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In a regular model let ${\ell }_N(\theta )$ be the log-likelihood, $U_N(\theta )=\nabla {\ell }_N(\theta )$ its score, $I_N(\theta )$ its total Fisher information, and $\hat{\theta }$ the unrestricted maximum-likelihood estimator. For a point null $\theta ={\theta }_0$ in a p-dimensional parameter space, the standard statistics areThese define the Wald test, score test and likelihood-ratio test. With an interior true parameter, increasing information, suitable differentiability and asymptotic normality, all have asymptotic chi-squared distribution with p degrees of freedom under the null. Observed information can replace expected information when it is asymptotically equivalent.

For q smooth independent restrictions $g(\theta )=0$, replace the Wald statistic by $g(\hat{\theta }{)}^T[G{I}_N(\hat{\theta }{)}^{-1}{G}^T{]}^{-1}g(\hat{\theta })$, where G is the derivative of g at the estimate; compare with chi-squared q. The likelihood-ratio statistic uses the difference between unrestricted and null-restricted maxima. The score statistic is computed at the restricted maximum and uses efficient information for the tested coordinates: with nuisance coordinates lambda this is $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$. Inverting the acceptance inequalities for point nulls produces approximate coverage-$1-\alpha$ confidence regions; for example the Wald region consists of theta satisfying $(\hat{\theta }-\theta {)}^T{I}_N(\hat{\theta })(\hat{\theta }-\theta )\le {\chi }_{p,1-\alpha }^2$. Analogous score and likelihood-ratio inequalities give their confidence regions.

For the Poisson family, put $Y_{-1}=1$ to include the initial observation, and defineMultiplying the initial and conditional Poisson probabilities gives, on the supported observation sequences,The parameter-free term is ${\sum }_{i=0}^n[Y_i\log Y_{i-1}-\log (Y_i!)]$, interpreting $0\log 0$ as zero. A positive count after a zero preceding count is outside the support.

For $S_n>0$, strict concavity gives unconstrained maximum $\tilde{\theta }=S_n/T_n$, and the upper endpoint restriction givesThere is an endpoint qualification in the printed question. With probability $e^{-\theta }$, $Y_0=0$ and all later counts are zero. Then $S_n=0$, $T_n=1$ and the likelihood is $e^{-\theta }$, which has no maximum on the stated space $0<\theta \le 1$. The formula produces zero only as an extended estimate on the closure $[0,1]$. This is a likelihood supremum at an excluded boundary, and zero cannot be called an actual maximum in the original space.

The observed negative second derivative is $S_n/{\theta }^2$. Conditional expectation gives $\mathbb{E}{Y}_0=\theta$ and $\mathbb{E}{Y}_i=\theta \mathbb{E}{Y}_{i-1}={\theta }^{i+1}$. Therefore for $0<\theta <1$,This is the bounded information in a subcritical Poisson branching process. Observing more generations of one family does not provide unbounded information. Indeed $\mathbb{P}(Y_n>0)\le \mathbb{E}{Y}_n={\theta }^{n+1}\to 0$, and zero is absorbing, so extinction occurs at a finite generation almost surely. The total observed progeny $S_{\infty }$ is finite; eventually $S_n=S_{\infty }$, $T_n=1+S_{\infty }$, and the extended estimate equals $S_{\infty }/(1+S_{\infty })$, a discrete nondegenerate random variable rather than a consistent increasingly precise estimate.

Here is a direct proof of the failure of a chi-squared Wald limit which does not rely solely on the information bound. Under any interior null ${\theta }_0$, the event $Y_0=1$, $Y_1=0$ has probability ${\theta }_0{e}^{-2{\theta }_0}>0$ and forces every later count to be zero. On it the genuine constrained estimate is $1/2$ for every $n\ge 1$, so the usual expected-information Wald statistic isThus it has a persistent atom tending to $4(1/2-{\theta }_0{)}^2$. The fact that atoms obstruct a continuous distributional limit rules out a chi-squared limit, including when ${\theta }_0$ equals one half and that atom is at zero. Any definition adopted for the separate all-zero sample event leaves this positive-count counterexample intact. Hence the Wald statistic does not have the usual asymptotic chi-squared null law for this increasing observation horizon.

For a regular iid model with a fixed-dimensional interior parameter ${\theta }_0$, identifiability, sufficient differentiability and moment domination, and nonsingular per-observation Fisher information $I({\theta }_0)$, a consistent maximum-likelihood estimator solves the score equation. Expanding it giveswhere $s_{\theta }={\partial }_{\theta }\log f_{\theta }$. The score has mean zero and covariance $I$ under the regularity assumptions. This proves the usual large-sample information bound and asymptotic efficiency within regular parametric inference. The observed curvature $j(\hat{\theta })/n$ and the empirical score covariance consistently estimate $I$ when the model is correctly specified.

For $r$ smooth constraints $h(\theta )=0$ with full-row-rank derivative $H$, the Wald test usesThe likelihood-ratio test and score test, evaluated using the constrained fit and nuisance-adjusted efficient information, have the same first-order chi-square law. A quadratic local likelihood expansion explains their asymptotic equivalence. Profile likelihood supplies inference about a parameter of interest with fitted nuisance parameters, and the corresponding chi-square cutoffs give approximate confidence regions.

These conclusions depend on the stated regularity: parameters at a boundary, nonidentifiability, singular information, changing dimension, or parameter-dependent support can give different rates and limits. Under misspecification, the fitted parameter generally approaches a pseudo-true value and its variance is a sandwich covariance matrix, not simply inverse Fisher information.

Converting size to a regression factor allows two unrestricted contrasts rather than imposing a linear trend on numerical size codes. The command relevel with a numeric second argument selects the second existing factor level as its reference level. For sorted codes $0,1,2$ this is medium, coded $1$. However, the displayed coefficient labels size1 and size3 instead correspond to codes $1,2,3$ with code $2$ as reference. The printed coding and coefficient labels are inconsistent. The actual factor levels and labels must be checked. Under the intended medium-size reference, the two reported contrasts compare small and large businesses with medium ones; with literal $0,1,2$ coding their printed names would be size0 and size2.

Using B's delayed-entry survival object, the Cox proportional-hazards model isHere $h_0$ is the unspecified baseline hazard for a medium-size village business and $c$ is the Cambridge indicator. The fitted hazard ratios are conditional on the other covariates and constant over business age under the proportional hazards assumption. The function fits regression coefficients through partial likelihood; its survival object must actually be B's version, rather than A's earlier object with the same name.

The location coefficient $-0.803$, with standard error $0.344$, gives $z=-2.335$ and a two-sided Wald test p-value about $0.02$. Its reported hazard ratio is $0.448$, with 95% confidence interval $(0.228,0.879)$. At the same age and size, a Cambridge business therefore has an estimated closure hazard about 55.2% lower than a village business. The reciprocal ratio is $2.23$. A hazard ratio is not a survival-probability ratio or a proportional increase in lifetime.

The intended small-versus-medium contrast is $-0.322$, with standard error $0.482$, $z=-0.669$ and p-value $0.50$. Its hazard ratio is $0.724$, with 95% confidence interval $(0.282,1.863)$. The intended large-versus-medium contrast is $-0.224$, with standard error $0.375$, $z=-0.598$ and p-value $0.55$; its hazard ratio is $0.799$, with interval $(0.383,1.666)$. Neither size contrast provides convincing evidence of an effect. The wide intervals also show why these results do not establish equality of size-specific hazards.

The sample has $60$ businesses. The likelihood-ratio test, Wald test and score test jointly test all three coefficients being zero; their statistics are $5.45$, $5.76$ and $6.01$ on three degrees of freedom, with p-values $0.142$, $0.124$ and $0.111$. None rejects at 5%. This is compatible with the significant one-coefficient location test: testing a specified one-dimensional contrast and testing three coefficients together are different questions. The location result offers some adjusted evidence, but the joint output does not demonstrate a strong overall improvement, and interpretation should acknowledge the uncertainty and the number of comparisons.

The reported $R^2=0.087$ is a Cox–Snell likelihood pseudo-R-squared, here $1-e^{-5.45/60}\approx 0.087$. It measures likelihood improvement rather than a literal fraction of variance in lifetimes explained. Its attainable maximum is reported as $0.979$. It does not validate the proportional hazards assumption or establish useful predictive performance. Adjusted survival predictions would require estimating the baseline survival function as well: $S(t\mid z)=S_0(t{)}^{e^{{\beta }^{T}z}}$. The coefficient summary alone does not supply adjusted five- or ten-year survival probabilities.

Before reporting results, check the size coding discrepancy, the precise sampling dates and the survival records. The stated January 2000–December 2005 span is six calendar years, despite being called five years. Check nonnegative ages, entry no later than exit, correctly recorded closure indicators and the same opening-based clock throughout. Establish whether inclusion and censoring are plausibly independent of lifetime given the covariates, and whether different opening cohorts or calendar environments can reasonably share the fitted lifetime model. Simply including delayed entry does not cure informative selection, omitted confounders or dependence between businesses.

Assess proportional hazards using Schoenfeld residuals against time and a proportional hazards assumption test, supplemented by groupwise log-minus-log survival plots. Investigate influential observations with coefficient-deletion diagnostics and deviance residuals, and inspect overall fit using Cox–Snell residuals. Consider a city-by-size interaction or time-varying effects if supported by the available events, rather than relying on the displayed three main effects. Finally, report the uncertainty and declining late risk sets—only six remain at the final listed event—and distinguish a conditional association from a causal explanation or a prediction for all future businesses.

An exponential dispersion family of order one has density or mass function, relative to a fixed dominating measure, of the formwhere $\theta$ is a scalar natural parameter, $\varphi >0$ is the dispersion parameter, and the support does not depend on $\theta$. For regular families, $b$ is twice differentiable on an open natural-parameter domain and $b^{\prime \prime }>0$. Known observation weights $w$ replace $\varphi$ by $\varphi /w$. “Order one” refers to the scalar canonical statistic $y$, rather than an arbitrary vector of canonical statistics.

Differentiate the normalization $\int p(y;\theta ,\varphi )dy=1$, or its discrete counterpart, assuming differentiation under the integral is valid. The first derivative gives $\mathbb{E}(Y-b^{\prime }(\theta ))=0$. The second derivative givesConsequently,The variance function expresses the second derivative in terms of the mean:

A generalized linear model specifies independent responses from an exponential dispersion family, a linear predictor ${\eta }_i=x_i^T\beta$, and a differentiable monotone link function relating it to the response mean by $g({\mu }_i)={\eta }_i$. The family, variance function, covariate design, link function and dispersion specification together define the model. The canonical link function identifies the linear predictor with the natural parameter:One advantage is that the likelihood equations become particularly simple. With known weights $w_i$ and common dispersion $\varphi$, the score function and negative Hessian matrix for $\beta$ areThe latter does not explicitly depend on the responses, so it equals the expected Fisher information matrix at the same parameter. With full-rank design and positive variances the likelihood is strictly concave in $\beta$, simplifying optimization. These statements concern the coefficient block at a fixed dispersion value.

Three explicit examples, taking unit weights, are as follows.

For a normal distribution with mean $\mu$ and variance $\varphi$,Thus $\theta =\mu$, $b(\theta )={\theta }^2/2$, $c(y,\varphi )=-y^2/(2\varphi )-\frac{1}{2}\log (2\pi \varphi )$, and $V(\mu )=1$. Its canonical link function is $\boxed{g(\mu )=\mu }$, the identity link.

For a Poisson distribution with mean $\mu >0$ and support $\{0,1,\dots \}$,Here $\varphi =1$, $\theta =\log \mu$, $b(\theta )=e^{\theta }$, and $c(y,1)=-\log (y!)$. Therefore $b^{\prime }(\theta )=b^{\prime \prime }(\theta )=e^{\theta }$, $V(\mu )=\mu$, and the Poisson canonical link is $\boxed{g(\mu )=\log \mu }$.

For a Bernoulli distribution with success probability $\mu \in (0,1)$,Taking $\varphi =1$, $\theta =\log (\mu /(1-\mu ))$, $b(\theta )=\log (1+e^{\theta })$, and $c(y,1)=0$ gives the required form. Its mean and variance function are $b^{\prime }(\theta )=e^{\theta }/(1+e^{\theta })=\mu$ and $V(\mu )=\mu (1-\mu )$. Its canonical link function is $\boxed{g(\mu )=\log (\mu /(1-\mu ))}$, the logit link. The fixed dispersion values in the Poisson and Bernoulli cases are part of those models; an arbitrary continuous dispersion is not being asserted for them.

To test $H_0:{\beta }_j=0$, first fit the unrestricted model and then the restricted model with that component fixed at zero, estimating all nuisance parameters, including dispersion when applicable, in each fit. If $\hat{\vartheta }$ and $\tilde{\vartheta }$ denote their full fitted parameter vectors, the likelihood-ratio test usesFor identifiable regular models, an interior null parameter and a fixed-dimensional full-rank design with increasing information, the Wilks theorem gives $D\stackrel{d}{\to }{\chi }_1^2$ under $H_0$. A level-$\alpha$ test rejects for $D$ greater than the $(1-\alpha )$-quantile of the chi-squared distribution with one degree of freedom.

Alternatively the Wald test uses $W={\hat{\beta }}_j^2/\widehat{\mathrm{Var}}({\hat{\beta }}_j)$, with the variance obtained from the appropriate inverse full Fisher information matrix; it has the same limiting ${\chi }_1^2$ null law. The score test fits only the restricted model. If $\lambda$ collects its nuisance parameters, its efficient information is $I_{jj\cdot \lambda }=I_{jj}-I_{j\lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda j}$, and the statistic is $U_j(\tilde{\vartheta }{)}^2/I_{jj\cdot \lambda }(\tilde{\vartheta })$, again asymptotically ${\chi }_1^2$. Accounting for the nuisance block is necessary rather than using the unadjusted diagonal information.

Set $\alpha =\log \{{\beta }^{(1)}/{\beta }^{(0)}\}$ and absorb the common multiplier into $h_0(t)=m(t){\beta }^{(0)}$. The model becomesIt is a Cox proportional-hazards model; the unknown baseline does not need a parametric form. The equality hypothesis is precisely $\alpha =0$. The absolute multipliers and $m$ have a scale ambiguity, but their hazard ratio $e^{\alpha }$ is identifiable when both groups supply information.

For an untied event at $a_k$, let $R_k$ be the risk set just before the event, and let $i_k$ be the subject who fails. The conditional event probability is proportional to that subject's hazard, so the common baseline cancels:Multiply these factors to form the Cox partial likelihood. One can maximize it and compare its maximum with its value at $\alpha =0$ using a partial likelihood-ratio test, or use the corresponding score test.

Explicitly, with group risk-set sizes $r_{k0},r_{k1}$ and observed event indicator $d_{k1}$ for group 1, the score at zero and its information areUnder the null the event label has exactly this Bernoulli distribution conditional on its current risk set. Hence $U^2/V$ is the usual one-degree-of-freedom log-rank test statistic, with an asymptotic chi-squared distribution when $V>0$ and enough informative events are observed. Censoring can differ between groups, provided it is independent of event time within the conditioning model. If there are recorded ties, use the appropriate tied-risk-set version: for $d_k$ events the null variance contribution is $d_k(r_k-d_k)r_{k0}{r}_{k1}/[r_k^2(r_k-1)]$.

Choose a parametric baseline hazard $h_0(t;\eta )$ and fitUnder independent right censoring, the full survival likelihood isEstimate $(\eta ,\beta )$ by maximum likelihood estimation and test $H_0:\beta =0$ with a likelihood-ratio test, Wald test, or score test. Equality of the two event-time distributions is exactly $\beta =0$ within this model.

For each observed event $i$ with $v_i=1$, let $R_i=\{j:x_j\ge x_i\}$ be its risk set. The Cox partial likelihood isMaximize it to obtain $\hat{\beta }$, estimate its variance from the observed partial information, and test $H_0:\beta =0$ using a partial likelihood-ratio test, Wald test, or score test. A positive fitted coefficient means the group with $z=1$ has the larger hazard.

Under a country-specific constant hazard survival model, let $D_k$ be the number of deaths and $Y_k$ the total person-time at risk in country $k$. The likelihood is proportional toso the maximum-likelihood estimator is ${\hat{\lambda }}_k=D_k/Y_k$. Test $H_0:{\lambda }_0={\lambda }_1$ by a likelihood-ratio test, score test, or Wald test; equivalently, give $D_k$ a Poisson distribution with mean ${\lambda }_k{Y}_k$ and test the country coefficient in a Poisson regression with offset $\log Y_k$.

For $K\sim \mathrm{Bin}(n,p)$, $n>0$, put $\hat{p}=K/n$ and let $z$ be the $1-\alpha /2$ quantile of the standard normal distribution. Inverting the approximate score test retains precisely the parameters satisfyingMultiplication and collection of terms gives the quadratic inequalityIts two roots are the displayed endpoints. The Wilson score interval stays within $[0,1]$ and remains nontrivial when $K=0$ or $K=n$, unlike the unmodified Wald confidence interval. Its coverage probability is approximate and varies with $p$; it is not a uniformly exact 95% guarantee. Separate pointwise intervals for several groups are not a simultaneous confidence statement. The method is named after Edwin Bidwell Wilson.^[ref]