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Here $\ell \sim \sqrt{ab}$, so the geometric condition is $b/a\ll 1$. Let $U$ be the signed roller velocity near its top, with magnitude $a|\Omega |$; choose the rotation sense making $U=a\Omega$ positive in the $x$ direction. The small-inertia condition iswith $|U|$ understood for a reversed rotation. The roller's slope and change of tangential speed are small in the gap region. The liquid is Newtonian, the roller imposes no slip and the free surface is shear-free; surface tension is neglected as in the gravity-only model.

For the leading flat film geometry, use $y=-d(x)$ as the roller boundary and $y=0$ as the free surface. The lubrication approximation gives $\mu u_{yy}=p_x$, $u(-d)=U$, $u_y(0)=0$, henceThe corresponding vertical velocity is determined by incompressibility, for example $v(x,y)={\int }_y^0{u}_x(x,s)ds$, so that $v(0)=0$ and $v(-d)=-U{d}^{\prime }$ by differentiating the constant-flux equation. This is the leading tangency condition at the roller. Steadiness makes $Q$ independent of $x$.

For the rotating roller under a nearly flat free surface, the excess pressure tends to zero in the outer fluid on both sides of the gap. ConsequentlySet $x=\sqrt{2ab}s$. The supplied recurrence gives $I_2=\pi /2$ and $I_3=3\pi /8$, soThe extension of the parabolic gap integrals to infinity is the usual matched thin-gap approximation; their dominant contribution comes from $x=O(\sqrt{ab})\ll a$. Taking the surface flat refers to this leading geometry, not setting its small hydrostatic displacement identically to zero in the pressure balance. An exactly flat stress-free surface at exactly constant pressure could not sustain this variable pressure gradient; the displacement needed to do so is checked in part (ii).

Use signature $(-,+,+,+)$, for which $\square =-{\partial }_t^2+\Delta$. With the Lorenz constraint in Proca theory, the Proca equation is $(\square -m^2)A^a=-{\mu }_0{J}^a$. A static solution therefore satisfies $(\Delta -m^2)A^a=-{\mu }_0{J}^a$.

For $R>0$ direct radial differentiation gives $(\Delta -m^2)(e^{-mR}/R)=0$. At the origin its $1/R$ singularity contributes $-4\pi {\delta }_0$: integrating the Laplacian over a small sphere gives limiting outward flux $-4\pi$, while the mass term's integral tends to zero. Thus $G_m(R)=e^{-mR}/(4\pi R)$ obeys $(\Delta -m^2)G_m=-{\delta }_0$. Convolution proves the printed Yukawa potential solutionThe integral requires suitable source decay, for example compact support. For conserved static sources, differentiating the convolution and integrating by parts gives ${\partial }_a{A}^a=0$ as well; its spatial divergence is the convolution with ${\partial }_i{J}^i=0$.

For the usual nonnegative mass parameter this is the decaying static solution, and $m=0$ gives the Coulomb kernel with Lorenz gauge imposed. If the arbitrary real parameter is negative, the printed kernel still gives a particular solution for compact sources, but grows at infinity; the decaying boundary condition selects $e^{-|m|R}/(4\pi R)$ instead. Static sources also allow homogeneous time-dependent solutions, so this construction establishes a static solution, not that every solution is static.

Take ${\partial }_b$ of the Proca equation. Commuting partial derivatives and antisymmetry make ${\partial }_b{\partial }_a{F}^{ab}=0$, while current conservation gives ${\partial }_b{J}^b=0$. Thus $m^2{\partial }_b{A}^b=0$. For $m\ne 0$, $\boxed{{\partial }_b{A}^b=0}$. This Lorenz constraint in Proca theory is a consequence of the massive equations, rather than a gauge choice.

The stated “arbitrary constant” includes a genuine exception at $m=0$. Then the divergence condition is not implied: with $J=0$ and $A_a={\partial }_a\chi$, $F=0$ solves the equations for any smooth $\chi$, whereas ${\partial }_a{A}^a=\square \chi$ need not vanish. In the massless theory the Lorenz condition can instead be imposed as a gauge fixing.

For compactly supported variations, antisymmetry of the electromagnetic field tensor gives $\delta (-F_{ab}{F}^{ab}/4)=-F^{ab}{\partial }_a\delta A_b$. Integrate by parts in the action to obtainArbitrariness of the variation gives the Proca equationtogether with the unchanged identity ${\partial }_{[a}{F}_{bc]}=0$ coming from $F=dA$. The signs follow from the source and mass terms as printed.

For the quadratic potential the characteristic curves are harmonic oscillations:The inverse is $x_0=x\cos t-v\sin t$, $v_0=v\cos t+x\sin t$. Therefore the Liouville equation has solutionThis formula works for smooth data or by transport for locally integrable data, and for measures by pushforward. Although the quadratic $V$ itself is not globally Lipschitz as in the initial sufficient hypotheses, its gradient is globally Lipschitz, which is the property actually needed for this global flow.

Formally the Jacobian determinant of the characteristic flow map satisfies $J^{\prime }=\mathrm{div}B({\Phi }_t)J=0$, $J(0)=1$, so $J=1$. Changing variables $z={\Phi }_t(z_0)$ givesTaking the $p$th root proves $\boxed{\Vert f(t){\Vert }_{L^p}=\Vert f_I{\Vert }_{L^p}}$ for $1\le p<\infty$. Equivalently multiply the equation by $p{f}^{p-1}$ and integrate the divergence term, with vanishing boundary terms. Volume preservation gives the result without needing such boundary decay for general integrable data.

Along a characteristic curve, the Liouville equation gives $d[f({\Phi }_t(z_0),t)]/dt=0$. HenceComposition preserves nonnegativity, so $f\ge 0$. For locally integrable data this formula defines the transported weak solution, and the Lipschitz invertible flow preserves null sets locally. Thus the same assertion holds almost everywhere without requiring pointwise smooth data.

Write $z=(x,v)$ and $B(z)=(v,-\nabla V(x))$. The characteristic flow map ${\Phi }_t$ solves $\dot{z}=B(z)$. Its vector field is globally Lipschitz continuous, so Picard-Lindelöf theorem gives a unique global flow with inverse ${\Phi }_{-t}$. Also ${\mathrm{div}}_{z}B=0$.

The proposed Dirac measure is the unit mass at ${\Phi }_t(z_0)$. For every compactly supported smooth test function $\psi$,This is the weak formulation of $f_t+\mathrm{div}(Bf)=0$, equal to the Liouville equation because $\mathrm{div}B=0$. The initial unit mass is recovered at $t=0$. The characteristic equations are precisely $\dot{\hat{x}}=\hat{v}$, $\dot{\hat{v}}=-\nabla V(\hat{x})$; the time-derivative dots, lost in the converted text, are present in the PDF.

Use Gaussian quadrature: choose the $n$ nodes as the roots of the degree-$n$ Legendre polynomial, which are distinct and inside $(-1,1)$. Retain the integrated Lagrange interpolation weights. With the standard normalization $P_n(1)=1$, they can be writtenThe orthogonality of the Legendre polynomial against all lower-degree polynomials makes this quadrature exact up to degree $2n-1$, twice the arbitrary-node degree apart from the endpoint count. This is the Gauss–Legendre choice of Gaussian quadrature nodes and weights.

Define the Lagrange interpolation polynomialsDistinct interpolation nodes ensure every denominator is nonzero and $L_j(x_i)={\delta }_{ij}$. For a polynomial of degree at most $n-1$, the difference $P(x)-{\sum }_{j}P(x_j)L_j(x)$ has degree at most $n-1$ and vanishes at $n$ distinct points, so it is zero. Integrating this Lagrange interpolation identity proves the required exact quadrature formula. This also covers $n=1$, when the empty product is one and its weight is two.

At the prescribed mean, $w_1-b$ has a normal distribution with mean $d=m-b$ and standard deviation $\sigma$. Integrating its negative part gives the expected losswhere $\phi ,\Phi$ are the standard normal density and distribution function. Differentiation at fixed $d$ yields $\partial L/\partial \sigma =\phi (d/\sigma )>0$; the expression extends continuously to $\sigma =0$. Therefore minimizing expected shortfall at a fixed mean chooses exactly the minimum-variance portfolio ${\theta }^{\ast }$ from part (a).

For $m\ge b$ it lies on the mean-variance efficient frontier. If “efficient frontier” has its usual undominated meaning, the printed claim needs this qualification: for $m<b$, the same optimizer lies on the lower minimum-variance branch and is dominated by the all-bank point $(b,0)$. It still solves the fixed-mean problem, but is not efficient in that unconstrained comparison. If the term instead means the full fixed-mean minimum-variance boundary, the claim holds for every feasible $m$.

When $a\ne 0$ and $n\ge 2$, the set of achievable mean-variance pairs isIndeed the decomposition in part (a) proves necessity. There is a nonzero $h$ orthogonal to $a$; scaling it makes $h^{\mathsf{T}}Vh$ any nonnegative number, proving sufficiency. For $n=1$ that orthogonal space is zero and only the boundary parabola is achievable, not its interior. If $a=0$, the set is the vertical half-line $(b,s^2)$, $s^2\ge 0$.

The mean-variance efficient frontier comprises undominated portfolios: no other portfolio has at least as high a mean and at most as large a variance, with one strict inequality. For $a\ne 0$ it is the upper branch $m\ge b$, $s^2=(m-b{)}^2/D$. Lower-branch portfolios have lower mean and larger variance than the all-bank portfolio, and points above the minimum-variance boundary are dominated at their own mean. On the upper branch increasing mean necessarily increases minimum variance, so no dominance is possible. If $a=0$, only the zero-variance all-bank point is efficient.

Put $b=(1+r)w_0$, $a=\mu -(1+r)p$ and $d=m-b$. The terminal mean and variance are $b+{\theta }^{\mathsf{T}}a$ and ${\theta }^{\mathsf{T}}V\theta$. A nonsingular covariance matrix is positive definite. If $a\ne 0$, define $D=a^{\mathsf{T}}{V}^{-1}a>0$. The Lagrange multiplier equations for the fixed-mean constraint give $2V\theta =\lambda a$, and thusFor a direct global proof, every feasible portfolio is ${\theta }^{\ast }+h$ with $a^{\mathsf{T}}h=0$, so its variance is $d^2/D+h^{\mathsf{T}}Vh$, minimized uniquely at $h=0$. If $a=0$, only $m=b$ is feasible and its minimum variance is zero, attained by the all-bank portfolio; no other required mean is achievable.

The Lehmann–Scheffé theorem states that an unbiased estimator which is a function of a complete sufficient statistic is the unique minimum-variance unbiased estimator; equivalently conditioning any unbiased estimator on such a statistic produces it.

Use the unbiased estimator $I=1_{X_1>x}$. Given $T_n=t$, the positive sample vector is uniform on the simplex ${\sum }_i{X}_i=t$: the joint exponential density is constant there, independently of $\lambda$. For $t<x$ the conditional probability is zero. For $t\ge x$, translating the first coordinate by $x$ changes the simplex size from $t$ to $t-x$; its $(n-1)$-dimensional volume ratio is $(1-x/t{)}^{n-1}$. Thusincluding $n=1$, when this is the indicator $1_{T_1\ge x}$ up to a null boundary. Conditional expectation preserves unbiasedness, and completeness from part (i) proves the displayed estimator is the unique minimum-variance unbiased estimator.

It does not attain the Cramér-Rao bound. Equality in that bound requires the centred unbiased estimator to be proportional to the score function, hence affine in $T_n$ at fixed $\lambda$. This estimator is zero on a positive-probability interval and positive elsewhere, so is not affine; the $n=1$ step function is not affine either.

For the last assertion, the remaining observations are independent of the first $k$. Conditional on $T_k$, the internal proportions of those first $k$ are independent of the remaining sum, soNow the tower property for the nested information $\sigma (T_n)\subseteq \sigma (T_k,T_n)$ givesConditioning first on both sums is essential; the sigma-fields generated by $T_k$ and $T_n$ alone are not nested.

Here $\varphi =e^{-\lambda x}$, so $\lambda =-(\log \varphi )/x$ and $d\lambda /d\varphi =-1/(x\varphi )$. The sample score function for $\lambda$ is $n/\lambda -T_n$, whose variance is $n/{\lambda }^2$. Transforming Fisher information givesFor one observation the information is the same expression without $n$.

The joint likelihood is ${\lambda }^n{e}^{-\lambda T_n}{\prod }_i{1}_{x_i>0}$, so the factorization criterion for sufficiency proves sufficiency. For any two positive sample vectors its likelihood ratio is $e^{-\lambda (T_n(x)-T_n(y))}$, independent of $\lambda$ exactly when their sums agree. The likelihood-ratio criterion for minimal sufficiency therefore makes $T_n$ minimal sufficient.

Its gamma distribution has density ${\lambda }^n{t}^{n-1}{e}^{-\lambda t}/\Gamma (n)$ on $t>0$. If $g(T_n)$ is integrable for every $\lambda$ and has expectation zero for every $\lambda$, thenBy uniqueness of the Laplace transform, $g(t)t^{n-1}=0$ almost everywhere. One may first multiply by $e^{-{\lambda }_{0}t}$ to obtain an integrable function and use transform parameters $\lambda >{\lambda }_0$; this handles the transform's integrability hypothesis. Since the gamma densities are positive on $(0,\infty )$, $g(T_n)=0$ almost surely for every $\lambda$. $T_n$ is complete as well as minimal sufficient.

Fix $M$. Each time the chain is in $\{1,\dots ,M\}$ immediately before a jump, its conditional chance to jump to zero is at least the $\delta$ from part (b). Count all these opportunities, including returns to the same state after an excursion; the $T_m$ in part (b) need not count those returns. By the Strong Markov property at successive such visits, the probability of surviving $r$ opportunities without absorption is at most $(1-\delta {)}^r$. Hence infinitely many such opportunities without hitting zero have probability zero.

A positive state cannot be occupied forever: its finite positive holding rate gives an almost surely finite holding time. For a nonexplosive continuous-time Markov chain, visits at arbitrarily late times to the finite set therefore require infinitely many of the counted opportunities. On the event of no absorption, the chain consequently eventually stays above $M$, almost surely. Intersect this probability-one assertion over integers $M$. The result is $X_t\to \infty$. If zero is ever reached it remains occupied forever, and for initial state zero this is immediate. Thus $\boxed{{\mathbb{P}}_i(E\cup F)=1}$ for every initial state.

This uses the usual assumption implicit in an $I$-valued chain defined for all $t\ge 0$: no explosion with loss to a cemetery state. The rate conditions alone do not ensure it. For example rates $q_{i,i+1}=i^3$, $q_{i0}=1$ allow an infinite upward path with probability ${\prod }_i{i}^3/(i^3+1)>0$ and a finite total holding time, since ${\sum }_{i}1/(i^3+1)<\infty$. The minimal chain then explodes; with a cemetery state the printed dichotomy would require an additional explosion alternative.

The chain is not an irreducible Markov chain, since zero is absorbing and cannot reach a positive state. Write $q_i=-q_{ii}={\sum }_{j\ne i}{q}_{ij}$. These are finite positive holding rates for the usual continuous-time Markov chain. Its first holding time in state $i\ge 1$ is exponential with rate $q_i$, and its jump destination has probabilities $q_{ij}/q_i$. ThereforeThe recursively defined $T_m$ are stopping times: after $T_m$, wait until the path first reaches a different state in the finite set. Set all later times to infinity if a previous time is infinite. On $A_m$, the Strong Markov property applies at $T_m$. Its next jump goes to zero with probability at least $\delta$; if it does, $T_{m+1}=\infty$. Other jumps may also prevent another qualifying visit. ThusStarting with $A_0$ certain gives $\boxed{{\mathbb{P}}_i(A_m)\le (1-\delta {)}^m}$.

For the natural filtration ${\mathcal{F}}_t=\sigma (X_s:s\le t)$, a stopping time $T$ takes values in $[0,\infty ]$ and satisfies $\{T\le t\}\in {\mathcal{F}}_t$ for every $t$. Its occurrence is decidable from information available by that time.

The Strong Markov property says that on $T<\infty$, conditionally on $X_T$, the post-$T$ process is independent of ${\mathcal{F}}_T$ and has the law of the original continuous-time Markov chain started at $X_T$. In particular ${\mathbb{E}}_i[f(X_{T+s})\mid {\mathcal{F}}_T]=P_{s}f(X_T)$ on $T<\infty$. The chain is an irreducible Markov chain if every state communicates with every other: for any $i,j$ some time has ${\mathbb{P}}_i(X_t=j)>0$, equivalently for finite rates there is a path of positive off-diagonal jump rates between them.

The antipodal map is the restriction of the smooth linear map $-I$, so is smooth. For $n\ge 1$ orient the sphere by the outward-normal convention. Its degree is $(-1{)}^{n+1}$: the ambient determinant is $(-1{)}^{n+1}$, and the outward normal is carried to the outward normal at the antipodal point.

Choose a regular value of $f$, available by Sard theorem; if it is outside the image, the degree is zero already. Otherwise its finite preimage is partitioned into distinct antipodal pairs. Differentiating $f\circ h=f$ shows that their local degree signs obeyFor even $n$, the signs in each pair cancel, so $\mathrm{deg}f=0$. For odd $n$, each pair contributes $2$ or $-2$, so $\mathrm{deg}f$ is even. In neither case can its degree equal the identity's degree one. Since the degree of a map between oriented manifolds is invariant under smooth homotopy, $f$ is not smoothly homotopic to the identity. For $n=0$, the symmetry makes $f$ constant on the two-point discrete sphere; a continuous homotopy into this discrete space cannot change either point's image, so the conclusion still holds.