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For a stable one-species linear noise approximation, the normalized stationary fluctuation-dissipation relation for a reaction network becomesTo see the factors explicitly, the linearized restoring rate is $a=H/\tau =4\beta m$, while the chemical noise intensity is the rate-weighted squared step sizeStationarity of the variance gives $2a\mathrm{Var}x=B$. Using $\lambda =2\beta m^2$ therefore yieldsThis is the stationary fluctuation result within the stated negligible-fluctuation/linear-noise approximation, not an exact identity for the boundary-corrected discrete chemistry.

Let $J_{+}=\lambda$ and $J_{-}=2\beta m^2$ be the production and loss fluxes, counted in molecules per unit time. At the steady state, both equal $J=\lambda$. The restoring elasticity and turnover lifetime areFlux-weighting the magnitudes of the reaction steps, one for production and two for loss, givesThis is an average over molecular fluxes, as requested, not an average over reaction-event counts; those are different when a reaction removes two molecules at once.

Apply the formal infinitesimal generator to the molecule count: its increment is $+1$ at rate $\lambda$ and $-2$ at rate $\beta x^2$. Thus the exact first-moment identity for the specified generator isNeglecting fluctuations gives the mean-field approximation $\dot{m}=\lambda -2\beta m^2$, so the positive steady mean isThere is a physical boundary qualification: the printed loss propensity is positive at $x=1$ and would produce a negative molecule count. Consequently it does not by itself define a molecule-count chain on all nonnegative integers. If losses are forbidden below two, the exact loss moment becomes $2\beta E[x^2{1}_{\{x\ge 2\}}]$; a mass-action two-molecule propensity instead uses $x(x-1)$. The calculation requested here uses the printed formal generator and its large-copy-number approximation, where such boundary corrections are negligible.

The given Riccati ansatz is $v^{\prime }=-i{v}^2+iz/6$. Differentiating it and substituting for $v^{\prime }$ showsIt therefore selects $C=i/6$ in the preceding scaled Painleve equation. Write $v=-i{\Psi }^{\prime }/\Psi$. The quadratic terms cancel in the ansatz, leavingThus every nonzero solution of the indicated Airy equation gives the local mKdV solutionon intervals where $\Psi \ne 0$. For example $\Psi (z)=A\mathrm{Ai}(-z/6^{1/3})+B\mathrm{Bi}(-z/6^{1/3})$, with constants not both zero, has the required equation. This branch is generally complex and meromorphic; the printed ansatz itself includes $i$. A real $v$ on a real interval would require simultaneously $v^{\prime }=0$ and $v^2=z/6$, which is impossible on an interval. Hence this particular ansatz does not furnish a nontrivial real-valued mKdV profile.

The scaling of the equation suggests $q=t^{-1/3}v(z)$, $z=x{t}^{-1/3}$, for $t>0$. Substitution givesIntegrating in $z$ gives the scaled Painlevé II equationwhere $C$ is an arbitrary integration constant. If the standard normalization $w^{\prime \prime }=2w^3+sw+\alpha$ is desired, set $z=3^{1/3}s$ and $v(z)=i{3}^{-1/3}w(s)$. Then $\alpha =-3iC$ andThe direct scaled equation avoids introducing imaginary variables when a real $v$ is required.

For a localized traveling wave put $q(x,t)=Q(\xi )$ with $\xi =x-ct-x_0$, and require $Q,Q^{\prime },Q^{\prime \prime }\to 0$ at infinity. The modified KdV equation integrates once to $Q^{\prime \prime }-cQ+2Q^3=0$. Multiplying by $Q^{\prime }$ and integrating again givesA nonzero decaying real solution requires $c>0$. Separating on the rising or falling half of the wave and using the indicated inverse-hyperbolic-secant integral yieldsThe signs and translations are arbitrary. Differentiating the sech profile verifies $Q^{\prime \prime }=cQ-2Q^3$ directly. The relevant antiderivative is the inverse hyperbolic secant, since $d(\mathrm{arcsech}x)/dx=-1/[x\sqrt{1-x^2}]$ for $0<x<1$.

Set $r=p/q$. If $r\notin [1,2]$, then $|\sqrt{2}-r|\ge \sqrt{2}-1>1/4\ge 1/(4q^2)$. If $1\le r\le 2$, irrationality of the square root of two implies that the integer $2q^2-p^2$ is nonzero. Factoring a difference of squares therefore givesThus the requested bound holds for every integer $p$ and every nonzero integer $q$, without using the Liouville approximation theorem. For completeness, irrationality follows because $p^2=2q^2$ in lowest terms would make $p$ even and then $q$ even, contradicting coprimality.

The Liouville approximation theorem states that if $\alpha$ is a real irrational algebraic number of degree $d$, there is a constant $C(\alpha )>0$ such that, for all integers $p,q$ with $q\ne 0$,Reduction of the fraction is not necessary: an unreduced denominator only weakens this bound. More generally the same type of bound holds for an algebraic number of degree $d$ whenever $p/q\ne \alpha$. The exception is essential for a rational $\alpha$, since an exact rational representation gives zero error. A rational number of degree one obeys the bound directly: if $\alpha =a/b$ in lowest terms and $p/q\ne \alpha$, then $|\alpha -p/q|=|aq-bp|/(|bq|)\ge 1/(|bq|)$.

The equalities are valid at a differentiable interior optimum under the intended concave formulation, but not under the printed assumptions alone. To derive the intended result, differentiate the finite conditional sum in the Bellman objective. For an interior portfolio fraction with positive invested wealth, its derivative givesThe consumption derivative gives $U^{\prime }(c_n^{\ast })=E_n[V_{n+1}^{\prime }(w_{n+1}^{\ast })G_{{\theta }_n^{\ast }}]$. Since $G_{\theta }=r+\theta (X-r)$, the first equality cancels its risky part and yields the second requested equality. The envelope derivative with respect to wealth gives $V_n^{\prime }(w_n^{\ast })=E_n[V_{n+1}^{\prime }(w_{n+1}^{\ast })G_{{\theta }_n^{\ast }}]$, giving the third. Thus, under these added hypotheses,The first prices the excess stock payoff using future marginal utility; the second equates marginal utility from consuming now to that from saving in the bank; the third identifies the marginal value of wealth with that common quantity. At portfolio or consumption boundaries these are replaced by one-sided optimality inequalities. Finiteness of the sample space justifies differentiating a finite sum; it does not ensure differentiability of a supremum or an interior optimum.

A concrete counterexample satisfies the printed convex assumptions and attains every supremum. Take one remaining decision, a one-point sample space, a deterministic stock return $X=x>r>1$, $U(c)=c^2$, and controls in $[0,w]\times [0,1]$. The optimum is $c^{\ast }=0,{\theta }^{\ast }=1$, since the objective $c^2+x^2(w-c{)}^2$ is convex in $c$ and its larger endpoint value is $x^2{w}^2$. Then $w_{n+1}^{\ast }=xw$ and $V_{n+1}^{\prime }=2xw$. The first claimed expectation is $2xw(x-r)>0$; the second compares $2rxw>0$ to $U^{\prime }(0)=0$; the third compares $2rxw$ to $V_n^{\prime }(w)=2x^{2}w$. All three fail. Consequently the requested equalities cannot be proved as printed; the derivation above supplies their precise intended interior version.

Here $U^{\prime }(x)=x^{-R}>0$ and $U^{\prime \prime }(x)=-R{x}^{-R-1}<0$: the specified utility is strictly concave, contrary to the word “convex” in part (i). Assume its expected values are finite on the admissible set. Define $G_{\theta }=r+\theta (X-r)$ and choose a constant portfolio fraction ${\theta }_{\ast }$ maximizing $E[U(G_{\theta })]$. Equivalently maximize $E[G_{\theta }^{1-R}]$ for $0<R<1$ and minimize it for $R>1$. An interior optimum satisfiesFor a constrained endpoint optimum the corresponding derivative has the appropriate one-sided sign. The objective is concave in $\theta$ and strictly concave if $P(X\ne r)>0$, so this condition or its endpoint version determines the optimum. Put $\kappa =E[G_{{\theta }_{\ast }}^{1-R}]$ and $\gamma ={\kappa }^{1/R}>0$.

Backward induction with independence of the returns gives $V_n(w)=A_n{w}^{1-R}/(1-R)$ and $A_N=1$. Maximizing the consumption expressiongives $c^{-R}=A_{n+1}\kappa (w-c{)}^{-R}$, henceWriting $b_n=A_n^{1/R}$ yields $b_N=1$ and $b_n=1+\gamma b_{n+1}$. Therefore the fully explicit horizon dependence isUse $b_n=(1-{\gamma }^{N-n+1})/(1-\gamma )$ when $\gamma \ne 1$, and $b_n=N-n+1$ when $\gamma =1$. The investor consumes this fraction, invests the remaining wealth with the same optimal portfolio mix at each date, and consumes all remaining wealth at $N$. If the stock is identical to the bank almost surely, every portfolio fraction is equivalent.

Let ${\mathcal{F}}_n$ represent the information at decision time $n$ and $E_n$ conditional expectation given it. $V_n(w)$ is the largest conditional expected utility from the remaining consumption opportunities with current wealth $w$; it may depend on the information state as well as $w$. The controls $c$ and $\theta$ are chosen using current information. In the usual allocation interpretation $0\le c\le w$ and $0\le \theta \le 1$, and next wealth is $(w-c)[r+\theta (X_{n+1}-r)]$. Other borrowing/short-selling rules require explicitly replacing this admissible set while preserving solvency.

The Bellman equation follows by separating current utility from future utility: after any current decision the best continuation has conditional value $V_{n+1}$, and maximizing over the current admissible decisions gives the stated recursion. At the last opportunity all wealth is consumed, because utility is increasing, giving $V_N(w)=U(w)$.

There are two printed qualifications. Strict convexity is not needed for this recursion, but it conflicts with the concave power utility in part (ii) and does not support the interior maximizing first-order conditions in part (iii). Also the displayed recursion starting at $n=0$ includes consumption at times $0,\dots ,N$, whereas the preceding sum starts at one. The same recursion represents that sum when begun at time one; alternatively relabel the consumption dates. Below the policy is given for the recursion's remaining horizon, so this indexing discrepancy does not leave the optimization unspecified.

Assume $I,J\ge 2$. The additive two-way model has dimension $1+(I-1)+(J-1)=I+J-1$; the null model, with no exam effects, has dimension $I$. Write row, column and grand means as ${\overline{X}}_{i\cdot },{\overline{X}}_{\cdot j},{\overline{X}}_{\cdot \cdot }$. Orthogonal projection onto the full model gives fitted values ${\overline{X}}_{i\cdot }+{\overline{X}}_{\cdot j}-{\overline{X}}_{\cdot \cdot }$, while the null fit is ${\overline{X}}_{i\cdot }$. The added exam sum of squares and full residual sum of squares areThe preceding orthogonal-projection result gives independent null chi-squares with $J-1$ and $(I-1)(J-1)$ degrees of freedom. ThereforeReject the no-exam-effect hypothesis for a large value. The printed alternative repeats the identifying constraint ${\sum }_j{\beta }_j=0$; a genuine alternative additionally has at least one nonzero ${\beta }_j$. No interaction term is being fitted; the independent equal-variance Gaussian error assumption is what makes this exact test valid.

Let $Q=P_1-P_0$ and $R=I-P_1$. For nested subspaces these are orthogonal projections onto ${\Theta }_1\cap {\Theta }_0^{\perp }$ and ${\Theta }_1^{\perp }$, of ranks $d_1-d_0$ and $d-d_1$. Under the null, both annihilate $\mu$. In an orthonormal basis adapted to ${\Theta }_0$, these two spaces and their complement, the coordinates of $X-\mu$ are independent $N(0,{\sigma }^2)$. Consequentlyand the two variables are independent. Equivalently the joint probability density of the two unscaled sums $a,b>0$ factors into gamma densities of respective shapes $(d-d_1)/2,(d_1-d_0)/2$ and common scale $2{\sigma }^2$.

The nuisance variance cancels inReject for a value above the upper $\alpha$ critical quantile to obtain a level-$\alpha$ test. Under a genuine alternative the numerator is noncentral chi-square with noncentrality $\Vert Q\mu {\Vert }^2/{\sigma }^2$, while the denominator retains its central law, explaining the upper-tail choice.

Complex conjugation permutes the irreducible characters, because the conjugate of an irreducible representation is irreducible. This involution fixes the real-valued characters and interchanges the other characters in pairs. If there are $r$ such pairs, then conjugating $A$ permutes its rows by $r$ transpositions. ThereforeIt follows that the determinant is real when $r$ is even and purely imaginary when $r$ is odd. It never vanishes, by the preceding part. Real-valued characters include both real and quaternionic representation types; the criterion concerns values of the characters, not those two representation types separately.

The matrix $A$ is square by completeness. Column orthogonality says $A^{\ast }A=\mathrm{diag}(|Z(g_j)|)$. Taking determinants givesIn particular its determinant is nonzero.

Apply completeness to the indicator $f_j$ of the conjugacy class of $g_j$. Its expansion coefficient is $⟨{\chi }_i,f_j⟩=\overline{{\chi }_i(g_j)}/|Z(g_j)|$, because that class has size $|G|/|Z(g_j)|$. Evaluating at $g_k$ givesThe complex conjugation is part of the printed formula; dropping it would not give the general column orthogonality relation.

The irreducible characters are an orthonormal basis of the space of complex class functions for the inner productThus the row orthogonality relation and completeness statement areIn particular the number of irreducible characters equals the number of conjugacy classes. In class-representative notation, the orthogonality relation is ${\sum }_j\overline{{\chi }_i(g_j)}{\chi }_k(g_j)/|Z(g_j)|={\delta }_{ik}$.

For the indicated cyclic Galois resolvent, reindexing its terms and using $g^m=1$ givesThere exists $w$ with $R(w)\ne 0$ because distinct field automorphisms are linearly independent as functions. A short proof is as follows. Take a nonzero relation with the fewest nonzero terms, normalize one coefficient, evaluate it at $uv$ and subtract a scalar multiple of the relation at $v$. Choose $u$ on which two participating automorphisms differ. This produces a nonzero relation with fewer terms, a contradiction.

Set $\alpha =R(w)\ne 0$. Then $g^j\alpha ={\zeta }^j\alpha$ are $m$ distinct conjugates. Also $g({\alpha }^m)={\alpha }^m$, so $a={\alpha }^m\in K$. The orbit has size $m$, giving $[K(\alpha ):K]=m$ and hence $K(\alpha )=M$. All the roots ${\zeta }^j\alpha$ belong to $M$, soThis also treats $m=1$ with a nonzero $w\in K$.

If $a=0$, the splitting field is $K$ and its Galois group is trivial. Otherwise choose a root $\alpha$ and a primitive $n$th root of unity $\zeta \in K$. Every root is ${\zeta }^j\alpha$, so $L=K(\alpha )$. Each automorphism is determined by $\sigma (\alpha )/\alpha \in {\mu }_n$. This is an injective group homomorphism, since the roots of unity are fixed by every automorphism. A subgroup of the cyclic group ${\mu }_n$ is cyclic. Thus $\mathrm{Gal}(L/K)$ is cyclic, with order dividing $n$.

This assertion is always true. Ordinal addition obeys $\alpha +\beta \ge \alpha$ and $\alpha +\beta \ge \beta$. Thus both summands are at most ${\omega }_1$. If $0<\beta <{\omega }_1$ and $\alpha <{\omega }_1$, both ordinals are countable, and their disjoint ordered union is countable, so their sum cannot be ${\omega }_1$. If instead $\alpha ={\omega }_1$ and $\beta >0$, appending a nonempty second block gives a sum strictly larger than ${\omega }_1$. These exhaust the possibilities with $0<\beta <{\omega }_1$; hence $\beta =0$ or $\beta ={\omega }_1$.