Move to a rotating frame. With detuning $\Delta ={\omega }_1-{\omega }_0+\omega$ (changing sign with the phase convention), the constant two-level Hamiltonian has generalized Rabi frequency $\Omega =\sqrt{{\Delta }^2+4v^2}$. Starting in state zero,The transition envelope is maximized on resonance, $\Delta =0$, namely $\omega ={\omega }_0-{\omega }_1$ for the matrix phases printed in the question.

First-order time-dependent perturbation theory givesThe lowest nontrivial transition probability is $|c_1(t){|}^2$, with errors of higher order in $V$.

The covering graph has at least one primitive cycle of lengths two and four and two inequivalent primitive cycles of length five. Thus the guaranteed minimum numbers of distinct orbits are respectively $1,1,2$.

Order a period-three orbit as $a<b<c$. Its images produce two subintervals $I_0,I_1$ with covering relations $I_0\to I_0,I_1$ and $I_1\to I_0$. Any closed walk in this directed graph gives, by the interval-covering lemma and the intermediate value theorem, a fixed point of the corresponding iterate. Primitive closed walks of every length exist, giving an orbit of every positive period; this is the period-three case of Sharkovsky theorem.

Since $y=u/x$, two real independent approximations are

The Liouville-Green approximation givesHereThe control ratio $|q\prime |q^{-3/2}=O(x^{-1/2})$ tends to zero, verifying consistency.

The standard removal of the first derivative, $y=u/x$, gives $u\prime \prime +q(x)u=0$ with $q(x)=x^{-1}-x^{-2}$.

After setting $z=1/x$, the transformed coefficients have a singularity beyond the regular-singular growth bounds at $z=0$. Thus infinity is an irregular singular point.

On the simplex define $f(x)=Ax/(1^{T}Ax)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By Brouwer fixed-point theorem, $f(x)=x$ for some $x$. Thus $Ax=\lambda x$, where $\lambda =1^{T}Ax>0$, and because $Ax$ is strictly positive, so is $x$.

The equivalent no-retraction theorem says there is no continuous retraction of the closed disk onto its boundary circle. A fixed-point-free map gives a retraction by projecting from $f(x)$ through $x$ to the boundary; conversely a retraction composed with the antipodal boundary map gives a fixed-point-free self-map.

The Brouwer fixed-point theorem says that every continuous map from a closed two-dimensional disk to itself has a fixed point.

Apply Cameron-Martin theorem to the absolutely continuous path $h(t)=xt/T$ up to $T$ and $h(t)=x$ thereafter. Its derivative is $x/T$ on $[0,T]$ and zero later, so the Radon-Nikodym factor is $\exp ((x/T)W_T-x^2/(2T))$, yielding the formula for every bounded measurable functional of the path after $T$.

The joint Gaussian density of the independent increments $\Delta W_i$ changes under shifts $c\Delta t_i$ byMultiplying the density and integrating the bounded cylinder function $g$ proves the identity.

The transformed process is centred Gaussian andThe Brownian strong law makes $t{W}_{1/t}\to 0$ almost surely as $t\downarrow 0$, giving continuity at zero; elsewhere continuity is immediate. Part (a) now applies.

A Brownian motion starts at zero, has continuous paths, and has independent increments with $W_t-W_s\sim N(0,t-s)$. The stated covariance makes every finite collection jointly Gaussian; disjoint increments have zero covariance and hence are independent. Their variances are interval lengths, proving all defining properties.

Neither proper gamma prior yields the condition in part (c). The risk in part (a) is unbounded as $\theta \to \infty$ because $\beta >0$. The risk in part (b) is analytic, tends to zero at infinity, and is not identically zero, so its prior average is strictly below its supremum for a gamma prior with positive density everywhere. Thus this sufficient condition proves neither estimator minimax for any $\alpha ,\beta >0$.

An admissible Bayes estimator whose Bayes risk equals its maximum risk is minimax. Indeed every estimator has maximum risk at least its prior-average risk, which is at least the Bayes risk; equality for the stated estimator attains this lower bound. Admissibility excludes a distinct estimator improving it everywhere.

Put $c=(\beta +1)/(\beta +2)$. The posterior mean of $e^{-\theta }$ is $\delta (X)=c^{\alpha +X}$. Using the Poisson probability-generating function, its risk is