predictor, so \(\theta_i=h(\eta_i)\). Differentiating the likelihood contribution with respect to \(\eta_i\) gives \((Y_i-b'(\theta_i))h'(\eta_i)/a_i(\phi)\). Its second derivative is
Consequently, the observed information contribution for \(\beta\) is
The second term depends on the observed residual. Its expectation is zero because \(E_\beta[Y_i]=b'(\theta_i)\), but it need not be zero for the observed data. With a canonical link, \(h(\eta)=\eta\), so \(h'=1\) and \(h''=0\); the residual-dependent curvature term vanishes identically. This pinpoints the mechanism without claiming that every noncanonical model must have unequal information in every special data set.
Common Mistake and Practical Scope
Common mistake: “The observed information uses the data, so it must differ from expected information; therefore Newton and Fisher scoring always take different steps.” The first statement is not a valid general argument. In a canonical-link GLM the likelihood Hessian for \(\beta\) is independent of the responses, so averaging it over the model changes nothing. The score still uses the observed responses, and both methods use that same score.
The fixed-dispersion qualification also matters. The result established here concerns the regression-coefficient block with \(\phi\) held fixed. If dispersion is estimated jointly, derivatives involving that additional parameter must be considered separately; the coefficient-block equality alone does not establish equality of the full information matrices. Similarly, a non-linear predictor in \(\beta\) can introduce additional second-derivative terms, so the canonical-link result for a linear predictor does not automatically apply.
Check Your Understanding
Use the derivations and examples above to answer these questions.
- Which part of the canonical-link likelihood derivative makes the observed Hessian independent of the response?
- For a Poisson log-link model, why can two different observed count vectors have different scores but the same information at a fixed \(\beta\)?
- What additional condition is needed to conclude that Newton and Fisher scoring produce the same coefficient increment?
- In the general-link expression, which term can make observed information differ from expected information?
- Does equality of observed and expected information imply that a full Newton step always increases the likelihood? Explain briefly.