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Let $p_{rs}(t)=P(X(u+t)=s\mid X(u)=r)$ and $P(t)=e^{Qt}$, where time homogeneity removes dependence on $u$. A recorded state at the next clinic visit contributes a transition probability; an exactly observed death contributes a statistical probability density, not the probability of being dead at that time. If the last recorded living state is $r$, the mixed panel and exact-death likelihood factor after an interval $t$ isThis sums over the unobserved living state immediately before death.

Conditioning on the recorded initial states, the contribution of the three displayed patient histories isIn this irreversible illness-death model, $p_{21}=0$, $p_{11}(t)=e^{-(a+b)t}$ and $p_{22}(t)=e^{-ct}$, simplifying it toThe factor $c$ is essential: the death time is known exactly. Replacing the final statistical probability density by $p_{23}(34.6)$ would instead model interval observation of death and give a different likelihood.

The assumptions are independent patient histories with common rates; the Markov property; constant rates over calendar/follow-up time in this model; the stated absence of recovery and absorption at death; accurate state labels and death times; and an observation/follow-up mechanism that is noninformative for the latent process given the observed history. Clinic dates are conditioned on. The displayed living endpoints contribute only the shown observations, with noninformative right censoring if they are follow-up endpoints. Initial state probabilities are omitted by conditioning on them. Progression between visits can be unobserved, which is precisely why panel-observed multi-state likelihood uses the matrix exponential rather than assuming a transition occurs at a visit.

The mean holding time from a transition intensity matrix is $-1/q_{rr}$. Apply this to the fitted exit rates and use monotonic inversion for each confidence interval:Thus the expected state durations are 86.96 months and 31.45 months, respectively. A confidence interval for a reciprocal rate reverses the endpoint order; the negative diagonal rates must first be converted to positive exit rates.

For the expected absorption time in an illness-death model, the time spent initially in the mild state is followed by an additional severe-state duration only if progression occurs before death. That probability is $0.0072/0.0115$. ThereforeThis is an unconditional mean including both possible paths to death, not a mean conditional on progression.

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.

Apply MAR standardization over a fully observed covariate. All first-year statuses are known, so the estimated probabilities of no use and use in year one are $58/102$ and $44/102$. Within those categories, missing at random lets the observed second-year probabilities represent the corresponding dropout outcomes as well. They are estimated by $7/37$ and $18/28$.

The law of total probability then givesEquivalently, impute expected drug-use counts $21(7/37)$ and $16(18/28)$ for the two dropout groups, add these to the 25 observed second-year users, and divide by 102. This uses both the complete records and the fully observed first-year information.

An ignorable missingness mechanism need not be absent from the data-generating process. It means that likelihood inference about the data-model parameters can omit the missingness factor, while still integrating over missing values.

Write the complete joint data statistical probability density as $f_{\theta }(y,x)$ and the conditional missingness probability as $g_{\psi }(r\mid y,x)$. All $y$ values and only $x_{\mathrm{obs}}$ are observed. Under missing at random, $g_{\psi }(r\mid y,x)$ is constant as $x_{\mathrm{mis}}$ varies with $(y,x_{\mathrm{obs}})$ fixed. The actual observed-data likelihood is thereforeThe assumed distinctness is understood as independent variation of $\theta$ and $\psi$. Maximizing over $\psi$ multiplies $L_{\mathrm{obs}}$ by a factor independent of $\theta$; likelihood ratios, scores and likelihood curvature for $\theta$ are therefore unchanged by omitting $g_{\psi }$. This proves likelihood ignorability.

For implementation of the observed-data likelihood with a missing covariate, a linear regression model for $Y\mid X$ must be accompanied by an appropriate model for the distribution of $X$. For independent individuals, write $f_{\beta }(y\mid x)$ for the regression statistical probability density and $g_{\eta }(x)$ for the age statistical probability density. Up to the ignorable factor,Ignorability does not authorize discarding missing-age cases or assuming their ages have the distribution seen in the complete cases. It removes the need to model the observation mechanism for likelihood inference under the stated conditions, not the need to handle the missing covariates.

For a proper continuous event time, the survival function is $S(t)=e^{-H(t)}$. The assumed invertibility of the cumulative hazard function gives, for $u\ge 0$,ThereforeThis is the cumulative hazard probability transformation; it applies also conditionally on a subject's covariates, using that subject's correctly specified cumulative hazard function.

The fitted transformed times are Cox–Snell residuals. They retain their event/right censoring indicators, so a right-censored residual represents an exponential observation known only to exceed its displayed value. Under the fitted model and independent right censoring, calculate the Kaplan–Meier estimator of residual survival and compare it with $e^{-y}$, or calculate the residual Nelson–Aalen estimator and compare its cumulative hazard with the diagonal $H(y)=y$. Systematic departures reveal model inadequacy; sparse extreme residual risk sets and parameter estimation require caution. Treating all censored residuals as observed event times would invalidate this diagnostic.

Without right censoring, the residual mean should be approximately one. With right censoring, use the modified Cox–Snell residualFor a true unit-rate exponential distribution, the memoryless property gives $E(U\mid U>c)=c+1$. Thus an event keeps its known transformed time, while a censored observation is replaced by the conditional expected event time. Under independent right censoring, iterated expectation makes the mean of these adjusted values one when the true hazards are used, and approximately one when fitted hazards are used. These mean-imputed values do not themselves have an exponential distribution, so the survival-curve diagnostic should still use the original censored residual dataset. Moreover fitting equations can force the adjusted sample mean to one, making its mean alone a weak diagnostic.

For the proposed mixture of a finite right censoring time $c\ge 0$ and no right censoring, $P(C<U)=\pi e^{-c}$ andConsequentlyWhen right censoring has positive probability this choice is unique; if $\pi e^{-c}=0$, no correction is needed and any $k$ has the same effect. Its independence from $c$ is the content of exponential memorylessness: the expected extra lifetime after any right censoring time is one. Conditioning on an arbitrary independent right censoring time proves the same correction beyond this special two-point mixture.

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.

When the event comes from the one-covariate subject, the Schoenfeld function for a three-person binary risk set isHenceThese are limits where necessary. At a very negative coefficient the model assigns almost no event probability to the observed one-covariate subject, producing the largest positive discrepancy. At zero coefficient all three subjects have equal hazard functions, so the expected event covariate is $1/3$ and the discrepancy is $2/3$. At a very positive coefficient the observed subject is predicted to have the event almost surely, so the discrepancy tends to zero. The function decreases strictly and stays positive at every finite coefficient: this one event alone favors increasing $\beta$, while the other events in the complete Cox partial likelihood determine its overall estimate.

Write $r={\theta }_A+{\theta }_B$, with ${\theta }_A\ge 0$, ${\theta }_B>0$ and a finite $\tau \ge 0$. The competing risks model with transient surgical mortality has survival functionIntegrating the disease cause-specific hazard against this survival function givesThe first term accounts for disease deaths while both causes act; the second includes survivors of that period who then face only disease mortality. The probability of eventually dying from surgery isTaking the limit in $F_B$ gives $F_B(\infty )={\theta }_B(1-e^{-r\tau })/r+e^{-r\tau }$, and thereforeThe ultimate-death conclusion uses the positive continuing disease hazard. If ${\theta }_B=0$, the surviving fraction $e^{-{\theta }_A\tau }$ would instead live indefinitely in this model; the asserted conclusion is not valid in that boundary case.

For mutually exclusive event types, let $T$ be the first event time and $J$ its type. The cause-specific hazard for cause $j$ isIt is a rate conditional on having had no event of any cause. The cumulative incidence function, also called the cumulative risk function, is the actual probability $F_j(t)=P(T\le t,J=j)$.

The total hazard function is $h(t)={\sum }_k{h}_k(t)$, giving $S(t)=\exp \{-{\int }_0^t{\sum }_k{h}_k(u)du\}$. Surviving every cause to time $u$ and then experiencing cause $j$ givesIn competing risks, $F_j$ is generally not $1-e^{-{\int }_0^t{h}_j}$, because competing events remove individuals before they can experience cause $j$. This formula does not require assuming independent latent failure times for the different causes.

Introduce nonnegative surplus variablesAt the proposed starting basic feasible solution, the basic variables are $x_1,r_1,r_2$ and the nonbasic variables are $x_2,x_3,r_3$. Its simplex dictionary, with objective $z=x_1+x_2+x_3$, isIncrease $x_3$ to decrease $z$. The simplex ratio test gives limits $1/5$ from $x_1$, $1/21$ from $r_1$, and $3/31$ from $r_2$. The smallest is $1/21$, so $x_3$ enters and $r_1$ leaves. After this single pivot the dictionary isAll objective coefficients of nonbasic variables are strictly positive. Thus the simplex method has reached its unique optimum:The feasible dual vector $y=(1/21,0,4/21)$ has the same objective, providing an independent weak duality certificate. Normalization givesThese are all the equilibria: the dual's strict slack in row two forces $p_2=0$, and the primal's strict slack in column two forces $q_2=0$. Equality of the two active row and column payoffs then fixes the displayed probabilities. The value is the first player's expected net loss; the second gains $4/5$.

For a fixed number choice, increasing one's own stake changes only the amount lost when one loses. It does not change the amount won, which is the opponent's stake, or the zero payoff of a tie. Thus doubling is either weakly dominated by retaining the original stake or payoff-equivalent to it. This establishes that there is no strategic advantage in doubling, but does not imply strict harm in every equilibrium.

For the first player, each of the two equilibrium choices loses with positive probability: choice one loses against the second player's four, and choice four loses against their one. Doubling either therefore gives a strictly smaller expected payoff than $-4/5$. By contrast, against the first player's support $\{1,4\}$, the second player's choices one and four either win or tie. Their own stake is never lost, so doubling it does not change their payoff.

More precisely, retain the first player's original-stake probabilities $(4/5,0,1/5)$. The second player may split their total probability $1/5$ on choice one, and $4/5$ on choice four, arbitrarily between original and double stakes. The first player's unused choice two has payoff at most $(1/5)2-(4/5)2=-6/5<-4/5$, even when the second player doubles their one stake. The doubled first-player choices are also worse. The second player's choice two, with either stake, is worse against the displayed first-player strategy. Hence these splits remain Nash equilibria of the enlarged zero-sum game.

The first player should not double; the second player is indifferent between doubling and retaining the original stakes on their equilibrium choices. This distinction is an example of weak domination does not exclude equilibrium strategies.

For a simple cooperative game, the Shapley value is the probability that a player is pivotal in a uniformly random ordering. Symmetry gives one value for permanent members and another for nonpermanent members.

A particular nonpermanent member is pivotal exactly when all five permanent members and exactly three of the other nine nonpermanent members precede them. The predecessor set then has size eight. There are $\left(\genfrac{}{}{0ex}{}{9}{3}\right)$ such sets, each giving $8!6!$ orderings. Their Shapley value is thereforeEvery ordering has exactly one pivotal member, since the empty coalition loses and the full coalition wins. This proves efficiency directly: $5\alpha +10\beta =1$, where $\alpha$ is the value of each permanent member. HenceThe vector has five entries $421/2145$ and ten entries $4/2145$. As a check, a permanent member is pivotal when they are last among the permanent members and occupy a position from nine to fifteen; counting those orderings gives the same $\alpha$.

Use the usual nonempty-clause convention and normalize repeated literals within a clause; tautologies are always satisfied. If empty clauses are admitted, discard them first: they contribute nothing to any assignment or to the optimum. Let $m$ be the resulting number of clause occurrences.

Assign independent fair truth values. A singleton clause is satisfied with probability $1/2$, a proper two-variable clause with probability $3/4$, and a tautology with probability one. By linearity of expectation, the expected number satisfied is at least $m/2$, hence at least $\mathrm{OPT}/2$.

To derandomize, use the method of conditional probabilities. After some variables have been fixed, let $F$ be the conditional expected number of satisfied clauses. For the next variable the two conditional expectations $F_0,F_1$ satisfy $F=(F_0+F_1)/2$. Fix the value with the larger expectation. This never decreases $F$. When all variables have been fixed, $F$ is the actual integer number of satisfied clauses, soCompute each conditional expectation by summing the probabilities of the clauses. Each has at most two variables, so its contribution is computed in constant time; scanning all clauses for each variable gives $O(nm)$ arithmetic operations. This is a polynomial-time approximation algorithm with approximation ratio $1/2$.

Apply the condition after the usual clause normalization, so each variable has at most one singleton clause. For a variable with such a clause, make its favored literal true with probability $p>1/2$; for other variables use a fair value. Make these choices independently. Then every singleton is satisfied with probability $p$.

Each literal of a proper two-variable clause is true with probability at least $1-p$. Independence bounds the probability that both are false by $p^2$, so the clause is satisfied with probability at least $1-p^2$. Tautologies have probability one. Thus the expected fraction satisfied is at least $\min (p,1-p^2)$. One term increases and the other decreases, so their intersection maximizes this bound:The method of conditional probabilities also works with these biased probabilities: before fixing a variable, the current expectation is the weighted average of its two conditional expectations, so choosing the larger cannot decrease it. Each clause contributes a constant-degree expression in $p$; since $p^2=1-p$, these expectations can be compared exactly in the fixed quadratic field $\mathbb{Q}(\sqrt{5})$ with polynomial bit complexity. The final deterministic assignment therefore hasThis is the golden ratio approximation for MAX-2SAT, under the stated restriction on normalized singleton clauses. Arbitrary conflicting singleton clauses do not admit the same independent-bias guarantee.

For a continuous local martingale $M$, its stochastic exponential isThe Itô formula gives $d{Z}_t=Z_{t}d{M}_t$ and $Z_0=1$, so $Z$ is a strictly positive local martingale. The finite-horizon Girsanov theorem says: if $Z$ is a true martingale on $[0,T]$ and $d\tilde{\mathbb{P}}=Z_{T}d\mathbb{P}$ on ${\mathcal{F}}_T$, then for every continuous local martingale $N$ under $\mathbb{P}$,For an infinite-horizon change of measure, assume $Z$ is a uniformly integrable martingale and use its terminal density $Z_{\infty }$. In particular, if $M=\int \theta dB$, the process $\tilde{B}=B-\int \theta ds$ is Brownian motion under the new probability measure, by the Lévy characterization of Brownian motion.

The quadratic variation cannot change. Adding the continuous finite-variation process $-[N,M]$ leaves $[\tilde{N}]=[N]$. Also the same pathwise partition sums defining $[N]$ have the same limit under an absolutely continuous change of measure; convergence in probability transfers by integrability of the density. Thus the bracket of the unchanged process $N$, now a semimartingale, is also unchanged. The local martingale property itself need not survive. This is quadratic covariation under an absolutely continuous measure change.

The union bound gives $g_k\le {\sum }_{x\in \partial {\Lambda }_k}{\mathbb{P}}_p(0\leftrightarrow x)$. Hence some boundary graph vertex has percolation two-point connection probability at least $g_k/(8k)$. The rotations and reflections of the square lattice let us choose such a graph vertex as $x=(k,j)$, $-k\le j\le k$. Reflection in the vertical line through $x$ sends $0$ to $e_{2k}=(2k,0)$ and fixes $x$. ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from $0$ to $e_{2k}$ reaches $\partial {\Lambda }_{2k}$, so $h_{2k}\le g_{2k}$. Taking roots givesBoth outside expressions have limit of a sequence $\gamma$, proving the even case. For $p>0$, the Harris-FKG inequality with the last horizontal edge gives $h_{2k+1}\ge p{h}_{2k}$, and also $h_{2k+1}\le g_{2k+1}$. The lower bound has rootThe upper bound has the same limit of a sequence. At $p=0$ all positive-distance connection probabilities vanish. Thus ${\lim }_{n\to \infty }{h}_n^{1/n}=\gamma$ for every $p\in [0,1]$.

Let $c_n$ count $n$-step self-avoiding walks on the square lattice from one fixed graph vertex. Splitting a self-avoiding walk after $m$ steps and dropping the avoidance constraint on its suffix gives $c_{m+n}\le c_m{c}_n$. The Fekete lemma therefore gives the connective constant $\mu =\lim c_n^{1/n}$. The planar dual graph of the square lattice is a translated copy of it, so its connective constant is also $\mu$.

Put $q=1-p$ and assume $q\mu <1$. Choose $a>\mu$ with $qa<1$. The definition of the connective constant provides a finite $K$ such that $c_j\le K{a}^j$ for every $j\ge 0$. A simple dual graph cycle of length $\ell$ surrounding $0$ has a graph vertex in a box of radius $\ell +1$: its diameter is at most $\ell$, and its coordinate ranges straddle the origin. Choose such a graph vertex as the starting point, orient the graph cycle, and omit its closing edge. The remaining self-avoiding walk has length $\ell -1$. Consequently the number $N_{\ell }$ of these graph cycles obeysfor fixed finite constants. A specified graph cycle is open in dual bond percolation, equivalently all its crossed primal edges are closed, with probability $q^{\ell }$. HenceTo make this tail argument valid for every $q\mu <1$, rather than only extremely small $q$, use a finite modification of Bernoulli percolation. Condition all primal edges within ${\Lambda }_N$ to be open. This event has positive probability for $p>0$. If the resulting percolation cluster of $0$ is finite, its outer boundary contains a simple closed dual graph cycle surrounding the whole box. Such a graph cycle has length tending to infinity with $N$ and crosses no forced-open edge. Under the conditioning its remaining edge states retain the original independent law, so its probability is still $q^{\ell }$. Choose $N$ so that the above tail is less than $1/2$. The conditional probability that $0$ belongs to an infinite percolation cluster is then at least $1/2$, and thus $\theta (p)>0$.

This connective-constant Peierls bound proves that every $p>1-{\mu }^{-1}$ is above or at the onset of positive percolation probability. Taking the infimum givesThe case $p=1$ is immediate. This proof uses the Peierls argument, without assuming the exact Harris-Kesten theorem.

The cubic lattice has graph vertices ${\mathbb{Z}}^d$ and an edge between $x,y$ when ${\sum }_i|x_i-y_i|=1$. In independent site percolation, every graph vertex is open with probability $p$ and closed with probability $1-p$, independently. The open subgraph contains only open graph vertices and the edges between them. Let $C_p(0)$ be the open connected component of a graph containing $0$, with $C_p(0)=\varnothing$ when $0$ is closed. The percolation probability isSince the cubic lattice is a locally finite graph, the König infinity lemma identifies an infinite open connected component of a graph with the existence of an infinite open graph ray from the origin. In particular, the origin itself must be open. The percolation critical probability is $p_c=\inf \{p:\theta (p)>0\}$.

Let $A_n$ be the event that the origin has an open graph path to the boundary of $[-n,n{]}^d\cap {\mathbb{Z}}^d$. The graph path can be stopped on its first boundary visit, so $A_n$ depends on only finitely many sites. Its probability $f_n(p)={\mathbb{P}}_p(A_n)$ is a finite sum of terms $p^a(1-p{)}^b$, and is therefore a continuous function of $p$. Also $A_{n+1}\subseteq A_n$, and the König infinity lemma gives ${\bigcap }_n{A}_n=\{0\leftrightarrow \infty \}$. By continuity from above of a measure,The monotone coupling of Bernoulli percolation described below shows that $\theta$ is nondecreasing. For $p<1$, its right limit of a function $\theta (p+)$ exists and is at least $\theta (p)$. For every fixed $n$,Taking the infimum in $n$ proves $\theta (p+)=\theta (p)$. This proves right continuity of percolation probability throughout $[0,1)$; at $1$ right continuity is understood relative to the parameter domain. The argument does not interchange two uncontrolled limits.