A cyclic rotation moves the first character to the end without changing the cyclic order of the remaining characters. Starting with the specified terminator-first row and repeatedly rotating left gives the following matrix; its row order is only a construction order, not a sorting order:Under lexicographic order, compare characters from left to right, using $\$<A<C<G$. The sorted cyclic rotation matrix is thereforeThe Burrows-Wheeler transform is the last column, read downwards, so the transformed string is

Expanding the squared Euclidean distances givesFor distinct centroids, chooseA point assigned to $C_1$ has the displayed distance difference at most zero and therefore $w^{\mathsf{T}}x+b\ge 0$; assignment to $C_2$ gives the reverse inequality. The separating hyperplane is the perpendicular bisector of the line segment joining the two centroids. Points tied between them lie on that hyperplane. This is the nearest-centroid decision hyperplane.

The requested nonzero $w$ needs the additional assumption $z_1\ne z_2$. It does not follow from the stated algorithm alone. In one dimension, take $C_1=\{-1,1\}$ and $C_2=\{0\}$, whose centroids are both zero. A tie rule that retains previous assignments can leave this configuration after an iteration. The inequalities at $-1,1$ imply $b\ge |w|$, while the inequality at zero implies $b\le 0$, forcing $w=0$. Thus the unconditional nonzero-separator claim has this degenerate exception.

Combining the first two parts gives the centroid sum-of-squares identityThe added term is nonnegative and, since $n_k>0$, vanishes exactly at $z={\bar{x}}_k$. Thus the unique minimizing centroid in this step of K-means clustering isAn empty cluster needs a separate algorithmic convention because its sample mean is undefined and its objective is identically zero.

The centred observations sum to zero:By linearity of the inner product,This cancellation holds for every $z$.

Write $n_k=|C_k|$ and suppose the cluster is nonempty. Expanding the Euclidean norm through its inner product givesSum over $x\in C_k$. The final term does not depend on $x$, so its sum is $n_k\Vert z-{\bar{x}}_k{\Vert }^2$, which proves the requested identity. This expansion is the starting point for the centroid sum-of-squares identity.

Direct independent sampling is preferable. The posterior distribution is the standard beta distribution $\mathrm{Beta}(9,38)$, for which exact posterior moments and standard random-number generators are available. For example, draw independent $U\sim \mathrm{Gamma}(9,1)$ and $V\sim \mathrm{Gamma}(38,1)$ with the same rate, and return $U/(U+V)$. This has the required beta distribution and avoids Markov chain Monte Carlo convergence problems. A correct, converged MCMC method could sample this target, but it is unnecessarily complicated here.

The trace has long, slowly changing stretches, indicating strong autocorrelation and poor exploration of the posterior distribution. The reported effective sample size of a Markov chain is only $36$, so the number of stored draws greatly exaggerates the information in the simulation. The potential scale reduction factor $\hat{R}=1.21$ is appreciably above one and does not support convergence between chains. The reported Monte Carlo standard error $0.04$ is also large relative to a posterior mean near $0.2$.

The exact beta distribution gives posterior mean $9/47\simeq 0.19149$ and standard deviationso the simulation's reported spread $0.11$ is another warning. Do not interpret its mean or interval estimates as reliable yet. Run multiple chains from dispersed starts, improve and tune the proposal or sampler, and use longer runs after adaptation. Recheck the traces, potential scale reduction factor, and effective sample size of a Markov chain. More iterations help only if the chain actually explores its target well; thinning alone does not fix poor mixing.

The two colonies with no observed cheaters are right-censored after seven days. Each contributes $\mathrm{Pr}(X>7)=(1-p{)}^7$, not the probability of an event on day seven or eight. The four observed first events contribute four factors of $p$ and $(6-1)+(7-1)+(5-1)+(5-1)=19$ failures. Including the two censored exposures givesThe beta distribution prior therefore yieldsIts probability density function is proportional to $p^8(1-p{)}^{37}$ for $0<p<1$: it vanishes at both endpoints, is unimodal, and has its mode at $8/45\simeq 0.17778$. The right panel of the sketch in question 5(b) shows this moderately right-skewed density. The calculation assumes daily observations detect the first event and that stopping the experiment after seven days is independent censoring.

For fixed prior parameters, divide numerator and denominator by $n$:under independent sampling from a geometric distribution with true success probability $p_0>0$. The law of large numbers gives $S/n\to 1/p_0$. In a large finite sample the posterior mean is close to $n/S=1/\bar{x}$, the maximum-likelihood estimate. The fixed prior's influence vanishes, and the estimate is consistent.

Using the expected value of a beta distribution, the posterior mean isEquivalently, it is the weighted averageof the prior mean and the maximum-likelihood estimate $n/S$.

The beta distribution prior density is proportional to $p^{a-1}(1-p{)}^{b-1}$. By Bayes' theorem, its product with the likelihood function gives the posterior densityThis is again a beta distribution, namelyThus the beta family is a conjugate prior family for the geometric distribution. The update is an instance of beta-geometric conjugacy: counts of observed successes and failures are added to the two beta parameters.

For independent geometric random variables supported on $1,2,\dots$, put $S={\sum }_i{x}_i$. Multiplication of their probability mass functions gives the likelihood functionThere are $n$ terminal successes and $S-n$ preceding failures.

Suppose $r>1$ and set $q=\log r/(r-1)$. Substitution of the positive equilibrium into the Jacobian matrix givesPart (b) showed $f(x)<0$ whenever $x\ne 1$. In particular,gives $\log r<r-1$, whereasgives $\det J_{\ast }>1$. The latter already violates the Jury stability criterion condition $1-\det J_{\ast }>0$. More directly, the product of the two eigenvalues is greater than one, so at least one has modulus greater than one. Therefore the positive equilibrium is unstable. This is the instability of the Nicholson-Bailey equilibrium; parasitoid feedback in this simple model does not stabilize coexistence.

The Jacobian matrix of the update map isIts eigenvalues are $r$ and zero. The extinction equilibrium is locally asymptotically stable for $0<r<1$ and unstable for $r>1$, by linear stability analysis for a discrete dynamical system. In the latter case, even a small perturbation along $P=0$ grows as $H_t=r^t{H}_0$.

At $r=1$, an eigenvalue equals one, so the strict linear criterion does not decide stability. Directly, $0\le H_{t+1}\le H_t$ and $0\le P_{t+1}\le b{H}_t$. Starting sufficiently close to extinction therefore keeps both coordinates close for all time, proving Lyapunov stability in the nonnegative state space. It is not asymptotically stable, since every nearby $(H,0)$ is another equilibrium. This distinction makes the statement “stable when $r>1$ fails” precise.

Since $a,b>0$, the coexistence value $P_{\ast }=(\log r)/a$ is positive exactly when $r>1$. Under that condition, $H_{\ast }>0$ as well. Therefore the required condition isFor $0<r<1$, the formal nonzero candidate has negative parasitoid density and is not biologically admissible; $r=1$ gives only the host-only line of equilibria.

At an equilibrium, $H=rH{e}^{-aP}$ and $P=bH(1-e^{-aP})$. Taking $H=0$ forces $P=0$, giving extinction. If $H\ne 0$, the first equation gives $e^{-aP}=1/r$ and henceThese are the only isolated candidates obtained from the equations. There is an exceptional parameter case: at $r=1$, $P=0$ and every $H\ge 0$ is an equilibrium. Thus the assertion of exactly two equilibria needs $r\ne 1$, and only the positive candidate belongs to the biological state space when $r>1$.

This is the Nicholson-Bailey model for a host-parasitoid interaction with discrete generations. The factor $r$ is the host multiplication factor in the absence of parasitism; $a$ measures effective attack or searching intensity per parasitoid; and $b$ is the number of next-generation parasitoids produced per attacked host, with any survival or unit conversion absorbed into it.

If the number of attacks on a host has a Poisson distribution with expected value $a{P}_t$, then the probability of no attack is $e^{-a{P}_t}$. Consequently $e^{-a{P}_t}$ is the surviving, unparasitized host fraction, and $1-e^{-a{P}_t}$ is the attacked fraction. The two update equations count host descendants from surviving hosts and parasitoid descendants from attacked hosts, respectively.

The function is strictly concave on $x>0$. Its derivative is positive for $0<x<1$, zero at $x=1$, and negative for $x>1$. Thus it rises from the limiting value $-1$ as $x\downarrow 0$, touches the horizontal axis at its unique maximum $(1,0)$, and then decreases to $-\infty$. It never crosses above the axis. The left endpoint is not included, and the slope tends to $+\infty$ there.

The left panel shows the required sketch; the right panel is used for question 6(e).

The product rule givesIn particular, the second derivative is negative throughout the domain.

The Poisson generalized linear mixed model assumes conditionally Poisson distributed counts with a log link. The fitted variance components are both zero, so this fit lies on a boundary and reduces to ordinary Poisson regression. It estimatesThus the fitted annual increase is about 6.6%, with approximate 95% confidence interval for the rate ratio $(e^{0.001},e^{0.128})\simeq (1.001,1.137)$. The age evidence is much weaker than in the Gaussian fit: the quoted p-value is $0.047$. Zero estimated group variances do not prove that real bird and year differences are absent.

The Poisson distribution forces conditional variance to equal the conditional expected value. Here fitted means are around two to three hatchlings, whereas the Gaussian model estimated a residual variance of only $0.188$. This suggests checking underdispersion and the bounded nature of hatching counts, rather than assuming a Poisson response solves the diagnostic problems. Inspect simulated count residuals and predictive distributions; if the number of eggs is observed, fit a binomial mixed model and check that model's dispersion too.

The displayed Gaussian and Poisson Akaike information criterion values should not simply be ranked as if they used the same likelihood convention: one evaluates a continuous density and the other a probability mass. Compare properly specified count models on the same observed response scale, with compatible likelihoods and predictive checks. A boundary fit, doubtful dispersion assumptions, and possible nonlinear age effects all warrant further work before a strong biological conclusion.