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Use the original PDF's value $\hat{p}=0.386$; the TeX aid misreads its last digit. In the fitted conditional mixture, $\hat{\theta }=0.575$ means that about 57.5% are assigned to the structural-zero, or modelled cured, component. Its asymptotic 95% confidence interval of $(0.30,0.81)$ shows substantial uncertainty about that fraction. An observed post-treatment zero alone is not proof of cure, since the non-cured Poisson distribution can also produce zero.

The parameter $p$ is a conditional allocation probability, not the ratio of post-treatment to pre-treatment means. That ratio among non-cured patients is $\beta =p/(1-p)$. HenceThe map $p\mapsto p/(1-p)$ is strictly increasing, so transform the endpoints of the supplied 95% confidence interval:No change in the non-cured mean is $\beta =1$, equivalently $p=1/2$, which lies in this interval. Thus the point estimate suggests a reduction, but this interval does not firmly establish one among the non-cured patients at the corresponding two-sided 5% level. It is not correct to describe $p=0.386$ itself as a 61.4% reduction in their mean count.

Under a full pre-conditioning mixture with cure independent of the baseline rate, the population mean ratio would be $(1-\theta )\beta$, whose plug-in value is about $0.425(0.629)=0.267$. This combines disappearance of counts in one component with a change of rate in the other; it is distinct from the conditional estimate's direct interpretation. The two marginal confidence intervals cannot simply be multiplied to obtain a calibrated joint 95% interval for this derived quantity.

Only 12 patients were observed, so the asymptotic intervals, particularly for a mixture parameter near a boundary, should be interpreted cautiously. The data describe one-minute counts and do not establish permanent cure. Nor does an uncontrolled before-and-after comparison alone prove a causal effect of the drug: natural count variability, secular changes and regression to the mean remain possible. The structural-zero interpretation and the reduction among non-cured patients are conclusions within the specified statistical model.

The central limit theorem states that for independent identically distributed variables with finite mean $m$ and positive finite variance $v$,Applying it to the exponential gaps givesInvert the approximate central probability $\mathbb{P}(-1.96\le (\lambda T_n-n)/\sqrt{n}\le 1.96)\simeq 0.95$. For an observed positive total $T_n$, an approximate 95% confidence interval isEquivalently its endpoints are $\hat{\lambda }(1\pm 1.96/\sqrt{n})$, with a nonpositive lower endpoint replaced by the parameter boundary. This is a large-$n$ approximation; the exact gamma law of $T_n$ permits an exact interval if needed. It is a two-sided parameter interval, not a 95th-percentile annual loss estimate.

Let annual total loss be $L={\sum }_{i=1}^{N_1}{X}_i$, with an empty sum equal to zero. The assumptions give a compound Poisson distribution. Conditioning on the count,The tower property and total-variance formula therefore giveThe $\lambda {\mu }^2$ term is the extra uncertainty in the number of defaults. Replacing it by zero would incorrectly treat that count as deterministic.

For sufficiently many annual events and an adequate normal approximation, the value at risk at confidence level $p$ is the compound-Poisson annual loss quantileThe supplied quantiles give $z_{0.90}=1.282$, $z_{0.95}=1.645$ and $z_{0.99}=2.326$. In particular a one-sided 95% annual loss quantile uses $1.645$, not the $1.96$ from the two-sided 95% rate interval. If the expected loss is covered separately, the approximate unexpected-loss reserve is the second term, rather than the full quantile.

Estimate the rate from a suitable historical count/exposure or gap sample, then insert $\hat{\lambda }$ into the loss formula. The interval in part (d) measures rate-estimation uncertainty; for nonnegative loss mean it can be propagated to lower and upper approximate quantiles because the displayed expression is increasing in $\lambda$. Severity parameters also require estimation, so the rate interval alone is not a confidence interval accounting for every source of uncertainty.

Mean and variance do not determine the exact loss quantile. The actual distribution isFor nonnegative losses it has a zero atom at least $e^{-\lambda }$, so if that mass exceeds $p$ its exact value at risk is zero. With a small rate or a strongly skewed severity distribution, a normal approximation can therefore be poor. A loan book with dependent defaults, rating migration or varying exposures also needs more than the constant-intensity independent-severity model. These are model qualifications, not deductions from the credit rating alone.

Use a longitudinal study with repeated measurements of the same individuals. Recruit people across several starting ages and measure the same ear, with a standardized anatomical definition and instrument, at baseline and subsequent visits over several years. Repeat measurements at each visit to estimate measurement error; where practical, conceal previous measurements from the observer. Record sex, body size, birth cohort and observer, and document reasons for loss to follow-up.

A simple random-intercept linear mixed model for ear length $Y_{ij}$ at elapsed follow-up time $t_{ij}$ isThe individual intercept ${\alpha }_i$ absorbs persistent differences between people. Alternatively, differences $Y_{ij}-Y_{i0}$ remove that intercept directly. Estimate the within-person regression coefficient $\beta$ and its confidence interval, accounting for the correlation of repeated measurements. Compare it with the cross-sectional slope rather than treating the two slopes as automatically identical.

This longitudinal study directly checks whether an individual's ears tend to lengthen as time passes. Several entry-age cohorts help assess whether the change varies with starting age. Calendar-period changes and selective dropout can still affect interpretation, so repeated observation alone does not eliminate every possible source of confounding or selection bias.

The association is strongly statistically significant under the reported regression assumptions, but age alone has only moderate predictive ability. The reported confidence interval for the age regression coefficient excludes zero by a wide margin. If it is the usual two-sided 95% interval from a normal linear model with an intercept and one predictor, there are $206-2=204$ degrees of freedom. Taking $t_{0.975,204}\simeq 1.97$ givesThus the two-sided p-value against zero slope is far below $0.001$. This conclusion concerns a statistical association in the sampled population; it does not establish an individual growth rate. Clustering by practitioner, nonconstant variance or nonindependent observations would require an appropriate standard error rather than blind reliance on this calculation.

The scatter plot has substantial vertical spread at each age. A narrow slope confidence interval describes uncertainty about the average trend, whereas a prediction interval for a new person's ear length must also include the residual variance. The reported interval even permits an approximate numerical assessment: write $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and let $s_e^2=\mathrm{RSS}/(n-2)$. In simple linear regression, $t^2={\hat{\beta }}^2{S}_{xx}/s_e^2$, while the fitted and residual sums of squares are ${\hat{\beta }}^2{S}_{xx}$ and $(n-2)s_e^2$. Hence the coefficient of determination isUnder these assumptions, age explains only about 27% of the sample variation; roughly 73% remains unexplained. The calculation is approximate because the printed slope and interval are rounded. For a new individual aged $x_0$, the usual prediction interval is based onThe leading one represents individual residual variation. The provided summaries do not determine $s_e$ and $S_{xx}$ separately, so they do not supply a numerical prediction interval.

Record sex, height or another measure of body size, ancestry, family characteristics and practitioner or measurement method. Use these covariates in a multiple linear regression, allowing scientifically justified nonlinear age effects or interactions, and assess prediction error on held-out observations. A guessed overall sex proportion cannot replace individual sex measurements or identify sex-adjusted effects.

There is a source inconsistency: the PDF's numerical ear-length summary is ten times the scale of its plotted measurements and fitted line. Indeed, the fitted line at the reported mean age gives about $67.7$ mm, whereas the printed mean is about ten times larger. Missing decimal points in the summary are a plausible explanation. The significance and approximate $R^2$ above use the mutually consistent slope, interval and graph; they do not treat the inconsistent summary as valid data.

For the log-transformed location-scale model, let $F_W$ denote the distribution function associated with $f$. The change-of-variables formula for a probability density gives the untruncated density and distribution functionConsequently its upper-truncated likelihood is ${\prod }_{i}p(t_i)/F(M{)}^n$ on the observed support, and the profile likelihood maximizes this over $\alpha ,\sigma ,q$. Restricting to the corresponding subfamilies gives the gamma distribution and lognormal distribution profiles. Write the plotted likelihood ratio asIt measures how well each endpoint fits relative to that model's own best endpoint, after nuisance refitting. It is not a probability density function for $M$ and need not integrate to one. Endpoints whose ratios are near one are relatively compatible with the observations; a rapidly falling profile indicates more information favoring endpoints near the maximum.

All three profiles favor the smallest admissible endpoint, approximately 14 days; their information about larger endpoints differs substantially. Reading the actual PDF legend and curves, the dotted gamma distribution profile falls initially but is nearly flat around $0.60$ from about 18 days onward. The solid generalized log-gamma profile declines much further, to roughly $0.20$ at 19 days and $0.18$ at 21 days. The dashed lognormal distribution profile falls most strongly, to roughly $0.14$ at 18 days and $0.08$ at 21 days. These are graphical readings, not recomputed fits.

Thus the gamma distribution fit gives weak discrimination between moderate and very large endpoints: much of its untruncated mass can already be below these endpoints, making further changes to the truncation have little effect. Over the displayed range, the lognormal distribution fit gives stronger relative evidence against large endpoints, and the more flexible generalized log-gamma fit is intermediate. The solid curve remains close to the dashed curve near the maximum but has appreciably more relative support farther out. The plot shows slowing declines rather than a sharply determined finite upper endpoint; it cannot establish a biological maximum.

Normalization is essential to this comparison. Although the generalized family includes the two special families and therefore has at least as large a raw maximized likelihood at any fixed $M$, its normalization constant can also be larger. Its normalized profile need not lie above the normalized profiles of its submodels. Their individual maxima have all been replaced by one, so this plot alone cannot compare overall goodness of fit or select a family while accounting for extra fitted parameters.

A set $\{M:R(M)\ge c\}$ is a relative-likelihood support set once $c$ is chosen. Turning it into a calibrated confidence interval requires an appropriate sampling argument, simulation or other justified method. In particular, a usual Wilks theorem chi-squared cutoff cannot simply be assumed for this changing-support endpoint problem. The original incubation observations are not supplied here, so numerical refitting or calibrated limits cannot be recovered from the graph alone.

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.

Define the difference as control minus intervention, positive for fewer deaths under intervention assignment. The observed difference per thousand isFor independent binomial distributions, its unpooled standard error per thousand isThe approximate 95% confidence interval is thereforeThe rare-event Poisson standard error $\sqrt{148+123}/30$ gives almost the same interval. A reduction of 1.5 per thousand is consistent with the data, since it lies inside the interval. Zero also lies inside, so a nonzero benefit is not established at the two-sided 5% significance level. This does not establish equivalence or prove the proposed effect size, and cannot by itself establish treatment contamination.

Set $Y_i=X_i^2$, $\overline{Y}=n^{-1}{\sum }_i{Y}_i$ and ${\tau }^2={\mathrm{Var}}_F(Y_1)={\mathbb{E}}_F{X}_1^4-{\theta }^2$. The fourth-moment assumption makes ${\tau }^2$ finite, so the variance of a sample mean is $v={\tau }^2/n$. Assume $n\ge 2$, as required for leave-one-out estimation.

The Jackknife variance estimator recomputes the original statistic on each sample with one observation removed. HereBy the definition of the Jackknife variance estimator,For completeness, expanding the centered sum givesConsequently $\mathbb{E}{\hat{v}}_{\mathrm{JACK}}={\tau }^2/n=v$. The jackknife variance estimator is unbiased for this statistic. This calculation uses only the finite fourth moment, not normality of the original observations.

Given the observed data, put $y_i=x_i^2$ and $\overline{y}=n^{-1}{\sum }_i{y}_i$. The empirical distribution assigns mass $1/n$ to each observed $x_i$, including repeated values with their multiplicities. A bootstrap sample consists of $n$ independent draws from this distribution, with replacement. The nonparametric bootstrap variance estimate is the conditional varianceIndependence of the bootstrap draws gives the exact conditional bootstrap variance of a sample mean:Thus its finite-sample expectation is $(n-1)v/n$, in contrast to the unbiased jackknife estimate. This is the jackknife and bootstrap variance of a sample second moment scaling difference.

To approximate it by Monte Carlo simulation, independently generate $B$ bootstrap samples. For replicate $b$, choose $n$ independent uniform indices $I_{b1},\dots ,I_{bn}\in \{1,\dots ,n\}$ and compute $T_b^{\ast }=n^{-1}{\sum }_j{x}_{I_{bj}}^2$. With ${\overline{T}}^{\ast }=B^{-1}{\sum }_b{T}_b^{\ast }$, useFor $B>1$ this is conditionally unbiased for the bootstrap variance and converges to it as $B\to \infty$. Uniform indices can be obtained by $I=\lceil nU\rceil$ for uniforms on $(0,1]$.

Let $G_{\ast }(t)=P_{\ast }(T^{\ast }\le t)$ be the conditional bootstrap distribution of the statistic. A nominal level $1-\alpha$ percentile bootstrap confidence interval isReuse the bootstrap replicates, sort them as $T_{(1)}^{\ast }\le \cdots \le T_{(B)}^{\ast }$, and take the order statistics with ranks $\lceil B\alpha /2\rceil$ and $\lceil B(1-\alpha /2)\rceil$ as empirical quantiles. These are quantiles of the statistic itself, rather than reflected quantiles of its centered error. Increasing $B$ reduces simulation error; it does not turn an approximate confidence interval into an exact finite-sample one.

An analytic alternative follows from the central limit theorem. If ${\tau }^2>0$, then $\sqrt{n}({\hat{\theta }}^n-\theta )\Rightarrow N(0,{\tau }^2)$, and the sample variance $s_Y^2=(n-1{)}^{-1}{\sum }_i(x_i^2-\overline{y}{)}^2$ converges to ${\tau }^2$ by the strong law of large numbers. The Slutsky theorem therefore gives the large-sample normal approximation intervalwhere $z_{1-\alpha /2}$ is the standard normal quantile. Its lower endpoint may be truncated to zero because the parameter is nonnegative. If ${\tau }^2=0$, $X_i^2=\theta$ almost surely and the statistic identifies the parameter exactly; the interval degenerates to that value. The nondegenerate normal interval is asymptotic, not an exact Student interval under arbitrary $F$.

A 95 percent confidence set is a measurable data-dependent set $C(X_1,\dots ,X_n)$ whose repeated-sampling coverage satisfies ${\mathbb{P}}_{\theta }\{\theta \in C(X_1,\dots ,X_n)\}\ge 0.95$ for every parameter value, with equality for an exact 95 percent procedure. The parameter is fixed; the set is random. This is not a posterior probability statement about the parameter after the data have been observed.

For independent normal random variables of known variance, $\overline{X}\sim N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The supplied quantile gives the exact confidence intervalIts deterministic length is $3.920\sigma /\sqrt{n}$. Requiring this to be at most $\epsilon$ givesAmong fixed-length intervals based equivariantly on the sample mean, the centred interval has greatest coverage: an interval of a given length captures most mass of a symmetric unimodal normal distribution when centred on its mean. Thus the width calculation is sharp for the usual fixed-length normal-mean construction.

Index city by $g\in \{0,1\}$ and age group by $j$, with the youngest group the reference level. For each of the 15 observed cells, the grouped-binomial logistic regression assumeswith independent cell counts. The response supplied to R is $C_{gj}/N_{gj}$, and the weights specify the binomial numbers of trials $N_{gj}$. They are not 15 arbitrary precision weights: the conditional variance of the proportion is ${\pi }_{gj}(1-{\pi }_{gj})/N_{gj}$. The logit link models log odds, and the two sets of factor coefficients use corner-point constraints.

There are nine parameters: one intercept, seven age contrasts and one city contrast. Hence the residual degrees of freedom are $15-9=6$, not the total population minus nine. The null model has only an intercept and $14$ residual degrees of freedom. Adding age uses seven more parameters and reduces the binomial deviance from $2330.46$ to $232.28$, a drop of $2098.19$. Adding city then uses one parameter and reduces it by $227.12$ to $5.15$. These are sequential likelihood-ratio tests, asymptotically compared with chi-squared distributions on seven and one degrees of freedom. Both provide extremely strong evidence of an effect; the printed age p-value $0.00$ is rounding, not an exactly zero probability. Because the table is incomplete, sequential main-effect sums need not be invariant to the order of fitting.

The intercept estimate $-11.69364$ is the log odds for the youngest Minneapolis–Saint Paul group. It implies baseline odds $e^{-11.69364}$ and a fitted probability about $8.35\times 10^{-6}$. Its standard error is $0.44923$; its reported Wald test tests log odds zero, or probability $1/2$, rather than a scientifically interesting comparison of groups.

The age coefficients are log odds ratios relative to the youngest group within the same city. Their exponentials are approximately $13.86$, $46.82$, $99.03$, $162.23$, $284.38$, $497.07$ and $484.57$ as age increases through the seven other groups. All these comparisons have large positive Wald statistics and small p-values. Thus the fitted probabilities are much higher in the older groups than in the youngest group. The slightly smaller coefficient in the oldest group than in the preceding group does not itself prove a real decline: that claim needs a contrast and its standard error, including the covariance of the two estimates.

The city coefficient $0.85492$ is the log odds ratio for Fort Worth relative to Minneapolis–Saint Paul, adjusted for age. Its standard error $0.05969$ gives $z=14.322$, so the null city effect is strongly rejected by the Wald test, consistent with the sequential likelihood-ratio test. The fitted constant-over-age city odds ratio and an approximate 95% confidence interval areThis is an odds ratio, not generally a probability ratio. The probabilities here are sufficiently small that the two ratios are close. The absence of an age-by-city interaction is an assumption of the fitted additive log-odds model.

The final binomial deviance is $5.1509$ on six degrees of freedom; its approximate upper-tail p-value is $0.525$. It gives no indication of substantial lack of fit against the saturated observed-cell model. The deviance residuals range from $-1.2830$ to $1.0820$, with median zero and no conspicuously large value in the five-number summary. Neither statement proves that every model assumption holds. In particular, the small case counts motivate checking the accuracy of the chi-squared asymptotic approximation, for example by a fitted-model parametric bootstrap.

The dispersion is fixed at one by the binomial distribution; it was not estimated as a free scale parameter. Four Fisher scoring iterations describe the numerical optimization, not four additional parameters. Residual and cellwise observed-versus-fitted comparisons would help assess adequacy and potential overdispersion.

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.

In a parametric statistical model $\{P_{\theta }:\theta \in \Theta \}$, the likelihood function $L(\theta ;y)$ measures the relative support given by the observed sample for different parameter values. Maximum likelihood estimation supplies a point estimate; the sampling distributions of suitable statistics calibrate significance tests and confidence intervals. A sufficient statistic retains all parameter-dependent information in the likelihood function, so one first seeks a reduction to such a statistic rather than discarding information arbitrarily.

The characteristic conditioning principle in Fisherian conditional inference is to calibrate inference conditional on the observed value of an ancillary statistic. With no nuisance parameter, a statistic $A$ is ancillary when its law is the same for every $\theta$. Its observed value can nevertheless describe the experimental configuration or precision. Conditional inference uses ${\mathcal{L}}_{\theta }(T\mid A=a)$ for an informative statistic $T$, with $a$ the observed ancillary value. Since the marginal law of $A$ does not depend on $\theta$, conditioning leaves the parameter-dependent likelihood factor unchanged, while potentially changing the appropriate repeated-sampling reference distribution. Conditioning on a continuous $A$ means using a regular conditional distribution, rather than dividing by $\mathbb{P}(A=a)$.

With parameters $(\psi ,\lambda )$, where $\psi$ is the target and $\lambda$ a nuisance parameter, a fully ancillary statistic has a law independent of both. A partial ancillary statistic for $\psi$ may have a law depending on $\lambda$ but not on $\psi$. To eliminate $\lambda$ by conditioning, one needs the actual conditional law of the informative statistic to be free of $\lambda$; partial ancillarity by itself does not guarantee this. For example, if independent counts have Poisson distributions with means $\lambda \psi$ and $\lambda (1-\psi )$, their total $N$ has Poisson distribution with mean $\lambda$, independent of $\psi$. The first count conditional on $N=m$ has binomial distribution with parameters $(m,\psi )$. Conditioning removes the nuisance intensity $\lambda$ exactly. More generally, nuisance-free pivotal quantities or conditional likelihoods can be used where available; there is no universal conditioning device that removes every nuisance parameter.

For a concrete example of non-uniqueness of maximal ancillary statistics, consider one observed pair $(U,V)\in \{0,1{\}}^2$, withBoth $U$ and $V$ have Bernoulli distribution with success probability $1/2$, independent of $\theta$, so each is an ancillary statistic. Their generated partitions are different and incomparable. Moreover, each is maximal: refining either two-point cell singles out a point whose probability is $\theta /2$ or $(1-\theta )/2$, which is parameter-dependent. Their joint statistic is not ancillary, since $P_{\theta }(U=V)=\theta$. Thus there is no common finer ancillary partition that reconciles the two choices. The conditional laws differ as labelled experiments: $P_{\theta }(V=U\mid U)=\theta$ and $P_{\theta }(U=V\mid V)=\theta$, but they condition on different observed information. This illustrates why “condition on a maximal ancillary” need not specify a unique procedure.

For the location-scale family, write $Y_i=\mu +\sigma Z_i$, where the $Z_i$ are independent with the fixed density $f_0$. Let $s_Z^2=n^{-1}{\sum }_i(Z_i-\bar{Z}{)}^2$. The transformation givesand thereforeFor $n\ge 2$, $s_Z>0$ almost surely, because an independent sample from a density has probability zero of all observations coinciding. The right side depends only on a sample from the fixed base law, so the entire vector $A$ has a law independent of $(\mu ,\sigma )$: $\boxed{A\text{ is ancillary}.}$ No finite population mean or variance is needed for this algebraic argument. With one observation the displayed standardization is undefined, so at least two observations are implicitly required.

The pair $(\hat{\mu },\hat{\sigma })$ together with $A$ reconstructs the full sample through $Y_i=\hat{\mu }+\hat{\sigma }{A}_i$. Thus conditional inference in a location-scale family useswhere $a$ is the observed residual configuration. Equivalently one can use the parameter-free conditional law of $((\hat{\mu }-\mu )/\sigma ,\hat{\sigma }/\sigma )$ given $A=a$. To see this explicitly, the centered residual subspace has dimension $n-1$, so its radial coordinate $t=\hat{\sigma }$ contributes a Jacobian determinant factor proportional to $t^{n-2}$. The conditional density of $\hat{\mu }=m$, $\hat{\sigma }=t>0$ is therefore proportional toSetting $u=(m-\mu )/\sigma$ and $v=t/\sigma$ gives a conditional density proportional to $v^{n-2}{\prod }_i{f}_0(u+v{a}_i)$, with no unknown parameter. The normalizing factor can depend on $a$, but not on $\mu$ or $\sigma$. These statements concern almost every attainable ancillary value.

Let the summands be independent and identically distributed random variables with mean $\mu$, variance ${\sigma }^2>0$, and standardized third cumulant ${\gamma }_1=\mathbb{E}[(Y_1-\mu {)}^3]/{\sigma }^3$. DefineAn Edgeworth expansion refines the central limit theorem using the higher cumulants of one summand. Some smoothness and moment assumptions are essential: finite variance alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose derivatives decay faster than every inverse power, together with $\mathbb{E}{e}^{\delta |Y_1|}<\infty$ for some $\delta >0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.

Write $\varphi$ and $\Phi$ for the standard normal density and standard normal distribution function. The first-order density Edgeworth expansion isThe remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu )/\sigma$ the cumulant expansion of its characteristic function givesExponentiating and using Fourier inversion yields the displayed correction, since the inverse transform of $(it{)}^3{e}^{-t^2/2}$ is $-{\varphi }^{(3)}(x)=(x^3-3x)\varphi (x)$. The Probabilists' Hermite polynomial convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.

The corresponding distribution function expansion isuniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\varphi (x)]=-H_3(x)\varphi (x)$. These are expansions of both the density and distribution function; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.

For a fixed $\alpha \in (0,1)$, write $z=z_{\alpha }={\Phi }^{-1}(\alpha )$. At leading order the proposed quantile expansion gives $\Phi (p_0(z))=\Phi (z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_{\alpha }=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the distribution function expansion. A Taylor expansion at $z$ givesSince $\varphi (z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first Cornish-Fisher expansion coefficients are

For the logistic distribution, set $Z_i=Y_i-\theta$. Its density $q(z)=e^z/(1+e^z{)}^2$ satisfies $q(-z)=q(z)$ and has exponentially decreasing tails. Symmetry and integrability give $\mathbb{E}{Z}_i=0$, and the supplied integral givesTherefore $\mathbb{E}{Y}_i=\theta$, $\mathrm{Var}(Y_i)={\pi }^2/3$, and the central limit theorem givesWriting $z_{-}=z_{\alpha /2}$ and $z_{+}=z_{1-\alpha /2}$, the condition $z_{-}\le T_n\le z_{+}$ is equivalent toConsequently the confidence intervalhas limiting coverage $\Phi (z_{+})-\Phi (z_{-})=1-\alpha$. Whether the finite endpoints are included makes no probability difference because the sample mean has a density.

The centered logistic distribution is symmetric, so ${\gamma }_1=0$. Its smooth density and exponentially decreasing derivatives satisfy the conditions stated above; in particular all moments exist and Fourier smoothing is available. The first correction in the Edgeworth expansion vanishes, leaving $F_{T_n}(z)=\Phi (z)+O(n^{-1})$ at each of the two fixed quantiles. Subtracting gives the stronger conclusionThe standardized law is independent of $\theta$, so this coverage-error conclusion holds with the same bound for every location parameter.

The mean ratio is $R=(\lambda e^{\beta })/\lambda =e^{\beta }$, so its logarithm is exactly $\beta$; the baseline coefficient $\alpha$ cancels. Under the stated asymptotic normal approximation, the Wald confidence interval for $\beta$ is $0.69\pm 1.96(0.15)=(0.396,0.984)$. Exponentiation is monotone and therefore preserves the interval's coverage event. HenceThe standard error of $\alpha$ is not needed for this confidence interval: the Poisson regression already supplies the standard error of the contrast $\beta$. It would be incorrect to combine the two reported standard errors as if the mean ratio depended on both coefficients independently.

For a normalized kernel for density estimation $K$, with $\int K=1$, the kernel density estimator isIt replaces each atom of the empirical measure by a translated, scaled copy of $K$: averaging these small probability-density bumps gives a smooth approximation to the sampling density. Integral one is part of the probability-kernel convention; without it, the estimator generally targets $f(x)\int K$.

The printed bandwidth assumption is insufficient for the claimed normal limit. The additional condition $n{h}_n\to \infty$ is needed at points with positive density. For a counterexample, take a normal distribution density $f$, $x=0$, the unit-width box kernel and $h_n=n^{-2}$. Then $n{h}_n^3\to 0$, but the probability that any observation lies in the window is at most $n\Vert f{\Vert }_{\infty }{h}_n\to 0$. On the complementary event the estimator is zero, and $\sqrt{n{h}_n}({\hat{f}}_{n,h_n}(0)-f(0))=-f(0)/\sqrt{n}\to 0$. The limit in probability is therefore zero, whereas the stated normal limit has positive variance.

We now prove the intended pointwise central limit theorem for a kernel density estimator under $h=h_n\to 0$, $nh\to \infty$ and $n{h}^3\to 0$. The first condition also follows from the last one. A mean value theorem bound, with $L=\Vert f^{\prime }{\Vert }_{\infty }$, gives the bias of a kernel density estimatorThus the bias multiplied by $\sqrt{nh}$ tends to zero.

The Lindeberg-Feller central limit theorem says that independent centered variables in each row of a triangular array have sum converging in distribution to $N(0,{\sigma }^2)$ if the sum of their variances tends to ${\sigma }^2>0$ and, for every $\eta >0$, the sum of their truncated second moments $\mathbb{E}[Z_{n,i}^2{1}_{\{|Z_{n,i}|>\eta \}}]$ tends to zero. Put $Y_{n,i}=K((x-X_i)/h)$ and $Z_{n,i}=(Y_{n,i}-\mathbb{E}{Y}_{n,i})/\sqrt{nh}$. The independence hypothesis holds within each row. A change of variable givesContinuity of $f$ at $x$ and the compact support and boundedness of $K$ justify the limit. Also $|Z_{n,i}|\le 2\Vert K{\Vert }_{\infty }/\sqrt{nh}\to 0$, so the Lindeberg condition is eventually identically zero for every $\eta$. The sum is exactly $\sqrt{nh}({\hat{f}}_{n,h}(x)-\mathbb{E}{\hat{f}}_{n,h}(x))$. Combining the centered limit with the negligible bias by the Slutsky theorem provesIf $f(x)=0$, the centered variance tends to zero, and the Chebyshev inequality proves the corresponding degenerate limit directly.

For $f(x)>0$, let $q_{\beta }$ be the supplied $\beta$-quantile of this limiting normal distribution. Inverting the limiting central probability gives the confidence intervalContinuity of the limiting distribution gives asymptotic coverage $1-\alpha$. Equivalently, its half-width is $z_{1-\alpha /2}\sqrt{f(x)\Vert K{\Vert }_2^2/(n{h}_n)}$ under the stipulated oracle quantiles. A degenerate limit at $f(x)=0$ does not by itself justify this nominal-coverage normal interval.

The fitted continuous-time multi-state model allows only adjacent fatigue-state jumps. Its four transition intensities, in reciprocal years, are shown here:

In state $2$, the total exit intensity is ${\lambda }_2=q_{21}+q_{23}=0.4684+0.3835=0.8519$. By the holding time law for a continuous-time Markov chain, the time until the next jump has an exponential distribution with this rate. ThereforeThe printed interval for $q_{22}=-{\lambda }_2$ gives the interval $(0.7219,1.005)$ for ${\lambda }_2$. Taking reciprocals reverses the endpoints, so the transformed $95\%$ confidence interval for the mean holding time isFor the next jump's destination, competing outgoing intensities giveIndeed the density of a first jump to $1$ at time $t$ is $q_{21}{e}^{-{\lambda }_{2}t}$; integrating gives the ratio above. This is a conditional next-jump probability, not a transition probability of being mild at a fixed later visit.