The first two mechanisms are standard independent, noninformative censoring. The third relabels the event time itself and therefore is not independent censoring for the original latent event time; it is more naturally interpreted as assigning a cause in a competing risks model. The detailed calculations follow in the three subparts.

Without censoring, every $v_i=1$ and $\mathbb{E}{T}_i=1/\lambda$. ThereforeThis is events divided by total person-time at risk.

If $U\sim \mathrm{Uniform}(0,1)$, the inverse transform sampling formulahas survival function $\mathbb{P}(T>t)=e^{-\lambda t}$ and hence the required exponential distribution.

With equal allocation and true alternative rates ${\lambda }_T,{\lambda }_C$, the asymptotic variance isThe normal approximation requires $|\log R{R}_A|/\sqrt{V_A}\ge z_{1-\alpha /2}+z_{1-\beta }$. Hencerounded upward. Writing ${\lambda }_T=R{R}_A{\lambda }_C$ expresses it using a specified control rate and the target rate ratio.

Since $X_j\sim \mathrm{Poisson}(n_j{\lambda }_j)$, the rate MLE has asymptotic variance ${\lambda }_j/n_j$. The delta method for the logarithm givesIndependence of the arms therefore yieldsUnder $H_0:{\lambda }_T={\lambda }_C=\lambda$, the asymptotic variance of the log rate ratio is

Targeting the Neyman ratio asymptotically minimizes the standard error for a fixed total sample size, so it cannot be worse than a fixed 1:1 design when the target probabilities are known. Targeting the ethical ratio can increase the standard error because it optimizes failures rather than information, although it may still outperform 1:1 for some probabilities.

In response-adaptive randomization, the probabilities are estimated during the trial. Delayed outcomes, time trends, unstable early estimates, and random final arm sizes complicate logistics and inference; naive Wald standard errors may also ignore adaptation.

Writing $v_H=p_H(1-p_H)$ and $v_L=p_L(1-p_L)$, Neyman allocation minimizes the variance $v_H/n_H+v_L/n_L$ at fixed total sample size and gives

Under the boundary null $p={\pi }_0=0.20$, rejection requires both $X_1>r_1$ and $X_1+X_2\ge r$. Independence of the two binomial distributions givesThe rejection event is increasing in $p$, so this boundary value is the supremum over the composite null $p\le {\pi }_0$.

While $I>0$, divide the susceptible equation by the recovered equation:Taking $R(0)=0$ and integrating givesAt the end of the epidemic $E_{\infty }=I_{\infty }=0$, so conservation gives $R_{\infty }=N-S_{\infty }$. Definingwe obtain the final size relation for an epidemic

For transition-intensity vector $q$ and generator $Q(q)$, the expected infectious occupancy over three months from $S$ is the Markov reward model quantityInsert the fitted intensities to obtain $m(\hat{q})$, evaluating the matrix exponential and integral numerically if necessary.

If $\eta =(\log q_{SE},\log q_{EI},\log q_{IR})$ has estimated covariance $V$, calculate the numerical gradient $g={\nabla }_{\eta }m(e^{\eta })$ at $\hat{\eta }$. The delta method gives estimated variance $g^{T}Vg$ and the approximate confidence intervalSimulation from $N(\hat{\eta },V)$ followed by transformation through $m$ gives a useful asymmetric alternative.

Let $P(t)=e^{Qt}$ be the transition probability matrix of the four-state continuous-time Markov chain, and put $N=\{S,R\}$ for a negative test and $P=\{E,I\}$ for a positive test. Starting susceptible at time zero, the likelihood contribution isEquivalently, with indicator diagonal matrices $D_N,D_P$ and the susceptible basis vector $e_S$ it isThis sums over every hidden state sequence compatible with the three test results.

The crude incidence-rate estimator isper person-month. This treats all 3000 person-months as susceptible time and assumes infection times are observed exactly, follow-up is complete, and each first infection is detected. In practice tests occur intermittently, so infection is interval-censored; latent, brief, or asymptomatic infections may also be missed.