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The unrestricted categorical MLE is ${\tilde{\pi }}_j=N_j/n$, so the multinomial combinatorial factor cancels from the likelihood ratio andAssume the true cell probabilities are positive and the null has identifiable dimension $d$ with full-rank information. With probability tending to one, all $N_j$ are positive. Put ${\delta }_j=n{\hat{\pi }}_j-N_j$. Regular estimation gives ${\delta }_j=O_p(\sqrt{n})$ and $N_j$ of order $n$. Expanding $-2N_j\log (1+{\delta }_j/N_j)$ yieldsThe linear terms cancel exactly because ${\sum }_j{\delta }_j=0$. ThereforeThe full probability simplex has dimension $K-1$, explaining the degrees of freedom. Existence and uniqueness of an MLE alone do not supply all Wilks regularity assumptions; positivity, smoothness and the local rank condition are also needed. The displayed approximation is not a finite-sample prescription for division by a zero observed count.

The likelihood-ratio statistic for nested parameter spaces isUnder the null, Wilks theorem gives $W_n\Rightarrow {\chi }_{d_1-d_0}^2$ when the full model and null submodel are smooth and identifiable, the true parameter is an interior regular point, Fisher information is nonsingular in the identifiable directions, and the usual differentiability and likelihood-asymptotic conditions hold. A level-$\alpha$ test rejects for $W_n$ exceeding the upper $\alpha$ quantile of that chi-square law. The approximation is for large samples; boundary parameters, singularities or nonidentifiability can give a different limiting distribution.

At a state with positive departure rate ${\lambda }_i=R-r_i$, the next-jump probabilities areIf at least two rates are positive, every state has positive departure rate and the normalizing constant $Z={\sum }_i{r}_i(R-r_i)=R^2-{\sum }_i{r}_i^2$ is positive. The jump-chain equilibrium law isIndeed ${\hat{\pi }}_i{\hat{p}}_{ij}=r_i{r}_j/Z$ is symmetric, proving both stationarity and reversibility. Jump observations weight the continuous-time equilibrium by the departure rate, so $\hat{\pi }$ is generally different from $\pi$.

If exactly one rate, $r_k$, is positive, state $k$ is absorbing and the denominator $R-r_k$ vanishes. There is no next jump from that state: the genuine embedded jump sequence terminates there. Under the usual extended-chain convention set ${\hat{p}}_{kk}=1$; every other state jumps to $k$ with probability one. Its equilibrium law is ${\delta }_k$ and it is trivially reversible there. The normalized $r_i(R-r_i)$ formula cannot be used in that absorbing case.

The stationary equation $\pi Q=0$ reads $r^T-R\pi =0$, because $\pi 1=1$. ThereforeIt is the unique stationary law and vanishes on the open classes. For distinct $i,j$, ${\pi }_i{q}_{ij}=r_i{r}_j/R={\pi }_j{q}_{ji}$. Thus the chain is reversible in its equilibrium law, including the reducible cases; on the closed class this is the usual reversible irreducible chain. This structure is the destination-dependent reset chain.

Let $S_{+}=\{j:r_j>0\}$, nonempty since $R>0$. Every pair of distinct states in $S_{+}$ communicates directly, and no transition leaves $S_{+}$ for a zero-rate state. Hence $S_{+}$ is the unique closed communicating class. Each zero-rate state forms an open singleton: it has total departure rate $R>0$, goes into $S_{+}$, and can never be revisited. This includes the case that $S_{+}$ itself is a single absorbing state.

For $R>0$, the chain has one communicating class exactly when every $r_j>0$. Sufficiency follows from the positive direct transition rate between each distinct pair of states. For necessity, if $r_j=0$ and there is another state, no other state can jump to $j$, either directly or as the final step of a path, so $j$ cannot communicate with them. The one-state case also satisfies the condition because $R=r_1>0$. A single class containing the entire state space is automatically closed.

When $R=0$ all rates vanish. Every state is absorbing and the communicating classes are the $s$ singletons, all closed. There are no open communicating classes. Every probability distribution is stationary in this degenerate case.

For $i\ne j$ the generator entry is $q_{ij}=r_j$. Its diagonal is the negative total departure rate, $q_{ii}=-(R-r_i)$. ThusEvery row sums to zero, as required for a continuous-time Markov-chain generator.

Let $U$ be one uniform random variable on $(0,1)$, and defineFor every $U>0$ the sequence is eventually zero, so $X_n\to 0$ almost surely. But $\mathrm{Pr}(|X_n|>1/2)=1/n$, whose sum diverges. This sequence is not completely convergent; the exceedance events are nested and dependent.

Almost sure convergence implies convergence in probability, so part (b) identifies the limit as a constant $c$ almost surely. For each $ϵ>0$, the events $A_n=\{|X_n-c|>ϵ\}$ are independent. Almost sure convergence gives $\mathrm{Pr}(A_n\text{ infinitely often})=0$. If their probability sum diverged, the second Borel-Cantelli lemma would instead give probability one. Hence the sum is finite for every $ϵ$, proving almost sure convergence implies complete convergence for independent sequences.

Under complete convergence of random variables, for each integer $k\ge 1$ the first Borel-Cantelli lemma gives $\mathrm{Pr}(|X_n-X|>1/k\text{ infinitely often})=0$. Intersecting these probability-one conclusions over the countably many $k$ shows that every tolerance is eventually satisfied. Therefore complete convergence implies almost sure convergence, without independence.

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first Borel-Cantelli lemma. For each fixed $N$, the tail of this subsequence lies in $\sigma (X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.

By Kolmogorov's zero–one law, every event in that tail sigma-field has probability zero or one. Consequently $\mathrm{Pr}(X\le r)\in \{0,1\}$ for every rational $r$. Since $X$ is a finite real random variable, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.

The first Borel-Cantelli lemma says that ${\sum }_n\mathrm{Pr}(A_n)<\infty$ implies $\mathrm{Pr}(A_n\text{ infinitely often})=0$, without an independence assumption. Indeed, for every $N$,The decreasing intersection of these tail unions is the infinitely-often event; continuity of probability from above proves the assertion. The second lemma says that for independent $A_n$, divergence of the probability sum implies $\mathrm{Pr}(A_n\text{ infinitely often})=1$.

At a non-planar point of a minimal surface, the shape operator is not zero. Minimality gives nonzero opposite principal curvatures, so $K<0$ nearby and $dN$ is invertible. The inverse function theorem makes the Gauss map a local diffeomorphism onto a patch of the sphere; part (ii) makes it conformal.

Choose stereographic coordinates $(u,v)$ on that sphere patch. Their metric is $4(d{u}^2+d{v}^2)/(1+u^2+v^2{)}^2$. Pulling them back by $N$, the conformal identity $N^{\ast }{g}_{S^2}=\lambda g_S$ givesThe positive coefficient shows that these are local isothermal coordinates. No general isothermal-coordinate existence theorem is needed: the nondegenerate conformal Gauss map supplies them explicitly.

Let $A=-dN$ be the self-adjoint shape operator. In an orthonormal principal basis, $⟨dN(v),dN(w)⟩=⟨Av,Aw⟩$, with diagonal coefficients $k_1^2,k_2^2$. The Gauss map is conformal with nonzero factor exactly when $k_1^2=k_2^2=\lambda >0$.

If the surface is minimal and has no umbilics, $k_2=-k_1$ and $k_1\ne 0$, so this equality holds. Conversely equality of the squares gives either $k_1=k_2$ or $k_1=-k_2$. The first possibility is excluded by the no-umbilic hypothesis. Hence $H=0$, proving minimality is equivalent to conformality of the Gauss map under the stated hypothesis. The factor is automatically positive because it equates positive definite inner products. The hypothesis is essential: a sphere has conformal Gauss map and nonzero mean curvature.

A minimal surface has zero mean curvature $H=(k_1+k_2)/2$; equivalently its area is stationary under compactly supported normal variations. Its principal curvatures satisfy $k_2=-k_1$, so its Gaussian curvature isEquality occurs precisely where its second fundamental form vanishes.

The element $ϵ=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a fundamental unit, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<ϵ$. Its conjugate is $\pm 1/u$, so $y=(u-\overline{u})/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(ϵ+1/ϵ)/(2\sqrt{10})=1$; for norm $+1$, $y<(ϵ-1/ϵ)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $ϵ$. For any positive unit, multiply by a suitable integer power of $ϵ$ to place it in $[1,ϵ)$; it must then equal one. HenceTo classify the norm-six elements, a principal ideal generated by such an element has norm 6. Unique prime-ideal factorization and part (b) force it to be either ${\mathfrak{p}}_2{\mathfrak{p}}_3=(ϵ+1)$ or ${\mathfrak{p}}_2{\overline{\mathfrak{p}}}_3=(ϵ-1)$. Therefore its generator is an associate of one of these two elements. Since their norms are respectively 6 and $-6$, the unit multiplying the first must have even exponent and the unit multiplying the second must have odd exponent. Combining the two cases gives all integer solutions:Conversely every displayed element is integral, since $ϵ$ is a unit, and has norm 6. This proves completeness of the norm-six equation in the quadratic field of square root ten.

The PDF's restriction to $n=0,1,2,\dots$ misses conjugate solutions. In particular $(x,y)=(4,-1)$ satisfies the equation but is not in that restricted family; it is the $n=-1$ member. A nonnegative-index version becomes correct by adding the conjugates of every displayed element, with both signs. The restriction as printed cannot describe all signed integer pairs.

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.

A canonical transformation preserves the symplectic form and therefore the form of Hamilton's equations for arbitrary Hamiltonians. In one degree of freedom,Thus the transformation is canonical exactly when $[Q,P]=1$. Equivalently the chain rule gives $[F,G{]}_{q,p}=[Q,P{]}_{q,p}[F,G{]}_{Q,P}$ for arbitrary transformed functions, so the bracket and Hamiltonian dynamics are preserved precisely with this factor one. The condition also makes the Jacobian nonzero and hence supplies local invertibility; a global canonical change of variables additionally requires the stipulated global bijectivity on its domain.