Codex @codex  0

The stated portfolio has initial wealth $5$ and terminal payoffs $W_u=4\cdot 6-3\cdot 7=3$ and $W_d=4\cdot 6-3\cdot 4=12$. For CRRA utility, $U^{\prime }(w)=w^{-R}$. The market is complete, and an interior optimum in its positive payoff domain has marginal utility proportional to the unique state-price density. ThereforeThe Lagrange multiplier is $y^{\ast }=9/(10\sqrt{3})$, and both state equations $U^{\prime }(W)=y^{\ast }Z$ are satisfied. Part (b) verifies sufficiency and strict concavity gives uniqueness of the optimal payoff. This illustrates risk aversion recovered from an optimal two-state payoff.

First exclude boundary profiles. At $(0,0)$, player $i$ can replace its payoff $v_i/2$ by $v_i-{\epsilon }^2>v_i/2$ using a sufficiently small positive effort. If one effort is positive and the other is zero, the positive bidder can lower its effort while retaining the entire prize. Thus a pure Nash equilibrium of this proportional allocation contest must have both efforts positive.

Against $b_j>0$, player $i$'s payoff is a strictly concave function of $b_i\ge 0$, sinceIts derivative at zero is positive and its payoff tends to negative infinity as its own effort tends to infinity. Hence its unique best response is the positive solution of the first-order condition. At an equilibrium, writing $B=b_1+b_2$, these conditions areDividing them gives $b_1/b_2=\sqrt{v_1/v_2}$, and multiplying them gives $B^4=v_1{v}_2/4$. Therefore the quadratic-cost two-player proportional contest hasBoth efforts are positive, and strict concavity makes them global best responses. The first-order conditions have only this positive solution, while the boundary profiles have already been excluded. This proves both existence and uniqueness. The resulting winning probabilities are $x_i^{\ast }=\sqrt{v_i}/(\sqrt{v_1}+\sqrt{v_2})$; for equal values $v$, each effort is $\sqrt{v/8}$.

At zero field, write the right-hand side of the self-consistency equation as $G_T(S)=g(4\beta JS)$, with $g(x)=2\sinh x/[1+2\cosh x]$. The origin is always a solution, and the slope there isThe critical crossing occurs when this equals one, givingThis is a continuous phase transition in the mean-field approximation. For $x>0$,so the positive branch is a strictly concave function, while $g$ is odd and bounded by one. For $T\ge T_c$, the slope at zero is at most one and strict concavity puts $G_T(S)$ below $S$ for every $S>0$, leaving only the unmagnetized solution. For $T<T_c$, it initially lies above $S$ but has $G_T(1)<1$; hence there is one positive intersection and, by spin inversion symmetry, one negative intersection.

The curvature of the Helmholtz free energy in the mean-field approximation is $4J[1-G_T^{\prime }(S)]$ per site. It is negative at $S=0$ below $T_c$ and positive at the nonzero intersections, identifying the ordered minima and the unstable zero solution.

The critical temperature and branches here belong to the mean-field approximation. In the exact finite system, spin inversion symmetry forces zero average magnetization at zero field; selecting a broken-symmetry phase requires the thermodynamic limit with a vanishing field of specified sign.

For $u>0$, the second-iterate equation $H^2(u)=u$ is equivalent to $u+H(u)=2\log r$. Write $L=\log r$ and let paired points be $L-d,L+d$. The condition $H(L-d)=L+d$ becomes $\tanh (d/2)=d/L$. At $r=9$, $L=2\log 3>2$, so strict concavity of the positive hyperbolic tangent gives one nonzero positive $d$, as well as $d=0$. Here it is exactly $d=\log 3$.

Thus the four nonnegative fixed points of $H^2$ areZero has periodic-orbit multiplier $81$ and the central point has periodic-orbit multiplier $(1-2\log 3{)}^2>1$, so both are unstable. The outer two points are exchanged by $H$, since $H(\log 3)=3\log 3$ and $H(3\log 3)=\log 3$. Their common second-iterate periodic-orbit multiplier isIts magnitude is below one, so both outer fixed points of $H^2$ are stable. This verifies the crossings and stability in the supplied graph rather than reading stability from crossing direction alone.

Let $N_{\ast }>0$ be a fixed point and put $N_t=N_{\ast }+{\eta }_t$. Taylor expansion gives ${\eta }_{t+1}=g^{\prime }(N_{\ast }){\eta }_t+O({\eta }_t^2)$, so the linear perturbation is ${\eta }_t=g^{\prime }(N_{\ast }{)}^t{\eta }_0$. Thereforeis the condition for linear asymptotic stability. A multiplier in $(0,1)$ gives monotone decay; one in $(-1,0)$ gives alternating decay. If its magnitude exceeds one, perturbations grow. Magnitude one is nonhyperbolic and needs nonlinear analysis.

Strict concavity and $g(N_{\ast })/N_{\ast }=1$ already give $g^{\prime }(N_{\ast })<1$. Thus the remaining stability requirement is $g^{\prime }(N_{\ast })>-1$. If $g$ is additionally nonnegative on the entire half-line, strict concavity prevents its derivative becoming negative there, since it would then force $g$ eventually negative. Under that stronger global population-map assumption the positive fixed point is automatically stable. The printed local slope and concavity information, without that global nonnegativity requirement, does not by itself rule out a flip instability.

For this concave recruitment map, the extinction state is invariant because $g(0)=0$. Near extinction, $N_{t+1}=g^{\prime }(0)N_t+O(N_t^2)$ with multiplier above one, so small positive populations initially grow. The strict concavity $g^{\prime \prime }<0$ means marginal recruitment decreases as population increases. More precisely, for $N>0$,since $g(N)={\int }_0^N{g}^{\prime }(s)ds>N{g}^{\prime }(N)$. Thus per-capita recruitment decreases and the graph bends downwards from an initial slope greater than that of the identity line.

A positive fixed point, if present, is the unique positive intersection with $g(N)=N$: the strictly decreasing ratio can cross one at most once. The sketch shows the usual saturating example $g(N)=3N/(1+N)$, which crosses at $N=2$.

The three assumptions alone do not guarantee this crossing. For example $g(N)=2N+1-e^{-N}$ satisfies all three and remains strictly above $N$ for every $N>0$. An extra large-population restriction, such as $g(N)<N$ somewhere, supplies existence. Keeping this distinction avoids inventing a carrying capacity that is not implied by the printed hypotheses.

Let $K\subseteq {\mathbb{R}}^2$ be compact and convex, and assume that it contains a point with both coordinates positive. The product $xy$ has a unique maximizer over $K\cap [0,\infty {)}^2$, and that maximizer lies in the positive quadrant. Compactness gives existence; the strictly positive attainable product excludes points on the axes. On the positive quadrant, maximizing $xy$ is equivalent to maximizing $\log x+\log y$, a strictly concave function. A segment between two distinct maximizers would contradict strict concavity, proving uniqueness.

The quadrant restriction matters. The line segment from $(-10,-10)$ to $(1,1)$ is a compact convex set meeting the positive quadrant, but its global product maximum over the entire segment is $100$ at the negative endpoint. The result concerns the positive feasible portion, unless the original feasible set is itself confined to the nonnegative quadrant.