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A reciprocal vector must have integral multiples of $2\pi$ as its dot products with lattice translations. The translations along the conventional axes require $G=2\pi (h/a,k/a,l/b)$, and the body-centering translation requires $h+k+l$ to be even. ThusThis is a face-centered tetragonal lattice, with conventional dimensions $4\pi /a,4\pi /a,4\pi /b$. The primitive unit cell volumes are $\boxed{a^{2}b/2}$ in direct space and $\boxed{2(2\pi {)}^3/(a^{2}b)}$ in reciprocal space. Their conventional cell volumes are $a^{2}b$ and $8(2\pi {)}^3/(a^{2}b)$, containing two and four lattice points respectively.

The listed plane is $x/a+z/b=1$, whose normal reciprocal vector is $\boxed{G_0=2\pi (1/a,0,1/b)}$. It is an allowed primitive vector in that direction; half of it has nonintegral indices. The plane spacing is $d=2\pi /|G_0|=(a^{-2}+b^{-2}{)}^{-1/2}=4a/5$ when $b=4a/3$. Bragg's law gives $2d\sin \theta =m\lambda$, soThese are approximately $18.21^{\circ },38.68^{\circ },69.64^{\circ }$. Here $\theta$ is the glancing Bragg angle; the angle between incident and scattered beams is $2\theta$.

Represent spin by $J_i=(\hslash /2){\sigma }_i$ in the basis $|\uparrow ⟩,|\downarrow ⟩$. The Pauli matrices are Hermitian and satisfy $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$, giving the general angular momentum commutation relations $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$. The particular representation has $J^2=3{\hslash }^{2}I/4$ and $J_3$ eigenvalues $\pm \hslash /2$, the features specific to spin one-half.

In the stated order of two-particle basis vectors, the operator isIts eigenvalue-one vectors are $|\uparrow \uparrow ⟩$, $|\downarrow \downarrow ⟩$ and $(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$, all symmetric under interchange. The eigenvalue-minus-three vector is $(|\downarrow \uparrow ⟩-|\uparrow \downarrow ⟩)/\sqrt{2}$, antisymmetric under interchange. These are the triplet state and singlet state, respectively.

Identical spin-half particles are fermions. With no other degrees of freedom to carry antisymmetry, only the singlet is allowed. Indeed, for total angular momentum $J=(\hslash /2)({\sigma }^{(A)}+{\sigma }^{(B)})$,This gives $2{\hslash }^2$ on the triplet and zero on the singlet, corresponding to total spin one and zero, as required by addition of angular momentum.

Differentiate the two linear equations in opposite orders. Their compatibility for a fundamental matrix of solutions requires $(U_y+UV)\Phi =(V_x+VU)\Phi$, hence the zero-curvature representationFor the displayed matrices, direct multiplication makes all off-diagonal entries zero and leaves the diagonal entries $-u_{xy}+e^u-e^{-2u}$, $u_{xy}-e^u+e^{-2u}$ and zero. Thus the nonlinear compatibility equation is the Tzitzeica equationUnder $x\mapsto cx$, $y\mapsto c^{-1}y$, the two derivative factors cancel, so the equation is unchanged. The positive-$c$ component is generated by $\boxed{x{\partial }_x-y{\partial }_y}$; negative $c$ adds a disconnected simultaneous reflection. Locally away from the axes, invariance makes $u=w(z)$ with $z=xy$. Since $u_{xy}=w^{\prime }(z)+z{w}^{\prime \prime }(z)$, the reduced equation isThis scaling-invariant reduction of the Tzitzeica equation describes local group-invariant solutions on each orbit sector; smooth extension across the axes imposes corresponding compatibility at $z=0$.

Completing the square givesSince $Q$ is positive-definite, the unique minimizer is $\boxed{a=-Q^{-1}Sx}$ and the minimum is $\boxed{x^T(R-S^T{Q}^{-1}S)x}$.

The Bellman equation is $V(n,x)=x^T{\Pi }_{0}x$ andSuppose $V(k+1,z)=z^T{\Pi }_{r}z+{\gamma }_{k+1}$, where $r=n-k-1$. Zero noise mean gives the expected quadratic $(Ax+Ba{)}^T{\Pi }_r(Ax+Ba)+\mathrm{tr}({\Pi }_r{N}_{k+1})$. Complete the square again, with $Q_r=Q+B^T{\Pi }_{r}B$ and $S_r=S+B^T{\Pi }_{r}A$. This yields the discrete Riccati recurrenceThe matrices $Q_r$ are positive definite. The minimum of the nonnegative stage-plus-future quadratic is nonnegative, so each ${\Pi }_r$ is symmetric positive semidefinite. Starting from ${\gamma }_n=0$, the constants obeyInduction proves $V(k,x)=x^T{\Pi }_{n-k}x+{\gamma }_k$, with optimal feedback control $U_k=K_{n-k-1}{X}_k$.

An arbitrage is a self-financing strategy with zero initial value and terminal value nonnegative almost surely and strictly positive with positive probability. If an equivalent martingale measure $Q$ exists, a self-financing portfolio in the stated constant-numeraire setting is a $Q$-martingale. Its terminal expectation therefore equals zero. Equivalence preserves the positive-probability event of positive profit, which would make that expectation strictly positive. This contradiction proves no arbitrage in this finite-time finite-state setting.

For the numerical tree use the bank account growing by $1+r=5/4$ and discount stock prices accordingly. The risk-neutral up probabilities solve $q{S}_u+(1-q)S_d=(5/4)S$, givingThus the four path probabilities are $1/16,5/16,25/48,5/48$, respectively.

The American put option has terminal payoffs $0,0,0,5$. At the upper time-one node, both immediate exercise and continuation are zero. At the lower node, exercise gives three while discounted continuation gives $(4/5)(1/6)5=2/3$. Hence the option values at time one are $0,3$, and backward induction givesThe lower node should indeed be exercised immediately, since $3>2/3$.

The seller's initial replicating hedge holds $-1/6$ share and four units in the bank account, costing $3/2$. At the bad node it is worth $-2+5=3$, enough for immediate exercise. If the buyer fails to exercise, the remaining European liability costs only $2/3$: hold $-5/6$ share and $32/3$ in the bank account, whose final payoffs are zero at stock price16 and five at stock price10. Switching to that hedge freesas certain profit at time one, worth $35/12$ if banked until time two.

An exponential family has density $h(x)\exp \{\varphi \cdot T(x)-k(\varphi )\}$ on a parameter-independent support. For an independent and identically distributed sample, its likelihood factorizes through ${\sum }_{i}T(X_i)$, so the Fisher-Neyman factorization theorem gives a sufficient statistic whose dimension is the fixed dimension of $T$, independent of sample size.

Expanding the normal log-density gives ${\varphi }_1=\mu /v$, ${\varphi }_2=-1/(2v)$, andThe mean-value parameter is $\nabla k=(\mathbb{E}X,\mathbb{E}{X}^2)$, henceFor $n\ge 2$ and positive sample variance, $\hat{\mu }=\bar{X}$ and $\hat{v}=n^{-1}{\sum }_i(X_i-\bar{X}{)}^2$, so ${\hat{\Phi }}_1=\bar{X}/\hat{v}$. For this normal natural-mean parameter ratio estimator, the central limit theorem gives asymptotic independent errors for $\hat{\mu }$ and $\hat{v}$, with variances $v/n$ and $2v^2/n$. Applying the delta method to $\mu /v$ yieldsIf $v=v_0$ is known, ${\Phi }_2=-1/(2v_0)$ is fixed and ${\hat{\Phi }}_1=\bar{X}/v_0$ has exact distribution $\boxed{N({\Phi }_1,1/(n{v}_0))}$. The variance-estimation contribution $2{\mu }^2/v^2$ disappears.