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A simple sufficient independent censoring assumption is that the potential observation-ending time $C$ and survival time $T$ are independent random variables. More generally, appropriate conditional independence given the covariates used in the analysis can suffice for analysis within those conditions. The requirement concerns the potential times, rather than independence of the observed duration $Y=\min (T,C)$ and event indicator $\delta$: those recorded quantities ordinarily depend on each other.

For an ordinary pooled Kaplan–Meier estimator, the observed risk set must represent the future event experience of everyone still event-free at each time. Conditional independence given a covariate does not alone justify ignoring that covariate in a pooled calculation. Also, there must be a positive probability of remaining observed through the horizons being estimated. If follow-up ends at a common closing time, it supplies no observations to identify the later survival function. Independent subjects with a common survival distribution and independent censoring form the basic model in Lu Tian's STAT331 notes, Unit 3, pages 1–4.

For example, subjects who stop attending because their condition is deteriorating can produce informative censoring. A predetermined common closing date can instead give administrative censoring, provided calendar entry and other relevant conditions do not select future survival times. The reason for stopping observation must be justified from study design and context; the recorded event indicators alone cannot verify independent censoring.

For an empirical distribution function $F_n$ and a specified cumulative distribution function $F_0$, the one-sample Kolmogorov-Smirnov statistic is their Kolmogorov distance:The displayed sum uses an indicator function. With this convention, $F_n$ is right-continuous. For distinct observations, its value and its left one-sided limit at the $i$th order statistic are $F_n(X_{(i)})=i/n$ and $F_n(X_{(i)}^{-})=(i-1)/n$.

For continuous $F_0$, both jump sides matter. Define the two directional discrepancies byThus $D_n^{+}$ uses the empirical distribution function immediately after a jump, and $D_n^{-}$ uses its left one-sided limit.^[ref] A supremum need not be attained.

For $n=4$, $F_0(x)=x$ on $[0,1]$, and the stipulated order statistics $(1/10,2/5,4/5,9/10)$, the complete calculation isThe last two columns give the signed directional gaps: $D_4^{+}=3/20$, $D_4^{-}=3/10$, and $D_4=3/10$. In contrast, checking only $|F_4(X_{(i)})-F_0(X_{(i)})|$ gives $3/20$ and misses the largest gap approached as $x\uparrow 4/5$. At $x=4/5$ itself, the gap has already fallen to $1/20$.

The finite formula follows from the behavior between jumps: the empirical distribution function $F_n$ is constant and the cumulative distribution function $F_0$ is monotone, so the largest directional gaps occur at the two interval ends, possibly as one-sided limits. The two tails introduce no larger gap. Checking every jump side therefore gives the exact supremum.

A common normalized Kolmogorov-Smirnov statistic is $\sqrt{n}{D}_n$, which equals $3/5$ in this example. Its large-sample reference law comes from the Kolmogorov-Smirnov theorem. The unscaled $D_n$ remains the largest difference of cumulative probabilities, not a p-value.

There is a useful exact exception to the usual warning about fitting statistical parameters. Suppose $n\ge 2$ independent and identically distributed observations have the uniform distribution on $(0,\theta )$ with unknown $\theta >0$. Let $M={\max }_i{X}_i$ and let $J$ be the label of its almost surely unique maximizing observation. Conditional on $M=m$ and $J=j$, the surviving labelled observations $(X_i{)}_{i\ne j}$ are independent with the uniform distribution on $(0,m)$; after division by $m$ they are independent and identically distributed uniforms on $(0,1)$. These are conditional distributions defined by probability density functions, since the event $M=m$ has probability zero.

To check this, the joint probability density function of $M$ and $J=j$ is $m^{n-1}/{\theta }^n$ for $0<m<\theta$. Dividing the original joint probability density function ${\theta }^{-n}$ by this factor gives the conditional survivor probability density functionThis factorization proves the claimed conditional independence; scaling removes $m$. Keep the surviving labels in a fixed order, or apply an independent random permutation. The sorted survivors instead have the joint law of uniform order statistics and exhibit statistical dependence; sorting does not affect their empirical distribution function. Delete the maximum itself, which would otherwise become a forced value $1$, and compute the Kolmogorov-Smirnov statistic using sample size $n-1$ and jump heights $i/(n-1)$. This gives an exact conditional Kolmogorov-Smirnov test under the uniform-endpoint statistical model. It does not justify fitting an arbitrary endpoint or other statistical parameters and using the ordinary $n$-sample calibration.

A Markov process is a stochastic process whose present state contains all the information in its observed past needed to predict its future. Precisely, if ${\mathcal{F}}_s$ is the natural filtration generated by observations up to time $s$, then, for $t>s$ and any measurable set $A$ of states,The equality is between conditional probabilities and holds almost surely. It expresses the Markov property: the past and future have conditional independence given the present. It does not assert independence of consecutive states. In the sunny/rainy example below, tomorrow's sunny probability is $0.8$ after sunshine but $0.4$ after rain, so knowing today's state matters.

The memory assumption concerns the chosen state space. If an equipment model's failure probability depends on its age, recording only “working” or “failed” may lose relevant information; recording age as part of the state can retain it. Similarly, a dependence on two previous observations can sometimes be represented by using their ordered pair as the current state. Whether such a description fits observations is a modeling question.

Transition Markov kernels describe the conditional distribution of a state at a later time. Successive kernels and the initial probability distribution determine every finite-dimensional distribution, through repeated conditional probability. The time-homogeneous Markov property is an additional assumption: the kernel depends on elapsed time rather than calendar time. The attraction is compression: prediction can proceed from a current state instead of an entire recorded history. Brownian motion and continuous-time Markov chains illustrate that a Markov process need not evolve in discrete time.

If D-separation and conditional independence are equivalent for a Directed acyclic graph, its moral graph is its conditional independence graph. After conditioning on all other vertices, an active path can only be a direct edge or a two-edge collider through a common child: two adjacent interior vertices cannot both be colliders. These are exactly the moral edges.

The printed formula expands to the main effects of school type, acceptance and college, together with school-type-by-college and acceptance-by-college interactions. It contains neither a school-type-by-acceptance interaction nor the corresponding three-way interaction. Its Poisson regression model is thereforeFor every college the mean table has odds ratioThus it tests conditional independence of school type and acceptance given college, while permitting both margins to vary freely across colleges. It does not merely assert a common, possibly nonzero association.

The maximum likelihood estimation score equations match every school-type-by-college and acceptance-by-college margin. Solving them gives the stratified two-by-two conditional independence model's fitted countsIndeed these counts factor by row and column and have exactly the observed margins. For nondegenerate tables there are three free mean parameters per college: one total and two independent margin proportions. There are $100$ cells and $75$ parameters, hence 25 residual degrees of freedom. The residual deviance iswith zero-count contributions interpreted as zero. The Poisson linear terms cancel because the fitted totals equal the observed totals. Under the null and suitable large-count regularity, compare $G^2$ with ${\chi }_{25}^2$. Sparse cells, structural zeros or degenerate margins require a different calibration, such as an appropriate conditional test or simulated reference distribution.

A large residual deviance is evidence of school-type/acceptance association in at least some colleges. A small one indicates compatibility with conditional independence, not proof of identical admission processes or absence of selection on qualifications. Residuals and individual college contrasts locate departures. Adding the school-type-by-acceptance term gives a common odds ratio, one extra parameter and 24 residual degrees of freedom. Adding the full three-way term permits each college's association and saturates the tables. The first comparison tests a common association; the second checks whether a common association is adequate. No frequencies are supplied here to evaluate these tests numerically.

A constant number star formation rate over $10\mathrm{Gyr}$ produces a uniform present-age distribution. The birth-age distribution under constant star formation therefore givesOutside this age interval the cumulative distribution function is zero or one. The census here retains stellar remnants as objects, as required by the population model.

Write $m=M/M_{\odot }$. The initial mass function is proportional to $m^{-3}$ on $m\ge 0.2$. Normalizing the tail integral gives the power-law initial-mass-function tail coordinate:For lower thresholds $X=1$. Because the birth mass and formation time have independence, the probability integral transform makes $(X,Y)$ uniform on the unit square. $X$ is an upper-tail coordinate, so larger masses have smaller $X$.

A red giant of current age $t$ lies in the stellar lifetime regions in a mass-age diagramThere are no red giants below $m=\sqrt{0.8}$ in this model. For $\sqrt{0.8}<m<1$ the upper boundary is the population's age limit, while for $m\ge 1$ both lifetime boundaries are present. A white dwarf lies above $t/\mathrm{Gyr}=10/m^2$.

In uniform probability coordinates the main-sequence boundary is $Y=20X$ and the end of the giant phase is $Y=25X$. The giant region is therefore $20X\le Y<25X$, clipped to the unit square: it is the triangle with vertices $(0,0),(1/25,1),(1/20,1)$. Its area, and hence the individual giant fraction, isThe white dwarf region $Y\ge 25X$ is the triangle with vertices $(0,0),(0,1),(1/25,1)$, givingThe remaining $97.5\%$ are on the main sequence. Boundaries have zero probability and do not affect these counts. Treating every post-giant object as a white dwarf is a stipulated toy-model assumption; real high-mass stars have other stellar remnants.

For the binary calculations, use a coeval binary population: components share their birth time even though their birth masses are independently drawn and their subsequent evolution is independent. Hence $(X_1,X_2,Y)$ is uniform in the unit cube. At a fixed age coordinate $Y=y$, a component is a giant when $y/25<X<y/20$, and a white dwarf when $0<X<y/25$. Define the conditional state fractionsComponents have conditional independence given $y$; averaging over the common age must be done after forming the conditional pair probabilities. The continuous-birth binary state fractions below are consequently volumes in this cube, not products of the marginal single-star fractions.

A credibility premium estimates a risk's expected future claims by combining its own observed experience with a collective or prior mean. In its basic form $Z\overline{X}+(1-Z)\mu$, the credibility factor $Z\in [0,1]$ is the weight assigned to individual experience. Larger credibility places less weight on pooling with the population. The amounts below are pure expected-claim premiums; no additional expense or profit loading is specified.

Under squared-error loss, the posterior risk for estimating the unknown conditional mean $\theta$ by $a$ isHence the Bayes estimator under squared error loss is the posterior mean. In the equal-year model, conditional independence makes the likelihood proportional to $\exp [-{\sum }_{j=1}^n(X_j-\theta {)}^2/(2{\sigma }_1^2)]$. Combining it with the normal prior and completing the square givesBecause $\mathbb{E}[X_{n+1}\mid \theta ]=\theta$, the required estimator is $m_n$; it also equals the predictive mean $\mathbb{E}[X_{n+1}\mid \mathbf{X}]$. The normal-normal credibility expression isThis is exact Bayesian credibility, not merely the best linear approximation to a possibly nonlinear posterior mean.

For unequal life counts use the usual exposure model: individual-life losses are conditionally independent with variance $v$, and the annual totals are conditionally independent given the same $\theta$. ThenThe dot denotes the stipulated normal approximation. The positive exposures $m_j$ are known. Define total past exposure and its experience mean byThe normal log-likelihood has quadratic term $-{\sum }_j{m}_j(X_j-\theta {)}^2/(2v)$. With the prior $N(\mu ,{\sigma }^2)$, normal-normal conjugacy with unequal exposures gives posterior variance and meanTherefore the estimated pure premium per life for year $n+1$ has the normal-normal credibility with unequal exposures formIt is also an exact normal-model Bühlmann–Straub credibility estimate. The individual experience is exposure-weighted: the unweighted mean of the yearly per-life observations would generally discard the different precisions.

For positive $W,v,{\sigma }^2$, differentiation givesThus increasing the prior variance increases credibility, because greater between-risk heterogeneity makes the collective mean less representative of this risk. Increasing the individual process variance decreases credibility, because a fixed amount of observed experience becomes noisier. The future exposure supplies no new observation, so it does not enter $Z_W$. It only scales the total premium:The independent-life variance scaling is part of the usual exposure interpretation. Without it, a known variance $v$ for one life does not determine the variance of a total: a common pairwise conditional covariance $c$ would add $m_j(m_j-1)c$ to $m_{j}v$. Similarly, cross-year conditional dependence would change the joint likelihood. Those assumptions are implicit in applying the standard credibility model to the final paragraph.

For the undirected representation of the pedigree graphical model, take the moral graph: remove arrowheads from all parent-child edges and join coparents. The added coparent edges areEvery Mendelian segregation factor involves a child and its two parents, so its variables form a clique in this graph. The product in the previous part is thus an undirected factorization. A separation of variables by a conditioning set in this graph implies the corresponding conditional independence.

This is a valid conditional independence graph representation of the joint genotype law. It need not be its minimal representation without a faithfulness of a directed acyclic graph assumption: some inheritance probabilities are zero, and special parameter choices can yield additional independences. It also differs from a graph of unconditioned correlations, since full siblings remain correlated through their parents when the parental genotypes are not conditioned on.

Use the shape-rate convention for the gamma distribution $\theta \sim \mathrm{Gamma}(\alpha ,\beta )$, with prior distribution probability density function proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$. The observations are independent conditional on the common parameter $\theta$. Unconditionally a common nondegenerate prior distribution induces covariance $\alpha /{\beta }^2$ between distinct observations, so the Bayesian inference model requires conditional independence.

For unit-exposure counts, the likelihood is proportional to ${\theta }^{{\sum }_i{x}_i}{e}^{-n\theta }$. Multiplying by the prior distribution givesThe future count has conditional expectation $\theta$, so posterior distribution averaging givesThis is an exact Bayesian credibility estimate, with credibility factor $Z=n/(n+\beta )$ and collective mean $\alpha /\beta$.

For the insurance exposure and claims inflation calculation put $c_j=c(1+r{)}^j$, assuming $c>0$, $1+r>0$ and positive policy counts $n_j$. Let $K_j$ be the total claim count in year $j$. Conditional independence of the policy counts gives $K_j\mid \theta \sim \mathrm{Pois}(n_j\theta )$, and the observed average amount satisfiesFor possible observations, the recovered $k_j$ are nonnegative integers. If they are not, the data have zero likelihood under the model and no conditional posterior distribution is defined for those impossible values.

Write $W={\sum }_{j=1}^n{n}_j$ and $K={\sum }_{j=1}^n{k}_j$. Since $Y_j$ and $K_j$ determine each other, conditioning on the observed average amounts is equivalent to conditioning on these counts. The independent yearly count likelihood isThe Poisson-gamma conjugacy with unequal exposures therefore givesFor the future year, $\mathbb{E}[Y_{n+1}\mid \theta ]=c_{n+1}\theta$; the number of future policies cancels because $Y_{n+1}$ is a per-policy average. HencewithThe collective mean $m$ is the prior distribution expected claim amount per policy in the future year's monetary units. The experience mean $m(\mathbf{y})$ first inflates each past per-policy amount to that same future year and then averages with policy-exposure weights. Equivalently, it estimates annual claim frequency by total observed claims divided by total past insurance exposure and converts that frequency to the future claim cost. This is inflation-adjusted Poisson-gamma credibility.

For fixed prior distribution parameters, $Z$ increases strictly with total past insurance exposure: $dZ/dW=\beta /(W+\beta {)}^2>0$, and $Z\to 1$ as $W\to \infty$. The prior distribution weight decreases to zero; $\beta$ acts as equivalent prior distribution insurance exposure. With no past insurance exposure the appropriate weight is zero and the prediction is $m$, without defining an empirical experience mean.

Condition on the claim count, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.

For each observation time, use only the currently susceptible risk set $S_t$, with new-infection indicators $Y_{it}$. PutThis is the spatial Bernoulli infection model, under conditional independence of the indicators. Distances must be positive. Already infected trees should not be included as new susceptible trials. The minus sign in the power kernel is present in the original PDF, although it is absent in the converted TeX.

The parameter $\alpha$ must be nonnegative: for $\alpha <0$ and positive exposure, $1-e^{-\alpha A_{it}}<0$ is not a probability. Therefore the stated unrestricted normal prior distribution cannot literally define the model. A coherent interpretation is a normal prior distribution restricted to $\alpha >0$,Here $\Phi$ is the standard normal distribution function. Use the same proper prior distribution in both kernel models. The given Gaussian prior distributions for $\beta$ and $\gamma$ are mathematically valid on the real line for finite positive distances, though restricting them to positive values would impose decaying kernels and would require the corresponding normalized prior distributions.

The full conditional kernels areNeither is generally normal or another elementary conjugate family: the infection factors contain $1-e^{-\alpha A}$, and $A$ depends nonlinearly on $\beta$. Thus a simple conjugate Gibbs sampler is unavailable. This does not make Gibbs sampling logically impossible; exact rejection or slice conditional draws could be used if implemented. A straightforward alternative is Metropolis–Hastings algorithm updates within each coordinate.

For example, put $\eta =\log \alpha$ and use symmetric Gaussian random-walk proposals for $(\eta ,\beta )$. The target in these coordinates isThe factor $e^{\eta }$ is the Jacobian determinant and must be retained. Accept according to the ratio of this target, either for a block or one coordinate at a time. Posterior averages and quantiles of $e^{\eta }$ and $\beta$ estimate the requested marginal posterior distributions.

To compare kernels, introduce $M\in \{1,2\}$ with positive model prior distribution probabilities $w_1,w_2$. Under model two replace $A_{it}$ by ${\sum }_{j\in I_t}{e}^{-\gamma d_{ij}}$, defining $L_2(\alpha ,\gamma )$. The joint model/parameter target iswhere $\zeta =\beta$ or $\gamma$ and $p_M$ is its normalized proper prior distribution. Within each model use the preceding random-walk updates. For a concrete model switch, keep $\alpha$, propose the other model's kernel parameter independently from its prior distribution, and use equal forward and reverse switch-selection probabilities. A move from $(1,\alpha ,\beta )$ draws $\gamma \sim p_{\gamma }$ and has ratioAccept with the minimum of this ratio and one. The auxiliary-coordinate swap $(\alpha ,\beta ,\gamma )\mapsto (\alpha ,\gamma ,\beta )$ has absolute Jacobian determinant one, so no additional factor occurs; unequal switch-selection probabilities would require their reverse/forward ratio. This is a valid model-switching Markov chain Monte Carlo construction even though the two active models have equal parameter dimension. Prior distribution proposals may be inefficient, so fitted proposals can be substituted provided their probability densities remain in the acceptance ratio.

After checking mixing, estimate posterior model probabilities by the fractions of retained iterations in each model. The Bayes factor in favor of the first kernel isUse the posterior distributions within each model as well as these odds to assess the competing distance kernels. Proper prior distributions and their model-dependent normalizing constants are essential for this comparison.

Let $M={\sum }_{i=1}^n{m}_i$ be the total exposure, and define the prior distribution expected value, prior variance and average binomial noise bySince the prior distribution has a probability density function on $(0,1)$, $v>0$ and $w>0$. All relevant moments exist because the parameter and observations are bounded. The binomial distribution givesThe law of total expectation, law of total variance and law of total covariance consequently yieldIn particular, conditional independence of the annual observations does not imply unconditional independence: they share the uncertain parameter.

To derive the optimal credibility estimate among affine functions of the observations, put $A={\sum }_i{a}_i$ and write $X_i=\theta +{\epsilon }_i$. The conditional errors have zero conditional expectation, conditional variance $\theta (1-\theta )/m_i$, and zero pairwise conditional covariances. They are also uncorrelated with every integrable function of $\theta$. Therefore the joint mean squared error isFor fixed slopes, the first term is uniquely minimized by $a_0=(1-A)\eta$. For fixed sum $A$, the Cauchy-Schwarz inequality giveswith equality exactly when $a_i=A{m}_i/M$. Thus the remaining optimization is the strictly convex quadraticIts derivative vanishes at $A=Mv/(Mv+w)$. Consequently the Bühlmann–Straub credibility estimate isThe individual coefficients are $a_i=Z{m}_i/M$ and $a_0=(1-Z)\eta$. The strict quadratic minimizations establish global optimality and uniqueness, not just necessary equations. Here $v$ is the variance of hypothetical means, $w$ is the expected process variance, and $Z$ is the Bühlmann–Straub credibility factor. More exposure gives more weight to the observed claim fraction. Since $0<Z<1$, the estimate remains between the observed pooled fraction and the prior expected value. If one separately permits a degenerate prior, $v=0$ leads to the constant estimate $\eta$, interpreted as $Z=0$.

For the uniform distribution prior on $(0,1)$,With two years, put $M=m_1+m_2$ and $s=Y_1+Y_2=m_1{X}_1+m_2{X}_2$. Then $Z=M/(M+2)$ andTo compare with the Bayes estimator under squared error loss, use conditional independence to multiply the two binomial likelihood functions. Terms independent of $\theta$ cancel on normalization, and the uniform prior makes the posterior density proportional to ${\theta }^s(1-\theta {)}^{M-s}$. Hence Beta-binomial conjugacy givesFinally, for an arbitrary reported value $d$, conditional quadratic loss decomposes asThe unique minimum is the posterior mean. Therefore the affine credibility estimate and the exact Bayesian estimate coincide in this case. The agreement illustrates exact beta-binomial credibility with unequal exposures; for a general prior distribution, minimizing over affine functions of the observations need not recover the unrestricted posterior mean.

Treat $G_i$ as the entire phase-known genotype of individual $i$ at the genetic loci being studied, and let $\overline{F}$ denote the nonfounders. Assume unrelated pedigree founders have independent genotype priors, each nonfounder's genotype depends only on its parents' genotypes, and the observed phenotypes are conditionally independent given the genotypes. These are the standard assumptions behind the pedigree likelihood; shared environmental effects or related pedigree founders would require additional factors.

The directed ancestry graph is acyclic, even when there is inbreeding. Order individuals with parents preceding offspring and apply the chain rule for probabilities. The parental conditional independence assumption givesConditional phenotype independence gives $P(X_1,\dots ,X_n\mid G_1,\dots ,G_n)={\prod }_{i}P(X_i\mid G_i)$. Multiplying these two expressions and summing over every possible latent genotype configuration provesThis is the law of total probability, not a claim that relatives have independent unconditional genotypes.

Let $A_n$ denote the event that at least one study is positive. Conditional independence gives $\mathbb{P}(A_n\mid T)=1-{\beta }^n$ and $\mathbb{P}(A_n\mid F)=1-(1-\alpha {)}^n$. Updating the prior odds by this event's Bayes factor givesThese are the odds under at-least-one-positive selection, not the odds using every study's actual result.

The model treats hospitals as exchangeable and assumes the same exponential mixing law for their total expected annual counts. Its shape is fixed: the prior standard deviation equals its mean. Real hospitals differ in size, patient mix, ascertainment and observation effort, so expected counts should not generally share one distribution without accounting for these predictors. Common regional conditions can also invalidate their assumed conditional independence. The resulting geometric distribution for observed counts is a restrictive distributional prediction that can be checked.

Interpret each mechanism for a fixed missingness pattern $r$: the observed and missing response components are the coordinates selected by that pattern, and the covariates $X$ are assumed fully recorded.

For missing completely at random, missingness depends on neither response values nor covariates:For covariate-dependent missing completely at random, response observation may depend on $X$ but not on the response after conditioning on $X$:This is conditional independence $R\perp Y\mid X$ and can still produce a marginal association between response and missingness.

For covariate-dependent missing at random, the observation probability may use observed response components and covariates, but not the values missing under that pattern:For missing not at random, there remains dependence on $Y^m$ after conditioning on $Y^o,X$; the displayed simplification is not valid. For example, refusal could depend on the unreported response itself, even among individuals sharing all recorded covariates. These mechanisms concern the reason for nonobservation, not whether the incomplete records merely look irregular. Their distinction is not generally identifiable from the observed responses alone.

The unit and item nonresponse distinction matters: embarrassment was not recorded for the 300 unit nonresponders. Conditional independence of unit response and virginity given embarrassment is therefore not enough unless the embarrassment distribution among unit nonresponders is also supplied.

One sufficient set of additional assumptions is that unit response is independent of the pair consisting of embarrassment and virginity. Equivalently for this estimation purpose, assume that unit nonresponders have the same embarrassment mixture as the other 750 individuals, and the same virginity probabilities within those strata. The former gives an embarrassed fraction $300/750=0.4$, and the latter gives probabilities $0.18$ and $0.08$. Their estimated positive proportion is consequently $0.4(0.18)+0.6(0.08)=0.12$, or 36 of the 300. Combining the groups givesThe item-response assumption from (ii) supplies the first 90; the added unit-response transport assumptions supply the final 36. A weaker sufficient assumption is simply that the unit nonresponders' marginal virginity proportion equals the estimated $12\%$ in the other group, but it is still untestable from these data.

Without any unit-response transport assumption, retain the item assumption and write the unidentified unit proportion as $p_U$. ThenThus the extrapolation is an assumption-dependent estimate, not a quantity determined by the observed table alone.

Assume that, conditional on the recorded embarrassment category, answering the item is independent of virginity status. Also assume that this category is assessed comparably for both groups and that both response strata have positive observation probabilities. This is covariate-dependent missing completely at random for the item response with embarrassment as the fully observed covariate; it supports MAR standardization over a fully observed covariate.

Use the responder probabilities $0.18$ and $0.08$ within strata, but the stratum sizes in the full group of responders plus item nonresponders: $300$ and $450$. This givesEquivalently, the estimated missing positive counts are $100(0.18)=18$ and $50(0.08)=4$, adding 22 to the 68 observed positives. The marginal estimate changes because the full group contains a larger proportion judged embarrassed. Merely copying the unstratified responder percentage would ignore that difference. The conditional independence assumption cannot be verified using the missing item responses.