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For ordered response categories, model each displayed conditional probability by a grouped-binomial logistic regression. The multinomial likelihood factors exactly into successive conditional binomial likelihoods, using the remaining category total as each stage's denominator. This is a different regression parameterization from baseline-category logits or cumulative odds. Separate cumulative binary indicators from the same person are dependent and cannot be treated as independent likelihood contributions.

For positive group totals, the improper prior $du/u$ on an independent Poisson baseline yields the multinomial likelihood kernel after integration. This prior is flat in $\log u$. A proper finite-variance log-rate prior generally introduces an additional coefficient-dependent factor.

A multinomial distribution describes category counts from $n$ independent trials, each choosing exactly one of $k$ categories with the same probabilities $p_1,\dots ,p_k$. Require $n\ge 0$ to be an integer, $p_i\ge 0$, and ${\sum }_i{p}_i=1$. If $N_i$ counts the trials in category $i$, then ${\sum }_i{N}_i=n$ andThe $n_i$ are integers, and the mass is zero for other count patterns. A category with $p_i=0$ has count zero almost surely, with $0^0=1$ in this formula. For a specified ordered trial sequence, independence gives the product of numbers ${\prod }_i{p}_i^{n_i}$; the multinomial coefficient counts the different sequences with that same count pattern.

For four trials with category probabilities $(1/2,3/10,1/5)$, the count pattern $(2,1,1)$ can occur in $4!/(2!1!1!)=12$ orders, each of probability $(1/2{)}^2(3/10)(1/5)=0.015$. Thus its total probability is $12(0.015)=0.18$. This is a constructed independent-trial example, distinct from drawing four objects without replacement from a ten-object population of composition $(5,3,2)$, where the multivariate hypergeometric distribution gives $2/7$.

The diagram shows all fifteen possible count patterns. Coordinates $(N_1,N_2)$ determine $N_3=4-N_1-N_2$; the triangle is a constraint on integer counts, not a continuous probability density. The mode is $(N_1,N_2,N_3)=(2,1,1)$, where the probability mass function is largest, but the expected values $(2,1.2,0.8)$ are not a possible count pattern.

Each single count has a binomial distribution, because recording category $i$ versus all other categories reduces every trial to a Bernoulli trial. Consequently,These statements about the joint count model are summarized in Aditya Guntuboyina's lecture notes.^[ref]

Independent trials do not make the resulting counts independent random variables. Within one trial the indicator random variables for two distinct categories cannot both equal one, so their product is zero and their covariance is $-p_i{p}_j$. Summing over independent trials gives $-n{p}_i{p}_j$. In the four-trial example, the variances are $(1,0.84,0.64)$ and the pairwise covariances are $-0.6,-0.4,-0.24$. Knowing that all four trials were in the first category fixes the other two counts at zero.

The attraction is that one formula describes any fixed-total list of category counts while preserving the constraint that they add up to $n$. It underlies multinomial likelihood calculations and uniform balls-in-bins allocation. The multinomial aggregation rule lets us merge categories without leaving this family of probability distributions. Allowing independent category counts with an unfixed total instead leads to a different model, related through Poisson-multinomial conditioning.

Use the index order in the probability specification: influence $i$, accommodation type $j$, contact $k$ and satisfaction $l$. Put $N_{ijk}=n_{ijk+}$, which is conditioned on. The independent multinomial likelihood isThus the relevant sample is a collection of conditional satisfaction distributions for fixed covariate strata; it is not a model for the marginal distribution of the three covariates. In the published dataset there are three influence, four accommodation, two contact and three satisfaction levels, giving 24 strata and 72 cells; these dimensions are documented in stat.ethz.ch/R-manual/R-devel/library/MASS/html/housing.html .

The missing output does not identify its two actual formulas. Two standard canonical models can nevertheless be derived explicitly, and the following general result identifies the sufficient statistics for any specified baseline-logit model. The first standard model has no response/covariate association: $p_{ijkl}=q_l$ for every stratum. Its likelihood factors asTherefore the pooled satisfaction totals $n_{+++l}$ are sufficient, with only two independent totals because the grand total is fixed. The maximum-likelihood estimator is ${\hat{q}}_l=n_{+++l}/1681$, and fitted counts are $N_{ijk}{\hat{q}}_l$. A Poisson surrogate formula with all covariate-stratum interactions and only a satisfaction main effect produces this same conditional model. It has two conditional parameters and 46 residual degrees of freedom when all 24 strata are present.

A direct grouped-binomial logistic regression can dichotomize satisfaction, for example high versus low-or-medium. In each covariate stratum use the high count as successes and the sum of the other counts as failures, with total $N_{ijk}$. This models $P(Y=\mathrm{high}\mid i,j,k)$ and is appropriate if that binary contrast is substantively intended, but discards distinctions among the merged categories. Another cutpoint gives a different binary question.

There is also an exact sequential binomial factorization preserving all three categories. With $p_1,p_2,p_3$ their probabilities in one stratum,Multiplying these two conditional binomial likelihoods gives the multinomial likelihood: the combinatorial coefficients multiply to $N!/(n_1!n_2!n_3!)$, and the probability factors reduce to $p_1^{n_1}{p}_2^{n_2}{p}_3^{n_3}$. Continuation-ratio logits can therefore fit two logistic regressions, using the remaining stratum total as the second denominator. These are conditional stage models, not two independent counts with the same fixed denominator. Their regression restrictions need not equal those of the additive baseline-category model.

Finally, an ordinal analysis may fit cumulative binary contrasts and examine whether their slopes can be shared in a proportional-odds model. The binary indicators at different cutpoints from the same person are dependent. Do not stack them as independent binomial observations and multiply their marginal likelihoods; fit the joint ordinal likelihood or use a method with the appropriate dependence correction. This distinction separates a convenient descriptive collection of binary fits from a coherent multinomial or ordinal model.

The response is the species categorical variable, while the four measurements are candidate predictors. The dot in the formula means all other columns of the data frame. The fitted object is a classification tree: it recursively partitions predictor space by binary threshold rules and assigns a species and class-probability vector to each terminal region. Sepal width was available but was not selected in the displayed tree.

At a node $t$, let $n_{tk}$ be the number of observations of class $k$ and $n_t={\sum }_k{n}_{tk}$. The multinomial likelihood is maximized by ${\hat{p}}_{tk}=n_{tk}/n_t$. The fitted class is one attaining the largest proportion, with a software convention resolving ties. The classification-tree deviance isA pure node has zero deviance. At each nonterminal node, consider candidate split points between distinct observed predictor values, and choose the split with the largest decrease $D(t)-D(t_L)-D(t_R)$. Repeating this local maximization constructs a binary decision tree; it does not solve a global optimization over all trees. Growth stops when nodes are pure or size/deviance controls prohibit further splitting. Mixed leaves with five or six observations can therefore remain. The node number $j$ has children $2j$ and $2j+1$, and a terminal-node marker means that no further split was retained.

The probability triples are in class order $(c,s,v)$, not $(s,c,v)$. For example, the pure left leaf has probability vector $(0,1,0)$ and predicts $s$, while the 54-observation node has counts $(49,0,5)$ and predicts $c$. The root counts are $(50,50,50)$, giving $D=300\log 3\simeq 329.584$ and a tied majority class. Its reported choice of $c$ does not indicate a more common class.

The complete derived decision rule is shown below. Each leaf includes its fitted class and the observed class counts, so both the branching logic and its remaining errors are visible. For the observed data, the printed strict inequalities leave no equality cases because the thresholds fall between measurement values. A consistent right-branch convention at equality extends the rule to new specimens.

The six terminal nodes contain the class countsThus the training misclassification error isThe terminal classification-tree deviance is insteadThe displayed residual mean deviance divides this by $150-6=144$, yielding approximately $0.1253$. This is the summary's reporting convention, not a normal-error residual mean square or an automatic chi-squared calibration with 144 statistical degrees of freedom. The two summary measures answer different questions: deviance depends on the fitted probability of each observed class, while misclassification counts only whether the majority-class prediction is correct. Both are training quantities. Cross-validation or independent test data, rather than the displayed training error, are needed to assess prediction on new specimens; cost-complexity tree pruning can select a less elaborate tree if predictive performance warrants it.

Put $N=n_0+n_1+n_2$ and $s=2x=2\theta (1-\theta )$. The usual recombination fraction range $0\le \theta \le 1/2$ gives $0\le s\le 1/2$, and the preceding likelihood function becomesTo see its relation to nonparametric linkage analysis, use Bayes' theorem and the constant disease probability $1/16$ to obtain the affected-pair marker-sharing probabilitiesTheir multinomial likelihood has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the Fisher-Neyman factorization theorem makes $B$ a sufficient statistic. The observed mean shared proportion is $\hat{p}=B/(2N)$, and the binomial likelihood kernel givesThe unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\hat{p}\le 1/2$, the constrained likelihood is maximized at the unlinked null. For $\hat{p}>1/2$, its log likelihood ratio iswhose derivative in $\hat{p}$ is $2N\log [\hat{p}/(1-\hat{p})]>0$. Hence it orders samples exactly by excess IBD sharing. Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information. This is allele-sharing sufficiency for recessive linkage, not a general equivalence for all disease models. Use limiting values for terms such as $0\log 0$, and assume $N>0$ for the estimators.

Take the affected sib pairs from independent families with the same disease model. Conditional on their marker sharing states, multiply the recessive affected-sib-pair likelihoods:Write $x=\sqrt{L_0}=\theta (1-\theta )$, so that $\sqrt{L_2}=1/2-x$ and $L_1=x(1/2-x)$. ConsequentlyFor an unordered count sample there can also be a multinomial likelihood coefficient, and the Mendelian probabilities of the observed marker sharing states supply further factors. All these factors are independent of $\theta$.

If genetic-study ascertainment explicitly conditions on both children being affected, the normalizing disease probability is constant:Thus the conditional sharing likelihood differs from the displayed likelihood function only by factors independent of the recombination fraction. Multiple pairs drawn from the same larger sibship would not, in general, justify treating these pair likelihoods as independent.

Set ${\eta }_{ik}={\beta }_k^T{\mathbf{x}}_i$, so ${\eta }_{i1}=0$. The multinomial logistic regression implies $p_{ik}=e^{{\eta }_{ik}}{p}_{i1}$. Normalizing givesMultiply the multinomial likelihoods for conditionally independent groups. The multinomial coefficients are constant in $\beta$, leavingThe reference constraint ${\beta }_1=0$ identifies the coefficients: a common shift of every category coefficient otherwise leaves all probabilities unchanged. The PDF places the full sum inside the numerator exponent; the TeX breaks that expression.

Put $a=y_4$ and $b=y_2+y_3$. Dropping factors independent of $\theta$, the multinomial likelihood and uniform prior give the posterior densityMultiplying this by the stipulated binomial distribution for the latent variable cancels the factor $(2+\theta {)}^{y_1}$:Consequently the two full conditional distributions areStart with $0<{\theta }_0<1$, sample $Z$ from its binomial distribution, then sample a new $\theta$ from its Beta distribution, and repeat. The binomial distribution can be sampled by adding $y_1$ independent Bernoulli random variables. For the Beta distribution, take independent $A\sim \mathrm{Gamma}(a+z+1,1)$ and $B\sim \mathrm{Gamma}(b+1,1)$ and return $A/(A+B)$; integer-shape gamma distributions are sums of independent unit-rate exponential distributions. Both shapes are positive even when some observed counts vanish.

Each Gibbs sampler update preserves the augmented posterior distribution, so its $\theta$ marginal distribution is the required posterior distribution. The positive full conditional distributions on the interior allow exploration of the entire support. These iterates are generally dependent; they are not the independent exact draws constructed in the later parts.

Write $y_{i+}={\sum }_j{y}_{ij}$ and ${\mu }_{i+}={\sum }_j{\mu }_{ij}$. Under the model in part (a),and $p_j$ is independent of the price category $i$. The Poisson-multinomial conditioning identity givesMore explicitly, the likelihood function for row $i$ factors asThe first factor is the likelihood of the row total and the second is the conditional multinomial likelihood. The free price effect ${\alpha }_i$ makes the first factor maximal at ${\hat{\mu }}_{i+}=y_{i+}$, while maximizing the remaining factors givesConsequently both fits have the same maximum-likelihood fitted valueHere every row total is $30$, the column totals are $31,28,31$, and $y_{++}=90$, so every fitted row isThis factorization is the Poisson trick.