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Take the event to mean three smoking-free weeks in succession, $Y_{i1}=Y_{i2}=Y_{i3}=1$. The initial cumulative count is zero. Along this path it equals $0,1,2$ in weeks one, two, three, respectively. In the cumulative-response logistic model, defineThe chain rule for probabilities then givesThe three conditional probabilities are approximately $0.19891,0.27015,0.35559$. If “stop during the first three weeks” instead means at least one smoking-free week by week three, its different event has probability $1-(1-{\pi }_0{)}^3\simeq 0.48590$: along the all-failure path the cumulative count stays zero. Stating the event resolves this wording ambiguity.

Under $H_0$, the seven free covariate coefficients all equal zero. The likelihood-ratio test statistic isThe larger fit has ten free statistical parameters, compared with three in the constant-rate fit, so the reference chi-squared distribution has seven statistical degrees of freedom. Its 95th percentile is $14.06714$. Therefore reject $H_0$ at 5%; the p-value is approximately $0.0043$, and the covariate model is preferred. The two constrained sex coefficients add no free statistical parameters.

For the dementia-to-death arrow, the hazard ratio for higher versus lower education is $e^{-1.490}=0.2254$, with approximate 95% confidence interval $(e^{-2.547},e^{-0.4327})=(0.0783,0.6488)$. Thus higher education is associated with a roughly 77.5% lower fitted death transition intensity, conditional on diagnosis age; the interval excludes one. This is an observational association and is not a direct multiplicative statement about death probability.

Each additional year of age at diagnosis multiplies the dementia-to-death transition intensity by $e^{0.07984}=1.0831$, with confidence interval $(e^{-0.007765},e^{0.1674})=(0.9923,1.1822)$. The estimate suggests an increase, but the interval includes one, so this individual age coefficient is not significant at 5%. Sex has hazard ratio one for this transition by the model's imposed constraint. The output supplies no estimated sex effect or test of that constraint for this arrow. The nonzero sex coefficient for $1\to 2$ must not be mistaken for a sex effect on $2\to 3$.

The fitted progressive illness-death model has three allowed arrows: $1\to 2$ with estimated transition intensity $0.1849$, $1\to 3$ with $0.01935$, and $2\to 3$ with $0.06143$, all in years${}^{-1}$. State 3 is an absorbing state, and state 2 has no return arrow.

Once in state 2, the holding time has an exponential distribution with rate $q_{23}$, so the mean holding time from a transition intensity matrix isApply a confidence interval for an inverse rate to the printed rate interval $(0.03552,0.1063)$: since inversion reverses order, the approximate 95% interval for the mean isThe time-homogeneous Markov property makes the future depend only on the current state; the exponential distribution also has the memoryless property. For a person currently in state 2,Equivalently, $p_{22}(2)=p_{22}(1{)}^2\simeq 0.9404163^2$. Thus the fitted two-year death probability is approximately 11.6%.

For independent binomial distributions in the two trial arms, the estimate of the log risk ratio isThe delta method gives $\mathrm{Var}(\log \hat{p})\simeq (1-p)/(np)$; replacing $p$ by the sample proportion gives $1/d-1/n$, where $d$ is the event count. Adding the independent arm contributions therefore givesThe standard error is $\sqrt{16/15}=1.0328$. Using the stipulated normal quantile two, the approximate confidence interval isThe point estimate of the risk ratio is $5/24\simeq 0.208$, corresponding to approximately 79% lower mortality risk in the transfusion arm. Exponentiating the endpoints gives an approximate 95% confidence interval for the risk ratio of $(0.0264,1.6437)$. It includes one, so the trial is compatible with no difference as well as substantial benefit or some harm. The point estimate favours transfusion, but the data are too imprecise to establish a difference at the 5% level. These are large-sample approximations, especially rough with only one event in an arm. A frequentist confidence interval describes repeated-sampling coverage, not a 95% posterior probability for this fixed parameter.

One method is trim and fill. Estimate the direction and extent of funnel asymmetry, temporarily trim the extreme studies on the overrepresented side to estimate a centre, and fill in mirror-image studies on the underrepresented side. Recompute the pooled effect using the observed and imputed studies. This estimates what the summary might be under a symmetry-based missing-study model. It can move an exaggerated effect towards the null, but genuine heterogeneity can violate its symmetry assumptions.

A second method is a selection model for publication bias. Specify how the probability of a result being available depends on quantities such as its $P$ value, direction or precision, and combine that mechanism with a model for the underlying study effects. If a study estimate has density $f(y)$ and availability probability $\pi (y)$, its observed density is proportional to $f(y)\pi (y)$, with a normalizing factor accounting for unobserved results. Fit the model, or vary the selection probabilities over plausible scenarios, to obtain selection-adjusted effects. Such models make assumptions about evidence that is missing, so their results are particularly useful as sensitivity analyses. Neither method removes publication bias without assumptions about the missing studies.

Label batches before consulting the random-number table, so their labels are independent of the test results. A uniformly random permutation of $1,\dots ,16$, restricted to a smaller label set, induces a uniformly random ordering of that set. This provides a simple rejection-and-restriction scheme:

With independent random digits instead, rejection sampling gives the same uniform choices: for A keep only digits $1,\dots ,6$; for C keep only $1,\dots ,5$. For B, use uniform two-digit numbers, retain labels $01,\dots ,15$, and reject repeats within a three-batch sample. Taking residues modulo six or fifteen from a table whose range is not divisible by that number would give unequal probabilities. These are respectively day-stratified simple random sampling, morning/afternoon stratified sampling, and a one-day design using cluster sampling. The printed word “rest” for A is interpreted as “test”, consistently with the surveillance task. If it instead meant leaving one batch untested, choose that omitted batch uniformly and test the other five; this alternative would test 25 batches per week, with inclusion probability $5/6$, and has a different testing budget.

Under the one-test-per-day interpretation, A spreads surveillance across all five days and is operationally straightforward, but uses only five tests per week and does not guarantee morning/afternoon representation. Within each day its uniform choice avoids systematically selecting a particular batch. B uses six tests, guarantees representation of both production periods, and spreads the selected batches over the week's production, though some weekdays can receive no test. C also uses six tests and may minimize the cost of collecting samples, but concentrates them on one randomly selected day.

If batches produced on the same day share contamination risks, C's six observations are positively correlated and provide less information than six dispersed batches. A simple equal-cluster-size approximation has design effect $1+(6-1)\rho$, where $\rho$ is within-day intraclass correlation; its effective sample size is consequently smaller than six when $\rho >0$. C can also miss intermittent problems occurring on other days.

I would prefer B for estimating and comparing batch contamination rates, assuming positive within-day correlation and no overriding collection-cost advantage for C. It combines slightly more testing than A with explicit production-period coverage and avoids C's concentration on a single day. A is a reasonable alternative when regular daily surveillance is the priority. All three are legitimate probability samples with the stated inclusion probabilities; the preference concerns precision and coverage, not a claim that C is intrinsically biased. Their relative efficiency is not universal without a model for production variability and costs.

For equal independent samples, take $p_B=0.0002$, $p_C=0.0006$ and $\bar{p}=0.0004$ in the two-proportion formula, with two-sided $\alpha =0.05$ and power $1-\beta =0.80$. It givesThus the usual normal planning estimate isThe combined approximate design therefore checks about eight blocks, with about eight positive batches from B and twenty-four from C at the rounded design.

Those expected counts are low enough for test discreteness to matter. If a genuinely exact two-sided Fisher exact test is specified, the null allocation of the positive batches between two equal-size suppliers is hypergeometric, conditional on their total. Under the proposed alternative, its power is obtained by summing the independent binomial probabilities over the exact rejection region:An independent calculation of the exact power of Fisher's exact test gives about $0.7750$ at $n=40000$ and $0.8728$ at $n=50000$. Consequently five whole 10,000-batch blocks per supplier suffice for at least 80% power with this conservative exact test. Four blocks are the intended normal-approximation answer, not an exact 80% guarantee irrespective of the chosen test. The distinction is a property of rare-event discreteness, not a change in the proposed effect size.

The Wald statistic has law $Z_1={\hat{\delta }}_1\sqrt{I_1}\sim N(\delta \sqrt{I_1},1)$. Write $C=\{Z_1\ge f_1\}$ for continuation. By the normal cumulative distribution function,The final sample size per arm is $N=n_1+n_2^{\ast }{1}_C$, soIt lies between $n_1$ and $n_2=n_1+n_2^{\ast }$, increasing with the treatment effect: a more promising treatment is more likely to reach full enrollment. Total expected recruitment over both arms is twice this expression. Equality at the futility boundary has probability zero under the normal model.

Assume independent censoring: the censoring mechanism contributes no factor involving the lifetime rate $\theta$. The exponential density and survival function are $f_{\theta }(x)=\theta e^{-\theta x}$ and $S_{\theta }(x)=e^{-\theta x}$. An observed event contributes the density; a right-censored lifetime contributes the probability of surviving its censoring time. Thus the likelihood for $\theta$, up to censoring factors independent of it, isFor $D>0$ and $T>0$, ${\ell }^{\prime }(\theta )=D/\theta -T$ vanishes at $D/T$, and ${\ell }^{\prime \prime }(\theta )=-D/{\theta }^2<0$ proves the maximum:Censored individuals add follow-up time to the denominator but no event to the numerator. If $D=0$ and $T>0$, the likelihood decreases for $\theta >0$ and has only a supremum as $\theta \downarrow 0$; zero is an extended boundary estimate, not a positive-rate exponential MLE. The derivation is for independent individuals entering at time zero; delayed entry would require conditional survival contributions and exposure measured from entry.

A loss network models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\ge 0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, andUnder fixed routing, each arriving call has a predetermined resource requirement. Take independent Poisson processes of rates ${\nu }_r$ and independent holding times with exponential distribution of rates ${\mu }_r$. A feasible arrival changes $n$ to $n+e_r$ at rate ${\nu }_r$; a departure changes it to $n-e_r$ at rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.

The product-form stationary distribution of a loss network isFor any feasible adjacent pair,These detailed balance equations prove stationarity and make the process a reversible Markov chain. Equivalently, Independent random variables with Poisson distributions of means ${\rho }_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.

By Poisson arrivals see time averages, a route-$r$ arrival sees acceptance probability $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking probability is $1-G(C-A_r)/G(C)$ and the expected value of its number in service is ${\rho }_{r}G(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.

The insensitivity of loss networks extends this occupancy formula to independent general holding-time distributions with the same expected values, under the usual fixed resource requirements and admission rule. Counts alone then need not be a Markov chain; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.

In the slotted ALOHA model, a station retains one packet until it successfully transmits. Conditional on $(S_t,N_t)=(S,N)$, take fresh independent trials with a Bernoulli distribution of parameter $1/S$ for the $N$ backlogged stations. Let arrivals be independent across slots and independent of these trials, with Poisson distribution of mean $\nu$. New arrivals join the next slot's backlog. An idle slot serves nobody, exactly one attempt serves one packet, and a collision serves nobody. Consequently the backlog changes by arrivals minus the success indicator.

Under these independence assumptions, $(S_t,N_t)$ is a time-homogeneous Markov chain: the feedback probabilities and the independent arrival law depend only on its current state. Merely specifying Poisson arrival marginals would not suffice. For example, take $S_t=1$, $a=b=c=0$, $N_0=0$, and arrivals $Y_0=H$, $Y_1=K$, $Y_2=H$ for independent random variables $H,K$ with Poisson distributions. Histories with $(N_1,N_2)=(0,2)$ and $(2,2)$ both have current backlog two, but the next backlog is respectively two and four. The usual model therefore includes fresh arrivals as an assumption.

Write $p_0=(1-1/S{)}^N$ and $p_1=(N/S)(1-1/S{)}^{N-1}$. Set $p_1=0$ for $N=0$, and use their exact binomial distribution probabilities at $S=1$. Away from the clipping boundary, the conditional drifts areNear $S=1$, the latter must use $\max (1,S+d)-S$ for each update $d$, rather than $d$ itself.

When both coordinates are large with $N/S\to \kappa$, the Poisson limit theorem approximates the attempt count, which has a binomial distribution, by a Poisson distribution of mean $\kappa$. Thus $p_0\to e^{-\kappa }$ and $p_1\to \kappa e^{-\kappa }$. Rescaling state by $K$ and slot time by $K$ motivates the fluid approximation of slotted ALOHAThis is an interior fluid approximation of slotted ALOHA, not the exact conditional drift at a clipped boundary.

Here is one explicit set of sufficient conditions, independent of the arrival rate within the subcritical range:Put $c_0=2/(e-2)$. ThenHence $g$ is negative below one and positive above one: the controller decreases the attempt denominator when offered contention is too small, and increases it when contention is too large.

Use the ratio time change for a homogeneous fluid model, $du/dt=1/s$. The equations becomeFor $\kappa \ge 1$,since $g(\kappa )\ge 0$ and $A(1+c_0)>1$. Also $h(1)=\nu -e^{-1}<0$. Therefore $h$ is negative on $[1,\infty )$, and by continuity on $[\bar{\kappa },\infty )$ for some $\bar{\kappa }<1$. Its value at zero is $\nu \ge 0$, so nonnegative backlog is preserved.

The ratio remains bounded by $K=\max (\kappa (0),1)$ and eventually enters $[0,\bar{\kappa }]$: on the compact interval $[\bar{\kappa },K]$, its derivative is bounded above by a strictly negative number. Once inside it cannot cross upward. The bounded ratio and smooth coefficients make the transformed equations exist for all $u\ge 0$. In this region $g(\kappa )\le g(\bar{\kappa })<0$, so $s(u)$ decreases at least exponentially in $u$. The original time satisfies $t(u)={\int }_0^{u}s(v)dv$ and has a finite limit $T$; boundedness of $\kappa$ gives $n(u)=\kappa (u)s(u)\to 0$ as well. Thus every nonnegative interior fluid trajectory drains to the origin in finite fluid time. At the origin the ratio equation is undefined; the usual stopped fluid trajectory is held there afterward. A state with $s=0<n$ enters the interior under the continuous limiting boundary drift $\dot{s}=c>0$; the same argument then applies.

This proof supplies sufficient conditions, not a characterization of every stabilizing triplet. It also does not assert the positive recurrent Markov chain property of the original stochastic chain solely from the heuristic ODE.

If $\nu >e^{-1}$, the ALOHA throughput bound givesNo choice of the three feedback increments can make this fluid backlog drain, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$. For the displayed controller, the supercritical fluid trajectory is global and grows instead of draining. At $\nu =e^{-1}$ the ray $n=s>0$ consists of stationary fluid states, explaining why the strict load inequality matters.

Give each nearest-neighbor edge of the cubic lattice an independent Bernoulli distribution state, open with probability $p$. The resulting product measure is denoted ${\mathbb{P}}_p$. In this bond percolation model, $C(0)$ is the percolation cluster of the origin, andThe uniform-label monotone coupling of Bernoulli percolation shows that ${\theta }_d$ is increasing.

Let $c_n(d)$ count $n$-step self-avoiding walks starting at the origin, with $c_0(d)=1$. Splitting a walk after $n$ steps, and discarding the avoidance constraint between the two pieces, gives $c_{n+m}(d)\le c_n(d)c_m(d)$. The Fekete lemma therefore gives the connective constantThe locally finite graph structure means that an infinite percolation cluster at the origin supplies an open self-avoiding walk of every length. Each specified walk has $n$ distinct edges and is open with probability $p^n$. The union bound givesFor $p\mu (d)<1$ the right side tends to zero. Hence the connective-constant lower bound for percolation is $p_c(d)\ge \mu (d{)}^{-1}$.

For the upper bound first work on the square lattice. A finite open percolation cluster has an outer boundary containing a simple closed graph cycle of dual edges, all crossing closed primal edges. Such a dual bond percolation circuit separates that cluster from infinity. Write $N_{\ell }$ for the number of simple dual circuits of length $\ell$ surrounding the origin. Each circuit meets the positive horizontal ray at distance at most $\ell$: its bounding box contains the origin and its diameter is bounded by its length. Choose such an intersection as an anchor and orient the circuit. Removing its last edge leaves a rooted self-avoiding walk of length $\ell -1$ in the translated square lattice. Consequently, for an absolute constant $K$,The exact constant and this possible overcount do not matter. If $(1-p)\mu (2)<1$, the root test givesA summable circuit count alone need not give a total sum below one. To use its tail correctly, let $B_R=[-R,R{]}^2\cap {\mathbb{Z}}^2$ and condition every edge internal to $B_R$ to be open. This finite event has positive probability. A closed dual circuit surrounding all of $B_R$ crosses no internal edge of $B_R$, and so its closed-edge probability remains $(1-p{)}^{\ell }$ under this conditioning. Its length tends to infinity with $R$. Choose $R$ so large that the union bound for all such circuits is below one. With positive conditional probability, none occurs.

On that event, $C(0)$ contains $B_R$ and cannot be finite: a finite cluster containing $B_R$ would have an enclosing closed dual circuit. Thus ${\theta }_2(p)>0$ whenever $p>1-\mu (2{)}^{-1}$. This is the connective-constant Peierls bound, proved by excluding short circuits through the open-box conditioning. An embedded coordinate plane in the cubic lattice has exactly the same bond percolation law as the square lattice, so $p_c(d)\le p_c(2)$. Together,There are $2d$ choices for the first step of a self-avoiding walk and at most $2d-1$ thereafter, because immediate reversal is forbidden. Hence $c_n(d)\le 2d(2d-1{)}^{n-1}$ and $\mu (d)\le 2d-1$. In particular $\mu (2)\le 3$. Substituting with the correct directions of the inequalities gives

The purpose of the RSW lemma is to turn local crossing information into control at every fixed shape and scale. Work with independent bond percolation on the square lattice. Let $H(a,b)$ be the event of an open left-to-right graph path in $[0,a]\times [0,b]\cap {\mathbb{Z}}^2$, and write $h_p(a,b)={\mathbb{P}}_p(H(a,b))$.

A useful precise uniform version of the RSW lemma is the following uniform RSW crossing estimate. If for a fixed parameter $p$ and some $\delta >0$,then for every fixed aspect ratio $\rho >1$ there is $c(\delta ,\rho )>0$, independent of $n$, such thatIn particular this bounds length $2n$ in terms of the uniform square bound at length $n$. The constants need not be sharp. The hypotheses used in the proof are planarity, translation and reflection symmetries, the Harris-FKG inequality and independence of unexplored edges. At a self-dual parameter the same argument for dual crossings gives an upper bound strictly below one as well. Neither an exact value of $p_c$ nor the existence of an infinite cluster is an assumption of this estimate.

Here is the gluing picture behind the RSW lemma. For increasing crossing or attachment events, the Harris-FKG inequality provides a lower bound on their joint probability. If an event of probability at least $\delta$ is a union of two reflection-related increasing alternatives $A_1,A_2$, the square-root trick for positively associated events givesIndeed their decreasing complements are positively associated too, so ${\mathbb{P}}_p(A_1^c\cap A_2^c)\ge {\mathbb{P}}_p(A_1^c{)}^2$. This prevents a square crossing from concentrating all its useful attachment locations on just one side.

The nontrivial first gluing step enlarges a square to aspect ratio $3/2$. Explore an extremal square crossing, revealing the edges on the explored side but leaving its other side unexamined. In that unexplored region the conditional law is still independent bond percolation. Compare possible attachments to the crossing with their reflected alternatives. Reflection symmetry and the preceding square-root estimate give a positive bound, depending only on $\delta$, for the required attachment after averaging over the explored crossing. Carry out the reflected construction at the other end and use the Harris-FKG inequality to combine the increasing attachment events with the original crossing. Planarity ensures that the relevant transverse paths actually meet. This RSW reflection extension lemma yields a lower bound $u(\delta )>0$ for the $3n/2$-by-$n$ crossing. The exploration is important: a reflection compares the laws of fresh configurations; reflecting the picture of an open path does not make the reflected edges open.

From that first extension, longer rectangles are obtained by a genuinely transverse gluing. Two $3n/2$-by-$n$ rectangles shifted by $n/2$ overlap in an $n$-by-$n$ square. Require a horizontal crossing in each long rectangle and a vertical crossing of the overlap square. Each horizontal crossing crosses that overlap from left to right, and therefore meets its vertical crossing. Their union crosses the $2n$-by-$n$ rectangle. The Harris-FKG inequality gives a lower bound $u(\delta {)}^2\delta$. Repeat a bounded number of times for any fixed $\rho$. Integer rounding uses neighboring lattice rectangles and bounded additional gluing steps; finitely many smallest scales can be absorbed into the constant. The constants deteriorate with $\rho$ but not with $n$. Merely requiring horizontal crossings in adjacent squares would not suffice, since their endpoints need not coincide; the overlap crossing solves that problem.

I apply the RSW lemma to the exact threshold for bond percolation on the square lattice. The planar duality for rectangle crossings says that an open horizontal crossing and a closed dual vertical crossing are complementary. At $p=1/2$, the dual edge states have the same law as the primal ones. For the balanced lattice rectangle with side lengths $n+1,n$, the rotated dual crossing rectangle has those same side lengths: its transverse lengths before rotation are $n,n+1$. Thus the exact self-dual rectangle crossing probability is $1/2$. Restricting a crossing of this rectangle to its first visit to the shorter vertical side givesThe one-unit balance avoids assuming an exact one-half probability for every finite vertex-square convention. The RSW lemma now gives scale-independent positive bounds for all fixed-aspect-ratio primal and closed dual rectangle crossings.

Arrange four appropriately overlapping long rectangles around a square ring. Closed dual crossings along the four sides, with transverse overlap crossings if needed, join to a closed dual graph cycle surrounding the inner square. The Harris-FKG inequality, applied to the closed dual states, and the RSW lemma give a constant $b>0$ for this circuit event, uniformly over ring size. Choose disjoint rings with radii increasing, for example, by a factor of four. Their circuit events depend on disjoint edge sets, so they are independent. An open graph path from the origin to infinity would have to avoid every one of these dual barriers. Its probability is at most $(1-b{)}^j$ after $j$ rings and therefore zero. HenceThis is the independent annular barriers for percolation argument. It is worth separating it from the other inequality: the absence of an infinite cluster at one parameter alone does not prove that every larger parameter percolates.

For the reverse inequality use sharpness of the percolation transition: below $p_c$, independent bond percolation on the cubic lattice has exponentially decaying connection probabilities. One can see why this is the relevant general ingredient through the finite-set criterion for percolation sharpness. For a finite set $S$ containing the origin putIf ${\phi }_p(S)<1$, split a long open self-avoiding walk at its first exit from $S$. Its internal connection, exit edge and subsequent connection have disjoint witnesses. The BK inequality gives a contraction by ${\phi }_p(S)$ each time distance decreases by the diameter of $S$ plus a fixed step. Iteration proves exponential decay of subcritical percolation at such $p$.

To identify this finite-set threshold with $p_c$, the Margulis–Russo formula expresses the derivative of $g_n(p)$ as the sum of pivotal-edge probabilities. Explore the cluster attached to the box boundary, and let $S$ be its complement. On failure of the origin-to-boundary event, $0\in S$; all edges from $S$ to the boundary cluster are closed, while the internal edges of $S$ remain fresh. A boundary edge is pivotal precisely when its endpoint in $S$ is connected to the origin inside $S$. Removing the factor $1-p$ for a closed pivotal edge yields the general differential inequalityFor $p>{\tilde{p}}_c$ every finite-set quantity on the right is at least one. Integrating from any $p_0\in ({\tilde{p}}_c,p)$ gives a positive lower bound for $g_n(p)$ independent of $n$, and taking $n\to \infty$ gives $\theta (p)>0$. Combined with the contraction below ${\tilde{p}}_c$, this proves $p_c={\tilde{p}}_c$ and the stated sharpness conclusion. This outline supplies the extra threshold argument rather than assuming the desired critical value.

If $p_c(2)>1/2$, sharpness would give $g_n(1/2)\le C{e}^{-cn}$. But a square crossing starting somewhere on the left side entails an open connection from that starting graph vertex to distance $n$. A union bound over the $n+1$ possible starting graph vertices givescontradicting exponential decay. Therefore $p_c(2)\le 1/2$, and the two directions establishThis is the Harris-Kesten theorem. The overall mechanism is that self-duality supplies a square crossing, the RSW lemma transports it between shapes and creates barriers at every scale, and sharpness converts the finite-scale crossing information into the exact threshold.

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.

For independent identically distributed observations $Y_1,\dots ,Y_n$, the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice ${\hat{\theta }}_n\in {\mathrm{argmax}}_{\theta \in \Theta }{L}_n(\theta )$, when a maximizer exists. It need not be unique. The one-observation score function is $s_{\theta }(y)={\nabla }_{\theta }\log f(y,\theta )$, and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives ${\mathbb{E}}_{\theta }{s}_{\theta }={\nabla }_{\theta }\int f=0$. Differentiating again gives the second information identity. Independent observations have total Fisher information $nI(\theta )$.

Assume fixed parameter dimension, an interior true parameter ${\theta }_0$, a positive-definite matrix $I({\theta }_0)$, the standard differentiability and integrability conditions, and statistical consistency of ${\hat{\theta }}_n$. Then the asymptotic normality of a maximum likelihood estimator isHere $I$ is the information per observation. Thus the leading covariance matrix of the estimator itself is $I({\theta }_0{)}^{-1}/n$.

To prove this, statistical consistency places the estimator in an interior ball about ${\theta }_0$ with probability tending to one, where its score function vanishes. Write $d_n={\hat{\theta }}_n-{\theta }_0$ and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give $J_n\stackrel{P}{\to }I({\theta }_0)$. The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so $J_n^{-1}\stackrel{P}{\to }I({\theta }_0{)}^{-1}$. Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since $I^{-1}I{I}^{-1}=I^{-1}$. Boundary parameters or singular information are outside this regular theorem.

The ordinary column Gram matrix is $G=X^{T}X$. Use the normalized empirical Gram matrixso that standard Gaussian entries give $\mathbb{E}\hat{\Sigma }=I_p$. This is the normalization needed for concentration around $\Vert \theta {\Vert }_2^2$.

The restricted isometry property of order $s$ with constant $0\le \delta <1$ means that the normalized map $X/\sqrt{n}$ approximately preserves the Euclidean norm of all vectors with at most $s$ nonzero coordinates:Equivalently, every principal block ${\hat{\Sigma }}_{MM}$ with $|M|\le s$ satisfies $\Vert {\hat{\Sigma }}_{MM}-I{\Vert }_{\mathrm{op}}\le \delta$. The least such $\delta$ is its restricted isometry constant. In the unnormalized definition apply this property to $X/\sqrt{n}$ itself.

For a standard normal variable $g$, direct Gaussian integration gives $\mathbb{E}{e}^{\lambda g^2}=(1-2\lambda {)}^{-1/2}$ for $\lambda <1/2$. Independence therefore yields the moment-generating function of a chi-squared distribution, centred here at its mean:The inequality follows from $-\log (1-u)-u={\sum }_{j\ge 2}{u}^j/j\le u^2/(2(1-u))$. For $t>0$, the Chernoff bound with $\lambda =t/(2(n+t))$ givesAt $t=0$ the trivial probability bound suffices. This proves the requested bound, with a stronger prefactor one.

For the lower tail, $\log (1+u)\ge u-u^2/2$ gives $\log \mathbb{E}{e}^{-\lambda Z}\le n{\lambda }^2$ for $\lambda \ge 0$. Taking $\lambda =t/(2n)$ yields $\mathrm{Pr}(Z\le -t)\le e^{-t^2/(4n)}$. Combining both tails gives the useful chi-squared concentration inequalityFor $z\ge 0$ set $t=4(\sqrt{nz}+z)$. A direct calculation showsThus the exponent is at least $z$, provingThe same threshold bounds the two-sided tail.

Finally fix a deterministic $\theta$ and put $r=\Vert \theta {\Vert }_2$. If $r>0$, independence of the Gaussian rows gives $X_i\theta /r\sim N(0,1)$ independently. ThereforeThe two-sided bound just proved suppliesFor $r=0$ the quadratic form is deterministically zero; the strict inequality makes the formula valid in that case too. When $r\le 1$, replacing the threshold by $4(\sqrt{z/n}+z/n)$ gives an absolute bound independent of $\theta$. In particular every fixed such direction concentrates at rate $n^{-1/2}$ for a fixed confidence level. The restricted isometry property requires a simultaneous statement over sparse directions; this fixed-direction calculation alone is not that stronger assertion.

Let ${\Delta }_n={\sup }_{\theta \in \Theta }|Q_n(\theta )-Q(\theta )|$. For any $\epsilon >0$, the setis compact. If it is empty there is nothing to prove. Otherwise continuity and the unique minimum give a strictly positive separation gapTake any measurable attained minimizer of $Q_n$. Its defining inequality givesConsequentlyThis proves argmin consistency under uniform convergence in probability. No continuity of $Q_n$ is needed once the minimizer exists; compactness and continuity concern the deterministic separation gap. The uniform convergence in probability assumption controls all candidate parameters, including the random minimizer.

For the estimating equation, fix $\epsilon >0$ and put $a={\theta }_0-\epsilon$, $b={\theta }_0+\epsilon$. The prescribed signs make $\eta =\frac{1}{2}\min \{-S(a),S(b)\}>0$. Pointwise convergence in probability at just these two points impliesand similarly $\mathrm{Pr}(S_n(b)\le 0)\to 0$. With probability tending to one, $S_n(a)<0<S_n(b)$. The intermediate value theorem then gives a zero inside $(a,b)$, and uniqueness identifies it with ${\hat{\theta }}_n$. HenceThis is consistency of a uniquely bracketed zero. It needs no monotonicity, no continuity of the limit $S$, and no uniform convergence of the $S_n$. The bracket interval must lie in the domain of $S_n$: read literally, the printed sign condition for every positive $\epsilon$ puts every real point into $\Theta$, so this requirement is satisfied. More generally it is enough to have an interval about ${\theta }_0$ and sign brackets arbitrarily close to it.