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For a normal sample of variance ${\sigma }^2>0$, define $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The exact chi-squared distribution identity $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ and the central limit theorem for independent squared standard normal random variables implyThe delta method for $g(v)=\sqrt{v}$, with derivative $g^{\prime }({\sigma }^2)=1/(2\sigma )$, therefore gives the displayed limit for $S$. Exact independence of the normal sample mean and sample variance also makes $S$ independent of $\overline{X}$. The resulting joint limit supports combining location and scale estimates when a model relates their parameters.

For independent $Y_{ij}\sim \mathrm{Bin}(m,p_i)$ with $k$ replicates per group and $\mathrm{logit}(p_i)=\mu +{\alpha }_i$, ${\alpha }_1=0$, let $S_i={\sum }_j{Y}_{ij}$ and $N=km$. The likelihood equations give ${\hat{p}}_i=S_i/N$, $\hat{\mu }=\log (S_1/(N-S_1))$, and ${\hat{\alpha }}_i=\mathrm{logit}({\hat{p}}_i)-\hat{\mu }$ when $0<S_i<N$. Boundary totals give infinite logits. With $0<p_1<1$ fixed and $N\to \infty$, the central limit theorem and delta method giveThe other groups do not increase information about the baseline logit when they have free offsets. For two groups, the information matrix for $(\mu ,{\alpha }_2)$ is $\left(\begin{array}{cc}{w}_1+w_2& w_2\\ w_2& w_2\end{array}\right)$ with $w_i=N{p}_i(1-p_i)$, whose inverse has first diagonal entry $1/w_1$.

For $\theta ={\beta }_1{\beta }_2$, use $\hat{\theta }={\hat{\beta }}_1{\hat{\beta }}_2$. Under regular regression assumptions, the delta method gives standard error $\hat{s}=(d^{\top }\hat{V}d{)}^{1/2}$, where $d=({\hat{\beta }}_2,{\hat{\beta }}_1{)}^{\top }$ and $\hat{V}$ estimates the covariance of the regression estimate. Resample the observational units, refit, and compute both ${\hat{\theta }}^{\ast }$ and ${\hat{s}}^{\ast }$. A percentile bootstrap confidence interval uses the .025 and .975 quantiles of ${\hat{\theta }}^{\ast }$. A bootstrap-t confidence interval uses quantiles of $({\hat{\theta }}^{\ast }-\hat{\theta })/{\hat{s}}^{\ast }$ and reverses them when solving for $\theta$. If both true coefficients vanish, the first derivative is zero and the standard first-order justification is unavailable; product-specific nonregular inference is then needed.

For independent Bernoulli random variables of fixed success probability $0<p<1$, put $\hat{p}=n^{-1}{\sum }_i{X}_i$ and $o=p/(1-p)$. The central limit theorem gives $\sqrt{n}(\hat{p}-p)\stackrel{d}{\to }N(0,p(1-p))$. Applying the delta method to the odds map $g(p)=p/(1-p)$, whose derivative is $(1-p{)}^{-2}$, givesThe ratio is undefined on the all-success event, whose probability is $p^n$. Any finite extension there has the same limiting probability distribution, since that probability tends to zero. This is an asymptotic statement and does not assert an exact finite-sample variance of the unextended ratio. It is useful for uncertainty in estimated odds when the success probability stays away from one.

For an independent statistical sample from the beta distribution with parameters $(\theta ,1)$, $\theta >0$, the probability density function is $\theta x^{\theta -1}$ on $(0,1)$. Its raw moments are $\mathbb{E}{X}^r=\theta /(\theta +r)$ for $r>-\theta$, by direct integration. Consequently the method of moments gives $T_1=\bar{X}/(1-\bar{X})$, whereas the maximum-likelihood estimator is $T_2=-n/{\sum }_i\log X_i$.

The second formula follows by setting the score function $n/\theta +{\sum }_i\log X_i$ to zero; its derivative is negative. Also $-\log X_i$ has an exponential distribution of rate $\theta$, since $\mathbb{P}(-\log X_i>u)=e^{-\theta u}$. The central limit theorem and delta method giveThus the asymptotic relative efficiency of $T_1$ relative to $T_2$ is $1-(\theta +1{)}^{-2}$. The transformed sum has a gamma distribution of shape $n$ and rate $\theta$, which also gives $\mathbb{E}{T}_2=n\theta /(n-1)$ for $n>1$ and $\mathrm{Var}(T_2)=n^2{\theta }^2/[(n-1{)}^2(n-2)]$ for $n>2$.

For independent observations from the exponential distribution with rate $\theta$, the log-likelihood is $n\log \theta -\theta {\sum }_i{X}_i$, soThe central limit theorem and delta method give

Let independent exposure sums $A,B$ have mean $N$ ADU at each detector pixel, with pure Poisson distribution noise and fixed gain $g$. The delta method gives $A/B\simeq 1+(A-N)/N-(B-N)/N$, so $\mathrm{Var}(A/B)\simeq 2/(gN)$ and $g\simeq 2/(N{E}^2)$. For nonuniform detector pixel means $N_i$, an unweighted spatial ratio variance is instead $(2/g)⟨1/N_i⟩$. This high-count approximation requires linear unsaturated data, bias subtraction and negligible read noise.

For the Kaplan–Meier estimator, the estimated variance isThe formula follows by summing approximate variances of the logarithms of the conditional survival factors and applying the delta method to their product. Its square root is the estimated standard error. The same expression uses the actual delayed-entry risk sets for the Kaplan–Meier estimator with delayed entry. When $0<\hat{S}(t)<1$ and every contributing $r_j>d_j$, an approximate log-scale confidence interval is $\exp \{\log \hat{S}(t)\pm z_{1-\alpha /2}\widehat{\mathrm{se}}(\hat{S}(t))/\hat{S}(t)\}$, with the upper endpoint capped at one. This interval can be inaccurate with very small risk sets.

The log odds ratio is the natural logarithm of an odds ratio. For two independent binomial groups with positive event and non-event counts $(a,b)$ and $(c,d)$, its estimate is $\log (ad/(bc))$, with first-order variance $1/a+1/b+1/c+1/d$ by the delta method. Zero cells need a method that handles boundary data; the displayed normal approximation is not valid there.

For two independent binomial distributions with event and nonevent counts $a,b$ and $c,d$, the odds ratio is $ad/(bc)$. The delta method applied to $\log (p/(1-p))$, whose derivative is $1/[p(1-p)]$, gives a group variance $1/[Np(1-p)]$. Substitution of the fitted proportion gives $1/a+1/b$ in the first group and $1/c+1/d$ in the second. Their independence gives the displayed sum. A normal approximation then gives a confidence interval on the log odds ratio scale; exponentiation produces an odds-ratio interval. Zero cells require another approximation or model.

For independent $N(\mu ,v)$ observations with unknown mean and variance, the first natural parameter is ${\varphi }_1=\mu /v$. Its maximum-likelihood estimator is $\bar{X}/[n^{-1}{\sum }_i(X_i-\bar{X}{)}^2]$. The delta method and independent asymptotic mean/variance errors give limiting variance $v^{-1}+2{\mu }^2{v}^{-2}$ for $\sqrt{n}({\hat{\varphi }}_1-{\varphi }_1)$. Knowing $v$ removes the second contribution and makes the estimator exactly normal with variance $1/(nv)$.

For a homogeneous continuous-time Markov chain, the holding time in a living state $r$ before the next transition has survival probability $e^{-{\lambda }_{r}t}$, where ${\lambda }_r={\sum }_{s\ne r}{q}_{rs}=-q_{rr}$. Indeed, conditional on still occupying that state, the probability of leaving in the next $dt$ is ${\lambda }_{r}dt+o(dt)$, so its survival function satisfies $S^{\prime }(t)=-{\lambda }_{r}S(t)$ and $S(0)=1$. Integrating gives the mean holding time from a transition intensity matrix:This is the mean time until any allowed exit from that state, not necessarily the patient's remaining lifetime.

For confidence limits, use the supplied standard error $s_r$ of the corresponding diagonal generator estimate. Changing its sign does not change its standard error, so it is also the standard error of ${\hat{\lambda }}_r$. The delta method, with derivative $d(1/\lambda )/d\lambda =-1/{\lambda }^2$, givesThe resulting approximate 95% Wald confidence intervals areThe diagonal standard errors already incorporate the covariance among outgoing rates. They should not be replaced by the square root of a sum of squared off-diagonal standard errors, which would silently assume those rate estimates independent.

Alternatively construct a positive rate confidence interval and apply the confidence interval for an inverse rate transformation. Here every normal rate interval has a positive lower endpoint, givingThese approximate intervals are, respectively, $(10.863,12.830)$, $(4.497,7.161)$, $(2.397,3.117)$ and $(1.548,2.631)$ years. They are asymmetric because inversion is nonlinear; positive log-rate or likelihood-based intervals could also be inverted. The symmetric delta intervals and the inverted-rate intervals are two legitimate first-order approaches and need not have numerically identical endpoints.

The death state has zero exit rate, so its holding time is infinite. There is no finite mean-sojourn estimate or ordinary finite confidence interval for this structurally absorbing state. All intervals above concern the mean parameter, not a prediction interval for one patient's realized holding time.

The response errors have standard deviations proportional to $|{\mu }_i|$. Thus, when fitted means approximate the true means, increasing absolute fitted values tends to accompany increasing residual spread. A residual-versus-fitted plot shows a fan for positive means, or spread increasing away from zero for means of either sign. Standardization by one common residual standard deviation, even with the usual regression leverage correction, does not remove this systematic heteroscedasticity.

Under independent errors the exact covariance matrix after fitting iswhere $k>0$ is the common proportionality constant. Hence the simple fan description is a diagnostic tendency, not an assertion that each fitted residual has exactly variance $k{\mu }_i^2$.

For positive responses, the intended variance-stabilizing transformation is the natural logarithm. The delta method givesUse $\log Y_i$ for positive data with modest relative error, then refit and check the transformed mean and residual structure. In particular, if the original mean was $X_i^T\beta$, the transformed mean is approximately $\log (X_i^T\beta )-k/2$, not automatically a linear function of the original covariates.

There is a domain qualification in the literal printed model: a nondegenerate normal distribution has negative support. For positive ${\mu }_i$, the probability of $Y_i\le 0$ is $\Phi (-1/\sqrt{k})$, so a real $\log Y_i$ is not defined throughout the stated sampling model. When $k$ is small this is a negligible-tail approximation, not an exact transformation to normal errors. If signed values must be retained and all ${\mu }_i\ne 0$, write $Y_i={\mu }_i(1+\sqrt{k}{Z}_i)$ with standard normal $Z_i$. ThenThe second term has a common distribution with finite variance, so log absolute value stabilization of a normal scale family is exact, but loses the sign of the mean and does not produce normal errors. Zeros and a zero mean are excluded from this transform. Alternatively retain the response scale and fit the stated variance model with weighted least squares weights proportional to $1/{\hat{\mu }}_i^2$ where the means stay away from zero. No arbitrary addition of a constant before taking logarithms exactly solves the given variance relation.

read.table(...,header=T) reads four grouped rows with column names. The displayed outcome counts give the observed risks $0.1519,0.2145,0.2611,0.5111$ in the indicated factor order. rbind in the first test pools over predisposition: the two rows represent no cannabis use and some use, with event/non-event counts $(341,1775)$ and $(82,238)$. Thus the pooled risks are $341/2116=0.1612$ and $82/320=0.2563$. This test concerns the marginal statistical association between use and outcome, not an association adjusted for predisposition.

Under independence, expected counts in the two-by-two contingency table are $E_{ij}=O_{i+}{O}_{+j}/N$. The ordinary Pearson chi-squared statistic is $17.5183$. The printed $16.8618$ is reproduced by the default two-by-two Yates continuity correction:All absolute differences here exceed one half. Comparison with a chi-squared distribution on one statistical degree of freedom gives $p=4.02\times 10^{-5}$. There is strong evidence of marginal association. The simplified output omits the usual continuity-correction wording, but its numerical value shows which statistic was used.

attach makes the data columns available by name, and tot=with+without obtains group sizes $1936,275,180,45$. The grouped binomial regression fits $Y_g\sim \mathrm{Bin}(n_g,{\pi }_g)$, with response $Y_g/n_g$ and prior weights $n_g$. These weights tell R the binomial denominator; treating the four proportions as four equally precise individual observations would be wrong. The default logit link isLet $A$ indicate some use and $B$ indicate predisposition. The additive logistic regression specifies $\eta =a+bA+cB$. Its log-likelihood, including the binomial constants, isDifferentiation gives score equations $X^T(Y-n\pi )=0$ and Fisher information matrix $X^{T}WX$, where $W_{gg}=n_g{\pi }_g(1-{\pi }_g)$. Fisher scoring solves these equations, and the inverse fitted Fisher information matrix gives the reported approximate coefficient variances. The number of scoring iterations records numerical convergence, not degrees of freedom or statistical evidence.

The intercept corresponds to no use and no predisposition and gives baseline fitted risk ${\mathrm{logit}}^{-1}(-1.73881)=0.1495$. The additive coefficients imply common adjusted odds ratios:They multiply the odds, not the risks. The Wald tests divide the estimates by their standard errors and use a standard normal distribution; both predictors have small two-sided p-values. Approximate $95\%$ confidence intervals for these odds ratios are $(1.296,2.266)$ and $(1.685,3.110)$, obtained by exponentiating the log-scale intervals. The intercept test against zero asks whether the reference risk equals one half, and is not a test of either exposure association. The binomial dispersion is assumed to be one because $\mathrm{Var}(Y_g)=n_g{\pi }_g(1-{\pi }_g)$; the output does not demonstrate independence or rule out extra-binomial variation.

There are four grouped probabilities and three fitted coefficients, leaving one residual degree of freedom. The fitted risks are approximately $(0.1495,0.2314,0.2869,0.4080)$. The binomial deviance compares this fit with the four unrestricted observed proportions:The deviance goodness-of-fit test against ${\chi }_1^2$ has p-value $0.0796$. The common-odds-ratio fit is not rejected at $5\%$, although the doubly exposed group's observed risk exceeds its fitted risk appreciably and suggests possible interaction.

The second formula adds $\delta AB$. Its four coefficients fit all four probabilities, making a saturated statistical model with zero grouped residual degrees of freedom. The near-zero deviance is roundoff from exact interpolation; it supplies no remaining grouped lack-of-fit test. Unlike Q1's unreplicated Gaussian fit, however, these grouped responses represent many independent Bernoulli trials, and their binomial variances are specified by their sizes and probabilities. Coefficient standard errors remain available from binomial information without estimating dispersion from the zero residual degrees of freedom.

Indeed, let ${\eta }_{ab}=\log [Y_{ab}/(n_{ab}-Y_{ab})]$ be the fitted cell logit link value. The saturated coefficients areThese calculations yield the printed estimates. The delta method gives $\mathrm{Var}({\hat{\eta }}_{ab})\approx 1/Y_{ab}+1/(n_{ab}-Y_{ab})$, and independent groups make the variance of $\hat{\delta }$ the sum of all eight reciprocal counts. Its square root is $0.37857$, reproducing the printed standard error.

The saturated logistic regression describes the conditional odds ratios as follows:Thus the main coefficients now concern effects at the other factor's reference level. Logistic interaction as a ratio of odds ratios gives $e^{\delta }=1.939$, with approximate $95\%$ confidence interval $(0.923,4.073)$. The interaction Wald test has $p=0.0802$, while the nested likelihood-ratio test uses deviance difference $3.0733$ on one statistical degree of freedom and has $p=0.0796$. The data suggest a larger odds association in predisposed subjects, but do not establish this interaction at the conventional $5\%$ level. This conclusion is specific to the odds scale; absolute-risk differences can vary even with no logit-scale interaction.

The Akaike information criterion is $-2\hat{\ell }+2k$. The additive fit has $k=3$ and AIC $31.766$; the interaction fit has $k=4$ and AIC $30.692$. The extra coefficient improves twice the log-likelihood by $3.0733$ but incurs penalty $2$, so AIC falls by only $1.0733$. It weakly favors the larger fit rather than decisively establishing an interaction, and AIC is not a p-value.

The marginal odds ratio $1.793$ and the fitted conditional odds ratios are different quantities. Predisposition is associated with both use frequency and outcome, so confounding is a substantive possibility; noncollapsibility of the odds ratio also means marginal and conditional odds ratios need not agree. This is an observational association analysis with only one adjustment factor. It does not by itself establish a causal effect, exclude other confounding, or validate the assumed independent binomial sampling.

Assume the two trial arms are independent samples, their cost and survival means have a valid joint normal approximation, and the supplied standard errors and correlation coefficient estimate the joint uncertainty in their mean differences. This could follow from a central limit theorem for patient-level cost/survival pairs, with finite second moments and suitably handled survival follow-up. The supplied covariance matrix for $(\widehat{\Delta C},\widehat{\Delta E})$ isThe off-diagonal entry has units pounds times days, and must not be discarded. Independence of trial arms does not mean cost and survival increments are independent: patient-level cost and survival are paired within each arm.

For a proposed ratio $r$, the null assertion is $\Delta C-r\Delta E=0$. Its estimated variance is $90000-4500r+225r^2$. Fieller's theorem inverts the associated two-sided normal tests. With $q={\Phi }^{-1}(0.975)\approx 1.96$, retain $r$ whenExpanding givesThe leading coefficient is positive, since $45>q(15)$; the effect increment is separated from zero at this level. With $q^2=3.841459$, the polynomial is approximatelyIt is nonpositive between its two real roots. The resulting approximate $95\%$ Fieller confidence set is therefore a bounded confidence interval:On the $365$-day life-year scale the endpoints are approximately £9,133 and £36,376. Exact coverage would need an exact pivot with its corresponding critical value; using estimated covariance and a central limit theorem makes this an asymptotic interval. In other data, a poorly determined denominator can give unbounded or disconnected Fieller confidence sets, which should not be replaced by an artificially bounded interval.

An alternative is a paired bootstrap: independently resample patients within each randomized arm, keeping each patient's cost and survival together; recompute both mean differences and their ratio in each replicate. The $2.5\%$ and $97.5\%$ empirical ratio quantiles give a percentile bootstrap confidence interval. The raw paired data are needed to calculate it, so the numerical endpoints cannot be obtained from this summary table alone. Near-zero bootstrap effect increments or sign changes warn that ordinary ratio-percentile intervals may be unreliable; one should retain the joint cost/effect uncertainty or invert a suitable bootstrap test instead.

A simpler first-order alternative is the delta method. For $r=\Delta C/\Delta E$, the gradient is $(1/\Delta E,-\Delta C/\Delta E^2)$, givingThe resulting approximate Wald confidence interval is $40\pm 1.96\sqrt{133.333}=(17.37,62.63)$ pounds per day. Its symmetric first-order form misses the denominator's nonlinear effect, explaining its marked difference from the Fieller confidence set; it is not the preferred ratio interval here.

The plug-in estimate isFor a concrete bootstrap construction, suppose the rows $(x_i,y_i)$ are independent and identically distributed, $\mathbb{E}(e_i\mid x_i)=0$, the design moment matrix is nonsingular, and the required moments exist. Use a paired bootstrap: sample $n$ rows with replacement, form $X^{\ast },Y^{\ast }$, refit by ordinary least squares, and compute ${\hat{\theta }}^{\ast }={\hat{\beta }}_1^{\ast }{\hat{\beta }}_2^{\ast }$. Repeat many times. Singular resampled designs require a suitable rank-preserving implementation or replacement replicate; with a regular continuous design their frequency vanishes asymptotically.

The bootstrap-t confidence interval method also needs a standard error in the original sample and every replicate. One asymptotically valid heteroscedasticity-robust covariance estimate isThe delta method for the coefficient product givesRecompute ${\hat{V}}^{\ast },d^{\ast },{\hat{s}}^{\ast }$ in each paired replicate. This is bootstrap inference for a product of regression coefficients.

With fixed rather than random design, resample centered residuals if errors are identically distributed with homoskedasticity, or use a properly specified wild bootstrap for heteroscedastic errors, keeping $X$ fixed and refitting after generating $Y^{\ast }$. The printed assumption $\mathbb{E}{e}_i=0$ alone does not establish independent observations or validity of a bootstrap confidence interval. The formulas below give asymptotic 95% coverage under the stated regularity conditions, not an exact finite-sample guarantee. In particular the ordinary first-order studentization needs $({\beta }_1,{\beta }_2)\ne (0,0)$ and a nonzero asymptotic variance; when both coefficients vanish the product has nonstandard behavior.

The logit link makes the probability constant within each group: write $p_i=\{1+\exp [-(\mu +{\alpha }_i)]{\}}^{-1}$, with ${\alpha }_1=0$. Let $S_i={\sum }_{j=1}^k{Y}_{ij}$ and $N=km$. Independence gives $S_i\sim \mathrm{Bin}(N,p_i)$. Up to terms independent of the parameters, the log likelihood isThe likelihood equations areTogether they also imply $S_1-N{\hat{p}}_1=0$. Thus, when all totals satisfy $0<S_i<N$,The group logits are unconstrained apart from their parametrization, and each binomial log likelihood is strictly concave in its logit, so these equations give the unique finite maximum. A total of $0$ or $N$ instead gives a boundary probability and an infinite logit; a finite MLE need not exist in that case.

For $n=2$, keep $0<p_1,p_2<1$ fixed and let $N=km\to \infty$. Since ${\hat{p}}_1=S_1/N$, the central limit theorem gives $\sqrt{N}({\hat{p}}_1-p_1)\stackrel{d}{\to }N(0,p_1(1-p_1))$. The derivative of the logit is $1/[p_1(1-p_1)]$, so the delta method gives the baseline logit estimator in a group-factor binomial model lawThe second group does not improve the baseline estimate because it has its own free offset. Equivalently, the Fisher information matrix for $(\mu ,{\alpha }_2)$ isIts first diagonal entry confirms the same asymptotic variance.

Let $Y_I,Y_C$ be attendance totals and $T_I,T_C$ the observed person-time at risk. Under independent Poisson distributions for the totals, the estimated incidence rate ratio is $\hat{R}=(Y_I/T_I)/(Y_C/T_C)$. Treating the exposures as known, the supplied large-count approximation, or the delta method, givesIndependence adds the log-count variances, while fixed exposure terms add no variance. Form the normal confidence interval on the log scale and exponentiate:This specifies the calculation without evaluating it. Individual randomization makes a child the assignment unit, but an independent Poisson count model remains an additional assumption about outcomes, not a consequence of randomization alone.

An exponential family has density $h(x)\exp \{\varphi \cdot T(x)-k(\varphi )\}$ on a parameter-independent support. For an independent and identically distributed sample, its likelihood factorizes through ${\sum }_{i}T(X_i)$, so the Fisher-Neyman factorization theorem gives a sufficient statistic whose dimension is the fixed dimension of $T$, independent of sample size.

Expanding the normal log-density gives ${\varphi }_1=\mu /v$, ${\varphi }_2=-1/(2v)$, andThe mean-value parameter is $\nabla k=(\mathbb{E}X,\mathbb{E}{X}^2)$, henceFor $n\ge 2$ and positive sample variance, $\hat{\mu }=\bar{X}$ and $\hat{v}=n^{-1}{\sum }_i(X_i-\bar{X}{)}^2$, so ${\hat{\Phi }}_1=\bar{X}/\hat{v}$. For this normal natural-mean parameter ratio estimator, the central limit theorem gives asymptotic independent errors for $\hat{\mu }$ and $\hat{v}$, with variances $v/n$ and $2v^2/n$. Applying the delta method to $\mu /v$ yieldsIf $v=v_0$ is known, ${\Phi }_2=-1/(2v_0)$ is fixed and ${\hat{\Phi }}_1=\bar{X}/v_0$ has exact distribution $\boxed{N({\Phi }_1,1/(n{v}_0))}$. The variance-estimation contribution $2{\mu }^2/v^2$ disappears.

Work at a regular admissible parameter value with $\varphi \ne 0$, so the score identities in the question apply. For one observation from the exponential dispersion family, the score function and second derivative areThe zero expected score immediately gives $\mathbb{E}Y=b^{\prime }(\theta )$. The equality between expected negative curvature and score variance then givesThusThe admissible density parameters necessarily make this variance nonnegative; the algebra does not require an unstated sign assumption beyond the regular family and $\varphi \ne 0$.

With independent observations and canonical parameters ${\theta }_i=x_i^T\beta$, the chain rule gives the vector score functionA finite interior maximum-likelihood estimator satisfies its score equation, provingDifferentiating again, using ${\nabla }_{\beta }{\mu }_i=b^{\prime \prime }(x_i^T\beta )x_i$, givesThis Hessian is independent of the observations, so taking expectation gives exactly the requested Fisher information matrix. In particular, $V$ is not the inverse observation-variance matrix: the linear predictor here is the canonical parameter, not the mean.

For the two-group Poisson regression, $\varphi =1$, $b(\theta )=e^{\theta }$, and ${\mu }_i=e^{x_i^T\beta }$. Put $a=e^{\alpha }$, $b=e^{\alpha +\nu }$, using $b$ here for the second group's mean. There are $2n$ observations, andHence the total Fisher information matrix and its inverse areThe theorem on asymptotic normality of a maximum likelihood estimator says that in a regular identifiable model with nonsingular information the estimator is approximately multivariate normal, centred at the true parameter and with covariance equal to the inverse total information. Here one can regard one observation from each group as an independent pair and apply its usual iid version to $n$ such pairs. ThereforeFor a direct check, the second score equation gives the fitted second-group mean ${\overline{Y}}_B$, and the first then gives the fitted first-group mean ${\overline{Y}}_A$. Thus the Poisson log-rate ratio estimate is $\hat{\nu }=\log {\overline{Y}}_B-\log {\overline{Y}}_A$. The delta method, using independent sample-mean variances $a/n$ and $b/n$, gives the same variance $(a^{-1}+b^{-1})/n$. If a group total is zero, its finite log-mean estimator fails to exist; with fixed positive true means this event has probability tending to zero exponentially as $n$ grows.