Start at any integer $x\ge 2$, for example two. Propose $x+1$ or $x-1$ with probability $1/2$ each; if the proposal is one, reject it immediately. Between distinct valid neighboring states the proposal is symmetric. The Metropolis–Hastings acceptance probability for an upward proposal isFor a downward proposal from $x\ge 3$, it isThese are exactly the ratios of the supplied negative binomial distribution probabilities; the factor $p^2$ cancels. The rejection at the boundary supplies a self-loop at two. All neighboring positive states communicate, so the chain is irreducible and aperiodic. Detailed balance gives the desired stationary law. Rejecting the invalid move, rather than forcing an upward move at two, preserves the proposal symmetry used in these ratios.

For the Metropolis–Hastings algorithm, choose a proposal transition $q(x,y)$ and initialize at a state with $\pi (x)>0$. From current $x$, propose $Y\sim q(x,\cdot )$ and accept it with probabilityOtherwise stay at $x$. The accepted off-diagonal transition satisfies detailed balance, since $\pi (x)q(x,y)\alpha (x,y)=\min \{\pi (x)q(x,y),\pi (y)q(y,x)\}$ is symmetric in $x,y$. Thus $\pi$ is stationary. For convergence from arbitrary starting states in the support, choose the proposal so the resulting chain is irreducible and aperiodic. Only ratios of target probabilities are needed, so an unknown normalizing constant cancels.

Count facial boundary walks with each edge-side counted once. Their total length is $2m$. Every face walk has length at least four: simplicity excludes loops and parallel-edge two-cycles, absence of triangles excludes length three, and a length-two face walk consisting solely of a bridge traversed twice could be the whole boundary only for the excluded connected graph on two vertices. Bridges may occur twice in a longer boundary walk and are still correctly counted. Thus $4f\le 2m$.

The Euler formula for a connected planar graph gives $2=n-m+f\le n-m+m/2=n-m/2$. Rearranging yields

The observed statistic $94.34$ exceeds the supplied $0.999$ quantile $13.82$ for two degrees of freedom. Thus reject independence, with $p<0.001$.

For one child of each sex among the first two, the three-child count is $3393$, below its independence expectation $3763.1$. Equivalently, the two-child count is $13079$, above its expectation $16472-3763.1=12708.9$. These departures have exactly the direction predicted by the suggestion. Thus the data support the proposed association: mixed-sex first pairs are associated with a greater tendency to stop at two children.

Let $N_{ij}$ be the observed cell counts, $N_{i+}$ the row totals, $N_{+j}$ the column totals and $N=33009$. The fitted expected frequencies under independence are $E_{ij}=N_{i+}{N}_{+j}/N$. For example the three-child, one-boy cell has expected count $7541\cdot 16472/33009=3763.1$.

The Pearson chi-squared statistic isUnder $H_0$ it has asymptotically a chi-squared distribution with $(2-1)(3-1)=2$ degrees of freedom, assuming the multinomial sampling model and sufficiently large expected counts. All expected counts here are large. The distribution is an asymptotic approximation, rather than an exact finite-sample identity.

Under the independence log-linear model for a two-way contingency table, the six cell probabilities factor asThe row distribution has one free parameter and the column distribution two, giving three free parameters under $H_0$. Under the unrestricted alternative, six cell probabilities sum to one, giving five free parameters under $H_1$. The difference is two, which is the degrees of freedom of the resulting chi-squared test.

The full column rank makes $X^{\mathsf{T}}X$ invertible. The least-squares normal equations from minimizing squared residuals giveThe errors have covariance matrix ${\sigma }^2{I}_n$ by independence and equal variance. Applying the covariance transformation rule to this linear estimator yieldsThe parameter estimator is unbiased because $\mathbb{E}Y=X\theta$.

Write $s_e=S_D/\sqrt{17}$. By symmetry of the $t_{16}$ distribution, $\mathbb{P}(Z>-1.746)=0.95$. Rearranging $Z=(\overline{D}-\mu )/s_e>-1.746$ gives $\mu <\overline{D}+1.746s_e$. Thus the interval printed in the question has endpointThere is a direction mismatch in the question. Since $D_i=X_i-Y_i$ is before minus after, a reduction corresponds to $H_1:\mu >0$. The upper interval $(-\infty ,b)$ excludes zero when $b<0$, which instead tests against a negative mean difference.

For the stated reduction alternative, use the lower 95% interval $(\overline{D}-1.746s_e,\infty )$. It excludes zero exactly whenThis is the level-$0.05$ one-sided paired t-test of $H_0:\mu =0$ against $H_1:\mu >0$. Both the requested upper endpoint and the correctly directed test are thus determined without silently reversing the difference convention.

Each before-and-after observation belongs to the same individual, so pairing removes individual baseline variation and allows dependence within a person's two readings. Under the stated independent normal model for the differences,The independence of the normal sample mean and sample variance theorem gives independence of $\overline{D}$ and $S_D^2$, andAlso $(\overline{D}-\mu )/(\sigma /\sqrt{n})$ is standard normal. Dividing it by the square root of the independent ${\chi }_{n-1}^2/(n-1)$ variable givesIts law contains neither unknown parameter, providing the pivotal quantity for a paired t-test.

Equality forces $\mathbb{E}X<\infty$ because its proposed value is at most $a$. The random variable $Y=X-a{1}_{\{X\ge a\}}$ is nonnegative and has expectation zero, so $Y=0$ almost surely. On $\{X<a\}$ it equals the strictly positive variable $X$, making that event null. On $\{X\ge a\}$ it equals $X-a$, forcing $X=a$ there. HenceThis is the equality in Markov's inequality under strict positivity. The reasoning even works without the discreteness assumption.

Let ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ and write ${\sigma }^2=\mathrm{Var}(X_1)<\infty$. Independence gives $\mathbb{E}{\overline{X}}_n=\mu$ and $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. Apply part (b) to the nonnegative variable $({\overline{X}}_n-\mu {)}^2$, which is the Chebyshev inequality:Thus ${\overline{X}}_n\to \mu$ in probability, proving the finite-variance weak law of large numbers.

Pointwise, $X\ge a{1}_{\{X\ge a\}}$, since $X$ is nonnegative. Taking expectations and dividing by $a>0$ proves the Markov inequalityIf the expectation is infinite, the inequality remains true in the extended-real sense.

The definition of convergence in probability isThe variables are defined on a common probability space so that these difference events make sense.