Let $D={\max }_{x,y}\rho (x,y)$ and let $\pi$ be the invariant distribution. Iterating the assumed Wasserstein contraction givesSince $\rho (x,y)\ge 1_{\{x\ne y\}}$, the coupling characterization of total variation distance impliesThe right side is at most $\epsilon$ whenwhich proves the claimed mixing bound.

For a path $\gamma =(v_0,e_1,v_1,\dots ,e_k,v_k)$, its $\ell$-length is ${\sum }_{j=1}^k\ell (e_j)$. The corresponding path metric isBecause every edge length is at least one, $\rho (x,y)\ge 1$ whenever $x\ne y$.

For a finite reversible lazy chain, the relaxation time is $t_{\mathrm{rel}}=1/\gamma$, where $\gamma =1-{\lambda }_2$ is the spectral gap. Its spectral profile isThe variational characterization of the spectral gap isEvery class $c_0^{+}(A)$ is contained in the class over which this last infimum is taken, so $\lambda (A)\ge \gamma$ and $\Lambda (r)\ge \gamma$. Therefore

The Dirichlet form of a Markov chain isWhen $P$ is reversible, this equals $⟨f,(I-P)g{⟩}_{\pi }$.

Let the energy of a flow be $\mathcal{E}(\theta )={\sum }_e\theta (e{)}^2$, with each unoriented edge counted once. Expanding as in part (i), dropping orientation signs, and summing over common edges givesuniformly in $r$.

Part (i), $\mathbb{E}{X}_r=1$, and the Paley-Zygmund inequality give a constant $a>0$ such that $\mathbb{P}(X_r>1/2)\ge a$ for every $r$. The preceding uniform expectation bound and Markov inequality allow a constant $L$ such thatOn this event, ${\theta }_{\omega }/X_r$ is a unit open flow from $o$ to $z_r$ with energy at most $4L$.

The events that there is such a bounded-energy unit flow from $o$ out of $B(o,r)$ are decreasing in $r$. Their intersection still has probability at least $a/2$. A diagonal compactness argument produces on this intersection a unit flow from $o$ to infinity, supported on its open cluster, with finite energy. The finite-energy flow criterion for transience makes that open cluster transient.

Thus a transient open cluster exists with positive probability. This existence event is a tail event: changing finitely many edges cannot destroy transience in every infinite component, because transience is invariant under finite graph modifications. The Kolmogorov zero-one law upgrades its probability to one.

Expanding the square and using the independence of the percolation edges givesThe ratio equals $p^{-N_r(\gamma ,{\gamma }^{\prime })}$, where $N_r$ is the number of common edges in the two truncated paths. It is at most $p^{-N}$ for $N=|\xi \cap {\xi }^{\prime }|$, under the convention in the hypothesis; in particular $N\ge 1$ because both paths contain $o$. For every integer $N\ge 1$,The tail-sum formula and the assumed exponential intersection tail therefore yield

Under the two-rate alternative, maximize the piecewise-exponential survival model likelihood separately to obtain the estimates in part (i). Under $H_0:{\theta }_1={\theta }_2=\theta$, there are 15 deaths and 2200 days of exposure, so $\hat{\theta }=15/2200$. Twice the maximized log-likelihood difference isUnder the null this likelihood-ratio test statistic is asymptotically ${\chi }_1^2$.

There are seven deaths by day 28. Their total time plus 28 days from each of the ten later observations gives first-interval exposureAmong the ten observations beyond day 28, eight are deaths, and their exposure beyond 28 is

For period survival analysis over a calendar window, patients diagnosed earlier enter at the window's start only if still alive. Their survival times are left-truncated at the elapsed duration from diagnosis to that date, while follow-up is censored at the window's end. This estimates survival using mortality experience in the chosen recent period without pretending that earlier survivors were under observation from diagnosis.

Conditioning on survival to $L_i$ changes the constant-hazard likelihood contribution toThereforeusing only person-time accumulated after entry.

In the Kaplan–Meier estimator, replace the ordinary risk set at event time $t$ byThe product-limit factor remains $1-d_j/|R(t_j)|$, but only individuals who have entered and not yet exited contribute to its denominator.

Left truncation means that an individual enters observation only after a delayed entry time $L_i$ and is included only if the event time exceeds $L_i$. Earlier failures are absent from the dataset, so analysis must condition on survival to entry.

Among the nine informative pairs, the active patient fails first once and the placebo patient fails first eight times. The conditional partial likelihood is proportional tosoUnder $H_0$, the number of active-first failures is $\mathrm{Binomial}(9,1/2)$. The exact two-sided sign test hasso the data reject at the 5% level and favor lower hazard under active treatment.

A pair is informative only when its first observed exit is an event while both members remain at risk; later events after one member has left contribute a factor one. The uninformative pairs are therefore pair 5, with tied events at $(17,17)$, and pairs 10, 11, and 13, whose earlier observations are censorings: $(63+,278)$, $(89+,1083+)$, and $(141,138+)$. At $(140,140+)$ both are in the risk set immediately before time 140, so pair 12 is informative.

Use each matched pair as a separate stratum with its own baseline hazard and fit one common treatment coefficient $\beta$. The Stratified Cox model partial likelihood multiplies the within-pair risk-set contributions, eliminating all pair-specific baselines. Test $H_0:\beta =0$ by the corresponding conditional score, likelihood-ratio, or Wald test.

Independent relabelling gives $\pi =q$, $\mathbb{E}v=1-q$, and leaves $X=T$ with mean $1/\lambda$. Its ratio is therefore $\lambda (1-q)$ rather than $\lambda$. To retain event hazard $\lambda$, simulate the common occurrence time with rateand label it as the target event with probability $1-q$. This is equivalent to independent competing clocks with target-event rate $\lambda$ and censoring-cause rate $q\lambda /(1-q)$; their minimum has rate $\lambda /(1-q)$.

For an independent $C\sim \mathrm{Exponential}(\mu )$, censoring is noninformative andThe observed time is exponential with rate $\lambda +\mu$, while $\mathbb{E}v=\lambda /(\lambda +\mu )=1-q$. Thus

Fixed administrative censoring $C=\tau$ is noninformative because it is unrelated to the random event time. The censored fraction isMoreover, $\mathbb{E}v=1-q$ andso the event-to-time ratio remains $\lambda$.

Neyman allocation optimizes information and may allocate more patients to the arm whose Bernoulli variance is nearer its maximum at $1/2$. The ethical allocation instead favors the arm with higher response probability. They coincide only for special parameter values; in general ethical gain is purchased with some loss of precision relative to Neyman allocation.

Minimizing the expected failures $n_H(1-p_H)+n_L(1-p_L)$ at a fixed variance gives the ethically optimal ratioIt allocates more patients to the dose with the larger response probability.