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Integrating the vector-field equations gives the one-parameter transformationsUnder this scaling, $v_t,v^2{v}_x,v_{xxx}$ have weights $\gamma -\beta ,3\gamma -\alpha ,\gamma -3\alpha$. Equality of all three weights givesThus the nontrivial Lie point symmetry generator can be normalized to $x{\partial }_x+3t{\partial }_t-v{\partial }_v$.

For the Miura transformation $u=v^2+v_x$, direct differentiation gives the operator identityFor example $u_x=2v{v}_x+v_{xx}$ and $u_{xxx}=6v_x{v}_{xx}+2v{v}_{xxx}+v_{xxxx}$; inserting them makes the $v_x{v}_{xx}$ terms cancel and leaves the right side. Hence $u$ solves $u_t-6u{u}_x+u_{xxx}=0$ whenever $v$ solves the mKdV equation. Both $v^2$ and $v_x$ have weight $-2\alpha$, so the corresponding Korteweg-De Vries equation symmetry is

For the WKB approximation, put $Q(x)=|x|-E$. Away from the turning points $x=\pm E$, the forbidden-region solutions have the form $Q^{-1/4}\exp (\pm \int \sqrt{Q}dx)$; in the allowed region put $k(x)=\sqrt{E-|x|}$ and use $k^{-1/2}\exp (\pm i\int kdx)$. The decaying tails selected by the boundary conditions areThe simple-turning-point Airy connection formula matches these toCompatibility requires the Bohr-Sommerfeld quantization conditionwith levels indexed from $n=0$. ThereforeThe cusp at zero does not alter the leading high-energy count. An exact matching check is available: on $x\ge 0$ the decaying solution is $\mathrm{Ai}(x-E)$, and parity at zero imposes ${\mathrm{Ai}}^{\prime }(-E)=0$ for even states or $\mathrm{Ai}(-E)=0$ for odd states. Their interlacing large negative zeros give the same combined leading quantization. Reindexing from one changes only the constant phase shift, not the requested growth law.

A state-price density is a strictly positive random variable $Z$ withOn the stock-state space this means a positive integrable function $z(s)$ satisfying ${\int }_0^{C}z(s)ds=C/R$ and ${\int }_0^{C}sz(s)ds=C$. Equivalently $dQ/dP=RZ$ is an equivalent probability measure with ${\mathbb{E}}_Q{S}_1=R$. Such measures exist because $0<R<C$, for example exponential tilts of the uniform law with mean tuned to $R$.

By the fundamental theorem of asset pricing, any arbitrage-free claim price $p$ is represented by such a density in the enlarged market. For payout $S_1^2$,with strictness because equivalence preserves the stock's nondegenerate distribution. Also $S_1^2<C{S}_1$ almost surely, so $p<C$. Hence $R<p<C$.

When $C=3R$, nonnegative wealth for all stock states requires $-1/2\le \pi \le 1$. At $\pi =1$ wealth is $S_1$. The tangent inequality for the strictly concave square root gives, for $s>0$,Uniform integration gives $\mathbb{E}\sqrt{S_1}=2\sqrt{C}/3$ and $\mathbb{E}{S}_1^{-1/2}=2/\sqrt{C}$, so the expected linear term vanishes exactly when $C=3R$. Strict concavity makes equality possible only for $\pi =1$. Thus the unique optimal holding is ${\pi }^{\ast }=1$, including this boundary optimum without assuming an interior maximizer.

The marginal utility price density isIt is integrable and satisfies both pricing constraints: $\mathbb{E}{Z}_{\ast }=1/R$ and $\mathbb{E}(Z_{\ast }{S}_1)=C/(3R)=1$. Therefore

A control policy specifies each control as a nonanticipating measurable function of the observed history, possibly with additional randomization. It may depend on time and the entire past; a stationary Markov policy uses only the current state. We use the standard admissibility assumptions allowing measurable choices and concatenation of continuation policies.

The dynamic programming operator is monotone: $f\le g$ implies $\mathcal{L}f\le \mathcal{L}g$. Nonnegative costs give $F_1\ge F_0=0$, so induction gives $F_{t+1}\ge F_t$. Thus $F_t$ increases pointwise to an extended nonnegative limit $F_{\infty }$. Backward induction identifies $F_t$ with the infimum of expected costs for the first $t$ periods: condition on the first control and next state, then optimize the continuation. Every infinite-horizon policy has at least that finite-horizon cost, so

For the Bellman equation, any infinite-horizon policy, conditional on its first action and next state, has continuation cost at least $F(X_1)$. Its total cost is therefore at least $\mathcal{L}F(x)$; infimizing gives $F\ge \mathcal{L}F$. For the reverse inequality at a state with finite $\mathcal{L}F(x)$, choose a first action within $\epsilon /2$ of its infimum and concatenate continuation policies whose costs are within $\epsilon /(2\beta )$ of $F(y)$ at each next state $y$. This produces a policy of cost at most $\mathcal{L}F(x)+\epsilon$. Such continuations are needed only at states of finite value, which have probability one under an action with finite continuation expectation. Let $\epsilon \downarrow 0$. If $\mathcal{L}F(x)=\infty$, the first inequality already gives $F(x)=\infty$. Therefore

Let $\Phi =\mathcal{L}\Phi \ge 0$, and use the permitted selector $u_{\ast }$. Under the stationary policy $u_{\ast }(X_t)$, repeated conditional expectation givesIf $\Phi (x)$ is infinite the desired comparison is immediate; otherwise these nonnegative quantities have finite expectations and the displayed recursion is legitimate. Drop the terminal term and apply monotone convergence theorem as $T\to \infty$. The policy's infinite-horizon cost is at most $\Phi (x)$, soThus the value is the least nonnegative Bellman fixed point. The argument works also for $\beta =1$ because costs and the dropped terminal term are nonnegative; it makes no unwarranted contraction or finite-value assumption.

In the normal form of a Bayesian decision problem, choose a measurable decision rule $\delta$ before seeing the data to minimize the prior Bayes risk $\mathbb{E}L(\Theta ,\delta (X))$. In the extensive form, after observing $X=x$, choose an action minimizing posterior expected loss. Conditional expectation makes these equivalent, subject to the usual measurable selection conditions. If ${\Pi }_x$ is the posterior distribution and $R_0(\Pi )={\inf }_a\int L(\theta ,a)\Pi (d\theta )$ is the best no-data loss, thenThe inequality follows by ignoring the data.

For a measurable region $S$, posterior or prior expected loss is ${\int }_S(c-\pi (\theta ))d\theta$. It is minimized by including exactly the locations where the integrand is negative; including equality points is harmless. Thus the highest density region $S^{\ast }=\{\pi \ge c\}$ is a Bayes act andFor $N(\mu ,{\sigma }^2)$, if $c\ge 1/(\sigma \sqrt{2\pi })$ the answer is zero. Otherwise put $z_c=\sqrt{2\log [1/(c\sigma \sqrt{2\pi })]}$. The optimal interval is $[\mu -\sigma z_c,\mu +\sigma z_c]$, andFor the specified prior and $c=1/2$, the prior density is everywhere below $c$, so $R_0=0$. Completing the square gives $\Theta \mid X=x\sim N(9x/10,1/10)$. Its peak is $\sqrt{10/(2\pi )}>1/2$, so a positive-measure region has posterior density strictly above $c$ and its optimal expected loss is strictly negative. That loss is independent of its mean, hence independent of $x$. Therefore $R_1<0=R_0$. Explicitly, with $z_c=\sqrt{2\log (2\sqrt{10/(2\pi )})}$, $R_1=z_c/\sqrt{10}-2\Phi (z_c)+1$.

For $0\le \alpha <1$, symmetry gives $\mathbb{E}{Y}_1=0$ andbecause the tail integrand is $O(x^{\alpha -2})$. The strong law of large numbers for independent identically distributed integrable variables therefore gives $S_n/n\to 0$ almost surely and hence in distribution. At $\alpha =0$, $Y_1$ is an equally likely sign, which is included in this argument.

For $\alpha =1$, $Y_1=X_1$ has the standard Cauchy distribution. Its characteristic function isFor $u>0$, close a contour in the upper half-plane; the pole at $i$ has residue $e^{-u}/(2i)$ and the semicircle integral vanishes. For $u<0$, symmetry gives the same value with $|u|$, and $u=0$ gives one. Independence now yields $\mathbb{E}{e}^{iu{S}_n/n}=\phi (u/n{)}^n=e^{-|u|}$. Thus $S_n/n$ is exactly standard Cauchy for every $n$. Consequently

For $0\le \alpha <1/2$, the variance is finite and equals $\mathbb{E}{Y}_1^2=2m(2\alpha )$, using the integral notation in the question. The central limit theorem for independent identically distributed variables with finite variance givesIn particular $m(0)=1/2$ gives variance one at $\alpha =0$. The invoked implications from almost-sure to distributional convergence and the characteristic-function uniqueness theorem are the standard probability results used above.

A meromorphic function on the Riemann sphere is rational: it has finitely many poles; subtract their principal parts, including a polynomial for a possible pole at infinity, and the remaining entire function is bounded at infinity, hence constant by Liouville theorem. A rational map of degree $d$ has $d$ preimages of a regular value, counted with multiplicity. An analytic isomorphism is bijective, so $d=1$, giving preciselyConversely these Möbius transformations have a Möbius inverse and extend holomorphically at their pole and infinity, so are analytic isomorphisms.

For a nonconstant analytic map $f:X\to Y$ between compact connected Riemann surfaces, the Riemann-Hurwitz formula isHere $g_X,g_Y$ are genera, $d$ is the degree (the sum of local multiplicities over any fiber), and $e_p$ is the local degree in coordinates where the map has the form $z\mapsto z^{e_p}$. Only finitely many terms are nonzero; these are the ramification points.

For a degree-two sphere map the formula gives total ramification two. Each local degree is at most two, so there are exactly two distinct ramification points, each of local degree two. Their branch values are distinct, because one fiber cannot have multiplicity four. Choose $T$ sending $0,\infty$ to the two ramification points, and $S$ sending their values to $0,\infty$. Then $SfT$ has its only zero at zero, of order two, and its only pole at infinity, of order two; it is $c{z}^2$ with $c\ne 0$. Rescale $S$ by $1/c$. Therefore $SfT(z)=z^2$, as required.