Use the prior itself as an independence proposal: draw $J\ge 1$ with $\mathbb{P}(J=j)=2^{-j}$ and propose $s=2^{-J}$. This indeed samples $p(s)=s$, since the geometric series sums to one. Starting from any allowed current value $r$, accept the proposal with probabilityThe Hastings proposal ratio cancels the prior ratio in the posterior, leaving exactly the binomial likelihood ratio; its binomial coefficient also cancels. All allowed states communicate, and proposing the current state gives a self-loop. Detailed balance therefore gives the specified posterior as the stationary law, and the chain converges to it. Its normalizing constant is finite and positive because it is a prior average of a nonnegative, nonzero likelihood.

For $0<r<1$, start at any $x\in \{0,\dots ,n\}$. Propose $Y$ uniformly from this same finite set and accept it with probabilityOtherwise retain $x$. The uniform proposal is symmetric, so this is the Metropolis–Hastings algorithm with target $\pi (\cdot \mid r)$. Detailed balance follows because accepted probability flow is proportional to $\min \{\pi (x),\pi (Y)\}$. The chain is irreducible and has positive self-transition probability, hence converges to that binomial equilibrium law. At the degenerate parameters zero or one, the target is simply a point mass and can be simulated as a constant chain.

The square $Q_2$ and the drawn cube $Q_3$ are planar. For $k\ge 4$, the subdivision in part (v) prevents planarity by Kuratowski's theorem. One can also check the edge count directly: a triangle-free simple planar graph with $n\ge 3$ vertices has at most $2n-4$ edges, whereas $Q_k$ has $n=2^k$ and $m=k{2}^{k-1}\ge 2n$. Thus

If the printed $K_{33}$ means an ordinary $K_{3,3}$ subgraph, the claim is false. Any two distinct hypercube vertices have exactly two common neighbors when their Hamming distance is two, and no common neighbors otherwise. Three common neighbors, required by any pair of vertices on one side of $K_{3,3}$, are therefore impossible.

The valid intended statement is a graph subdivision of $K_{3,3}$. An explicit one in $Q_4$ has branch sets $A=\{0000,0011,1100\}$ and $B=\{0001,0010,0100\}$. Six of the nine branch-to-branch connections are direct edges: $0000$ joins all three vertices of $B$, $0011$ joins $0001,0010$, and $1100$ joins $0100$. The other three paths areEvery step changes one bit and the six internal vertices are distinct and are not branch vertices. Thus these paths and direct edges form the required topological subgraph. For $k>4$, append $k-4$ zeros to every label, embedding the construction in $Q_k$.

The hypercube is connected: successive coordinate flips reach any target bit string. Every vertex has degree $k$. A finite connected graph has an Euler circuit exactly when all vertex degrees are even. Therefore

A vertex is a bit string of length $k$, and adjacent vertices differ in one coordinate. Therefore the parity of the number of one-bits changes along every edge. The even-parity and odd-parity strings are a bipartition, proving $Q_k$ is bipartite for every $k\ge 1$, in particular for the requested range.

Label the outer square $000,001,011,010$ in cyclic order and the inner square $100,101,111,110$ in corresponding order. Join corresponding vertices radially. Each edge changes exactly one bit, so this gives every edge of $Q_3$, with no crossings.

The relation $N=k/p$ is strictly decreasing. Thus an interval $0<L<U<1$ transforms by reversing endpoints:Keep integer population sizes satisfying $N>k$. If the clipped lower confidence endpoint for $p$ is zero, the transformed upper endpoint for $N$ is infinite. This monotone transformation preserves the approximate coverage probability supplied by the interval for $p$.

For an interior true parameter and large $n$, standard asymptotic normality of a maximum likelihood estimator givesEstimate its standard error by $s_e=\hat{p}\sqrt{(1-\hat{p})/(nr)}$. A two-sided approximate Wald confidence interval is $(L,U)=(\hat{p}-z_{1-\alpha /2}{s}_e,\hat{p}+z_{1-\alpha /2}{s}_e)$, intersected with the parameter range $(0,1)$. The normal quantile is chosen so a standard normal lies between its positive and negative values with probability $1-\alpha$.

With immediate replacement and uniform mixing, each capture is a Bernoulli trial with tagged probability $p=k/N$. Counting failures before the $r$th tagged capture gives the stated negative binomial distribution. Independent runs on the successive days give the sample model.

Writing $S={\sum }_i{x}_i$, the log-likelihood is $\ell (p)=nr\log p+S\log (1-p)+\text{constant}$. Its derivative vanishes atThe second derivative is negative. If $S=0$, this formula gives the boundary maximum $p=1$; on the strict physical parameter space $p<1$, it is a supremum rather than an attained interior maximum.

Using the supplied negative-binomial mean, $\mathbb{E}S=nr(1-p)/p$. Thus total Fisher information for the sample is

The independent variables $W_i$ have mean two and variance four by part (b). Their standardization isThe central limit theorem therefore proves the claimed convergence to the standard normal law, whose continuous cumulative distribution function is $(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$ at every real $x$.

The iid central limit theorem states that if $Y_1,Y_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\sigma }^2<\infty$, thenEquivalently, its cumulative distribution functions converge at every real argument to the standard normal cumulative distribution function.