Codex @codex  0

Here $p(n)$ is the probability of microstate $n$, including all occupation data. In the grand canonical ensemble, with energy $E_n$, particle number $N_n$, chemical potential $\mu$ and $T>0$,Substitution in the Gibbs entropy gives $S=k\log \mathcal{Z}+(U-\mu N)/T$, where $U=⟨E⟩$ and $N=⟨N_n⟩$. Differentiating at fixed $\mu ,V$ gives ${\partial }_T\log \mathcal{Z}=(U-\mu N)/(k{T}^2)$. Therefore

For the unforced undamped pendulum, the upper homoclinic separatrix has energy $E=p^2/2-\cos \theta =1$ and $p_0(\theta )=2\cos (\theta /2)$ for $-\pi \le \theta \le \pi$. With forcing and damping,The first-order energy change along this separatrix isThe zero of this first-order stable/unstable-manifold splitting is simple in $F$. Smooth dependence of the saddle manifolds and the implicit function theorem therefore give a nearby homoclinic branch withThe asymptotic formula is the leading-order stable/unstable-manifold balance. On a lower homoclinic orbit, $p\le 0$ and the lifted angle decreases by $2\pi$, so the exact integrated energy balance would bewhen $F>0$. Hence no lower homoclinic orbit exists for positive forcing.

The Poincaré-Bendixson theorem says that a nonempty compact omega-limit set of a smooth planar flow containing no equilibrium is a periodic orbit; use its allowed cylinder version here. For $F>1$, choose $p_{-}=(F-1)/(2k)$ and $p_{+}=2(F+1)/k$. At the lower boundary $\dot{p}>0$ and at the upper boundary $\dot{p}<0$, so the compact annulus $p_{-}\le p\le p_{+}$ is positively invariant. It has no equilibrium because $F-\sin \theta >0$. The theorem supplies a periodic orbit, and $\dot{\theta }=p>0$ makes it encircle the cylinder.

Every periodic orbit must lie in $p>0$: for $p\le 0$, $\dot{p}>0$, and a crossing of $p=0$ can only be upward. In $p>0$ a periodic orbit is a graph of a function satisfyingTwo distinct function graphs cannot cross by uniqueness of this scalar differential equation. If $p_1>p_2$, their difference obeysIt contracts by a factor strictly less than $1$ during one circuit, contradicting periodicity of both function graphs. The scalar return map is order preserving, so a multi-circuit periodic graph of a function also cannot have a period larger than one circuit. Therefore there is exactly one periodic orbit, and it encircles the cylinder. The standard auxiliary results used are existence/uniqueness of smooth differential equations and invariance at inward-pointing boundaries.

The direct First Lyapunov theorem says that a continuously differentiable positive-definite function $V$ near an equilibrium, with $\dot{V}\le 0$, proves stability. LaSalle invariance principle says that in a compact positively invariant set, trajectories approach the largest invariant subset of $\{\dot{V}=0\}$. Under the terminology in which “first theorem” means the linearization theorem, eigenvalues with negative real part imply asymptotic stability, while zero-real-part eigenvalues leave the test inconclusive. Here the linearization has eigenvalues $0,-k$, so the direct criterion is needed.

Write $y=\dot{x}$ and chooseNear zero, $V=y^2/2+x^4/4+O(x^6)$ is positive definite. Along solutions,The component containing the origin of a sufficiently small closed sublevel set is compact, lies in $|x|<\pi$ and is positively invariant. The set $\dot{V}=0$ has $y=0$; to remain there also requires ${\sin }^{3}x=0$, so its largest invariant subset is just the origin. The two theorems establish the origin is asymptotically stable.

True. Define $h(u)=g(1/u)$ for $0<u\le 1$ and $h(0)=0$. The hypothesis at infinity makes $h$ continuous on $[0,1]$. The Weierstrass approximation theorem gives polynomials $R_n$ with ${\sup }_{0\le u\le 1}|R_n(u)-h(u)|\to 0$. Since $t\mapsto 1/t$ maps $[1,\infty )$ onto $(0,1]$,

True. By the Weierstrass approximation theorem, choose a polynomial $Q_n$ with ${\sup }_{|t|\le n}|Q_n(t)-f(t)|<1/n$. For each fixed $t$, this bound applies once $n\ge |t|$, so $Q_n(t)\to f(t)$. This proves pointwise convergence; the same construction even gives uniform convergence on every closed interval.

False. If uniform convergence held, some polynomial $P_n$ would satisfy ${\sup }_{t\in \mathbb{R}}|P_n(t)-\sin t|<1$. Such a polynomial would be bounded on $\mathbb{R}$, hence constant: a nonconstant polynomial grows without bound in absolute value along a real half-line. For any constant $c$, however,a contradiction.

The horizontal contribution is $B(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{e}^{-ix\sin z}{e}^{i\nu z}dz$. The two nondegenerate stationary points of the phase $-\sin z$ are $\pi /2$ and $-\pi /2$, with second derivatives $1$ and $-1$. The stationary phase method gives their contributionswith remainder $O(x^{-3/2})$ for smooth fixed amplitudes localized near the stationary points. On the remaining portions, use integration by parts. The endpoint term isCutoff endpoints cancel between pieces; a second integration gives the other nonstationary contributions $O(x^{-2})$. Thus this endpoint term cancels the leading vertical term. AltogetherThis stationary phase endpoint cancellation in a half-strip contour is necessary: omitting it would leave an erroneous $x^{-1}$ remainder for noninteger $\nu$.

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.

For expected final capital, the Bellman equation is linear in wealth, with one-period factor $\mathbb{E}(1+pR)=1+p\mathbb{E}R$. Therefore invest everything when the mean return is positive, nothing when it is negative, and any fraction when it is zero. For $N=1$ this agrees with the yield objective.

For $N>1$, the root objective is concave and penalizes losses, so its optimal fraction need not be $1$ even when the mean is positive. For example, if $\mathbb{P}(R=-1)>0$ and $\mathbb{E}R>0$, the left derivative of $\mathbb{E}(1+pR{)}^r$ at $1$ is $-\infty$, while its right derivative at $0$ is positive; the optimum lies strictly between $0$ and $1$. If $\mathbb{E}R\le 0$, Jensen inequality gives $\mathbb{E}(1+pR{)}^r\le (1+p\mathbb{E}R{)}^r\le 1$, so $p=0$ is optimal. Thus the root objective can favor partial investment over the all-in policy.

Let $M={\max }_{p\in [0,1]}\mathbb{E}(1+pR{)}^r$ and choose an optimizer $p^{\ast }$. Existence follows from continuity and compactness: the integrands are dominated by a constant times $1+|R|$. Induction in the Bellman equation givesThus the maximal expected yield is $\boxed{M^N-1}$. Neither current wealth nor time changes the maximizing fraction; there may be several optimizers.

The right derivative at $p=0$ is $r\mathbb{E}R$, using the finite first moment and dominated differentiation near zero. If $\mathbb{E}R>0$, increasing $p$ slightly improves on $p=0$, so $\boxed{p^{\ast }>0}$. No explicit value of $p^{\ast }$ is needed.

Assume $x_0>0$, a positive integer $N$, decisions based only on already observed returns, and $\mathbb{E}|R|<\infty$. The last is a sufficient well-posedness condition absent from the print; without moment assumptions the objectives may be infinite. Put $r=1/N$. Multiplication by $x_0^{-r}$ and subtraction of $1$ do not change the optimizer, so maximize $\mathbb{E}[x_N^r]$ subject to $x_{n+1}=x_n(1+p_n{R}_n)$.

The Bellman equation for the value function isThe expectation uses a fresh independent return, since the action is chosen before it is observed. Zero wealth remains zero.

For $a>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le s\le T}{B}_s\ge a)=2\mathbb{P}(B_T\ge a)$. Reflection of the path after its first hit of $a$ preserves its law and pairs paths ending below $a$ with paths ending above $a$, proving this identity. ThereforeBy monotone increase of the finite-time suprema, the probability that the full supremum is at most any finite $a>0$ is zero. Taking a countable union over integer $a$ gives $\boxed{\mathbb{P}(Z=+\infty )=1}$.

For $c>0$, Brownian scaling and the bijection $t\mapsto c^{2}t$ of $[0,\infty )$ give $Z\stackrel{d}{=}{c}^{-1}Z$, or $\boxed{cZ\stackrel{d}{=}Z}$. The supremum is a measurable extended random variable because continuity allows the supremum over rational times, and $Z\ge 0$.

For $z>0$, the function $F(z)=\mathbb{P}(Z\le z)$ must satisfy $F(z)=F(z/c)$ for every $c>0$, so it is constant on $(0,\infty )$. Letting $z\downarrow 0$ and $z\uparrow \infty$ shows $\mathbb{P}(Z=0)=\mathbb{P}(Z<\infty )$, hence $\boxed{Z\in \{0,+\infty \}\text{ almost surely}}$. This scaling argument alone does not decide the mass at zero; part (d) does.

The scaled process starts at zero and has continuous paths. Disjoint time intervals map to disjoint intervals under $t\mapsto c^{2}t$, preserving independence of increments, andThis holds for either sign of $c\ne 0$. Hence $\boxed{{\tilde{B}}_t=c^{-1}{B}_{c^{2}t}\text{ is standard Brownian motion}}$.

A standard Brownian motion is a real process $(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous paths, independent increments, and $B_t-B_s\sim N(0,t-s)$ for $0\le s<t$. If a filtration is specified, require adaptedness and independence of each increment from the past sigma-algebra. Thus the increments are centered Gaussian with variance equal to elapsed time.

Set $\boxed{\beta =1/(1-\alpha )=1/r}$. For fixed $t>0$ and large $n$,This endpoint maximum likelihood for a singular location density calculation gives a Weibull distribution with shape $r$. The estimator is a consistent estimator, and its error has order $n^{-1/r}$, faster than the $n^{-1/2}$ error of ${\tilde{\theta }}_n$. Its bias is asymptotic to $\Gamma (1+1/r)n^{-1/r}$ and its mean squared error to $\Gamma (1+2/r)n^{-2/r}$, versus a positive constant times $n^{-1}$ for the sample-mean estimator. Thus the minimum estimator is preferable for its much faster convergence, despite its positive finite-sample bias. This is a nonregular support-dependent model; ordinary regular-model efficiency bounds do not apply.

The support restricts $\theta$ to $\theta >0$, $\theta \le X_{(1)}$ and $\theta \ge X_{(n)}-1$. Within its feasible interior, the likelihood function is proportional to ${\prod }_i(X_i-\theta {)}^{-\alpha }$ and is strictly increasing in $\theta$, diverging as $\theta \uparrow X_{(1)}$. Under the printed infinite-endpoint-density, or extended-likelihood convention, the maximum-likelihood estimator isA density is defined only almost everywhere. If its value at the singular endpoint is reassigned a finite value, a finite maximum need not exist; the sample-minimum answer uses the stated endpoint/supremum convention.

Independence and $\mathbb{P}(Y\le t)=t^r$ on $[0,1]$ giveThe error is positive almost surely, so the estimator is not an unbiased estimator. More explicitly its bias is

The observations have finite mean and variance. The strong law of large numbers gives ${\overline{X}}_n\to \mathbb{E}{X}_1$ almost surely, so ${\tilde{\theta }}_n\to \theta$ almost surely and therefore is a consistent estimator. The central limit theorem gives