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The authoritative definition here is $Y_n=|X_n|$. On the conditioning event, $X_{n-1}=0$. Since $p_{0,1}=0$, the additional condition $Y_n=1$ then forces $X_n=-1$. The Markov property therefore givesAs usual, this conditional probability is asserted when the conditioning event has positive probability.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.

For a finite graph $G=(V,E)$, let $o(\omega )={\sum }_{e\in E}\omega (e)$, and let $k_G(\omega )$ count connected components of the spanning open subgraph, including isolated graph vertices. The random-cluster model with $p\in [0,1]$ and $q>0$ isThe partition function $Z_{G,p,q}$ is the sum of these weights over all configurations, making the expression a probability measure.

For the lattice box, take all nearest-neighbor edges with both endpoints in ${\Lambda }_n$. A random-cluster boundary condition $\xi$ is a partition of the boundary graph vertices. Vertices in one block are identified, or wired together, before counting components; the identifications do not add random edges. Let $k_{\xi }(\omega )$ be the number of components of the resulting quotient open graph. ThenThe free random-cluster boundary condition has singleton blocks, while the wired random-cluster boundary condition has one boundary block. Arbitrary partitions are allowed; no assumption that the partition itself has a planar realization is needed for the monotonicity statement.

Write $\xi ⪯\zeta$ if every block of $\xi$ is contained in a block of $\zeta$, so $\zeta$ makes at least as many identifications. The precise boundary monotonicity of the random-cluster measure isEquivalently, the expectation of every real-valued order-preserving function is larger under the more wired measure. This is stochastic domination of probability measures on the coordinatewise configuration order.

To prove it, condition on every edge except $e=\{u,v\}$. If $u,v$ are already connected using those open edges and the boundary wiring, opening $e$ does not change $k_{\xi }$, so its conditional open probability is $p$. If they are not connected, opening it reduces $k_{\xi }$ by one. The ratio of the open weight to the closed weight is then $p/[q(1-p)]$. Thus the random-cluster single-edge conditional probability isFor $q\ge 1$ the second number is at most the first. Adding open edges or making the boundary partition coarser can only turn a disconnected pair into a connected one. Therefore these conditional open probabilities are increasing in both the exterior configuration and the amount of wiring.

Run a heat-bath Markov chain for each boundary partition. At every step choose the same uniformly sampled edge in both chains, sample the same independent uniform variable $U$, and set that edge open if $U$ is below its conditional open probability. Start both chains at the all-closed configuration. By the preceding inequality, the two configurations remain ordered at every update. Each marginal chain has its corresponding random-cluster measure as its stationary distribution: resampling one coordinate from its conditional distribution preserves that law. Because $0<p<1$ and $q>0$, every update gives both possible states positive probability; the finite chain is an irreducible Markov chain with aperiodicity, and hence converges to its unique stationary distribution. Taking expectations of any order-preserving function and passing to the limit proves the claimed stochastic domination of probability measures. At $q=1$ the two conditional probabilities agree, and the boundary condition has no effect on the independent bond percolation law.

For the final identity fix a plane embedding of the finite planar graph, and include the unbounded face in its planar dual graph. In the dual configuration ${\bar{\omega }}^{\ast }$, a dual edge is open exactly when its crossed primal edge is closed. Let $m=o(\omega )$. The spanning open primal subgraph has $|V|$ graph vertices, $m$ edges, and $k(\omega )$ components. The Euler formula for a connected planar graph, applied componentwise with the common exterior face accounted for, gives its number of faces asDeleting a closed primal edge merges its incident original faces precisely when its dual edge joins distinct dual components. Consequently the faces of the open primal subgraph correspond exactly to components of the open complementary dual subgraph. This remains true for bridges and loops, with their dual loops and bridges, and for a disconnected primal graph. Thus the planar cluster-count identity isAt the self-dual parameter of the random-cluster model, put $s=\sqrt{q}$ andFor fixed $G$ the factor $(1-p_{\mathrm{sd}}{)}^{|E|}$ is independent of $\omega$. The weight is therefore proportional toAbsorbing the remaining graph-dependent factor into the normalization gives the symmetric cluster weight at the self-dual parameter:Counting the outer dual face and all isolated primal graph vertices is essential for the constant in the Euler identity. No assumption that $G$ is self-dual is required for this proportionality.

Let $Y$ denote the full data and $R$ the pattern of indicators, with $R_j=1$ for an observed component and $R_j=0$ for a missing one. For each fixed pattern $r$, split the data as $y=(y_{\mathrm{obs}}(r),y_{\mathrm{mis}}(r))$. The missing at random condition isHere $\psi$ parameterizes the missingness mechanism. Equivalently its conditional probability, with the observed data fixed, is constant over all possible completions of the missing data. The restriction is pattern-specific because the observed components depend on $r$. Missing at random allows missingness to depend on observed values; missing completely at random imposes independence from all the data.

Under missing completely at random, the observed second-year outcomes form a random subsample, so the direct pooled complete-case analysis estimate isIt is valid under missing completely at random because dropout no longer changes the marginal distribution of second-year use. Pooling is simpler than reporting separate conditional probabilities, but MCAR alone does not make this estimate more efficient than the estimate in the previous part. The fully observed first-year variable carries useful outcome information and can improve precision even when dropout is completely random.

To make the qualification explicit, write $X=Y_1$, $m(X)=P(Y_2=1\mid X)$, $\theta =E\{m(X)\}$ and $\rho =P(R=1)$ for the constant response probability. For $N$ patients the leading variance of the pooled estimator is $\mathrm{Var}(Y_2)/(N\rho )$. The standardization estimator in the preceding part has leading varianceIndeed its first-order centered contribution is $m(X)-\theta +R\{Y_2-m(X)\}/\rho$: the two terms have zero covariance, and their variances give that expression. The law of total variance therefore yieldsThe inequality is strict when there is dropout and first-year use predicts second-year use, as the distinct conditional rates suggest here. Thus the requested universal efficiency claim needs qualification. In the unrestricted joint binary-outcome model, maximizing the observed-data likelihood under either missing at random or missing completely at random gives the same standardization estimate $0.384889$: the all-patient first-year proportion and the two observed conditional second-year proportions maximize its factored outcome likelihood. Restricting the distinct missingness parameters to a common response probability affects their factor, not this estimate. The pooled value is the simple valid MCAR answer; retaining the first-year information gives the efficient MCAR answer and does not require replacing the previous estimate.

Conditional on an event and the immediately preceding history, probabilities are proportional to the three instantaneous hazards. Put $w=e^{{\beta }_0}$. The two zero-covariate subjects each have weight one, and the one-covariate subject has weight $w$. The common baseline hazard cancels. ThusEach individual with zero covariate has probability $1/(2+w)$; the first boxed probability is their combined probability. Conditioning on an event at a specified continuous time can be understood by the limiting conditional event probabilities in a short interval.

The hazard-weighted covariate mean is $w/(2+w)$, so the Schoenfeld function at the true coefficient isMultiplying by the two conditional probabilities givesThis verifies the score-centering property directly for this risk set.

Let $U_1=e^{-iH(t-t_1)/\hslash }$ and $U_2=e^{-iH(t_2-t)/\hslash }$ describe unitary time evolution. For an ideal projective measurement with the Lüders rule, its unconditioned outcome probability is $p_j=\Vert P_j{U}_1|\psi ⟩{\Vert }^2$. After that outcome, the normalized state is $P_j{U}_1|\psi ⟩/\sqrt{p_j}$. The Born rule probability of successful postselection is then $|⟨{\psi }^{\prime }|U_2{P}_j{U}_1|\psi ⟩{|}^2/p_j$. Multiplying gives the joint probabilityConditional probability therefore gives the Aharonov-Bergmann-Lebowitz rule:The denominator must be positive; otherwise the selected subensemble does not occur. Define the forward-evolved ket $|a⟩=U_1|\psi ⟩$ and backward-evolved ket $|b⟩=U_2^{†}|{\psi }^{\prime }⟩$. The numerator becomes $|⟨b|P_j|a⟩{|}^2$, which is unchanged by interchanging $a,b$. Equivalently, with ${\rho }_a=|a⟩⟨a|$ and ${\rho }_b=|b⟩⟨b|$,This expresses the boundary-state symmetry explicitly. It follows from the ordinary time-asymmetric preparation, Born rule, and state update; it does not posit an additional backward dynamical collapse.

Restore the post-selected vector omitted entirely from the TeX aid by reading the original PDF. Write $D=N^2-N+1$ andIts norm is one because $N+(N-1{)}^2=D$. For the uniform prestate and $H=0$, put $c=1/\sqrt{(N+1)D}$. The individual transition amplitudes areIn experiment $E_i$, the complement amplitude is $c-c=0$. Thus for every $i\le N$,The successful postselection rate in this experiment is $c^2=1/((N+1)D)$.

In the fully resolved experiment $E_0$, the ABL rule squares the individual amplitudes before summing. Their squared sum is $D{c}^2=1/(N+1)$, givingThese probabilities sum to one. For $N=1$ they reduce to a single certain outcome in either quantum measurement. For $N\ge 2$, the N-box pre- and post-selection paradox is that each separate binary question can be answered affirmatively with certainty, although the fully resolved quantum measurement cannot give all those outcomes at once.

There is no inconsistency. In $E_i$, the unresolved complement preserves coherent cancellation between its basis contributions under the Lüders rule. In $E_0$, those alternatives are resolved, so their squared amplitudes add instead. The different projective measurements disturb the state differently and have different postselection success rates. Merely merging the recorded $E_0$ outcomes afterwards does not reproduce $E_i$. This is the context dependence of pre- and post-selected measurements; certainties in mutually alternative experiments do not describe simultaneous measurement-independent properties.

Put $a=P(A)$, $b=P(B)$ and $j=P(A\cap B)$. Because $a,b>0$, the conditional probability inequality $P(A\mid B)>P(A)$ is equivalent to $j>ab$. But that same inequality gives $j/a>b$, or $\boxed{P(B\mid A)>P(B)}$. Thus positive association of two events is symmetric.

For the complement of an event, $P(B^c)=1-b>0$, andTherefore $B^c$ strictly repels $A$; equivalently $P(B^c\mid A)=1-j/a<1-b=P(B^c)$. Both directions are stated because the wording switches the grammatical roles of attraction and repulsion. The substantive conclusion is the same: the conditional probabilities involving the complement are strictly smaller than their unconditional counterparts.

For $k\in {\mathcal{N}}_i$, put $a_k^{(i)}={\sum }_{j\in {\mathcal{N}}_k\setminus \{i\}}{s}_j$ and $H_i={\sum }_{j\in {\mathcal{N}}_i}{s}_j$. Removing all factors independent of $s_i$ from the posterior distribution gives, for $u\in \{-1,1\}$,A stable implementation uses the difference of the log weights. The conditional log odds areso ${\pi }_i(+1\mid s_{-i},x)=1/(1+e^{-L_i})$, a logistic function. For very large $|L_i|$, evaluate this logistic expression using the sign of $L_i$ to avoid exponential overflow.

For Gibbs sampling for a finite hidden spin field, initialize any spin configuration. At every step choose $i\in D$ uniformly, draw a fresh uniform variable, set spin $i$ to $+1$ with the probability above and to $-1$ otherwise, and leave all remaining spins fixed. This Random-scan Gibbs sampler has the stated posterior distribution invariant by detailed balance of a random-scan Gibbs sampler. All conditional probabilities are strictly between zero and one. For the one-site case, the conditional probabilities are both $1/2$.

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

Let $C_{12}=\mathbb{E}[S_1{S}_2]\ge 0$, using the supplied correlation property of the Curie–Weiss model. Since $1_{\{S_i=1\}}=(1+S_i)/2$ and each spin has zero expected value,Divide by $\mathbb{P}(S_1=1)=1/2$ to obtainThe expression assumes $n\ge 2$, as required for $S_2$ to exist. It makes the connection between the spin correlation and the requested conditional probability explicit.

Reading the arrows in the original diagram gives parents $\mathrm{pa}(A)=\mathrm{pa}(B)=\varnothing$, $\mathrm{pa}(C)=\{A,B\}$ and $\mathrm{pa}(D)=\{B\}$. The Bayesian network factorization isFor binary variables, each conditional probability table has one free probability per parent configuration, since its two entries sum to one. There are one each for $A$ and $B$, four for $C$, and two for $D$, givingThis is the dimension of the unrestricted binary Bayesian network family; extra statistical parameter equalities or deterministic relationships would describe a smaller model and are not imposed by the diagram.

By conditional probability and continuity of event times,When $\tau >0$ and $q={\theta }_A+{\theta }_B>0$, the common factor $1-e^{-q\tau }$ cancels, givingBefore $\tau$, the constant cause-specific hazards divide the total event intensity in these proportions, regardless of how long the interval is. If $\tau =0$ or both intensities are zero, the conditioning event has probability zero and this conditional probability is undefined, not a ratio assigned by convention.

Write $\mathcal{E}$ for the directed edges from parent to child and $x_v(i)$ for the state at site $i$. The global Markov property for an undirected graph on this tree, together with the specified parent transitions, gives the complete-data likelihood functionOne can obtain the factorization by successively removing terminal subtrees: after conditioning on their parent, each subtree is independent of the rest. The root factor and transition products also show that different sites are independent. Only ancestral states are latent variables.

At the current Markov kernel $K^{(o)}$, define expected transition countsThe E-step of the expectation-maximization algorithm isMaximize separately for each row under $K(a,b)\ge 0$ and ${\sum }_{b}K(a,b)=1$. For $N_a={\sum }_b{N}_{ab}>0$, the Lagrange multiplier equation is $N_{ab}/K(a,b)=N_a$, with zero-count entries set to zero at a boundary maximum. Thus the EM transition-count update on a tree isIf $N_a=0$, that row does not occur in $Q$; any row probability distribution, including its previous value, maximizes the objective.

The expected counts can be computed exactly by the Felsenstein pruning algorithm and an outward belief propagation pass. For one site, let $L_v(a)$ be the conditional probability of observations in the subtree below $v$, given state $a$. For a leaf $v$ with observed state $d_v$, $L_v(a)=1_{\{a=d_v\}}$. For an internal vertex,Define $A_v(a)$ as the joint mass of state $a$ at $v$ and all observations outside its subtree. The root has $A_{v_0}(a)=1/4$. For a child $w$ of $v$, putThe edge posterior probability required above isMultiply the inside and outside factors to obtain the joint mass of the observations and endpoint states; division by the site likelihood function proves this expression. Sum $\xi$ over edges and sites to form $N_{ab}$. The two passes take $O(k|\mathcal{E}|4^2)=O(kn)$ operations on a binary tree. Scaling messages or computing them logarithmically avoids underflow.

A strictly positive initial Markov kernel ensures $Z_i>0$ for every observed leaf pattern, so every E-step is defined. At a later boundary iterate, require positive likelihood for the actual data and restrict the conditional distribution to its support. EM likelihood monotonicity proves that the update cannot decrease observed-data likelihood function. It is a maximum-likelihood iteration, not a guarantee of finding the global maximum-likelihood estimate; different initial kernels may lead to different stationary points.

The common-age conditional probabilities giveHere $GW$ includes both possible component orders. Hence the binary star fractions have the concise ratioAs a check on the role of shared age, $p_{GG}=4p_G^2/3$ rather than $p_G^2$. This is covariance induced by a shared latent variable: the red giant indicator random variables have covariance equal to the variance of $g(Y)$, namely $1/97200$.

For a starburst galaxy with age $t$, put $y_b=t/(10\mathrm{Gyr})$. There is no age averaging. The component masses remain independent, so the fraction of systems with two red giants isThe last comparison is ambiguous in the printed question. Literally, the fraction of individual red giants in the first galaxy is $1/180$. Since $p_{GG}^{\mathrm{burst}}<1/8100<1/180$, no allowed starburst age satisfies that literal comparison. If the intended comparison is instead between systems containing two red giants in the two galaxies, compare with $p_{GG}=1/24300$; the answer isFor completeness, the individual red giant fraction in the burst is $g(y_b)=t/(900\mathrm{Gyr})$, which exceeds $1/180$ precisely for $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. This is a third, different comparison; it does not change the computed fraction of two-giant systems.

Use conditional independence at the common age $Y=y$. A system with a red giant and a second evolved component is either a pair of red giants, with conditional probability $g(y{)}^2$, or a red giant and a white dwarf, with conditional probability $2g(y)w(y)$. The factor two counts which component is the white dwarf; a pair of red giants is counted only once. Since $w=9g$, the unconditional probability of this event isThis event is contained in the event of at least one red giant. Therefore the requested conditional probability, among systems selected for containing a red giant, isIn particular the denominator is a fraction of systems, not the individual-star fraction $1/180$.