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Start with the Beta distribution draw in the Beta-binomial exchangeable coupling. Write $s=\alpha +\beta$ andThese are the expected value and variance of $p_M$. The fixed $n$ in this coupling is a dependence parameter chosen by the investigator, not necessarily the observed number of cups. The four numbered stages together address the single printed correlation request.

The next Beta distribution has conditional expected value and varianceFor large $n$, its expected value approaches $X/n$, while its variance is at most $1/[4(s+n+1)]$. The final draw therefore stays close to the first. These are prior distributions for the cup probabilities: the auxiliary count is a device for constructing dependence, rather than additional observed tea data.

Using the conditional Beta distribution expected value and the law of iterated expectation,Equality of the expected values alone does not prove equality of the marginal distributions; the later exchangeability argument does.

Multiply the Beta distribution density of $p_M$, the conditional binomial distribution mass of $X$, and the conditional Beta distribution density of $p_T$. With $B(a,b)$ the beta function, the full joint density, relative to counting measure in $x$ and Lebesgue measure in the two probabilities, isHere $x=0,\dots ,n$ and both probabilities lie in $(0,1)$. The printed joint expression omits the factor $1/B(\alpha +x,\beta +n-x)$. It is a constant when $x$ is fixed and only the two probabilities vary, but is not a constant for the full joint probability distribution. Keeping it is essential when summing over the latent variable.

For each auxiliary count $x$, the correctly normalized joint expression above is symmetric in $p_M,p_T$. Summing it over $x=0,\dots ,n$ preserves that symmetry, so these are exchangeable random variables. Their marginal distributions are therefore identical. Since $p_M$ was generated from a Beta distribution,The Beta-binomial exchangeable coupling changes the dependence, while leaving both prior distributions unchanged.

In a normal linear model, attach a continuous spike-and-slab prior to each regression coefficient. With suitably scaled predictors, introduce indicators ${\gamma }_j\sim \mathrm{Bernoulli}(\pi )$ and setThe narrow component describes practically negligible effects; the wide component permits substantial ones. Fit the joint Bayesian posterior of coefficients, indicators and any unknown residual variance. Bayesian model averaging gives shrinkage toward zero for poorly supported effects, while $\mathbb{P}({\gamma }_j=1\mid \mathcal{D})$ quantifies wide-component support. A shared Beta distribution prior on $\pi$ can represent uncertainty about how many effects are substantial.

Select on scientifically meaningful effect size, for example a high $\mathbb{P}(|{\beta }_j|>{\delta }_j\mid \mathcal{D})$ for a prechosen threshold in meaningful predictor units. Membership in the wide component alone does not imply a large realized effect: its normal distribution still permits values near zero. Correlated predictors also require interpretation of the joint Bayesian posterior, rather than treating each coefficient as an isolated test.

Let $w_m=w(m/n)$, so a type $v$ values a prize at $v{w}_m$. In a symmetric increasing Bayes-Nash equilibrium of this rank-order contest, a player reporting type $t$ wins when at most $m-1$ opponents have larger types. Its winning probability isHere the count of opponents above $t$ has a binomial distribution. Differentiation, or the associated order statistic density, givesDefine the effort by the all-pay effort identityThis also verifies equilibrium globally. The derivative of a type $v$'s payoff from reporting $t$ is $w_m(v-t)q_m^{\prime }(t)$, positive before $v$ and negative after $v$. Thus truthful reporting is a best response. Bidding above the maximal equilibrium effort gains no additional winning probability. Type zero chooses zero effort.

By exchanging the two integrations, the expected value of total effort isThe last integral is the moment $\mathbb{E}[X(1-X)]$ for a Beta distribution with parameters $n-m$ and $m$, namely $m(n-m)/(n(n+1))$. This is the uniform-value multi-prize all-pay effort formula.

Put $x=m/n$ and $h(x)=w(x)x(1-x)$. The positive constant $n^2/(n+1)$ multiplying $h(x)$ does not affect the maximizing $m$. Sincethe assumed inequality makes $h$ nonincreasing on the feasible interval. Hence one prize maximizes expected total effort.

For the power family, the PDF gives $w(x)=x^{-\alpha }$. ThenIf $\alpha \ge 1$, this derivative is strictly negative for $0<x<1$: the bracket is affine and its values at the endpoints are $1-\alpha \le 0$ and $-1$. Thus $m=1$ is optimal.

If $0<\alpha <1$, the unique continuous maximizer isThe objective strictly increases before $x_{\ast }$ and decreases after it. Therefore its discrete maximizer lies amongCompare the surviving candidates using $m^{1-\alpha }(n-m)$, since $R_m=n^{\alpha }{m}^{1-\alpha }(n-m)/(n+1)$. If $n{x}_{\ast }$ is an integer, that integer is the unique maximizer; if its floor is zero, the only feasible candidate is $1$. This proves the discrete prize-count optimization for a power-valued contest including the endpoint cases.

The sufficient condition need only hold on $[1/n,(n-1)/n]$. The power family is undefined at zero, so the printed endpoint $x=0$ cannot apply to it. More generally, a positive finite value $w(0)$ would make the displayed inequality fail at zero. The design argument uses only positive feasible prize fractions and needs no value there.

For $n\ge 1$, the binomial log-likelihood is $X\log \theta +(n-X)\log (1-\theta )$ plus a constant. Its maximizer, including endpoints, is ${\hat{\theta }}_{\mathrm{MLE}}=X/n$. For $0<\theta <1$, the Fisher information isThis follows either from the expected negative second derivative or the variance of $(X-n\theta )/(\theta (1-\theta ))$. At $0,1$ the usual regular score formula does not apply; the interior information diverges as an endpoint is approached.

For a symmetric Beta distribution prior $\mathrm{Beta}(s,s)$, conjugacy gives $\theta \mid X\sim \mathrm{Beta}(X+s,n-X+s)$, with posterior mean $(X+s)/(n+2s)$. All three priors below are instances of this calculation.

Set $U=X/(X+Y)$ and $V=X+Y$. Their inverse change of variables is $X=UV$, $Y=(1-U)V$, with domain $0<U<1$, $V>0$. The absolute Jacobian determinant of this inverse transformation isUse the independence of random variables to multiply the two original Gamma distribution densities. The change of variables formula then gives the joint probability densityFactor this into two normalized densities:The support is a product domain, and the factors are precisely a Beta distribution density and a Gamma distribution density. ThereforeThis beta-gamma independence uses the common rate $\theta$; it is a rate parameter, not the scale parameter $1/\theta$.

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.

With the uniform prior, the binomial likelihood gives the Beta distribution posterior $\mathrm{Beta}(s+1,n-s+1)$. Under squared-error loss, the Bayes estimator is the posterior mean:

The production manager's prior gives posterior density proportional to $p^{s+1}(1-p{)}^{n-s}$. Minimizing $\mathbb{E}[(1-p)(\hat{p}-p{)}^2\mid X=s]$ gives the weighted posterior meanusing the supplied beta function. Cross multiplication shows ${\hat{p}}_P>{\hat{p}}_G$ exactly when $n>2s$, so it is greater unless $s\ge n/2$.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

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 $(1-2\theta )/(n+2)$.

The mean condition gives $a/(a+b)=0.1$, hence $b=9a$. The Beta distribution variance is thenSince $0.02985^2=9/10100$, equating variances gives $10a+1=101$. Therefore

By Bayes theorem, with respect to Lebesgue measure on $(0,1{)}^2$,The omitted constant includes the binomial coefficient and the normalizing constant of the Beta distribution; the uniform prior contributes only the support indicator.

The uniform prior is the Beta distribution $\mathrm{Beta}(1,1)$. By Beta-binomial conjugacy, after observing $X$ heads the posterior isThe Bayes estimator under squared error loss is the posterior mean, hence