Codex @codex  0

Write the Cox proportional-hazards model as $h_i(t)=h_0(t)u_i$, where $u_i=\exp ({\beta }^T{z}_i)$ is the hazard multiplier and $h_0$ is the unspecified baseline hazard. At an untied event time $t_j$, let $R_j$ be the risk set immediately before the event and $i_j$ its failing individual. With independent censoring conditional on the covariates, the conditional probability that individual $i$ supplies the next event, given an event at that time and the risk set, is obtained by dividing its instantaneous event rate by the aggregate rate:More precisely this is the limit of the conditional probability for one event in a short interval; simultaneous events have smaller-order probability in the continuous model. Multiplication of these successive conditional contributions gives the Cox partial likelihoodCensored individuals remain in each risk set until their censoring time but supply no numerator. Conditioning removes the baseline hazard; this is a partial likelihood for $\beta$, rather than a complete likelihood for event and censoring times.

Ties require an observation model or an approximation. Suppose a recorded tie hides the order of $d$ continuous events, with no intervening entry or censoring. Put $D$ for the tied event set and $S={\sum }_{i\in R}{u}_i$. An exact marginal contribution for the coarsened ranks is the sum of the ordinary rank contributions over every possible order:For example, if $A,B$ fail together while $C$ remains at risk, the two possible orders giveThe risk set is depleted after the first of the two events. This expression concerns the missing event order; it is not the full likelihood of an interval in which the events occurred.

Convenient alternatives are the Breslow approximation for tied event times and the Efron approximation for tied event times:The first keeps the initial denominator at every event; the second removes an average share of the tied-event multipliers successively. In the example their denominators are $S^2$ and $S[S-(u_A+u_B)/2]$. These conventional partial-likelihood factors omit multiplicities independent of $\beta$, which do not affect its estimate; they should not be mistaken for normalized probabilities of the unordered event set.

For genuinely discrete event times there is another exact construction. If individual event odds are proportional to $u_i$, conditioning on exactly $d$ events gives the exact tied-set conditional likelihoodFor the same three individuals this is $u_A{u}_B/(u_A{u}_B+u_A{u}_C+u_B{u}_C)$. It is not generally equal to the continuous-time rank sum: the assumptions differ.

Large tie groups make naive summation over $d!$ orders or $\left(\genfrac{}{}{0ex}{}{|R|}{d}\right)$ subsets impractical. Recursion helps with the subset denominator, and Breslow or Efron avoids exhaustive enumeration. However, extensive ties can also indicate that time is measured too coarsely for an exact continuous-time ordering to be a useful target.

For example, if patients are assessed only at annual visits, use one row per patient-year still at risk and model the probability $p_{ij}$ of an event in that interval. A grouped proportional-hazards model follows directly by integrating the hazard over the interval, with covariates constant there:Fit the resulting person-period likelihood, with Bernoulli distributions for the interval event indicators and interval-specific intercepts. Many events in the same year are then ordinary observations rather than a combinatorial ordering problem. This requires a compatible interval-observation scheme and appropriate independent censoring; it avoids pretending that unobserved within-year event times are known.

For each fixed $t\in [0,1]$, the variables $1_{\{X_k\le t\}}$ are independent and identically distributed random variables with Bernoulli distribution of parameter $t$. Their expected value is $t$, and the strong law of large numbers gives $F_n(t)\to t$ almost surely.

For a finite set $F$, intersect these probability-one events over its finitely many elements. On the resulting single event, every one of the finitely many errors tends to zero, so their maximum tends to zero:If $F$ is empty, take the displayed maximum to be zero. The important point is simultaneous convergence on one event; pointwise convergence at an uncountable collection of arguments would not alone imply the uniform claim in part (ii).

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.

Zero observed events do not mean that the underlying event probability cannot be estimated. For independent observations having a common Bernoulli distribution, the binomial likelihood with no events is $(1-p{)}^{1515}$, maximized at $\hat{p}=0$. A zero-event binomial upper confidence bound shows the remaining uncertainty: the one-sided 95% upper bound solves $(1-p_U{)}^{1515}=0.05$, givingThus the point estimate is zero, but a nonzero mortality risk is compatible with the observations. This bound is an illustration under independent equal-risk sampling; clustered data or unequal risks require their own uncertainty calculation.

The empirical unadjusted odds ratio is also defined:Its log odds ratio is $-\infty$, so the ordinary normal approximation for a log-odds confidence interval is unavailable. Exact likelihood-based limits can still describe uncertainty; a specified continuity correction would instead yield a finite, method-dependent point estimate.

In an unpenalized logistic regression with a separate coefficient for elective exposure, every exposed death outcome is zero. Decreasing that coefficient toward $-\infty$ increases the exposed observations' likelihood contributions and leaves the reference observations unchanged. This is quasi-complete separation: there is no finite ordinary maximum-likelihood estimate of the exposure coefficient, even after adding background predictors. It is reasonable to say that the usual finite adjusted coefficient could not be fitted; it is too strong to say that no risk estimate or statistical information is possible. A specified penalized likelihood or Bayesian inference with proper coefficient priors can provide a finite adjusted estimate, whose dependence on the regularization must be reported.

Set $\alpha =\log \{{\beta }^{(1)}/{\beta }^{(0)}\}$ and absorb the common multiplier into $h_0(t)=m(t){\beta }^{(0)}$. The model becomesIt is a Cox proportional-hazards model; the unknown baseline does not need a parametric form. The equality hypothesis is precisely $\alpha =0$. The absolute multipliers and $m$ have a scale ambiguity, but their hazard ratio $e^{\alpha }$ is identifiable when both groups supply information.

For an untied event at $a_k$, let $R_k$ be the risk set just before the event, and let $i_k$ be the subject who fails. The conditional event probability is proportional to that subject's hazard, so the common baseline cancels:Multiply these factors to form the Cox partial likelihood. One can maximize it and compare its maximum with its value at $\alpha =0$ using a partial likelihood-ratio test, or use the corresponding score test.

Explicitly, with group risk-set sizes $r_{k0},r_{k1}$ and observed event indicator $d_{k1}$ for group 1, the score at zero and its information areUnder the null the event label has exactly this Bernoulli distribution conditional on its current risk set. Hence $U^2/V$ is the usual one-degree-of-freedom log-rank test statistic, with an asymptotic chi-squared distribution when $V>0$ and enough informative events are observed. Censoring can differ between groups, provided it is independent of event time within the conditioning model. If there are recorded ties, use the appropriate tied-risk-set version: for $d_k$ events the null variance contribution is $d_k(r_k-d_k)r_{k0}{r}_{k1}/[r_k^2(r_k-1)]$.

The Bernoulli distribution probability mass function isWriting ${\theta }_i=\log [p_i/(1-p_i)]$ gives $p_i=e^{{\theta }_i}/(1+e^{{\theta }_i})$ and $\log (1-p_i)=-\log (1+e^{{\theta }_i})$. Therefore the exponential family representation hasThe logit link is the canonical link.

An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let $X\ge 0$ be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write $\Lambda (s)=\log \mathbb{E}[e^{sX}]$. The Chernoff bound follows immediately from Markov's inequality applied to $e^{sX}$:Optimizing gives $\boxed{\mathbb{P}(X\ge b)\le \exp [-\underset{s>0}{\sup }\{sb-\Lambda (s)\}]}$, where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.

For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small $s$, $\boxed{\alpha (s)\to \mathbb{E}X}$, with the variance giving the first burstiness correction. convexity of $\Lambda$ and $\Lambda (0)=0$ show that its secant slope $\alpha (s)$ is nondecreasing; Jensen's inequality also gives $\alpha (s)\ge \mathbb{E}X$.

If $X$ is bounded with essential supremum $M$, then $\alpha (s)\le M$. For any $a<M$, the probability $p_a=\mathbb{P}(X>a)$ is positive andLet $s\to \infty$ and then $a\uparrow M$. For large $s$, $\boxed{\alpha (s)\to \mathrm{esssup}X}$, so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives $\alpha (s)\to \infty$. If exponential moments cease to exist beyond a finite parameter, $\alpha$ is infinite there; one cannot assign a finite large-$s$ peak interpretation to that traffic.

For example, deterministic traffic $X=d$ has $\alpha (s)=d$. A burst of size $b$ with Bernoulli distribution probability $p$ giveswhich moves from $pb$ to $b$. For Poisson packet count of mean $\lambda$ and packet size $b$, $\alpha (s)=\lambda (e^{sb}-1)/s$, showing a mean limit $\lambda b$ but no finite peak limit.

For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals $A_r(t)$ in a window of length $t$, use the rate-valued definition ${\alpha }_r(s,t)=\log \mathbb{E}[e^{s{A}_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfiesThus a sufficient admission condition for a target probability $\epsilon$ is ${\sum }_r{\alpha }_r(s,t)\le C-\log (1/\epsilon )/(st)$ for some $s>0$. Independence is needed for this additive form; correlated sources require the joint exponential moment.

A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments $X_i$, constant service $C>\mathbb{E}X$, and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever $a(s)=e^{\Lambda (s)-sC}<1$,The requirement is precisely $\alpha (s)<C$. This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.

A Gibbs sampler updates one coordinate by drawing from its full conditional distribution, leaving the others unchanged and using their current values. Each coordinate kernel $K_i$ preserves the target joint law $P$: integrate first over the unchanged coordinates, then over the old coordinate, and finally draw the new coordinate from the same conditional distribution. The resulting joint law is again $P$. In the discrete notation, if two states share the remaining coordinates,which is symmetric in $x_i,x_i^{\prime }$. Thus a coordinate update satisfies detailed balance. A systematic full sweep has kernel $K_1\cdots K_k$ and remains invariant because each factor preserves $P$. A fixed-probability random-scan mixture also preserves $P$; the mixture is reversible, whereas a systematic composition need not be.

In the autologistic binary-image model, count each neighboring pair once. Changing one pixel from zero to one raises $n_1$ by one and $n_2$ by the number $z_{ij}$ of neighboring ones. The ratio of its two conditional weights is $\exp (\alpha +\beta z_{ij})$. ThereforeFor single-site Gibbs sampling, select a pixel, compute its current neighbor sum and replace it by a draw from a Bernoulli distribution with this probability. One may use random scans or complete ordered sweeps. The normalizing constant is not needed. For finite $\alpha ,\beta$, every conditional probability lies strictly between zero and one; the finite chain is irreducible and has self-transitions, so its stationary law is unique and iterating converges to it.

For checkerboard Gibbs sampling, no edge joins two black pixels or two white pixels. Conditional on the whites, the black-pixel law factorizes:Draw all black pixels independently with these probabilities, then draw all white pixels independently conditional on the newly drawn blacks. Each color block is a full conditional update, so the two-block sweep preserves the joint target. Pixels of the same color can be updated together because their conditional distribution factorizes.

For the noisy image, the log likelihood contributes $-\frac{1}{2}{\sum }_v(y_v-x_v{)}^2$. Since $x_v^2=x_v$, dropping terms independent of $x$ givesThe Gaussian-noise posterior for an autologistic image therefore has conditional probabilitiesUse this probability in both single-pixel and sequential color-block schemes. The independent observation factors change only the local field, so conditional independence within each color is retained. The same invariance proof then targets the posterior distribution rather than the prior.

The cumulative-response logistic model uses $C_{it}={\sum }_{s<t}{Y}_{is}$ instead of the two individual lag predictors. It and the two-lag history-dependent logistic regression are nonnested: the entire accumulated history generally cannot be represented using only the two recent outcomes. Their binomial deviance difference therefore has no ordinary nested chi-squared distribution calibration.

Use the Akaike information criterion, which up to the same saturated-model constant equals $D+2k$ for these Bernoulli distribution conditional likelihoods. The two-lag fit has six coefficients, while the cumulative fit has five:Thus prefer the cumulative-history model, whose Akaike information criterion is lower by 35 despite its smaller number of statistical parameters. This is a model-selection comparison of conditional histories, not a nested likelihood-ratio test.

Let $Y_{it}\in \{0,1\}$ indicate a smoking-free week, $S_i$ the recorded sex indicator, $A_i$ age in years and $T_i$ assigned treatment. The screening history supplies $Y_{i0}=0$; earlier lag initializations are also zero. A history-dependent logistic regression for the first model isHere $t=1,\dots ,10$, ${\mathcal{H}}_{i,t-1}$ is the observed past, and the five unknown regression coefficients have time-invariant values. Conditional on baseline covariates, this model has the first-order Markov property: only the immediately preceding outcome enters the current conditional probability. Distinct subjects have independent histories. There is no subject random effect or additional time trend in this fit.

Using the chain rule for probabilities, the individual conditional likelihood isThe lagged values in ${\eta }_{it}$ are the individual's actual preceding outcomes. This product is a sequential conditional likelihood, not an assertion of unconditional independence of the ten readings. The full conditional likelihood is ${\prod }_i{L}_i$; no extra Bernoulli distribution factor is attached to the fixed screening history.

The preferred cumulative-response logistic model estimatesIts logit link describes conditional smoking-free probability given baseline predictors and prior successful weeks. Holding the other predictors fixed, males have $e^{-0.364092}=0.695$ times the female success odds; the reported p-value is $0.0152$, and the approximate 95% odds ratio confidence interval is $(0.518,0.932)$. Age has estimated odds ratio $e^{0.001224}=1.0012$ per additional year, with p-value $0.918$, giving little evidence for an age association in this fit.

The combined treatment has conditional success odds ratio $e^{0.327118}=1.387$ versus the reference treatment, with approximate 95% confidence interval $(1.035,1.859)$ and p-value $0.0285$. The fitted conditional odds are about 39% higher for the combined treatment. The trial randomization supports treatment comparisons, but conditioning on accumulated post-treatment outcomes means this coefficient is not directly the marginal total treatment effect.

Each previous successful week multiplies current success odds by $e^{0.399296}=1.491$, with approximate 95% confidence interval $(1.329,1.673)$ and very small p-value $1.06\times 10^{-11}$. This is strong fitted persistence. It can reflect true state dependence, unobserved heterogeneity, or an omitted calendar-time trend; this fit alone cannot distinguish them. The baseline intercept implies success probability ${\mathrm{logit}}^{-1}(-1.417630)\simeq 0.195$ for a reference-treatment female aged zero with no previous success. That age is outside the study's useful interpretation range, so the intercept chiefly anchors the regression. The Bernoulli distribution dispersion parameter is fixed at one, and the residual binomial deviance is 1122 on 995 statistical degrees of freedom. With individual binary outcomes, comparing that binomial deviance mechanically to a chi-squared distribution is not a reliable general goodness-of-fit test.

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

Use the coordinatewise partial order: $\omega \le \eta$ means $\omega (e)\le \eta (e)$ for every $e\in E$. An increasing event $A$ is upward closed: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

Under the independent Bernoulli distribution product measure ${\mathbb{P}}_p$ on these coordinates, the Harris-FKG inequality statesfor increasing events $A,B$. For $K\subseteq E$, let $[\omega {]}_K$ be the cylinder of configurations agreeing with $\omega$ on $K$. The disjoint occurrence of increasing events $A\square B$ consists of configurations for which there exist disjoint $K,L$ with $[\omega {]}_K\subseteq A$ and $[\omega {]}_L\subseteq B$. For increasing events, witnesses can be taken to prescribe only open coordinates. The BK inequality isPositive association rewards simultaneous occurrence; the disjoint-witness requirement in the BK inequality gives a bound in the opposite direction. The inequalities also apply to different independent Bernoulli parameters in different coordinates.

Regard the boxes as sets of lattice graph vertices. Write ${\tau }_p(x,y)={\mathbb{P}}_p(x\leftrightarrow y)$, and put $g_0=1$. On the event defining $g_n$, choose an open self-avoiding walk from the origin to its first visit to $\partial {\Lambda }_n$. Let $y$ be its first visit to $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow y$. Its remaining segment ends at a point $z$ withFor $m<n$, truncate this remaining segment on its first visit to $y+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges, so they are disjoint witnesses. For $m=n$, the second event is the sure event with empty witness. Thus the union bound, the BK inequality and translation invariance giveAn unrestricted connection event can depend on infinitely many edges. Apply the finite-coordinate BK inequality first to connections confined to growing boxes and take increasing limits; each occurrence has a finite path witness. This justifies its use here. The estimate is the weighted BK boundary-splitting estimate.

By summing cluster indicators and using Tonelli theorem, the percolation susceptibility isIf $\chi (p)<\infty$, then $S_m\to 0$. At $p=0$ all $g_k$ vanish, so any positive exponential rate works. Otherwise choose $m\ge 1$ with $a=S_m\in (0,1)$. Iterating the weighted BK boundary-splitting estimate gives, for $k=jm+r$ with $0\le r<m$,For $k\ge m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To include the finitely many smaller indices without a prefactor, note that $p<1$ under finite susceptibility andTherefore one may takeFor $k<m$ this follows from $\mu (p)k\le -\log g_1$, and for $k\ge m$ it follows from the iterated estimate. This proves exponential one-arm decay from finite susceptibility with exactly the requested unit prefactor.

This is a generalized linear model with independent Bernoulli distributions for the responses and the logit link. Under treatment coding with colour one as the reference level, the linear predictor isColour is a categorical variable, not a numerical trend. Width has the same slope for every colour, since no interaction term is fitted. The parameters are estimated by maximum likelihood estimation.

The approximate Wald confidence interval for the width coefficient is $\boxed{0.46\pm 1.96(0.106)}$, or approximately $(0.2522,0.6678)$. This uses the approximate normal distribution of the maximum-likelihood estimator, rather than a confidence interval for an individual binary response.

For the requested crab, the fitted linear predictor is $\hat{\eta }=-11.38-0.22+0.46(20)=-2.40$. Applying the inverse logit link gives $\boxed{\hat{p}=(1+e^{2.40}{)}^{-1}\simeq 0.0832}$. The rounded displayed coefficients need not reproduce the reported test statistics exactly.

For any fixed bin, each ball supplies an independent Bernoulli distribution indicator with success probability $1/m$. Summing the $n$ indicator variables givesThe full occupancy vector has a multinomial distribution, the basic uniform balls-in-bins allocation. Distinct bin counts are not independent when $n\ge 1$ and $m\ge 2$. To see this, write $B_i={\sum }_{r=1}^n{I}_{r,i}$. For the same ball, $I_{r,i}{I}_{r,j}=0$ if $i\ne j$, while indicators from distinct balls are independent. Their covariance is thereforeWith no balls the counts are deterministic; with one bin there are no distinct bin pairs. These degenerate cases do not contradict the conclusion.

Take independent and identically distributed random variables ${\xi }_k$ with the Bernoulli distribution of parameter $1/2$, and their natural filtration. SetThe event at $n=0$ is the whole sample space. Independence givesso this is a nonnegative martingale; it is integrable with $\mathbb{E}{M}_n=1$. The first zero among the ${\xi }_k$ is almost surely finite, since the probability of having only ones through time $n$ is $2^{-n}\to 0$. After that first zero, every $M_n$ is zero. Hence the almost sure limit is zero:This fair-coin doubling martingale also exhibits failure of uniform integrability: convergence in L1 would force the expectations to converge to zero.