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Assume $\alpha ,\beta ,\kappa >0$, with the given gamma using shape and rate. PutThe joint posterior density before normalization isComplete each square and integrate over all ${\theta }_i$. The Gaussian integral contributes ${\tau }^{-n/2}$, givingThis hierarchical form completely specifies the normal-gamma distribution of the posterior. For comparison, $\tau \mid \theta ,X$ has shape $\alpha +n$ and rate $\beta +\frac{1}{2}{\sum }_i[{\theta }_i^2+\kappa (X_i-{\theta }_i{)}^2]$; confusing this conditional shape with the marginal shape would lose the Gaussian-integration factor.

Marginally $\theta$ has a multivariate Student distribution with $2A=2\alpha +n$ degrees of freedom and scale matrix $B/[A(1+\kappa )]I_n$. The coordinates are conditionally independent given $\tau$, but are generally dependent after integrating their shared precision. This is conjugate shrinkage with uncertainty in the overall noise scale.

Complete the square in the independent normal prior and likelihood:Thus, conditional on the full observations, the hospital parameters remain independent andIf $j$ is the observed minimum index and $X_j=x$, its conditional distribution is exactly this formula with $X_j=x$. No extra truncation is introduced because the selection index is already a function of the conditioned data. Its posterior mean is shrunk toward zero, so the worst measured performance is not simply identified with the worst true performance.

For level $k$, let $H_k$ be the horizontal generator: neighboring horizontal sites exchange at rate $1/6$, with reflecting endpoints. It is symmetric and has zero row sums, so the uniform row vector $u_k=(1/(k+1),\dots ,1/(k+1))$ satisfies $u_k{H}_k=0$. Removing paths as soon as they jump vertically gives the killed horizontal generator $H_k-(2/3)I$. ThereforeConditional on any elapsed waiting time without a vertical jump, the horizontal position remains uniform. This proves the exit-position uniformity as well as independence of that position from the vertical holding time.

Now consider the upward rate block $U_k$ from level $k$ to $k+1$. For $k>0$ each of its $k+2$ destination columns has sum $1/3$: an endpoint receives its single doubled edge jump, and each other site receives two jumps of rate $1/6$. HenceLikewise every destination column of the downward block $D_k$ has sum $1/3$, so $(u_k{D}_k{)}_j={\mu }_k/k$. Thus, conditional on either next vertical level, the arrival position is uniform on that new level. At the corner the two upward rates are equal, giving uniform arrival on level one directly.

Starting with the single site at level zero, induction now proves the uniform conditional law for a triangular-wedge walk: conditional on every attainable past-level sequence, arrival is uniform; the killed semigroup makes departure uniform; and the rate-block column sums make arrival at the following level uniform again. This proves both equalities of the requested property and justifies the averaging in part (b). The statement does not condition departure on the future level, which would in general bias the exit positions.

For $k>0$, an interior horizontal position has total upward and downward rates $1/3$ each. At either edge the upward rate is $1/3+1/6=1/2$ and the downward rate is $1/6$. Thus every position has total vertical-jump rate $2/3$. Averaging over the uniform conditional horizontal law givesAt zero there are two upward jumps of rate $1/3$, so ${\lambda }_0=2/3$ and ${\mu }_0=0$, agreeing with these formulas. Dividing by $2/3$ yields the two embedded transition probabilities in part (a).

The constant vertical hazard makes every vertical holding time exponential with rate $2/3$. Conditional uniformity at a new level, together with preservation of uniformity during horizontal motion, makes the next direction depend only on that level. Consequently the vertical process has the Markov property in its own history and the generator in part (a). Part (c) proves the preservation assertion and the required conditioning on past levels, rather than assuming strong lumpability for every possible horizontal starting point.

The embedded birth-death chain has upward probability $p_k=(k+2)/(2(k+1))$ and downward probability $q_k=k/(2(k+1))$. To calculate a return probability, solve the harmonic difference equation on $\{0,\dots ,m\}$ with boundary values one at zero and zero at $m$. Successive differences have ratio $q_k/p_k=k/(k+2)$, so their weights areTheir total infinite sum is $2$. Letting $m\to \infty$ gives the probability of ever hitting zero from $k$:In particular the chain leaves zero to one and returns with probability $1/2<1$, so it is transient. This finite-interval calculation selects the minimal nonnegative hitting-probability solution, rather than merely observing that $h_k$ is harmonic.

The continuous-time chain has the same embedded transitions and constant total exit rate $2/3$, so it is nonexplosive and has exactly the same return events. It too is transient, neither positive recurrent nor null recurrent.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.

The null condition follows by expansion:It vanishes exactly when $E=G$ and $F=0$; the immersion hypothesis ensures their common value is positive. Thus it characterizes isothermal coordinates for an immersion.

With complex coordinate $\zeta =u+iv$, mixed derivatives commute andand similarly for $y,z$. By the Cauchy-Riemann equations, each ${\phi }_j$ is holomorphic exactly when the corresponding coordinate is harmonic. Combining with the preceding part gives an isothermal immersion is minimal exactly when all three ${\phi }_j$ are holomorphic. Without immersion the null condition could also describe a degenerate map, so the nondegeneracy hypothesis matters.

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.

The prime factorization is $561=3\cdot 11\cdot 17$, so it is composite and squarefree. Moreover $3-1=2$, $11-1=10$ and $17-1=16$ all divide $560=561-1$. The Korselt criterion therefore proves that 561 is a Carmichael number.

By the Korselt criterion, the prime factors are distinct. There cannot be just one prime factor, since a Carmichael number is composite. If there were exactly two, write $n=pq$ with $p<q$. Then $q-1\mid pq-1=p(q-1)+(p-1)$, so $q-1\mid p-1$, impossible because $0<p-1<q-1$. Hence at least three distinct prime factors are necessary.

On the standard two-dimensional representation, $t_{\theta }$ has eigenvalues $e^{i\theta },e^{-i\theta }$. Its $m$th symmetric power has monomial basis $X^{m-j}{Y}^j$, $j=0,\dots ,m$, of weights $m-2j$. The character is thereforeThe quotient is interpreted by continuity at $0,\pi$, where its values are $m+1$ and $(-1{)}^m(m+1)$.

Apply the Weyl integration formula for SU2 to the character product:The last equality follows directly by product-to-sum integration. A finite-dimensional representation of a compact group decomposes into irreducibles, and the squared character norm is the sum of squared multiplicities. Since this norm is one, there is exactly one constituent with multiplicity one. Thus every symmetric power is irreducible, and the different powers are pairwise inequivalent.

Normalize Haar measure on $\mathrm{SU}(2)$ to total mass one. For a continuous class function $F$, the Weyl integration formula for SU2 isEquivalently it is ${\pi }^{-1}{\int }_0^{2\pi }F(t_{\theta }){\sin }^2\theta d\theta$. For arbitrary continuous $F$, replace $F(t_{\theta })$ on the right by its conjugacy average ${\int }_{\mathrm{SU}(2)}F(h{t}_{\theta }{h}^{-1})dh$.

Here is a direct proof. Identify $\mathrm{SU}(2)$ with unit quaternions, the sphere $S^3$. Multiplication by a unit quaternion is a transformation given by an orthogonal matrix of ${\mathbb{R}}^4$, so normalized round volume is bi-invariant and is Haar measure. Write a quaternion as $\cos \theta +\mathbf{n}\sin \theta$, $0\le \theta \le \pi$, $\mathbf{n}\in S^2$. The round volume element is ${\sin }^2\theta d\theta dA(\mathbf{n})$, and the total volume is $2{\pi }^2$. Conjugation rotates the unit imaginary vector $\mathbf{n}$, so a class function depends only on $\theta$. Integrating over the sphere of area $4\pi$ gives the factor $4\pi /(2{\pi }^2)=2/\pi$. The endpoint degeneracies have measure zero. Conjugacy averaging and Haar invariance prove the more general version.

The positive roots can be writtenwhose squares are $2+\sqrt{3}$ and $2-\sqrt{3}$ and whose product is one. Therefore all roots lie in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Conversely $a+b=\sqrt{6}$ and $a-b=\sqrt{2}$, giving $\sqrt{3}=\sqrt{6}/\sqrt{2}$ in the splitting field. HenceThe quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are distinct, so the compositum has this degree. Their independent sign changes are four commuting automorphisms, each nonidentity of order two. Thus

Let $a=\sqrt{2+\sqrt{5}}>0$ and $b=i/a$, so $b^2=2-\sqrt{5}$. The roots are $\pm a,\pm b$. The element $2+\sqrt{5}$ is not a square in $\mathbb{Q}(\sqrt{5})$: its field norm is $-1$, whereas the field norm of a square is a square of a rational number and hence nonnegative. Thus $[\mathbb{Q}(a):\mathbb{Q}(\sqrt{5})]=2$, and $[\mathbb{Q}(a):\mathbb{Q}]=4$. This also proves irreducibility of the quartic.

The splitting field is $K=\mathbb{Q}(a,i)$, since $b=i/a$ and $i=ab$. Because $\mathbb{Q}(a)\subset \mathbb{R}$, adjoining $i$ doubles the degree:The Galois group acts faithfully on the four roots and preserves the partition into the two opposite pairs $\{a,-a\}$ and $\{b,-b\}$, possibly interchanging the pairs. The full permutation group preserving this partition has order eight and is the dihedral group of a square. Since the Galois group has order eight, it is exactly that group. Equivalently its generators can be chosen as $r:a\mapsto b,b\mapsto -a$ and complex conjugation $s:a\mapsto a,b\mapsto -b$, with $r^4=s^2=1$ and $srs=r^{-1}$.

The Fisher information isUnder standard interior, full-rank and large-sample regularity conditions, the maximum-likelihood estimator is asymptotically normal with mean $\beta$ and covarianceThereforeThe sums already depend on sample size; no extra factor $1/n$ should be inserted. The variance is estimable by substituting fitted means. If all $z_i$ are the same, $ac-b^2=0$ and the two coefficients are not separately identifiable.

Put $t_i={\beta }^T{x}_i=1/{\mu }_i>0$. Up to constants independent of $\beta$, the log-likelihood isThe score function and its derivative areThus an interior maximum-likelihood estimator solvesFor a full-rank design the Hessian is negative definite on the parameter region, so there is at most one interior solution and it is the maximum. A Newton method, also Fisher scoring here because the observed Hessian equals its expectation, isRecompute the means at every step and use a line search if necessary to keep all $t_i>0$ and increase the log-likelihood. Existence, identifiability and nonsingularity are regularity conditions, not consequences of the density alone.

The density is that of a Gamma distribution with shape $\nu >0$ and rate $\nu /{\mu }_i$, requiring ${\mu }_i>0$. Direct integration givesfor $\nu +r>0$. Using $\Gamma (\nu +1)=\nu \Gamma (\nu )$ and $\Gamma (\nu +2)=\nu (\nu +1)\Gamma (\nu )$,

The continued-fraction algorithm givesafter which the process repeats. ThusThe first two convergents are $3/1$ and $10/3$, of field norms $-2$ and $1$ respectively. The even period gives the fundamental solutionIt is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding Pell equation group law givesIndeed $3970^2-11\cdot 1197^2=1$.

For the continued-fraction assertion, take the standard Pell equation hypothesis $N>0$ nonsquare. If $(x,y)$ is a solution with $x,y>0$, thenThe convergent criterion for continued fractions therefore makes $x/y$ a convergent to $\sqrt{N}$. Write its periodic expansion as $[a_0;\overline{a_1,\dots ,a_{\ell }}]$, and index convergents by $p_j/q_j$ with $p_0/q_0=a_0$. The norm-one convergents occur at $j=k\ell -1$ with $k\ell$ even. In particular the fundamental positive solution comes at $\ell -1$ if $\ell$ is even, and at $2\ell -1$ if it is odd. If $\epsilon =x_1+y_1\sqrt{N}>1$ is this least norm-one unit, all integer solutions are represented by $\pm {\epsilon }^m$, $m\in \mathbb{Z}$.

The abelian group law comes from multiplication of the norm-one expressions:Their field norms multiply, so the result remains a solution. Multiplication also proves associativity and commutativity; the identity is $(1,0)$ and the inverse of $(x,y)$ is $(x,-y)$. Cubing givesThe last form uses $N{y}^2=x^2-1$.

For completeness, the assertion about all powers follows by dividing any positive unit by the largest power of $\epsilon$ not exceeding it: the remainder is a norm-one unit in $[1,\epsilon )$, and minimality forces it to be $1$. Sign changes and inverses give the remaining solutions. If the phrase “nonsquare integer” includes negative $N$, real continued fractions do not apply: $x^2+|N|y^2=1$ has only $(\pm 1,0)$ except for $N=-1$, when $(0,\pm 1)$ also occur. The same multiplication law still works in those finite groups.

For subhorizon pressureless matter during radiation domination, expansion is too rapid for significant self-gravitating growth. Neglecting the small matter-gravity term gives ${\ddot{\delta }}_m+t^{-1}{\dot{\delta }}_m\simeq 0$, hence a constant plus a logarithm of time; the exact coupled radiation problem supplies the detailed amplitudes. This is the Mészáros effect. Modes entering the horizon during radiation domination are suppressed relative to modes entering later, while after equality ordinary matter growth resumes.

The matter power spectrum consequently has a turnover near the equality horizon scale $k_{\mathrm{eq}}\sim a_{\mathrm{eq}}{H}_{\mathrm{eq}}$ in units with $c=1$. Shorter-scale modes retain the radiation-era suppression. This statement concerns cold pressureless matter; baryons coupled to photons also show acoustic effects before recombination.