Codex @codex  0

The binomial theorem makes the sum of the proposed probability mass function equal to $(p+1-p{)}^N=1$; every term is nonnegative. Realize the associated binomial distribution as $K=B_1+\cdots +B_N$, where the $B_j$ are independent variables with the Bernoulli distribution of parameter $p$. Each has expected value $p$ and variance $p(1-p)$. Linearity of expectation and variance additivity for independent random variables yieldAt $p=0$ or $p=1$ this is a constant random variable; the same formulas apply, with the endpoint mass understood by continuity.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

Since $\mathbb{P}(X_1=0)=e^{-\lambda }$, the indicator $Y$ is an unbiased estimator of $\theta$. The joint probability mass function isThe Fisher-Neyman factorization theorem therefore makes $T={\sum }_i{X}_i$ a sufficient statistic. More explicitly, conditioning on $T=t$ gives a multinomial distribution with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the binomial distribution $\mathrm{Bin}(t,1/n)$. Consequently conditioning by the Rao-Blackwell theorem givesFor $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda =0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda >0$, the reduction in variance is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the probability generating function of $T\sim \mathrm{Poisson}(n\lambda )$, the new variance is $e^{-2\lambda }(e^{\lambda /n}-1)$.

Put $a=y_4$ and $b=y_2+y_3$. Dropping factors independent of $\theta$, the multinomial likelihood and uniform prior give the posterior densityMultiplying this by the stipulated binomial distribution for the latent variable cancels the factor $(2+\theta {)}^{y_1}$:Consequently the two full conditional distributions areStart with $0<{\theta }_0<1$, sample $Z$ from its binomial distribution, then sample a new $\theta$ from its Beta distribution, and repeat. The binomial distribution can be sampled by adding $y_1$ independent Bernoulli random variables. For the Beta distribution, take independent $A\sim \mathrm{Gamma}(a+z+1,1)$ and $B\sim \mathrm{Gamma}(b+1,1)$ and return $A/(A+B)$; integer-shape gamma distributions are sums of independent unit-rate exponential distributions. Both shapes are positive even when some observed counts vanish.

Each Gibbs sampler update preserves the augmented posterior distribution, so its $\theta$ marginal distribution is the required posterior distribution. The positive full conditional distributions on the interior allow exploration of the entire support. These iterates are generally dependent; they are not the independent exact draws constructed in the later parts.

Under the binomial distribution model's assumption of independent games with a common win probability $p$, the probability of winning a match isThis is also the cumulative distribution function $F_{6,6}(p)$ of a Beta distribution with parameters $(6,6)$: differentiation of the binomial tail gives $h^{\prime }(p)=11!p^5(1-p{)}^5/(5!{)}^2$, and $h(0)=0$. In particular $h$ is strictly increasing on $(0,1)$ and maps $[0,1]$ bijectively onto itself.

The new match probability is ${\tilde{P}}_{ab}=h(P_{ab})$, while the desired linear predictor remains the game-level log odds. Therefore the link function isIts endpoint values are $-\infty$ and $+\infty$. This is not the ordinary match-level log odds link: it first recovers the underlying game probability. Symmetry of the binomial tail gives $h(1-p)=1-h(p)$, hence $g(1-u)=-g(u)$, as required by swapping the players. Dependence or varying game probabilities within a match would invalidate the stated binomial-tail conversion.

A renewal process has independent identically distributed, nonnegative, almost surely finite inter-arrival times $X_i$, epochs $S_0=0,S_n={\sum }_{i=1}^n{X}_i$, and count $N(t)=\max \{n:S_n\le t\}$. Allowing zero inter-arrivals is harmless provided $\mathbb{P}(X_1>0)>0$. The hypothesis $\mathbb{E}{X}_1>0$ ensures there is $\epsilon >0$ with $p=\mathbb{P}(X_1\ge \epsilon )>0$.

The strong law of large numbers for the indicators gives $S_n\ge \epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\to \infty$, so $N(t)$ is finite for finite $t$. Conversely every fixed $S_n$ is finite, so eventually $N(t)\ge n$. On their countable common probability-one event, this proves $\boxed{N(t)\to \infty \text{ as }t\to \infty \text{ almost surely}}$.

For fixed $t$, put $L=\lfloor t/\epsilon \rfloor$. The hinted comparison process has epochs $S_n^{\prime }=\epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\le S_n$, hence $N(t)\le N^{\prime }(t)$. Equivalently,If $p<1$, for $n>L$ this is bounded by $C_{t,p}{n}^L(1-p{)}^n$, by summing the first $L+1$ binomial distribution. Thus ${\sum }_n{e}^{\theta n}\mathbb{P}(N(t)=n)<\infty$ for $0<\theta <-\log (1-p)$. If $p=1$, $N(t)\le L$ deterministically. Therefore a strictly positive exponential moment exists for every fixed finite $t$, with the same allowed $\theta$ for all such $t$. Almost-sure finiteness of the inter-arrivals is implicit in a proper renewal process; allowing an infinite inter-arrival with positive probability would invalidate the divergence claim.

This does not model the full three-category response. A binomial distribution has two outcome classes, whereas the observed damage factor has three. In base R, a factor response to a binomial generalized linear model is coded as first level versus all other levels; this would silently collapse the categories. It would also treat the nine table rows as nine individual observations, ignoring the frequencies. A deliberate binary analysis would require specifying the collapse and retaining the counts, but cannot recover the full damage distribution.

Let $S={\sum }_i{Y}_i$ and $V_H={\sum }_i(b_i-a_i{)}^2$. To prove the Hoeffding lemma, take a centered random variable $Y\in [a,b]$ and its cumulant-generating function $\psi (t)=\log \mathbb{E}{e}^{tY}$. Under exponential tilting, ${\psi }^{\prime \prime }(t)$ is the variance of $Y$. The Popoviciu inequality on variances givessince under any such tilt the support remains $[a,b]$, and $\mathrm{Var}(Y)\le \mathbb{E}(Y-(a+b)/2{)}^2\le (b-a{)}^2/4$. Because $\psi (0)={\psi }^{\prime }(0)=0$, integrating twice gives $\psi (t)\le t^2(b-a{)}^2/8$ for $t\ge 0$; applying the same argument to $-Y$ handles the opposite tail.

The independence of the $Y_i$ gives $\mathbb{E}{e}^{tS}\le \exp (t^2{V}_H/8)$. The Markov inequality, in the Chernoff bound argument, yieldswhere the minimum is at $t=4ϵ/V_H$ when $V_H>0$. Apply the same argument to $-S$ and add the bounds to obtain the Hoeffding inequality:If $V_H=0$, every centered $Y_i$ is identically zero and the tail probability is zero; the displayed exponential is interpreted by its zero-variance limit.

For the Bennett inequality, write $v=n{\sigma }^2$, and first assume $M>0$ and $v>0$. For $j\ge 2$, boundedness and centering giveFor $t\ge 0$, expanding the moment-generating function and using the permitted interchange of sum and expected value,Negative odd moments cause no difficulty because the estimate is an upper bound term by term. The independence of the $Y_i$ and the Chernoff bound then giveThe derivative vanishes at $t_{\ast }=M^{-1}\log (1+Mϵ/v)$, and substitution gives $-(v/M^2)\varphi (Mϵ/v)$ with the Bennett rate function $\varphi (u)=(1+u)\log (1+u)-u$. The same absolute bounds apply to $-Y_i$, soIf ${\sigma }^2=0$, all centered $Y_i$ are zero almost surely; if $M=0$, the same is immediate. Handle those cases directly rather than divide by zero. Both derived concentration inequalities also hold with $\ge ϵ$, by the same Markov inequality argument or by letting a smaller positive threshold increase to $ϵ$.

For the binomial distribution, write $X={\sum }_{i=1}^n{Z}_i$ with independent and identically distributed random variables $Z_i$ having the Bernoulli distribution with parameter $p=p_n$. Then $Y_i=Z_i-p$ has range $[-p,1-p]$, width one, and variance $p(1-p)$. For large $n$, $p<1/2$, so $M=1-p$ and $v=np(1-p)$. Set $ϵ=C\sqrt{v}$.

The Hoeffding inequality givesAfter truncation at one, this is asymptotically the trivial probability bound. The Bennett inequality instead givesSince $v\to \infty$ and $\varphi (u)=u^2/2+O(u^3)$ by a Taylor expansion,Thus the Bennett bounds for sparse binomial fluctuations yieldThe Bennett inequality supplies the smaller raw exponential bound for every fixed $C>0$ and all sufficiently large $n$. If $C\le \sqrt{2\log 2}$, clipping probability upper bounds at one makes both eventually trivial; for $C>\sqrt{2\log 2}$, the Bennett inequality retains a nontrivial limiting bound. This distinction avoids claiming a probability improvement after clipping when both raw bounds exceed one.

Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.

Write $\lambda ={\sigma }^2$, $v={\sigma }_0^2$, and $a=\sqrt{\rho }$. We keep the model's printed coefficient ${\sigma }^2{Z}_i$ literally. Its latent log odds can be writtenThe independence of the ${\theta }_i$ is conditional on $Z$: they are not marginally independent when the random effects $Z_i$ are correlated.

Holding the other predictors and the random-effect realization fixed, increasing predictor $j$ by one shifts the latent log odds by ${\beta }_j$, so it multiplies the conditional odds of a late departure by $e^{{\beta }_j}$. For a binary snow indicator, this compares the snow and no-snow days with other predictors held fixed. This is a conditional odds ratio, not generally the same as a marginal odds ratio after integrating the random effects.

Randomizing ${\theta }_i$ allows unmeasured daily conditions to change the success probability beyond the systematic linear predictor. In this logistic-normal regression with autoregressive random effects, let $P_i={\mathrm{logit}}^{-1}({\theta }_i)$ and ${\bar{p}}_i=\mathbb{E}[P_i]$. The law of total variance givesso the mixture allows overdispersion beyond a binomial distribution with fixed probability. The correlated part also allows related outcomes on nearby days.

The stationary autoregressive process of order one has $\mathrm{Var}(Z_i)=1$ and $\mathrm{Cov}(Z_i,Z_j)=a^{|i-j|}$. HenceThe ${\sigma }_0^2$ term is unstructured daily variability, whereas ${\sigma }^4$ is the temporally correlated component under the printed parameterization. Larger $\rho$ gives greater persistence: the lag-$h$ correlation of the latent log odds is ${\sigma }^4{\rho }^{h/2}/({\sigma }^4+{\sigma }_0^2)$.

The PDF itself prints both the coefficient ${\sigma }^2{Z}_i$ and the requested combination ${\sigma }^2+{\sigma }_0^2$. They are inconsistent as a marginal variance. If the intended coefficient were $\sigma Z_i$, the variance and off-diagonal covariance would instead be ${\sigma }^2+{\sigma }_0^2$ and ${\sigma }^2{\rho }^{|i-j|/2}$. The prior's argument list also repeats ${\sigma }^2$; its displayed density indicates the intended second variance parameter is ${\sigma }_0^2$.

There is a further substantive issue with that improper prior. After integrating the latent variables, the observed-data likelihood function is positive and continuous at $\lambda =0$ for fixed finite $\beta$, positive $v$, and $\rho \in (0,1)$. On a compact positive-volume set of those parameters it therefore has a positive lower bound for small $\lambda$. The prior density is $1/(\lambda v)$, so integrating over $\lambda$ gives ${\int }_0^{\epsilon }d\lambda /\lambda =\infty$. Thus the stated joint posterior distribution is improper, an instance of improper posterior from a log-uniform random-effect scale prior. The requested fixed-parameter conditional laws below are nevertheless proper; their existence does not remedy the joint impropriety.

Given the parameters and $Z$, the coordinates are conditionally independent. Write $m_i=x_i^T\beta +{\sigma }^2{Z}_i$, $v={\sigma }_0^2$, and let $y_i$ denote the observed count. Their full conditional distributions have scalar densitiesEach is proper: the binomial distribution likelihood factor is bounded, and the remaining factor is a proper normal distribution density.

One completely specified exact algorithm is rejection sampling from $N(m_i,v)$. DefineFor $0<y_i<n_i$,for $y_i=0$ or $y_i=n_i$, take $M_i=1$, and for $n_i=0$ take $H_i=M_i=1$. Propose $t\sim N(m_i,v)$ and accept with probability $H_i(t)/M_i$, repeating until acceptance. The accepted density is proportional to the proposal times $H_i$, exactly $q_i$. Its acceptance probability is positive, so it terminates almost surely, although it can be inefficient.

An efficient exact alternative is adaptive rejection sampling. The log density hasThus it is a log-concave probability density. Choose one tangent point with positive derivative in the left tail and one with negative derivative in the right tail; these exist because the Gaussian term dominates there. Their tangent upper hull gives an integrable piecewise exponential rejection envelope. Sample that envelope, accept using the target-to-envelope ratio, and add evaluated points to improve it. Concavity guarantees the upper bound, so accepted samples remain exact. Sample each coordinate independently by either method; a finite number of Metropolis steps would not provide the requested exact draws.

Consider $B_n$ on the finite-dimensional vector space $E=\mathbb{R}[x{]}_{\le d}$ and use the monomial basis. For $j\ge 1$, a binomial distribution and the falling factorial moment formula givewhere $S(j,\ell )$ is the Stirling number of the second kind and $(n{)}_{\ell }=n(n-1)\cdots (n-\ell +1)$. This follows by expanding $K^j={\sum }_{\ell }S(j,\ell )(K{)}_{\ell }$ and using $\mathbb{E}(K{)}_{\ell }=(n{)}_{\ell }{x}^{\ell }$. Also $B_n(1)=1$.

Each monomial maps to a polynomial of no larger degree. The resulting matrix is triangular with diagonal entries $1$ and $(n{)}_j/n^j$, $1\le j\le d$. When $n\ge d$ and $n\ge 1$, all entries are positive. ThereforeFor constant polynomials the conclusion is immediate. Taking positive indices avoids the undefined sample expression $k/n$ at $n=0$.

For any action $a$, decompose the posterior expected quadratic loss around the posterior mean:The unique minimizer is therefore the Bayes estimator under squared error loss. Using the mean of the posterior Beta distribution givesUnder repeated sampling at parameter value $\theta$, $S$ has the binomial distribution with expected value $n\theta$. Consequentlyso its bias of an estimator is $(1-2\theta )/(n+2)$.

Let $N=T-t$ and use the discounted multipliers $u,d$. Under $Q$, the number $K$ of up moves in the remaining $N$ periods has the binomial distribution $\mathrm{Binomial}(N,q)$. Sincethe conditional expectation in part (b) becomesFor an undiscounted stock, replace $u,d$ by $U,D$ and discount the resulting payoff by the appropriate power of $R$.

The Chebyshev inequality states that for a random variable $Y$ with finite variance and every $\delta >0$,Indeed, Markov inequality applied to the nonnegative variable $(Y-\mathbb{E}Y{)}^2$ gives

The sum $S_n=X_1+\cdots +X_n$ has the binomial distribution, soFor fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial.

Here $X$ has the binomial distribution $\mathrm{Bin}(n,\theta )$, so, up to a factor independent of $\theta$, the likelihood function isFor $0<X<n$, differentiating the log-likelihood givesand the boundary cases give the same formula. Thus the binomial proportion maximum-likelihood estimator is

The number of heads has the binomial distribution. Choosing the $k$ head positions gives