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Write $e=y-g_{\theta }(x)$ and $h_{\theta }(x)=\partial g_{\theta }(x)/\partial \theta$. Holding the nuisance parameter $\eta$ fixed, the parametric score function is the derivative with respect to $\theta$ of the one-observation log-likelihood. The joint probability density function isBecause $\partial e/\partial \theta =-h_{\theta }(x)$, differentiating givesFor instance $E_{\eta }{h}_{\theta }(X{)}^2<\infty$ ensures a square-integrable score function. Its expected value is zero by independence and the centered normal distribution of the error. This is the Gaussian regression score.

Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.

Use the normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design matrix $X$, independent errors ${\epsilon }_i\sim N(0,{\sigma }^2)$, and ${\sigma }^2>0$. Its log-likelihood isFor fixed ${\sigma }^2$, differentiating with respect to $\beta$ gives $X^T(Y-X\hat{\beta })=0$. Since $X^{T}X$ is positive-definite, the unique least-squares estimate isLet $\mathrm{RSS}=\Vert Y-X\hat{\beta }{\Vert }^2$. The variance score equation is $-n/(2{\sigma }^2)+\mathrm{RSS}/(2{\sigma }^4)=0$, givingThis is the maximum-likelihood estimator, whereas the residual standard error reported by linear-regression software is $s=\sqrt{\mathrm{RSS}/(n-p)}$. The latter uses the unbiased residual estimate of Gaussian noise variance. These formulas assume $n>p$ and positive RSS; exact interpolation gives a variance-boundary degeneracy rather than an interior maximum.

For the Poisson regression, the log-likelihood isSince ${\mu }_i=e^{x_i^T\beta }$, the score function and its derivative areConsequently a finite maximum-likelihood estimator satisfiesThese nonlinear equations are normally solved by iteratively reweighted least squares. If $X$ has full column rank, the Hessian is negative definite and any finite solution is the unique maximizer. Full rank does not by itself rule out a boundary solution at infinite coefficients for every conceivable data set. The Poisson generalized linear model has dispersion one, so there is no separate scale parameter in this likelihood.

The saturated model estimates each Poisson mean separately as $Y_i$, permitting the boundary value zero when $Y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson deviancewith $0\log (0/{\hat{\mu }}_i)=0$. A zero observation contributes $2{\hat{\mu }}_i$. If the model includes an intercept, the score equation gives ${\sum }_i{Y}_i={\sum }_i{\hat{\mu }}_i$, so the summed linear terms cancel; the unsimplified expression applies generally.

For goodness of fit, the null hypothesis is that the specified independent Poisson mean model is correct; the alternative is an unrestricted collection of means. When a quadratic likelihood approximation is valid, for example with fixed $n$ and large expected counts, $D$ has approximately a chi-squared distribution with $n-p$ statistical degrees of freedom. Reject for a large upper-tail deviance. Merely increasing the number of sparse observations does not make this saturated-model reference reliable: the alternative dimension also increases. With small fitted means, replication-based checks or a parametric bootstrap are more defensible. This qualification concerns the absolute goodness-of-fit deviance; a fixed-dimensional nested likelihood-ratio comparison can have a better chi-squared approximation.

The likelihood-ratio statistic isThe null hypothesis $\pi =0$ is at the boundary of the permitted mixture weights. Assuming the remaining parameters are identifiable and interior, including $\tau >0$, an efficient normalized score for $\pi$ is asymptotically $Z\sim N(0,1)$ under the null hypothesis. The locally unconstrained quadratic log-likelihood maximizer is proportional to $Z$, while the constrained maximizer truncates it at zero. Thus $G^2\Rightarrow (\max (0,Z){)}^2$, giving the one-sided boundary likelihood-ratio limitHere ${\delta }_0$ is a point mass at zero. For a positive observed statistic the upper-tail probability is half the ordinary ${\chi }_1^2$ upper tail. The 5% critical value is the 90th percentile of ${\chi }_1^2$, about 2.7055, displayed as 2.71. Hence the reported likelihoods give a very marginal rejection of $\pi =0$ at 5%, with $p\simeq 0.04955$.

Using the ordinary 95th-percentile cutoff 3.84 would be inappropriate for this one-sided boundary test. The conclusion is close enough to the threshold that likelihood rounding matters: since each log-likelihood is printed to two decimals, the unrounded statistic can differ from 2.72 by nearly 0.02. The intended calculation rejects using the reported values, but an exact decision from the fitted data requires unrounded likelihoods. If the nuisance dispersion is also on its boundary or the mixture is not identifiable, the stated half-and-half null distribution needs reconsideration.

Write $m_i=\mu +{\alpha }_i$. Apart from a constant, the log-likelihood isFor each group,Thus maximum likelihood estimation sets ${\hat{m}}_i={\overline{Y}}_i$. The corner-point constraint ${\alpha }_1=0$ identifies $m_1=\mu$, givingThe baseline mean is the first group mean, rather than the grand mean, because of the chosen identifiability constraint.