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Deleting $q$ fitted mean parameters improves the Akaike information criterion exactly when the displayed condition holds, since the criterion difference is $n\log ({\mathrm{RSS}}_{\mathrm{reduced}}/{\mathrm{RSS}}_{\mathrm{full}})-2q$. For one parameter, the equivalent partial F-test statistic is below $(n-p)(e^{2/n}-1)$, where $p$ is the full mean-model dimension. This is not the usual fixed-level significance-test cutoff. Fits must use the same observations and response transformation.

The corrected Akaike information criterion adds a finite-sample penalty to the Akaike information criterion, where $k$ counts fitted parameters and $n>k+1$. The displayed correction is commonly used for small samples; its exact unbiasedness depends on the model family.

Both candidate models fit one parameter, but their parameter symbols have different meanings. For the exponential distribution in model B,Its second derivative is $-N/{\theta }_B^2<0$, so this is its maximum-likelihood estimate. The maximized log-likelihoods, including their normalization terms, areUse the corrected Akaike information criterion because $N=4$ is small relative to the number $k=1$ of fitted parameters:The correction is $2$ for each model, givingPrefer model A, whose corrected Akaike information criterion is smaller by $7.67252$. The ordinary Akaike information criterion also gives that ranking because the two models receive equal penalties. This compares the two proposed models; it does not establish that either is a correct description of the process. The usual corrected Akaike information criterion formula is a small-sample model-selection convention, rather than an exact finite-sample guarantee for these density families.

The Poisson generalized linear mixed model assumes conditionally Poisson distributed counts with a log link. The fitted variance components are both zero, so this fit lies on a boundary and reduces to ordinary Poisson regression. It estimatesThus the fitted annual increase is about 6.6%, with approximate 95% confidence interval for the rate ratio $(e^{0.001},e^{0.128})\simeq (1.001,1.137)$. The age evidence is much weaker than in the Gaussian fit: the quoted p-value is $0.047$. Zero estimated group variances do not prove that real bird and year differences are absent.

The Poisson distribution forces conditional variance to equal the conditional expected value. Here fitted means are around two to three hatchlings, whereas the Gaussian model estimated a residual variance of only $0.188$. This suggests checking underdispersion and the bounded nature of hatching counts, rather than assuming a Poisson response solves the diagnostic problems. Inspect simulated count residuals and predictive distributions; if the number of eggs is observed, fit a binomial mixed model and check that model's dispersion too.

The displayed Gaussian and Poisson Akaike information criterion values should not simply be ranked as if they used the same likelihood convention: one evaluates a continuous density and the other a probability mass. Compare properly specified count models on the same observed response scale, with compatible likelihoods and predictive checks. A boundary fit, doubtful dispersion assumptions, and possible nonlinear age effects all warrant further work before a strong biological conclusion.

For a normal linear model with $n$ fixed observations and $p$ mean coefficients, maximizing over the error variance gives ${\hat{\sigma }}^2=\mathrm{RSS}/n$. The maximized negative twice log-likelihood is $n\log (\mathrm{RSS}/n)$ plus a constant common to the candidate models. Counting the variance as another parameter adds the same penalty two to every model, so the stated criterion is equivalent for comparing them:The Akaike information criterion balances fit against parameter count; smaller values are preferred. It does not compare models using different response scales or different sets of observations without further adjustment.

Stepwise selection by the Akaike information criterion compares allowed one-term additions and deletions, takes a change lowering the criterion, and repeats until no allowed change improves it. The search direction, initial fit and lower/upper scopes matter. A displayed no-change row represents the current model. The term degrees of freedom show how many coefficients a change removes, the residual sum reflects its fit, and the AIC column determines the selected move. The search is greedy, not a guarantee of the best subset over the entire model space. The software's scope and direction conventions are documented in stat.ethz.ch/R-manual/R-devel/library/MASS/html/stepAIC.html .

Deleting $q$ coefficients changes the criterion byThus the AIC deletion threshold in a normal linear model isFor a single coefficient, its partial F-test statistic must be below $(n-p)(e^{2/n}-1)$ for AIC to favor deletion. At the full 47-observation, 16-coefficient fit this is approximately $1.348$, corresponding to $|t|<1.161$. Therefore AIC selection is not equivalent to retaining only coefficients with a conventional five-percent p-value.

The final model should be described by its retained terms, signs, conditional effect sizes, uncertainty, residual diagnostics and prediction behavior. A coefficient that survives selection need not identify a causal mechanism, and its ordinary final-model p-value ignores the preceding search. Stability under bootstrap resampling and predictive cross-validation are useful. The absent search trace and final model prevent identifying the actual deletion sequence, final predictors, coefficients, RSS or AIC. The given starting name does not specify any of those facts.

For minimization of an energy $E(\vartheta )$, simulated annealing at temperature $t>0$ targets the Boltzmann distribution ${\pi }_t(\vartheta )\propto e^{-E(\vartheta )/t}$ relative to a specified parameter reference measure. With proposal distribution $q({\vartheta }^{\prime }\mid \vartheta )$, accept withAt each temperature run the chain long enough to explore that law, then cool and retain the best state seen. For symmetric proposals every improvement is accepted, while a worsening move is accepted with probability $e^{-\Delta E/t}$. Equilibrium laws concentrate near global minima when their normalizing integrals are finite and the minima are suitably isolated. A particular rapid cooling schedule does not automatically inherit this guarantee.

Likelihood maximization uses energy $E=-\ell$, not $E=\ell$. The source's later wording “$f(p)$ equal to the log-likelihood” conflicts with its preceding minimization convention. The intended maximum-likelihood Boltzmann distribution is proportional to $e^{\ell /t}=L^{1/t}$. With the opposite sign, likelihood maxima would be disfavored; for binomial data with both successes and failures the density often would not even be integrable near the endpoints as $t$ decreases.

For the maximum-likelihood estimates, write $s={\sum }_{j=1}^m{x}_j$, $\bar{x}=s/m$ and $S={\sum }_j(x_j-\bar{x}{)}^2$. In the binomial distribution model with known positive integer $n$, the log-likelihood isDifferentiating gives $s/p-(mn-s)/(1-p)=0$, hence $\hat{p}=s/(mn)$. The second derivative is negative in the interior. If $s=0$ or $s=mn$, the maximum is at the corresponding endpoint. This model requires integer observations in $\{0,\dots ,n\}$; otherwise its likelihood is zero.

For the normal distribution model write $v={\sigma }^2$. Its log-likelihood isFor fixed $v$, the unique maximizing mean is $\bar{x}$. Substituting it and differentiating in $v$ gives $v=S/m$, so, assuming $S>0$,If $S=0$, the normal density likelihood grows without bound as $v\downarrow 0$ with $\mu =\bar{x}$; there is no finite interior maximum or ordinary finite AIC for that fit.

For Gaussian likelihood Gibbs annealing, use $L_N^{1/t}$ relative to $d\mu dv$. Completing the square and identifying an inverse-gamma distribution gives the Gibbs sampler conditionalsIntegrating out $\mu$ contributes a factor proportional to $v^{1/2}$, so the marginal is $v\sim \mathrm{InvGamma}(m/(2t)-3/2,S/(2t))$. The joint target is proper for $0<t<m/3$ when $S>0$; starting the Lebesgue-based annealing above this range would require a proper reference factor or bounded parameter domain. For sufficiently small $t$, its moments satisfyThus equilibrium draws concentrate at $(\bar{x},S/m)$ as $t\downarrow 0$.

There is also a direct convergence proof for the actual successive Gibbs sampler updates here. Choose deterministic $0<t_r\le m/10$ with $t_r\to 0$, start at a finite deterministic $v_0>0$, draw ${\mu }_r$ from the first conditional using $v_{r-1}$, then draw $v_r$ from the second using ${\mu }_r$. Conditional normal moments give $E[({\mu }_r-\bar{x}{)}^2\mid v_{r-1}]=t_r{v}_{r-1}/m$ and $E[({\mu }_r-\bar{x}{)}^4\mid v_{r-1}]=3t_r^2{v}_{r-1}^2/m^2$. The permitted inverse-gamma moments therefore yieldThe coefficient on the preceding first moment is at most $1/6$, and that on the preceding second moment is at most $1/8$. The remaining coefficients are uniformly bounded, so both moment sequences are bounded by iteration. Since $t_r\to 0$, the recursions then give $E[v_r]\to S/m$ and $E[v_r^2]\to (S/m{)}^2$. Also $E[({\mu }_r-\bar{x}{)}^2]=t_{r}E[v_{r-1}]/m\to 0$. HenceThis proves convergence in mean square, and thus in probability, to the normal maximum-likelihood estimates for this explicit cooling/update scheme, rather than merely assuming equilibration.

For binomial likelihood-power annealing, normalizing $p^{s/t}(1-p{)}^{(mn-s)/t}$ on $(0,1)$ givesIts mean is $(s+t)/(mn+2t)$, and its variance isThe mean tends to $s/(mn)$ and the variance tends to zero, proving concentration at the binomial maximum-likelihood estimate, including the endpoint cases. Independent exact draws from these beta laws are a simple annealing implementation within this model.

For model selection, the Akaike information criterion is $2d-2\hat{\ell }$ with $d_B=1$ and $d_N=2$. ThusUse the endpoint convention $0\log 0=0$. For a trans-dimensional annealing for penalized likelihood construction, target the disjoint model spaces with unnormalized densitiesrelative to $dp$ and $d\mu dv$, respectively. The energies are half the unoptimized AIC, $d-\ell$; their joint minimizer is the fitted model with smaller AIC. Keep constants such as $C_B$ and the normal density normalization, which do not cancel across models. The within-model conditional draws are unchanged by these constant penalties.

Here is a complete dimension-changing proposal. Select a binomial-to-normal move with probability $b>0$ and a normal-to-binomial move with probability $d>0$, both fixed; other steps can use the within-model updates. From $B,p$, draw $u\sim N(0,1)$ with density $h(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$ and setThis bijection from $(p,u)\in (0,1)\times \mathbb{R}$ onto $({\mu }^{\prime },v^{\prime })\in \mathbb{R}\times (0,\infty )$ has inverse $p=(1+e^{-{\mu }^{\prime }}{)}^{-1}$, $u=\log v^{\prime }$. Its absolute Jacobian determinant is $v^{\prime }/[p(1-p)]$. The Binomial-normal reversible-jump annealing acceptance ratio is thereforeFor a reverse proposal from $N,\mu ,v$, compute $p^{\prime }=(1+e^{-\mu }{)}^{-1}$ and $u^{\prime }=\log v$ deterministically and accept withThese ratios are reciprocal at inverse states and enforce detailed balance on the dimension-matched augmented spaces. They specify the proposal density, model-selection probabilities and Jacobian explicitly. The map is a valid example, not a claim of optimal mixing. Positive-temperature beta draws stay in the open interval even when the limiting estimate is an endpoint. At each temperature ensure adequate model switching; retain the best fitted penalized likelihood encountered. If minima tie, limiting concentration need not choose a unique model.

There is a material measurement qualification. A binomial probability mass and an unrounded normal density use different observation reference measures. Their raw AIC values are not an invariant, scientifically calibrated comparison of the same recorded data. The expressions and jump algorithm above implement the formal density comparison suggested by the question, with its units fixed. For integer-recorded measurements, a compatible normal model instead assigns probabilities $P_{\mu ,v}(x-1/2<Z\le x+1/2)$, suitably adjusting boundaries or truncation if required. Then both models describe the same recorded outcomes relative to counting measure. Use those masses in the normal likelihood and the same reversible-jump formula; its simple normal MLE and conjugate Gibbs updates will generally no longer apply. This observation-model correction cannot be silently combined with the continuous-normal formulas.

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.

For minimizing an energy $F(\theta )$, simulated annealing at temperature $T>0$ uses the Boltzmann distribution ${\pi }_T(\theta )\propto e^{-F(\theta )/T}$. At each temperature run a Metropolis–Hastings algorithm or Gibbs transition preserving that law, then reduce $T$. With a symmetric proposal, accept ${\theta }^{\prime }$ with probability $\min \{1,e^{-[F({\theta }^{\prime })-F(\theta )]/T}\}$: downhill moves are accepted and uphill moves remain possible. Keep the best state encountered. As temperature decreases, equilibrium mass favors global minima, provided the target is proper and the relevant concentration hypotheses hold. An arbitrary finite cooling run does not by itself prove global convergence.

Let $s={\sum }_{i=1}^m{x}_i$, $\bar{x}=s/m$ and $S={\sum }_i(x_i-\bar{x}{)}^2$. For the Poisson distribution model the observations must be nonnegative integers. Its log-likelihood isFor $s>0$, differentiation gives ${\ell }_P^{\prime }=-m+s/\lambda$ and ${\ell }_P^{\prime \prime }=-s/{\lambda }^2<0$. If $s=0$, the likelihood is decreasing and the maximizer is the boundary value zero. For the normal distribution model write $v={\sigma }^2>0$, givingFor each $v$, this is maximized at $\mu =\bar{x}$, and differentiating there gives $v=S/m$. HenceThe normal variance estimate has denominator $m$, not $m-1$. A finite positive normal maximum requires $S>0$; if all observations coincide, the likelihood is unbounded as $\mu =\bar{x},v\downarrow 0$. The normal annealing calculation below assumes $S>0$.

To maximize likelihood, choose the energy $-\ell$, so the annealing law is $L^{1/T}$, not $L^{-1/T}$. This sign is important in the Poisson request: literally inserting a positive log-likelihood into a minimizing Boltzmann law would give a nonnormalizable probability density function growing like $e^{m\lambda /T}$ at infinity. We use likelihood-power annealing with the maximizing sign.

For the normal distribution model, take base measure $d\mu dv$. The joint annealing probability density function isCompleting the square gives the Gibbs sampling conditional for $\mu$. Matching the conditional $v$ probability density function to the supplied inverse-gamma distribution convention givesIntegrating out $\mu$ contributes a factor $\sqrt{v}$, so the marginal law isThus this joint Boltzmann law is proper for $T<m/3$. Its variance marginal and the normal conditional both concentrate at the stated estimates as $T\downarrow 0$. For example, for $T<m/7$ its marginal mean and variance areAlso ${\mathbb{E}}_T(\mu -\bar{x}{)}^2=(T/m){\mathbb{E}}_{T}v\to 0$.

There is a direct convergence proof for the actual changing-temperature Gibbs algorithm, beyond convergence of its equilibrium probability densities. Start at a finite deterministic $v_0>0$, choose deterministic $0<T_n\le m/10$ with $T_n\to 0$, draw ${\mu }_n$ conditional on $v_{n-1}$, then draw $v_n$ conditional on ${\mu }_n$ using the two boxed distributions. Conditional normal moments giveThe inverse-gamma distribution first and second moments consequently yieldandIn the first recursion the coefficient of the preceding mean is at most $1/6$; in the second the coefficient of the preceding second moment is at most $1/8$. The means therefore stay bounded, and then so do the second moments. Letting $T_n\to 0$ in these formulas gives $\mathbb{E}{v}_n\to S/m$ and $\mathbb{E}{v}_n^2\to (S/m{)}^2$. Together with $\mathbb{E}({\mu }_n-\bar{x}{)}^2=T_n\mathbb{E}{v}_{n-1}/m\to 0$, this provesThis Gaussian likelihood Gibbs annealing convergence uses the explicit conditional structure and does not assume that one Gibbs sampling sweep is an exact equilibrium draw.

For the Poisson distribution model, the likelihood-power law with respect to $d\lambda$ has the normalized probability density functionsoThe mean tends to $\bar{x}$ and the variance to zero, proving convergence to a point mass at the Poisson maximum-likelihood estimate, including the zero-boundary case.

The Akaike information criterion is $2k-2\ell (\hat{\theta })$, counting one parameter for Poisson and two for normal. With $0\log 0=0$ and $S>0$, the two forms areandChoose the smaller criterion. Likelihood constants that differ between models, including the factorial and $2\pi$ terms, cannot be discarded.

For a fully specified trans-dimensional annealing for penalized likelihood, let $M=P$ or $N$ and defineUse the combined target, relative to counting measure on models and Lebesgue measure on their parameters,where $w_P,w_N>0$ are fixed reference weights. For $S>0$ and $0<T<m/3$ both model components are integrable. The factor $2T$ is a convenient temperature convention keeping the within-model laws exactly those already derived. The extra factor $e^{-k_M/T}$ makes the zero-temperature objective AIC rather than unpenalized likelihood.

Use, for example, probability $1/2$ for a model-change attempt in either model and probability $1/2$ for a within-model update. Within the Poisson distribution model draw a fresh gamma value from its boxed law; within the normal distribution model perform the boxed Gibbs sampling sweep. For a Poisson-to-normal proposal, draw $u\sim q(u)=N(0,{\tau }^2)$ with a fixed ${\tau }^2>0$, and setThe inverse normal-to-Poisson proposal is $\lambda =v$ with reverse auxiliary $u=\mu -v$. This dimension matching uses one auxiliary variable to augment the one-dimensional Poisson parameter space. The absolute Jacobian determinant isIf the forward model-jump selection probability is $b$ and the reverse probability is $d$, the reversible-jump Markov chain Monte Carlo acceptance ratio isAccept with probability $\min (1,R_{P\to N})$. For the reverse move compute $\lambda =v,u=\mu -v$ and useFor the suggested equal selection probabilities, $b=d=1/2$ and those factors cancel. The probability density function $q$ must still be included, despite symmetry of its normal distribution: it is an auxiliary draw in only the forward dimension-increasing proposal. Pairing a move with its inverse makes the accepted forward and reverse probability flows equal after the change of variables, proving detailed balance for each fixed-temperature model-switch kernel.

Cool through proper temperatures, allow enough model and parameter exploration at each stage, and retain the smallest observed $E_M$. The minima of the two energies are exactly the two AIC values above. Low-temperature equilibrium mass concentrates on the better criterion and its fitted parameters; if they tie, the minimizing set contains both models. Unlike the direct normal Gibbs argument, this multimodel exploration claim needs an appropriate cooling/mixing justification and is not guaranteed by an arbitrary short run.

Finally, an exact statistical comparison must put both likelihoods on the same observable and reference-measure convention. A Poisson distribution model is a probability mass function on counts, whereas the displayed normal distribution model is a continuous probability density function. The formulas above are the question's formal Gaussian working-likelihood comparison. A coherent exact normal approximation for count observations can instead use probabilities of rounding intervals; that changes its likelihood and generally loses the simple analytic normal estimates. This distinction matters when interpreting the numerical AIC difference, while leaving the dimension-matching and annealing construction applicable to properly specified likelihoods.

Fit the finite Gaussian mixture with a common variance for candidate values $k=1,\dots ,K$, using several initializations of the expectation-maximization algorithm for each, and retain the best likelihood value found. Maximized likelihood alone cannot select the number of components: a smaller mixture embeds in a larger one by giving an added component zero weight, so its likelihood cannot decrease.

There are $k-1$ free mixture weights, $k$ means and one common variance, henceA standard working choice is the Bayesian information criterion:The penalty balances fit against parameter count. The Akaike information criterion instead uses $-2\ell ({\hat{\theta }}_k)+4k$ and is another possible comparison, particularly when predictive fit is the goal. The parameter count would differ if component variances were separate, but that is not this model.

Likelihood fitting must first be made well posed. Even a shared variance does not prevent all degeneracy: if $k=N$, choose ${\mu }_j=x_j$, ${\alpha }_j=1/N$ and let the common variance tend to zero. Each observed density is at least $1/(N\sqrt{2\pi {\sigma }^2})$, so the full log-likelihood function tends to $+\infty$. The same construction works when $k$ reaches the number of distinct observed values. Thus one must not feed infinite fitted likelihoods into an information criterion. Restrict candidate sizes to a sensible range below that degeneracy, or impose a fixed positive variance floor and use that same constrained model in every comparison. Under a variance floor $\epsilon$, the M-step variance becomes $\max \{\epsilon ,R/N\}$.

There is also nonregular mixture model selection: at a missing component its weight is on a boundary and its mean is unidentified, while permuting labels leaves the distribution unchanged. Consequently an ordinary chi-squared calibration for a likelihood-ratio test between successive component counts is not generally justified. Information criteria are practical comparisons, not an automatic universal proof that the true number has been recovered. Predictive cross-validation or a suitably constructed parametric bootstrap comparison can supplement them. Select the component count by a penalized, well-defined mixture fit, rather than by unpenalized likelihood increase.

Write $q=n-p$. In this normal linear model, the ordinary least squares criterion is $\Vert Y-Xb{\Vert }^2$. Differentiating with respect to $b$ gives the normal equations for linear least squares, $X^{T}Xb=X^{T}Y$. Full column rank of the design matrix makes $X^{T}X$ invertible, and completing the square proves that the unique minimum isSince $\hat{\beta }=\beta +(X^{T}X{)}^{-1}{X}^Tϵ$, its sampling distribution is the multivariate normal distributionLet $H=X(X^{T}X{)}^{-1}{X}^T$ be the hat matrix. It is symmetric and idempotent with rank $p$, and $HX=X$. The regression residual vector is $(I-H)Y=(I-H)ϵ$, soAn orthogonal projection of a Gaussian vector onto the residual space gives $\mathrm{RSS}/{\sigma }^2\sim {\chi }_q^2$. The residual projection and fitted projection have zero cross-covariance, hence are independent because they are jointly normal. In particular, $s^2$ is an unbiased estimator of ${\sigma }^2$ and is independent of $\hat{\beta }$. The maximum-likelihood estimator of ${\sigma }^2$ instead divides by $n$ and is not unbiased.

For the null hypothesis ${\beta }_1=0$, put $v_{11}=((X^{T}X{)}^{-1}{)}_{11}$. Under that null hypothesis, ${\hat{\beta }}_1/(\sigma \sqrt{v_{11}})$ is standard normal and independent of $q{s}^2/{\sigma }^2$. Therefore the Student t-test usesA two-sided level-$\alpha$ test rejects when $|T|>t_{q,1-\alpha /2}$. The denominator is the estimated standard error of the regression coefficient.

The fitted linear regression includes an intercept and three covariates:Thus $n=25$, $p=4$ and the residual degrees of freedom are $21$. The fitted conditional mean isEach slope measures a conditional association with the response while the other two covariates are held fixed. Increasing wind speed by one mile per hour corresponds to a fitted decrease of $0.527410$ in the concentration units used. Its standard error is $0.224904$, giving $t=-2.345$ and a two-sided p-value of $0.0289$: evidence of a negative conditional association at the 5% level. The temperature slope is $0.124991$ per degree Fahrenheit, with standard error $0.075173$, $t=1.663$ and p-value $0.1112$. The insolation slope is $-0.005259$ per langley per day, with standard error $0.006637$, $t=-0.792$ and p-value $0.4370$. Neither of these last two conditional effects is significant at 5%; that does not establish that either true slope is zero. These are associations, not a demonstration of causal effects.

The intercept estimates the mean at zero values of all three covariates. Its value $3.784916$ has a large standard error $7.979647$, giving $t=0.474$ and p-value $0.6402$. The zero-covariate combination may be outside the observed range, so this test may have little practical relevance. Every coefficient test uses a Student t-distribution with $21$ degrees of freedom.

The reported residual standard deviation $s=1.844$ measures unexplained variation in concentration units; the subsequent table gives $\mathrm{RSS}=71.428$, consistent with $s=\sqrt{71.428/21}$ after rounding. The five-number residual summary describes the spread, from $-2.3052$ to $3.4033$, with median $-0.4990$. An intercept forces the residual sum to zero, not the residual median. These five numbers alone cannot establish normal errors, constant variance or independence; residual-versus-fitted and quantile-quantile plots would be more informative, and successive days also warrant checking serial dependence. The coefficient of determination $R^2=0.6533$ means that 65.33% of the sample response sum of squares about its mean is explained by this fit. Its adjusted version iswhich accounts for the number of estimated regression coefficients.

The stepwise selection by the Akaike information criterion searches between the intercept-only model and the specified model with all three main effects. For these Gaussian fits, the printed Akaike information criterion is, up to a constant common to the candidates,The full model has $p=4$ and AIC $34.246$. Removing insolation increases the residual sum of squares by $2.136$ but decreases AIC to $32.982$, so that removal is chosen. Removing temperature or wind instead would give AIC $35.337$ or $38.060$ and is worse. In the next step the candidate models include deletions and restoration of insolation. Keeping wind and temperature has the smallest AIC: removing temperature gives $33.389$, restoring insolation gives $34.246$, and removing wind gives $37.960$. Hence the search stops, givingThe accompanying F-tests compare a larger model with a one-parameter reduction: $F=\Delta \mathrm{RSS}/s^2$, where $s^2$ is estimated from the larger model. For example, removing insolation gives $2.136/(71.428/21)=0.628$, equal to the corresponding squared test statistic. Deletions from the second-stage model use $73.564/22$ in the denominator, whereas restoring insolation uses the full-model denominator. These tests describe the comparisons; they do not determine the AIC choice. In particular, temperature is retained despite its deletion-test p-value $0.15016$. AIC balances fit and complexity, and a stepwise optimum need not be a global optimum over a richer model class. Ordinary coefficient inference after choosing a model does not account for the selection itself.

For positive responses, the Box–Cox transformation is $T_{\lambda }(y)=(y^{\lambda }-1)/\lambda$ for $\lambda \ne 0$, and $T_0(y)=\log y$. Its profile likelihood refits the linear regression and variance at each $\lambda$ and includes the transformation Jacobian. The graph peaks at roughly $\lambda =0.3$–$0.4$. The approximate 95% likelihood-ratio confidence interval contains values a little below zero through approximately one, using the cutoff ${\ell }_{\max }-{\chi }_{1,0.95}^2/2$. Thus a logarithmic transformation ($\lambda =0$) and a square-root transformation ($\lambda =1/2$) are plausible, while leaving the response on its original scale ($\lambda =1$) is at or very near the upper confidence boundary. The broad profile supports considering a mild power transformation; it does not determine a sharply estimated power. The raster does not justify more precise endpoints or a decisive assertion about which side of the boundary $\lambda =1$ lies.

Use the population as exposure regardless of whether a population-size category is also a covariate. Define a demographic factor with levels $\mathrm{Low}$ for at most 20%, $\mathrm{Medium}$ for $(20,30]$%, and $\mathrm{High}$ above 30%; define $H_{it}=1_{\{P_{it}>250\}}$ if following the suggested population dichotomy. Treat village-size codes as a categorical variable, not equally spaced areas.

A candidate multivariate Poisson generalized linear mixed model retains the random intercept and slope from part (c) and usesThis specifies the suggested village-size-by-population interaction while keeping its main effects. One may add selected time-by-covariate interactions to explain slope heterogeneity, provided information from 32 clusters supports them. Compare continuous demographic and population formulations as a sensitivity analysis: categorization loses information, and the coefficient of a size category measures association beyond the already included exposure effect.

Fit models with the same likelihood convention and conditional distribution when comparing Akaike information criteria; use joint fixed-effect likelihood-ratio tests, boundary-aware parametric bootstrap tests for random effects, residual diagnostics and village-level cross-validation to justify a parsimonious final model. Preserve interactions' main effects and report interval estimates from the final multivariate model, rather than interpreting separate univariate fits. For example, the adjusted inland-versus-coastal rate ratio is $e^{{\theta }_I}$, while a medium-versus-small village ratio is $e^{{\gamma }_2+{\kappa }_{2}H}$; a Wald interval for that contrast uses the full coefficient covariance matrix. Conditional time effects and standardized marginal predictions should be distinguished.

The displayed rows cannot identify this final model: population exceeds 250 in every displayed village-year, no large village is shown, and inland status is confounded with the identity of the two displayed villages. The complete 160-row file is needed for fitted effects, model selection, all-village plots and uncertainty. No numerical best model or evidence of the specified covariate associations can honestly be reported from this excerpt.

Finally, even a significant adjusted association with young-male percentage would not show that young males were involved in most incidents. This is the ecological fallacy, specifically that aggregate composition does not identify offender composition: the same village counts can be assigned to different individual age and sex groups without changing any observed aggregate predictor. Individual incident-level age, sex and involvement records are required to answer that last question. The aggregate dataset cannot establish the claimed majority involvement.

Let $Y_{ijkl}$ be the flight distance in millimetres for paper level $i$, launch-angle level $j$, design level $k$ and replicate $l$, where $i,j,k\in \{1,2\}$ and $l\in \{1,2\}$. The full factorial design is fitted by the normal linear modelThe assumptions are independent errors with a common variance and normal distribution, and the displayed factor-combination mean structure. The corner-point constraints areand $(PAD{)}_{ijk}=0$ whenever at least one index equals $1$. Thus the eight free mean coefficients are $\mu ,P_2,A_2,D_2,(PA{)}_{22},(PD{)}_{22},(AD{)}_{22},(PAD{)}_{222}$. These treatment coding coefficients are comparisons at the other factors' reference levels, rather than necessarily averaged main effects.

To test the three-factor interaction term, compare the full model with the model retaining all main effects and all two-factor interactions. The null hypothesis is $(PAD{)}_{222}=0$, against a nonzero value. Removing it increases the residual sum of squares by $21025$ and removes one parameter. With eight residual degrees of freedom, the nested-model F-test givesThere is no evidence that this three-factor interaction is needed. This does not establish that all two-factor interactions vanish: the paper-design interaction is substantial.

The stepAIC command performs stepwise selection by the Akaike information criterion. Its upper scope is the full factorial model and its lower scope is the intercept-only model. Starting from the full model, it compares admissible additions and deletions, preserving the hierarchy of interaction terms, and chooses changes that reduce the Akaike information criterion. With the supplied scope its default search permits both directions. For these Gaussian fits the displayed criterion, up to constants common to all models, is $16\log (\mathrm{RSS}/16)+2k$, where $k$ counts mean coefficients. The first deletion changes it from $226.72$ to $224.76$: the small increase in residual sum of squares is outweighed by removing a parameter. This is a greedy model selection procedure, not an exhaustive search or a sequence of fixed-level hypothesis tests. The command's arguments are described in the MASS documentation.

For the selected model, let $p=1_{\{i=2\}}$ and $d=1_{\{k=2\}}$. Thenindependently between flights. It has no angle term. The maximum-likelihood estimates of its mean coefficients are also obtained by ordinary least squares:For example, the four fitted means are just the four paper-design cell means, each averaging four flights over the angle levels. With $s^2=\mathrm{RSS}/12$, the variance-covariance matrix of the coefficient estimates is $s^2(X^{T}X{)}^{-1}$. Its diagonal gives the displayed standard errors; their ratios are $s/2,s/\sqrt{2},s/\sqrt{2},s$, with $s=892.6$ mm.