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Allow each subject to have a separate baseline and slope through a correlated random-intercept and random-slope model:The subject random effects are independent of all measurement errors, and subjects are independent. The within-person covariance between times $s,t$ is ${\tau }_0^2+(s+t)\rho {\tau }_0{\tau }_1+st{\tau }_1^2$, with an additional ${\sigma }^2$ when the same reading is used twice.

Prefer the model with a random slope. The maximum likelihood estimation refits improve twice the log-likelihood by $227.9604$ while adding two covariance statistical parameters. Both the Akaike information criterion ($5165.779$ to $4941.819$) and Bayesian information criterion ($5183.367$ to $4968.200$) strongly favour it. The printed nominal likelihood-ratio test is overwhelming. The usual reference chi-squared distribution with two statistical degrees of freedom is not an exact regular calibration, since zero slope variance is a boundary and the intercept–slope correlation is then unidentified; a design-specific parametric bootstrap could calibrate it. This qualification does not undermine the substantial descriptive improvement shown by both information criteria.

The preferred fit estimates population mean baseline strength ${\hat{\beta }}_0=61.09286$ kg and weekly gain ${\hat{\beta }}_1=2.83143$ kg/week. The between-person baseline standard deviation is ${\hat{\tau }}_0=15.785582$ kg, and the between-person slope standard deviation is ${\hat{\tau }}_1=2.738666$ kg/week. Their estimated correlation $\hat{\rho }=0.117$ is weakly positive, giving random-effect covariance about $5.058$ kg${}^2$/week; its uncertainty is not supplied. The measurement-error standard deviation is $\hat{\sigma }=9.923280$ kg. The printed $-0.038$ instead describes the correlation between the estimated fixed effects, not between the subject random effects. These statistical parameter estimates come from restricted maximum likelihood; the model comparison uses the separate maximum likelihood estimation fits.

The population mean fitted trajectory is $61.09286+2.83143t$ kg. HenceUsing the reported slope standard error, an approximate 95% confidence interval for the population mean gain is $28.3143\pm 1.96(2.984464)=(22.46,34.16)$ kg; a Student t confidence interval with 499 statistical degrees of freedom is almost identical.

Successful individuals can improve far more than the average. Their latent ten-week gains have fitted normal distributionA clear estimate for an upper-performing group uses a between-person slope quantile: the 95th percentile is $28.3143+1.645(27.38666)\simeq 73.4$ kg, and the 97.5th percentile is about $82.0$ kg. The upper 5% of fitted underlying gains begin around 73 kg. These are person-to-person performance quantiles, not confidence intervals for the population mean. Identifying the best observed trainee would require that person's data or fitted subject random effects; the aggregate output cannot identify a literal maximum. If performance means the observed difference of endpoint readings, add the measurement-error variance $2{\sigma }^2$ to the latent gain variance.

Let $y_i$ be each trial's estimated log risk ratio, with estimated sampling variance $s_i^2$. A fixed-effect meta-analysis models the independent estimates as $y_i\simeq N(\mu ,s_i^2)$, with one common true log risk ratio $\mu$. Differentiating the Gaussian log-likelihood, or minimizing ${\sum }_i(y_i-\mu {)}^2/s_i^2$, gives inverse-variance weights $w_i=1/s_i^2$ andThe variance follows directly by adding independent variances of the weighted estimates: ${\sum }_i{w}_i^2{s}_i^2/({\sum }_i{w}_i{)}^2=1/{\sum }_i{w}_i$.

Using the printed rounded estimates and standard errors, ${\sum }_i{w}_i=50.2923$, soThe pooled risk ratio is about $0.902$. An interval using the same quantile two is $(-0.3848,0.1792)$ on the log scale, so it includes no effect. Recomputing all trial estimates from the event counts would change the last digits because the printed log estimates and errors are rounded.

The key assumptions are independent trials, a common true treatment effect on this chosen log scale, approximately unbiased and approximately normal trial estimates, and suitable sampling-variance estimates. The common-effect assumption is stronger than merely studying the same named treatment: systematic population, treatment or design differences may produce genuine heterogeneity. For an unbiased interpretation of the pooled evidence, inclusion of studies must also not depend selectively on their results. Treating the plug-in variances as fixed is the usual approximation in the displayed variance formula.

Choose candidate variables from subject-matter knowledge, measurement quality and the target question before screening on outcomes. For an adjusted-effect analysis, retain prespecified confounders and essential design variables even when their individual $P$ values are large. For prediction, consider reliable predictors available at the intended prediction time; do not include future information. Handle missingness transparently, using a justified multiple imputation strategy where appropriate rather than changing the analyzed population unpredictably as variables enter.

Encode a categorical variable with $K-1$ indicators relative to a stated reference category, and assess it as a group. Check multicollinearity, sparse categories and plausible interactions. The relevant information is chiefly the number and distribution of events, not merely the number of enrolled subjects: hundreds of correlated or rarely varying predictors cannot be supported by a small event count.

Compare prespecified nested models by a likelihood-ratio test based on the partial log-likelihood and the change in parameter dimension. Penalized methods such as ridge regression or Lasso regression applied to the Cox likelihood can stabilize many-variable models; choose tuning by suitable validation and account for mandatory variables. The Akaike information criterion or a limited, documented model-selection procedure can help balance fit and complexity. Unrestricted stepwise searches and repeated univariable screening invite unstable choices, omit joint confounding effects, and make naive post-selection intervals misleading. Prefer a defensible, validated covariate set to a collection chosen only for small $P$ values.

The Type II maximum likelihood estimator maximizes the Bayesian model evidence after integrating out the breed parameters:For positive sample sizes define $S={\sum }_i\log (m_i/\beta )\ge 0$. Up to an additive constant, the log-likelihood isDifferentiation gives the interior score equationAlso ${\ell }^{\prime \prime }(\alpha )=-I/{\alpha }^2+{\sum }_i(\alpha +n_i{)}^{-2}<0$. If $S>0$, the derivative decreases from infinity to $-S$, so a unique finite maximum exists. When all sample sizes equal $n$, the equation becomes $In/[\alpha (\alpha +n)]=S$, givingFor $S=0$ no finite maximum exists. Plugging this hyperparameter estimate into the breed posterior distributions is an Empirical Bayes method.

Positivity means $u>0$ almost everywhere, since the real logarithm must belong to $L^1$. A positivity-preserving operator gives $s=Tu\ge 0$. For fixed $x$, the scalar shifted Poisson data fidelity is $f_x(s)=s-g(x)\log (1+s)$, withComposition with the linear operator $T$ proves convexity in $u$, and strict convexity holds along pairs whose forward images differ on a set of positive measure. The admissible class itself is convex: the concavity of the logarithm gives $\log (\theta u+(1-\theta )v)\ge \theta \log u+(1-\theta )\log v$, controlling its negative part, while ${\log }^{+}w\le w$ controls its positive part.

Because $|\Omega |=1$ and $\log (1+s)\ge 0$, the Jensen inequality gives the requested bound:The final equality uses nonnegativity and the unit area. Since $t-G\log (1+t)\to \infty$, this controls the forward-image norm on energy sublevels.

The printed strictly positive problem has no minimizer. This is nonattainment under strict positivity for shifted Poisson fidelity, rather than a failure of the coercivity calculation. Indeed, with $0<g<1$ and $s>0$, $\log (1+s)<s$ implies $f_x(s)>0$. Also $Tu$ cannot vanish identically for a strictly positive $u$. To see this, let $E_n=\{u\ge 1/n\}$. If $Tu=0$, positivity and $0\le {\chi }_{E_n}\le nu$ give $T{\chi }_{E_n}=0$. But ${\chi }_{E_n}\to {\chi }_{\Omega }$ in $L^1$ and continuity of $T$ would imply $T{\chi }_{\Omega }=0$, contradicting the hypothesis. Thus every admissible $u$ has positive fidelity and hence positive total energy.

Conversely, the constants $u_{\epsilon }=\epsilon {\chi }_{\Omega }$ are admissible, have zero total variation, and satisfyTheir logarithms are integrable for each $\epsilon >0$, but the limit is excluded. ThereforeA bounded minimizing sequence and bounded-variation compactness do not repair a nonclosed positivity/logarithm constraint. In particular, a literal existence or uniqueness proof for that domain is impossible.

The natural correction is to minimize over $BV(\Omega )$ with $u\ge 0$, omitting the unnecessary $\log u\in L^1$ condition: the fidelity only contains $\log (1+Tu)$, which is already integrable. Here is the full existence for nonnegative shifted Poisson regularization argument, also valid for any bounded nonnegative data $g$. Let $G=\Vert g{\Vert }_{\infty }$, $m_G={\inf }_{t\ge 0}\{t-G\log (1+t)\}>-\infty$, and take a minimizing sequence of energy at most $C$. The displayed bound gives $\Vert T{u}_n{\Vert }_1\le C_1$ and $\alpha \mathrm{TV}(u_n)\le C-m_G$. Write $c_n={\int }_{\Omega }{u}_n\ge 0$. The Poincaré inequality for total variation and mean control for positive imaging operators yieldThe denominator is nonzero, so the full $BV$ norm is bounded. Bounded-variation compactness gives $u_n\to u$ in $L^1$ along a subsequence, with $u\ge 0$. Continuity gives $T{u}_n\to Tu$ in $L^1$. On $s\ge 0$, $|f_x^{\prime }(s)|\le 1+G$, so the fidelity converges in the integral; lower semicontinuity of variation completes the direct method in the calculus of variations.

For the actual printed data $0<g<1$, the corrected problem has the unique minimizer $u=0$, even if $T$ is not injective. Zero attains energy zero. Any other zero-energy candidate would have both $Tu=0$ and $\mathrm{TV}(u)=0$; on the connected square, zero variation makes $u$ a nonnegative constant, and $T{\chi }_{\Omega }\ne 0$ forces that constant to vanish. For general positive bounded data, an injective $T$ is a sufficient uniqueness condition, because its fidelity is strictly convex; injectivity is not a necessary condition in every instance.

In the finite-dimensional interpretation, let ${\lambda }_{ij}=1+(Tu{)}_{ij}$. Independent Poisson observations with these intensities have negative log-likelihood ${\sum }_{ij}[{\lambda }_{ij}-g_{ij}\log {\lambda }_{ij}]+C(g)$. Removing the constant $\sum 1$ gives precisely the stated fidelity. Thus the model is Poisson counting noise with a unit background intensity, or an approximate version of it for rescaled/continuous grey values. Literal Poisson counts are integers; the constraint $0<g<1$ is a grey-value normalization, not a literal unscaled count sample.

For independent identically distributed observations $Y_1,\dots ,Y_n$, the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice ${\hat{\theta }}_n\in {\mathrm{argmax}}_{\theta \in \Theta }{L}_n(\theta )$, when a maximizer exists. It need not be unique. The one-observation score function is $s_{\theta }(y)={\nabla }_{\theta }\log f(y,\theta )$, and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives ${\mathbb{E}}_{\theta }{s}_{\theta }={\nabla }_{\theta }\int f=0$. Differentiating again gives the second information identity. Independent observations have total Fisher information $nI(\theta )$.

Assume fixed parameter dimension, an interior true parameter ${\theta }_0$, a positive-definite matrix $I({\theta }_0)$, the standard differentiability and integrability conditions, and statistical consistency of ${\hat{\theta }}_n$. Then the asymptotic normality of a maximum likelihood estimator isHere $I$ is the information per observation. Thus the leading covariance matrix of the estimator itself is $I({\theta }_0{)}^{-1}/n$.

To prove this, statistical consistency places the estimator in an interior ball about ${\theta }_0$ with probability tending to one, where its score function vanishes. Write $d_n={\hat{\theta }}_n-{\theta }_0$ and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give $J_n\stackrel{P}{\to }I({\theta }_0)$. The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so $J_n^{-1}\stackrel{P}{\to }I({\theta }_0{)}^{-1}$. Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since $I^{-1}I{I}^{-1}=I^{-1}$. Boundary parameters or singular information are outside this regular theorem.

For the normal linear model, the log-likelihood isThe full column rank of the design matrix makes $X^{T}X$ invertible. Differentiating in $\beta$ gives the normal equations $X^{T}X\hat{\beta }=X^{T}Y$. With the minimized residual sum of squares substituted, differentiation in ${\sigma }^2$ givesThese are the maximum-likelihood estimators almost surely; $RSS>0$ almost surely when ${\sigma }^2>0$ and $n>p$. The linear transformation of the multivariate normal distribution yieldsDefine the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. The fitted values are $\hat{Y}=HY=X\hat{\beta }$, the regression residuals are $e=Y-\hat{Y}=(I-H)Y$, and $RSS=e^{T}e$. The orthogonal projection $I-H$ has rank $n-p$ and annihilates $X\beta$. By Cochran's theorem,Consequently $E({\hat{\sigma }}^2)=(n-p){\sigma }^2/n$, with bias $-p{\sigma }^2/n$. The unbiased estimator isIts square root, the reported residual standard error, estimates $\sigma$; the assertion of unbiasedness applies to the variance, not generally to its square root.

For the paper-strength analysis, let $h_i$ be the measured percentage and $z_i=h_i-\overline{h}$. Both normal linear models use independent errors of common variance ${\sigma }^2$. Their mean functions are ${\beta }_0+{\beta }_1{z}_i$ for lm1, and ${\beta }_0+{\beta }_1{z}_i+{\beta }_2{z}_i^2$ for lm2, with separately fitted coefficients. The reported residual standard errors are $11.82$ and $4.42$, respectively.

For lm1, the estimated conditional expectation at a new percentage $x$ isThe original data mean $\overline{h}$ must be used for centering the new percentage; its numerical value is not supplied in the excerpt. Because ${\sum }_i{z}_i=0$, the two columns of this design matrix are orthogonal, and $\mathrm{Cov}({\hat{\beta }}_0,{\hat{\beta }}_1)=0$. Hence the coefficient standard errors in the output giveThis estimates uncertainty in the mean. Predicting an individual future batch would additionally require the new-error variance, estimated by $11.82^2$.

To compare the nested normal linear models, test $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$ within the quadratic model. The Student t-test statistic isIts two-sided p-value is $1.89\times 10^{-8}$. Equivalently $T^2$ is a partial F-test statistic with null distribution $F_{1,16}$. Reject the linear restriction and prefer lm2 at the 5% level. Its residual spread is much smaller; the increase in the coefficient of determination from $0.3054$ to $0.9085$ supports the same conclusion, although the test is the relevant complexity-adjusted comparison.

For regression diagnostics, inspect regression residuals against fitted means and hardwood percentage for omitted curvature, a scale-location plot for nonconstant variance, and a quantile-quantile plot against the normal distribution for departures from the error assumption. Check unusual observations using regression leverage and Cook's distance, and examine residuals in collection or batch order if dependence is plausible. Independence and a common variance require substantive justification as well as these plots; a small p-value for a polynomial term does not itself check the error model.

Introduce a latent indicator $Z_i=1$ for a structural zero. In this parametrization $P(Z_i=1)={\pi }_i$, and conditional on $Z_i=0$ the response has a Poisson distribution with mean ${\mu }_i$. The zero-inflated Poisson regression usesIts observed probability mass function is ${\pi }_i+(1-{\pi }_i)e^{-{\mu }_i}$ at zero and $(1-{\pi }_i)e^{-{\mu }_i}{\mu }_i^{y_i}/y_i!$ at a positive count. The zero-mixing probability here is explicitly ${\pi }_i$; it is distinct from the weight of the count component.

For the expectation-maximization algorithm, start with positive ${\mu }_i$ and $0<{\pi }_i<1$. At iteration $k$, the E-step computesThis distinguishes structural zeros from Poisson-generated zeros. Up to terms independent of the new parameters, the expected complete-data log-likelihood isIt separates into a fractional-response logistic fit and a weighted Poisson fit. Because treatment is binary, each component is saturated over its two treatment groups, so the M-step has closed forms.

For $j=0,1$, let $I_j=\{i:x_i=j\}$, $n_j=|I_j|$, $S_j={\sum }_{i\in I_j}{\tau }_i$, and $C_j={\sum }_{i\in I_j}{y}_i$. The M-step score equations giveHere $\sum (1-{\tau }_i)y_i=C_j$, since every positive count has ${\tau }_i=0$. The explicit coefficient updates areIterate the E- and M-steps until the observed log-likelihood and parameter estimates stabilize. This is the EM algorithm for zero-inflated Poisson regression with explicit binary-group updates; merely naming two regression routines would not supply these expressions. Both treatment groups must be represented for both contrasts to be identifiable. Zero fitted group means or endpoint mixing probabilities are boundary solutions, interpreted through limits of the log or logit coefficients. As with other mixture models, multiple starts help distinguish competing stationary solutions; EM increases the likelihood but does not guarantee a global maximum from an arbitrary start.

The Bradley-Terry model assigns comparison probability $q_i={\theta }_i/({\theta }_i+{\theta }_{i+1})$ to each observed edge. Up to factors independent of the parameters, its likelihood function isThe comparison graph is a path graph, hence a tree. After fixing ${\theta }_n=1$, its edge ratios are unconstrained positive coordinates: every collection $r_i={\theta }_i/{\theta }_{i+1}>0$ determines uniquely ${\theta }_i={\prod }_{k=i}^{n-1}{r}_k$.

For a convenient strict-concavity calculation, use the edge log-ratios in a Bradley-Terry comparison tree ${\eta }_i=\log r_i$. The log-likelihood separates asEach derivative is $m(p_i-q_i)$, where $q_i$ is the logistic function of ${\eta }_i$, and each second derivative is $-m{q}_i(1-q_i)<0$. Because $0<p_i<1$, there is a unique finite global maximum atConverting back gives the Bradley-Terry maximum-likelihood estimate on a path:Equivalently, recurse backwards using ${\hat{\theta }}_i={\hat{\theta }}_{i+1}{p}_i/(1-p_i)$. All estimates are finite and positive. The sample size $m$ changes the curvature of the likelihood but not this maximizer; absence of comparison cycles is what permits every empirical edge proportion to be fitted simultaneously.

For a finite Gaussian mixture with a common variance, introduce independent latent labels $Z_i\in \{1,\dots ,k\}$ with probabilities ${\pi }_j>0$, ${\sum }_j{\pi }_j=1$, and conditional responses $Y_i\mid Z_i=j\sim N({\mu }_j,{\sigma }^2)$, with common ${\sigma }^2>0$. The observed density and log-likelihood areThe expectation-maximization algorithm replaces the difficult log of sums by an expected complete-data objective. Choose positive initial weights and variance and separated initial means. At iteration $r$, the E-step computes the mixture responsibilitiesThe common normalizing factor cancels. For numerical stability the probabilities can be evaluated by subtracting the largest log weight before exponentiating.

With the old responsibilities held fixed, the expected complete-data log-likelihood, up to irrelevant constants, isLet $N_j={\sum }_i{\tau }_{ij}^{(r)}$. A Lagrange multiplier for ${\sum }_j{\pi }_j=1$, weighted least squares for the means, and differentiation in the common variance give the M-step:The variance update uses the new means and old responsibilities, with denominator $n$, not a residual degrees-of-freedom adjustment: this is maximum likelihood estimation. Repeat the E- and M-steps until the observed log-likelihood and parameters stabilize.

By EM likelihood monotonicity, exact updates do not decrease the observed likelihood. They need not reach its global maximum, so use several starting configurations and keep the best converged fit, checking for empty or nearly empty components and vanishing variance. The labels are interchangeable; sorting means after fitting supplies an interpretable labeling. Distinct starting means do not guarantee that all fitted components remain distinct. With a common variance and the usual fixed small $k$ relative to distinct observations the model avoids the individual-component variance-collapse pathology of unrestricted Gaussian mixtures, but degenerate data or too many components still require attention.

The statistician first models the age- and gender-dependent mean by linear regression. Its fitted value is $27.01675-0.10483\mathrm{age}+1.47031\mathrm{gender}$. At fixed gender, a year of age is associated with a decrease of $0.10483$ BMI units; the group coded gender $1$ has a fitted mean $1.47031$ units above the group coded $0$ at the same age. This question does not state which gender receives which code. The intercept refers to age zero in the reference group, outside a typical adult-patient range, so its substantive interpretation is limited. Both slope Student t-tests have small p-values, and the overall F-test supports an age/gender-dependent mean. The coefficient of determination is only $0.18$, leaving considerable individual variation to investigate.

The residual diagnostics check whether mean adjustment is plausible and whether a single normal error law is adequate. The residual-versus-fitted plot has no strong curved mean trend, although its smooth curve is not perfectly flat. The scale-location plot suggests some decline in spread as fitted BMI increases, so common residual variance is a working approximation rather than an established fact. The normal Q-Q plot has systematic departures, notably shorter tails than the normal reference, suggesting the residual law is not exactly normal. The regression leverage plot does not display an obviously extreme leverage point, but the marked observations and Cook's distance should still be checked for influence. Independence between patients cannot be diagnosed from these four plots alone.

Next the statistician extracts the regression residuals and fits an intercept-only normal model as a baseline residual distribution. With an intercept in the original ordinary least squares model, the residuals sum to zero by the normal equation; the near-zero fitted residual mean and its $p=1$ are therefore automatic, not new evidence of good fit. The $2.698$ standard error in this refit uses $499$ degrees of freedom, whereas the original $2.704$ uses $497$ after estimating three regression coefficients. The reported log-likelihood for the single-normal residual model is $-1205.25$.

The histogram suggests a shape worth exploring beyond one normal density, with broad shoulders and some asymmetry, without showing unambiguous separated clusters. The statistician fits two- and three-component finite Gaussian mixtures with a common variance by the expectation-maximization algorithm, starting from separated means and positive weights. Successive likelihoods increase and the displayed final iterations stabilize, as expected from EM likelihood monotonicity. This is evidence of numerical convergence from those starts, not proof of a global maximum.

The two-component fit assigns weights $(0.3317,0.6683)$, residual means $(-2.7844,1.3819)$ and common variance $3.4175$. The three-component fit assigns weights $(0.2051,0.3926,0.4023)$, residual means $(-3.7306,-0.4417,2.3329)$ and common variance $2.1448$. The code correctly passes the square roots of those variances to dnorm and overlays the resulting weighted densities. Both curves broadly follow the histogram and are very similar. In particular, a two-component mixture need not have two distinct visible modes.

These fits explore residual mixture clustering: groups differ in BMI relative to the same age/gender-adjusted mean, rather than simply in raw BMI. Membership can be summarized by the fitted mixture responsibilities, retaining uncertainty instead of asserting a certain label for each patient. The common-variance assumption is economical, but should be checked; the common regression slope assumption and constancy of mixture proportions across age and gender are also substantive. A residual mixture alone does not establish genuine biological subpopulations; a flexible continuous distribution, missing predictors, nonlinear mean effects or heteroscedasticity could explain a similar marginal shape.

There is also a two-stage residual mixture fitting limitation. Fitted regression residuals are not exactly independent identically distributed observations: even under a normal regression, their covariance matrix is ${\sigma }^2(I-H)$, and estimating the mean adjustment introduces uncertainty. With $500$ subjects and three initial regression coefficients, treating them as independent is a useful approximation for exploring shape, but ordinary mixture likelihoods omit the first-stage uncertainty. A joint mixture regression with shared slopes, or a suitable bootstrap of the whole analysis, would give a firmer basis for parameter uncertainty and cluster selection. Examine component membership versus age/gender and repeat the algorithm from multiple starts before making substantive clustering claims.

The three-component finite Gaussian mixture with a common variance has the largest displayed maximized log-likelihood, but it also has more fitted parameters. For $k$ components there are $k-1$ free weights, $k$ means and one common variance, making $d=2k$ residual-distribution parameters. Applying the nominal Akaike information criterion and Bayesian information criterion to the three residual fits givesBoth criteria favour the two-component model. It improves the likelihood substantially over one normal component; the third component gains only $1.5550$ in log-likelihood at a cost of two further parameters. The Akaike information criterion difference between two and three components is small, about $0.89$, so that criterion alone gives only a slight preference; the Bayesian information criterion preference for two is stronger. Their nearly indistinguishable overlaid curves also support retaining the simpler two-component fit.

These calculations are model-selection aids, not an exact test. Ordinary chi-squared calibration of a likelihood-ratio test for the number of mixture components is invalid in general: under a smaller-component null, some weights lie on the boundary and extra-component parameters are unidentified. This is nonregular mixture model selection. A parametric bootstrap or predictive cross-validation is preferable for a formal comparison, and should account for the mean-adjustment stage. The table uses the provided residual likelihoods and their nominal parameter counts; first-stage regression uncertainty and possible local EM maxima remain qualifications. On the supplied evidence, select two components without claiming that two distinct biological groups have been established.

With the onset means ${\mu }_k(\theta ,\lambda ,\kappa )$ defined by the discrete back-calculation of infection incidence, the independent Poisson observation model yields the product likelihoodFor positive means its log-likelihood is ${\sum }_k\{y_k\log {\mu }_k-{\mu }_k-\log (y_k!)\}$. A zero mean assigns probability one to a zero count and zero to a positive count. Any unknown pre-observation infection history must be parameterised or supplied as well; it cannot silently be excluded merely because the observation series starts at $t_0$.

For $n\ge 1$, the binomial log-likelihood is $X\log \theta +(n-X)\log (1-\theta )$ plus a constant. Its maximizer, including endpoints, is ${\hat{\theta }}_{\mathrm{MLE}}=X/n$. For $0<\theta <1$, the Fisher information isThis follows either from the expected negative second derivative or the variance of $(X-n\theta )/(\theta (1-\theta ))$. At $0,1$ the usual regular score formula does not apply; the interior information diverges as an endpoint is approached.

For a symmetric Beta distribution prior $\mathrm{Beta}(s,s)$, conjugacy gives $\theta \mid X\sim \mathrm{Beta}(X+s,n-X+s)$, with posterior mean $(X+s)/(n+2s)$. All three priors below are instances of this calculation.

Writing $y_i$ for the observed counts, independence of the Poisson distributions gives the log-likelihoodIts score function and negative second derivative areUnless all $x_i$ vanish, $J(\beta )>0$, so the log-likelihood is strictly concave. Use Newton method, starting from a finite ${\beta }_0$, withStop when the score function or the change in $\beta$ is small; a reduced step can ensure that the log-likelihood increases. Any finite zero of the score function is the unique maximum-likelihood estimator.

Existence needs a qualification. If the nonzero $x_i$ include both signs, $U$ decreases from $+\infty$ to $-\infty$, so a finite root exists. If all nonzero $x_i$ are positive, a finite root exists exactly when ${\sum }_i{x}_i{y}_i>0$; otherwise the supremum is approached as $\beta \to -\infty$. The corresponding condition for all negative $x_i$ is ${\sum }_i{x}_i{y}_i<0$, with a supremum at $+\infty$ otherwise. When all $x_i=0$, the parameter is unidentifiable. These cases explain when the iterative algorithm actually has a finite estimate to find.