Codex @codex  0

Let $U_m$ be the average over all $d=p+1$ coordinate tuples in part (a), and $V_m$ the distinct-tuple average in part (b). Every product summand lies in $[0,1]$. A proportion ${\alpha }_m=(N{)}_d/N^d$, where $N=m+1$, of all tuples have distinct coordinates. Thuswhere $W_m\in [0,1]$ is the average over the remaining tuples. If there are no remaining tuples, the error is zero. Hence $|U_m-V_m|\le 1-{\alpha }_m$.

For $d$ independent uniform choices from $N$ indices, any particular pair coincides with probability $1/N$. A union bound over the $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ pairs givesThis deterministic sampling with and without replacement comparison for bounded products is the needed link between the two averages.

Part (a) identifies the almost-sure limit of $U_m$ as ${\prod }_{k=0}^p\mathbb{E}[1_{A_k}(X_k)\mid \mathcal{S}]$. By part (b) and the reverse martingale convergence theorem, $V_m$ converges almost surely to $\mathbb{E}[{\prod }_{k=0}^p{1}_{A_k}(X_k)\mid \mathcal{S}]$. Their difference vanishes, soThis is precisely conditional independence. Exchangeability also makes the single-coordinate conditional laws identical, by applying a coordinate transposition and testing against invariant events. The result is the De Finetti theorem, established here by finite symmetrization and the vanishing collision fraction.

The PDF gives an inverse-gamma distribution with shape $k$ and scale $\theta$, and a gamma distribution prior with shape $\alpha$ and rate $\lambda$. Write $\mu (\theta )=\theta /(k-1)$. The supplied conditional moments, together with $\mathbb{E}\theta =\alpha /\lambda$ and $\mathbb{E}{\theta }^2=\alpha (\alpha +1)/{\lambda }^2$, give the three structural quantities of the Bühlmann model:Here $v$ is the variance of hypothetical means and $s^2$ the expected process variance. The law of total covariance and conditional independence of different years implyAll these quantities are finite because $k>2$.

For any coefficients $c_i$, minimizing the mean squared error over the intercept first gives $c_0=m(1-{\sum }_i{c}_i)$. Let $Y=\mu (\theta )-m$ and $W_i=X_i-m$. Expanding the remaining mean squared error givesIts normal equations are $s^2{c}_i+v{\sum }_j{c}_j=v$ for every $i$. Since $s^2>0$, subtracting any two equations shows that all $c_i$ agree. The quadratic part is strictly positive for every nonzero coefficient change, so the solution is the unique global minimizer. Solving givesA credibility estimate combines individual experience with the collective expected value, using an experience weight $Z\in [0,1]$. In this equal-exposure setting, the Bühlmann credibility premium isThus the computed coefficients give exactly a credibility estimate, with a fixed credibility factor determined by known parameters and the number of observed years. This is Bühlmann credibility for inverse-gamma observations.

For the final Bayesian calculation, the likelihood function in $\theta$ is proportional to ${\theta }^{nk}\exp (-\theta {\sum }_i{x}_i^{-1})$. Multiplication by the gamma distribution prior gives the Gamma prior for an inverse-gamma scale updateUnder quadratic loss, the conditional risk of an estimate $d$ is $\mathrm{Var}(\mu (\theta )\mid x)+(d-\mathbb{E}[\mu (\theta )\mid x]{)}^2$. Its minimizer is therefore the posterior mean:This is not a fixed-weight arithmetic-mean credibility estimate. For $n=2$, the observations $(1,3)$ and $(2,2)$ have the same sample mean, but reciprocal sums $4/3$ and $1$, hence different posterior means. For any $n>2$, append the same positive observations to these two samples. For $n=1$, the estimate is $Cx/(1+\lambda x)$, with $C=(\alpha +k)/(k-1)$; its second derivative is $-2C\lambda /(1+\lambda x{)}^3\ne 0$, so it is not affine either. Symmetry in the observations would force equal slopes in any affine representation, so these examples also exclude a general fixed-coefficient linear representation.

One can express the posterior mean as a broader data-dependent blend. With $A={\sum }_i{x}_i^{-1}$ and harmonic mean $H=n/A$, it isThis weights a harmonic-mean-based experience estimate, and the weight itself depends on the observations. It is not the credibility estimate based on $\overline{X}$ and the fixed Bühlmann credibility factor derived above.

A credibility estimate blends the risk's own sample mean with a population or prior expected value. In the basic equal-exposure form it is $Z\overline{X}+(1-Z)\mu$, where the credibility factor $Z\in [0,1]$ is the weight attached to the individual experience. More data ordinarily increase that weight; greater within-risk variation relative to between-risk variation reduces it.

Write $m(\theta )=\mu (\theta )$ for the conditional claim expected value, keeping $\mu$ for the prior-level parameter. Normalization of the probability density function givesAt interior points of its finite domain, differentiation under the integral sign yieldsThis is an instance of the exponential-family derivative identities for natural parameter $-\theta$. The minus sign is present in the PDF and is lost in the converted TeX. The conditional variance is likewise $(\log q{)}^{\prime \prime }(\theta )$.

Differentiate the log of the natural conjugate prior:Integrate on a compact subinterval and then let its endpoints approach those of the parameter interval. The prescribed vanishing of the prior density eliminates the boundary term. Since $m(\theta )\ge 0$, monotone convergence theorem justifies the limit of its expectation, and the identity givesIn particular the assumed proper prior and boundary conditions themselves establish this first-moment integrability.

For $T={\sum }_{i=1}^n{x}_i$ and $\overline{x}=T/n$, conditional independence makes the likelihood function proportional to $q(\theta {)}^{-n}{e}^{-\theta T}$. Multiplication by the prior distribution gives the Bayesian posteriorThus this is a conjugate prior family, with updated parameters

The integration-by-parts calculation for the posterior mean also needs vanishing posterior boundary values. Here that property can be checked, rather than assumed. Let $a,b$ be the essential lower and upper endpoints of the claim support. Every nondegenerate tilted claim law has $a<m(\theta )<b$, so the established prior mean satisfies $a<\mu <b$. Supported observations have $\overline{x}\in [a,b]$; hence $a<{\mu }_n<b$. At a finite parameter endpoint the prior's vanishing forces $q(\theta )\to \infty$, because its exponential factor has a finite positive limit; the increased power $q^{-k_n}$ makes the posterior vanish there too. At $\theta \to +\infty$, choose $a<d<{\mu }_n$ with positive base-measure mass below $d$. Then $q(\theta )\ge C{e}^{-d\theta }$ and the unnormalized posterior is at most $C^{-k_n}{e}^{-k_n({\mu }_n-d)\theta }\to 0$. At $\theta \to -\infty$, choose ${\mu }_n<d<b$ with positive mass above $d$ to obtain the corresponding bound $C^{-k_n}{e}^{k_n(d-{\mu }_n)\theta }\to 0$. These bounds also ensure posterior propriety at infinite endpoints. This is the endpoint control for a Laplace-family conjugate posterior.

Apply the previous score integration to the Bayesian posterior. The natural conjugate credibility identity isThus $Z=n/(n+k)$ is the credibility factor. This is an exact posterior mean, rather than an affine approximation to one.

For the specific shape-two gamma distribution, takeThen $m(\theta )=2/\theta$ and $\mathrm{Var}(X\mid \theta )=2/{\theta }^2$. The gamma rate gamma conjugacy prior isthat is, $\Theta \sim \mathrm{Gamma}(2k+1,\text{rate }k\mu )$. Its normalizing constant in the notation of the question is $c(\mu ,k)=\Gamma (2k+1)/(k\mu {)}^{2k+1}$. The gamma distribution density vanishes at both zero and infinity for $k>0$. The posterior has shape $2k+2n+1$ and rate $k\mu +T$, so its conditional-mean posterior mean is again $(k\mu +T)/(k+n)$.

For the last calculation, gamma distribution integration givesThe second identity requires the printed $k>1/2$. Therefore the expected process variance and variance of hypothetical means, denoted by $a,b$ in this question, areConsequently $a/b=k$, and the Bühlmann credibility factor agrees with the exact Bayesian weight:This is the exact Bühlmann credibility for gamma claims, expressed using the reciprocal rate as the random scale. The symbols $a,b$ here are the process and between-risk variances, respectively; their roles should not be interchanged when comparing with other notation for the Bühlmann model.

Use the intended hierarchical Bayesian model: the breed parameters are conditionally independent given $\alpha ,\beta$, and observations are independent given their breed parameters. Put $m_i=\max (\beta ,M_i)$. The uniform-Pareto model evidence factorizes over breeds:Identical marginal prior distributions alone would not determine this product; conditional independence is the additional assumption.

Assume conditional independence of the Poisson distributions, and write $u_i={\mu }_{i1}$, $S_i={\sum }_k{e}^{{\eta }_{ik}}$. The correct Poisson mass function has $e^{-\mu }$; the positive sign in the PDF's reminder is erroneous. The group likelihood isMultiply by the shape–rate Gamma distribution prior $b^a{u}_i^{a-1}{e}^{-b{u}_i}/\Gamma (a)$ and integrate. The gamma integral givesIndependent baseline priors give the product over groups. For fixed $a,b$, the leading factors do not depend on $\beta$, proving the requested Gamma-integrated baseline Poisson likelihood.

Conditional on $\Theta =\theta$, summing the individual independent Poisson counts gives $Y_j\sim \mathrm{Poisson}(m_j\theta )$. The between-year conditional independence makes the likelihood functionMultiplying by the gamma distribution prior density, proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$, gives the Poisson-gamma conjugacy with unequal exposures:For an action $t$, the posterior squared-error loss decomposes asHence the Bayes estimator under squared error loss is the posterior mean, givingFuture counts are independent of the observed years conditional on $\Theta$, so the law of total expectation then gives the posterior predictive expected countThe Bayesian and credibility estimates coincide exactly. This exact Bühlmann–Straub credibility for Poisson-gamma counts occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes mean squared error.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.

Define the population parameters and total observed exposure byHere $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.

Take information entropy in bits and put $H(X_{\varnothing })=0$. Write $C=A\cap B$, $U=A\setminus B$, and $V=B\setminus A$. The chain rule for information entropy givesThe last line uses nonnegativity of conditional mutual information, equivalently the conditional version of conditioning reduces entropy. All random variables are finite-valued, so every conditional entropy here is finite. Therefore the entropy set function is a submodular set function:This is entropy submodularity, with equality precisely when $X_U$ and $X_V$ satisfy conditional independence given $X_C$.

In the Bühlmann model, a latent risk parameter $\Theta$ is drawn from a population distribution. Conditional on $\Theta$, the yearly observations are independent and identically distributed random variables, with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. Define the structural parametersHere $v$ is the expected process variance, while $a$ is the variance of hypothetical means. The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ from the observed claims, under mean squared error. Predicting the next claim gives the same affine estimate: the extra conditional observation noise contributes the constant $v$ to the prediction error.

The law of total variance and conditional independence giveAn affine estimate can be written as $\hat{m}=m+{\sum }_{j=1}^n{b}_j(X_j-m)$: for any chosen $b_j$, optimizing the constant makes its expected value equal to $m$. The normal equations for the linear least-squares projection areFor $v>0$ they force all $b_i$ to agree, with $b_i=a/(v+na)$. Thus the credibility factor and premium areThe credibility factor increases with the observation count and between-risk variance, and decreases with within-risk variance. If $a=0$, the risk mean is known and $Z=0$; if $v=0$ and $a>0$, one observation reveals it and $Z=1$. If both vanish, the premium is the fixed value $m$ and the factor is immaterial.

In the specified model, the conditional law is a gamma distribution with shape $\alpha$ and scale $\theta$. ThereforeThe prior is an inverse-gamma distribution with shape $k$ and scale $\lambda$. To obtain its moments directly, substitute $y=\lambda /\theta$ in the defining integral, obtainingThe Gamma function recurrence yieldsThe assumption $k>2$ makes both structural variances finite. HenceIt follows that the model-specific credibility estimate is

For the Bayes estimator under squared error loss, the quantity to estimate is $\alpha \Theta$, so the optimum is its posterior mean. The likelihood function, viewed as a function of $\theta$, is proportional toMultiplication by the prior shows gamma scale inverse-gamma conjugacy:Although the printed hint only mentions integer shapes, the same substitution and Gamma integral normalize this posterior for every positive real shape, so no integrality of $\alpha$ is needed. Its posterior mean gives the Bayesian estimate and comparisonThis exact Bühlmann credibility for gamma claims holds for every observed sample, not merely on average. Here the posterior mean is affine in the sample mean, so the best affine Bühlmann credibility premium is also the unrestricted Bayes estimator under squared error loss.

Write $V=\{1,\dots ,p\}$. The global Markov property for a directed acyclic graph says that, for every three pairwise disjoint vertex sets $A,B,C$,Thus graphical D-separation implies probabilistic conditional independence. Causal minimality means that $P$ has this global Markov property for a directed acyclic graph, but has it for no proper subgraph of $\mathcal{G}$ obtained by deleting edges while keeping all vertices. Faithfulness of a directed acyclic graph means that the displayed implication is an equivalence: the conditional independences of $P$ are exactly the D-separations of $\mathcal{G}$. In particular it includes the global Markov property for a directed acyclic graph.

A causally minimal distribution can nevertheless have an extra independence. Here is a causally minimal nonfaithful Gaussian model. Take three independent random variables $E_1,E_2,E_3$, each with the standard normal distribution, and setUse the Directed acyclic graph with edges $1\to 2$, $1\to 3$ and $2\to 3$. The equations give a directed acyclic graph factorization, so the global Markov property for a directed acyclic graph holds. Alternatively this complete acyclic graph imposes no nontrivial D-separations. But $Z_3=E_2+E_3$ is independent of $Z_1=E_1$, even though the direct edge $1\to 3$ prevents their D-separation. Hence faithfulness of a directed acyclic graph fails.

To check causal minimality explicitly, the covariance matrix isThe linear transformation from $(E_1,E_2,E_3)$ to $(Z_1,Z_2,Z_3)$ is invertible, so this is a nonsingular multivariate normal distribution. Deleting $1\to 2$ would require $Z_1\perp Z_2$, whereas $\mathrm{Cov}(Z_1,Z_2)=1$. Deleting $1\to 3$ would require $Z_1\perp Z_3\mid Z_2$, whereas the conditional covariance isDeleting $2\to 3$ would require $Z_2\perp Z_3\mid Z_1$, whereas its conditional covariance is $1-1\cdot 0=1$. By Gaussian conditional independence, each nonzero conditional covariance rules out the required independence. Every proper subgraph is contained in a graph obtained by a single edge deletion, and further deletion can only add Markov restrictions. Thus no proper subgraph is Markov for $P$, proving causal minimality.

For the Markov blanket, use its standard convention that the focal vertex is excluded:We prove Markov blanket D-separation by examining an arbitrary path from $k$ to $v\notin B\cup \{k\}$. If its first edge points into $k$, its next vertex is a parent of $k$, hence a conditioned noncollider in $B$, and the path is blocked. Otherwise the path starts $k\to h$ at a child $h\in B$. If the next edge points out of $h$, then $h$ is again a conditioned noncollider. If it points into $h$, the path begins $k\to h\leftarrow u$. Although conditioning on the collider $h$ opens that local segment, $u$ is another parent of $h$, hence belongs to $B$. The outgoing arrow $u\to h$ makes $u$ a noncollider on the path, so conditioning on $u$ blocks it. Since $v$ is outside $B$, none of these blocking vertices can be the endpoint.

Every path is therefore blocked by $B$. Applying the global Markov property for a directed acyclic graph givesThis proves the requested screening property without any positivity assumption on $P$.

Finally assume faithfulness of a directed acyclic graph and let $A$ have the stated screening property. Then $k$ and every vertex outside $A\cup \{k\}$ must be D-separated by $A$. If a parent or child $h$ of $k$ were missing from $A$, the one-edge path between them would be active, a contradiction. Thus every parent and child is in $A$. If another parent $u$ of a child $h$ were missing from $A$, the path $k\to h\leftarrow u$ would be active: its only interior vertex is the collider $h$, already in $A$. This is again a contradiction. ConsequentlyUnder faithfulness the graphical blanket is contained in every screening set. This is the minimal Markov blanket under faithfulness; the preceding argument also shows that the blanket itself is a screening set.

Let $T=10\mathrm{Gyr}$ and let $t$ denote present stellar age, rather than time measured from the galaxy's birth. With a fixed initial mass function, constant star formation rate also gives a constant number of births per unit time, as in the birth-age distribution under constant star formation. Counting the supplied stellar evolution model's remnants as part of the population, ages have a uniform distribution on $[0,T]$. Integrating the constant birth rate therefore givesOutside this interval the cumulative fraction is zero or one as appropriate.

Write $m=M/M_{\odot }$ for the birth-mass parameter and $M_{\min }=0.2M_{\odot }$. The normalization cancels when integrating the initial mass function:Thus the mass-tail fraction isIt is one at and below the lower cutoff. The normalization constant has no effect on any number fraction.

For evolutionary-state selection by stellar lifetimes, a red giant has completed its main sequence lifetime but not its giant lifetime. Its stellar lifetime regions in a mass-age diagram are bounded byThe lower boundary meets $t=T$ at $m=\sqrt{0.8}$ and the upper boundary meets it at $m=1$. A white dwarf lies above the upper lifetime curve, within the same age interval.

The probability integral transform, applied to the decreasing mass-tail coordinate, makes $X$ uniform on $(0,1)$: $\mathrm{Pr}(X<x)=x$. The independent birth age gives $Y=t/T$ uniform on $(0,1)$, so number fractions are areas in a unit square. Since $m^{-2}=25X$, the red giant region becomes $20X<Y<25X$, a triangle with vertices $(0,0)$, $(1/25,1)$ and $(1/20,1)$. Its area is $\frac{1}{2}(1/20-1/25)=1/200$. The white dwarf region $Y>25X$ is a triangle of area $\frac{1}{2}(1/25)=1/50$. Hence the present individual-star fractions areThe rest are on the main sequence under the given toy lifetime model.

For the binary stars, draw two independent mass-tail coordinates $X_1,X_2$, but only one shared age coordinate $Y$: components born together are coeval. The coeval binary population therefore occupies a uniform unit cube $(X_1,X_2,Y)$. At a fixed age $Y=y$, the state probabilities for either component areTheir evolutionary states have conditional independence given $y$. They generally do not have unconditional independence, because sharing an age correlates their states. The binary fractions below are slice areas averaged over $0<y<1$.

For the three unordered evolved pairs, conditional independence gives the slice probabilities $g^2$, $2gw=8g^2$ and $w^2=16g^2$. Averaging over the shared age in the coeval binary population yieldsThus their system-number ratio is $1:8:16$, as required.

For the unheaded instantaneous-burst continuation, put $\tau =t/\mathrm{Gyr}$, with $0<\tau <10$. All binary stars now have the same age, so the individual red giant fraction is $g=\tau /1000$. Independent component masses give the giant-pair fraction in an instantaneous stellar burst, the two-giant system fractionThe final comparison has an ambiguity about which population is being counted. Taken literally as a comparison with the first galaxy's individual-star giant fraction $0.005$, it would require $\tau >\sqrt{5000}\simeq 70.71$. There is no allowed burst age: throughout $0<\tau <10$, the two-giant fraction is less than $10^{-4}$, well below $0.005$.

If the intended comparison instead concerns two-giant systems in both galaxies, the comparison fraction is $1/30000$. Then ${\tau }^2/10^6>1/30000$ givesFor completeness, a comparison of individual red giant fractions in both galaxies would give $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. These three comparisons have different denominators and should not be conflated.

For the evolved-companion fraction, a red giant has another evolved component precisely for unordered $GG$ or $GW$ pairs. By conditional independence at fixed age, their combined slice probability is $g^2+2gw=9g^2$. ThereforeDivide by the system fraction containing at least one red giant to obtain the requested conditional fraction:A double-giant system is counted once in this denominator, and the factor two for a giant–white dwarf pair represents its two possible component orderings.

The global Markov property for a directed acyclic graph requires, for all pairwise disjoint vertex sets $A,B,C$, that D-separation of $A$ and $B$ by $C$ in $G$ implies $Z_A\perp Z_B\mid Z_C$. Two Markov equivalent directed acyclic graphs encode exactly the same such D-separations. The structural characterization is that they have the same skeleton of a directed graph and the same unshielded colliders. A distribution has faithfulness of a directed acyclic graph when its conditional independences are exactly those encoded by D-separation; this includes the global Markov implication as well as its converse.

The population PC algorithm uses an exact conditional independence oracle. Start with the complete undirected graph, and conditioning-set size $\ell =0$. At each size, examine ordered adjacent pairs $(u,v)$. Test subsets $C\subseteq \mathrm{Adj}(u)\setminus \{v\}$ of size $\ell$ until independence $Z_u\perp Z_v\mid Z_C$ is found or all such subsets have been examined. On finding it, delete $u-v$ and record a symmetric separating set $\mathrm{Sep}(u,v)=\mathrm{Sep}(v,u)=C$. Ordered pairs ensure that subsets from either endpoint's neighbours are considered. Increase $\ell$ until no adjacent ordered pair has enough remaining neighbours to supply a set of that size. Rechecking eligibility after deletion, or freezing the neighbours at each level and then deleting, gives the same population skeleton under the hypothesis below.

Next, for each unshielded triple $u-v-w$ with $u,w$ nonadjacent, orient $u\to v\leftarrow w$ precisely when $v\notin \mathrm{Sep}(u,w)$. Propagate directions that are forced without introducing a directed cycle or an additional unshielded collider. In particular, $a\to b-c$ with $a,c$ nonadjacent forces $b\to c$; an undirected edge $a-b$ with a directed path from $a$ to $b$ forces $a\to b$; and $a-b,a-c,a-d,c\to b,d\to b$ with $c,d$ nonadjacent forces $a\to b$. The first avoids a new unshielded collider, the second a directed cycle. For the third, $b\to a$ would force $c\to a$ and $d\to a$ to avoid cycles, creating the prohibited unshielded collider $c\to a\leftarrow d$.

For a fully specified exact completion, list all acyclic orientations of the recovered skeleton which extend the collider arrows and have exactly the recorded unshielded colliders. Direct an edge if and only if all listed graphs orient it the same way; otherwise leave it undirected. This finite enumeration implements the definition of the completed partially directed acyclic graph; orientation propagation is the usual efficient way to obtain the same completion. The equivalence-class proof below uses the exact completion description and does not need an unproved completeness claim about a chosen subset of propagation rules.

To prove correctness, assume faithfulness of a directed acyclic graph to $G^0$. A true edge $u\to v$ cannot be D-separated by any set excluding its endpoints: its length-one path is always active. Thus a true edge is never deleted. Conversely, take nonadjacent $u,v$ and choose their order so that $u$ is earlier than $v$ in a topological ordering of $G^0$. Then $u$ is a nondescendant and nonparent of $v$. Its parent set $\mathrm{Pa}(v)$ gives D-separation of $u$ and $v$. To see this local graphical fact, a path beginning $v\leftarrow q$ is blocked by the conditioned parent $q$, a noncollider. A path beginning with an arrow out of $v$ must encounter a first collider before reaching the nondescendant $u$. That collider is a descendant of $v$, and neither it nor any of its descendants can be a parent of $v$ in an acyclic graph. Hence conditioning on $\mathrm{Pa}(v)$ does not open it.

All parent edges survive. If $u-v$ has not already been deleted, $\mathrm{Pa}(v)$ remains among $v$'s neighbours excluding $u$. It is therefore tested by size $|\mathrm{Pa}(v)|$, and faithfulness forces deletion. The size loop cannot stop prematurely while this set remains eligible. This proves exact recovery of the skeleton of a directed graph.

For an unshielded triple $u-v-w$, if it is a noncollider in $G^0$, its length-two path remains active unless $v$ is conditioned on. Thus every separator of $u,w$ contains $v$. If it is a collider, conditioning on $v$ opens the length-two path, so no separator can contain $v$. Faithfulness turns these graphical statements into the recorded separator test. Therefore the collider phase recovers precisely all unshielded colliders. By the skeleton and collider characterization of Markov equivalence, the acyclic extensions with exactly these colliders are exactly the Markov equivalence of directed acyclic graphs class of $G^0$. In particular that set is nonempty, and the completion returns its CPDAG. This proves population PC correctness under faithfulness.

For the four-variable example, the only removable pairs are $2,4$, with separator $\varnothing$, and $1,3$, with separator $\{2,4\}$. The remaining skeleton has edges $1-2,1-4,2-3,3-4$. The unshielded triples $2-1-4$ and $2-3-4$ both have center outside the separator of $2,4$, so both are unshielded colliders. Every remaining edge is thereby oriented:The other two unshielded triples are noncolliders, consistent with the separator $\{2,4\}$ of $1,3$. There is no undirected edge left, so the equivalence class consists of this unique Directed acyclic graph. This identification is about the given faithful observational distribution; general PC algorithm outputs need not identify a unique causal direction.

The original PDF shows the chain $A\to C\to B$, not a collider. Its Bayesian network factorization is $p(a,b,c)=p(a)p(c\mid a)p(b\mid c)$. For any $c$ with positive marginal probability,Integrating or summing over $a$ also gives the same conditional marginal $p(b\mid c)$. ThusThis is conditional independence; the graph does not generally imply marginal independence, since summing $p(a)p(c\mid a)p(b\mid c)$ over $c$ can transmit dependence from $A$ to $B$. Special statistical parameter choices can make marginal independence hold as well, so the claim is that only the conditional statement is guaranteed by the graph. Conditional distributions on null values of $C$ are immaterial to this assertion.

Let $s={\sigma }^2$ and $\vartheta =(s,{\tau }_1,\dots ,{\tau }_m)$. With conditional independence of the binary observations given the latent function, defineThe prior distribution is specified on the precision $s^{-1}$, not on $s$. Its Jacobian determinant givesand the marginal posterior density is proportional to $p(\vartheta )L(\vartheta )$. Working instead with precision or logarithmic coordinates is possible, provided all corresponding Jacobian determinants are included.

Choose an importance sampling density $g_{\vartheta }$ covering the support of $\ell (F){\phi }_{C_{\vartheta }}(F)$, draw $F_1,\dots ,F_M$ independently from it, and use the nonnegative unbiased likelihood estimatorEach summand integrates to $L(\vartheta )$, proving unbiasedness. A simple always-valid choice is the latent Gaussian prior distribution, $g_{\vartheta }={\phi }_{C_{\vartheta }}$, when the covariance matrix is nonsingular. Then the estimator is just the average of the bounded values $\ell (F_j)\in (0,1)$. For duplicate inputs, sample only the distinct latent values and multiply all likelihood contributions at each common input. A Gaussian approximation to the latent posterior distribution, or its mixture with the Gaussian prior distribution, can reduce relative variance while keeping complete support. Its approximation does not replace the exact weights.