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Put $x=\beta ϵ$. The one-particle canonical partition function isand independence gives $Z=z^N$. Therefore the independent symmetric three-level system hasand, using part (c),As $T\to \infty$, $x\to 0$, all three levels become equally likely andAs $T\to 0^{+}$, every particle occupies the nondegenerate ground level $-ϵ$, and

Substitution of the canonical probabilities into the Gibbs entropy givesHence

For the Gibbs entropyfirst maximize subject only to ${\sum }_n{p}_n=1$ over $\Omega$ accessible microstates. A Lagrange multiplier givesso every $p_n$ is equal. Normalization yields the microcanonical ensemble

For the canonical ensemble, impose both normalization and the mean-energy constraint ${\sum }_n{p}_n{E}_n=E$. Stationarity ofgives $p_n=C{e}^{-\beta E_n}$. The canonical partition function fixes $C$, soThermodynamic consistency identifies $\beta =(k_{B}T{)}^{-1}$. This is the maximum-entropy derivation of equilibrium ensembles.

The microcanonical ensemble describes an isolated system with fixed energy, volume, and particle number. It assigns equal probabilities to the accessible microstates in the chosen narrow energy shell. The canonical ensemble describes a system in thermal contact with a large heat reservoir: its temperature, volume, and particle number are fixed, while its energy can fluctuate.

Using the states from part (a),The Born rule therefore givesAfter the equal-sign outcomes the state is respectively $|\uparrow \uparrow ⟩$ or $|\downarrow \downarrow ⟩$. These are triplet states, hence eigenstates of ${\mathbf{S}}^2$ with eigenvalueThe opposite-sign product states are each a nontrivial linear combination of the singlet and the $m=0$ triplet, so they are not eigenstates of ${\mathbf{S}}^2$.

Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. ThenThus $|\psi ⟩$ is unchanged. It is the spin-one-half singlet state, so joint rotations leave it invariant. Equivalently, direct use of the Pauli matrices showsIt is therefore an eigenstate of every Cartesian component of the combined spin, with eigenvalue zero.

Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$,Its normalized eigenstates areand

An interval map $F:I\to I$ has a horseshoe if there are two closed subintervals $J_0,J_1\subseteq I$ with disjoint interiors such thatIt is chaotic in Glendinning's sense if some positive iterate $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a 3-cycle and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. Ifthen the intermediate value theorem givesConsequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If insteadthenand again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

Recall that positivity and asymptoticity meanfor each fixed $j$. For $0\le k\le n$, compare the $k$th term of
${\chi }_{n+1}$ with the $k$th term of ${\chi }_n$:For the remaining term, compare with the final term of ${\chi }_n$:Since ${\chi }_n$ is positive and dominates every comparison term, summing these finitely many bounds givesThus $({\chi }_n)$ is an asymptotic sequence.

Apply the bounded-differences inequality with every
$L_i=2\beta +M/n$. With probability at least $1-\delta$,Using part (c) and substituting the definition of $F$ gives

Let $Z=(X,Y)$ be an independent test example and write $Z_i=(X_i,Y_i)$. Exchangeability of $Z$ and $Z_i$ givesThe datasets $D_i(Z)$ and $D$ differ in one coordinate, so stability givesAveraging over $i$ and subtracting the empirical loss proves

Replace the $i$th example of $D$ to obtain $D^{\prime }$. The population-loss term changes by at most $\beta$ by the assumed uniform stability of a learning algorithm. In the empirical average, each of the $n-1$ unchanged examples contributes a change at most $\beta$. For the replaced term, first change the hypothesis, costing at most $\beta$, and then change the evaluated example; boundedness of the loss costs at most $M$. Thus the empirical term changes by at mostCombining the two terms gives

A function has the bounded-differences property with constants
$L_1,\dots ,L_n$ when replacing only coordinate $i$ changes its value by at most $L_i$. The Bounded differences inequality states that, for independent inputs,and the analogous lower-tail bound also holds.

For a symmetric matrix,
$\mathrm{range}V=(\ker V{)}^{\perp }$. Since $q\notin \mathrm{range}V$, there is
$\eta \in \ker V$ with ${\eta }^{T}q\ne 0$. For any target excess mean $h$, setThen ${\theta }^{T}q=h$ andThus the required minimum is zero for every $m,x$.

Choose the sign of $\eta$ so that ${\eta }^{T}q>0$, buy the risky portfolio $\eta$, and finance it by borrowing its time-zero cost in the bank. Its initial wealth is zero, while its terminal excess payoff has variance zero and positive mean ${\eta }^{T}q$, so it is a strictly positive constant almost surely. This is an arbitrage.