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.

With frequency-dependent transmission parameter $\beta$, the nonlinear SEIR equations are

The $S\to E$ transition must be redefined. A constant $q_{SE}$ gives each susceptible the same external infection pressure even when no infectious person remains. For influenza, the force of infection should depend on the current infectious prevalence. The progression $E\to I$ and recovery $I\to R$ can retain rates $q_{EI}$ and $q_{IR}$.

At an interior maximum, $\dot{E}(t)=0$, equivalentlyThusThis is positive for distinct positive rates. Its continuous limit when $q_{EI}=q_{SE}=q$ is $1/q$.

Substitute part (ii) into $\dot{E}+q_{EI}E=q_{SE}{S}_0{e}^{-q_{SE}t}$. Multiplication by the integrating factor $e^{q_{EI}t}$ givesSince $E(0)=0$, integration and rearrangement yield

The separable equation $\dot{S}=-q_{SE}S$ with $S(0)=S_0$ has solution

The linear SEIR model with a constant susceptible-to-exposed intensity isAdding the four equations confirms conservation of population: $S+E+I+R=N$.

If $[M{]}_T\le b$, the Novikov condition holds for $(M-M_0)/2$ becauseIts stochastic exponential is therefore a true martingale with expectation one. Using the factorization from part (i),

Factor the square root of the stochastic exponential asThe stochastic exponential in this expression is a positive local martingale and hence a supermartingale, so its expectation is at most one. If $[M{]}_T\ge a$ almost surely, then

Reversing the finite summations givesThis is a Riemann sum for the continuous functional $f\mapsto {\int }_0^{1}f(t)dt$ evaluated at the interpolated random walk; the interpolation error tends to zero in probability. The continuous mapping theorem and part (b) yield

The evaluation map $f\mapsto f(1)$ is continuous on $C[0,1]$ in the uniform norm. The continuous mapping theorem applied to part (c) therefore givesThis recovers the central limit theorem from the functional version.

Change variables $x={\lambda }_{n}y$ and use the profiles from part 5:Since $v_n\to Q{e}^{i\gamma }$ in $L^2$, the difference made by replacing $|v_n{|}^2$ with $|Q{|}^2$ tends to zero against the bounded function $\varphi ({\lambda }_n\cdot )$. For each fixed $y$, continuity gives $\varphi ({\lambda }_{n}y)\to \varphi (0)$, and dominated convergence then givesThus the mass measures $|u(t_n,x){|}^{2}dx$ converge weakly to the Dirac delta function $\Vert Q{\Vert }_2^2{\delta }_0$.

Choose a smooth radial cutoff ${\theta }_R$ that vanishes on $|x|\le R$, equals one on $|x|\ge 2R$, and satisfies $|\nabla {\theta }_R|\le C/R$. The first localized virial identity and the estimate from part 2, applied to $w=u(t)$ and $\psi ={\theta }_R$, givebecause the mass and energy are conserved.

For fixed $R$, the strong profile convergence from part 5 and ${\lambda }_n\to 0$ implyIntegrating the flux estimate from $t$ to $t_n$ and then letting $n\to \infty$ yieldsuniformly for $0\le t<T$. Taking $R$ sufficiently large proves the claim.