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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.

The regular level set theorem says that a regular level of a smooth map from an $m$-manifold to an $r$-manifold is an $(m-r)$-manifold. The corresponding sublevel-set theorem says that $\{F\le c\}$ is a manifold with boundary $F^{-1}(c)$ when $c$ is a regular value of a smooth real-valued function, with the interior $F<c$.

Apply this to $F(x)=|x{|}^2$ on ${\mathbb{R}}^{n+1}$. On $F^{-1}(1)$ its gradient $2x$ is nonzero. Thus $S^n$ is an $n$-manifold and $B^{n+1}$ is an $(n+1)$-manifold with boundary $\boxed{\partial B^{n+1}=S^n}$. Locally the implicit function theorem supplies $F-1$ as one coordinate; the ball occupies its nonpositive side, which is a half-space after reversing that coordinate. This also covers $n=0$: the sphere consists of two points and the ball is the interval with those endpoints.

A regular value $y$ of a smooth map $f:X\to Y$ is one for which the differential $d{f}_x:T_{x}X\to T_{y}Y$ is surjective at every $x\in f^{-1}(y)$. A value outside the image is regular vacuously. The condition concerns all preimages, not just one.

An embedded smooth $n$-dimensional manifold $X\subseteq {\mathbb{R}}^N$ has, around every point, a parametrization by an open subset of ${\mathbb{R}}^n$ which is a homeomorphism onto a relative neighbourhood, has rank $n$, and has smooth coordinate changes. Equivalently, after smooth ambient coordinates it is locally a coordinate $n$-plane. For a manifold with boundary use relatively open subsets of ${\mathbb{H}}^n=\{u_n\ge 0\}$ instead, with parametrizations and transitions smoothly extendible across the boundary. Points corresponding to $u_n=0$ form the boundary. Interior charts are ordinary manifold charts; a boundary point is not a corner.

A singular point would satisfy $p=0$, $p_w=2w=0$ and $p_z=-3z^2+2z+1=0$. Thus $w=0$ and $z=1$ or $-1/3$. But $p(1,0)=1$ and $p(-1/3,0)=-5/27$, both nonzero. The curve is smooth everywhere.

The complex implicit function theorem says that near a zero of a holomorphic function with a nonzero partial derivative, the corresponding variable is a uniquely defined holomorphic function of the others. Where $p_w\ne 0$, use $z$ as a local coordinate with $w=w(z)$; where $p_z\ne 0$, use $w$ with $z=z(w)$. On overlaps the coordinate transitions are holomorphic with nonzero derivatives. These charts give $Z$ a Riemann surface structure, and both coordinate projections are holomorphic in them.

A ramification point of a nonconstant holomorphic map is a point where in local coordinates the map has the form $u\mapsto u^e$ with $e\ge 2$; its ramification, or branching, order is $e-1$. For $g=w$, take $(z,w)=(1,i)$. Here $p_w=2i\ne 0$, so $z$ is a local coordinate. Differentiating $w(z{)}^2=z^3-z^2-z$ gives $w^{\prime }(1)=0$ and $2i{w}^{\prime \prime }(1)=4$. ThereforeThe nonzero quadratic coefficient gives local degree $\boxed{2}$ and branching order $\boxed{1}$. If “branching order” denotes local degree rather than its excess in a different convention, the corresponding value is two.

A rational function is a quotient $r=P/Q$ of polynomials with $Q\ne 0$. Cancel common factors. It determines a map on the Riemann sphere by assigning infinity at poles and assigning at infinity the limit in the sphere. Equivalently homogenize $P,Q$ to the same degree $d=\max (\mathrm{deg}P,\mathrm{deg}Q)$ and set $f([Z:W])=[P_h(Z,W):Q_h(Z,W)]$. They have no common projective zero: finite ones would contradict coprimality, and at infinity at least one has a nonzero leading coefficient. This also proves the map is holomorphic.

The degree of the nonconstant rational function is $d$. The degree of a holomorphic map of compact Riemann surfaces is the number of preimages of any target point, counted with local mapping multiplicity. For target $[a:b]$, the homogeneous equation $b{P}_h-a{Q}_h=0$ is a nonzero homogeneous polynomial of degree $d$. Over $\mathbb{C}$ it factors into $d$ linear factors, with multiplicity, giving exactly $d$ projective zeros. Locally its zero order is precisely the map's multiplicity, since the other target-coordinate denominator is nonzero. The two degrees are equal.

There is a Möbius transformation sending any ordered triple of distinct points to $0,1,\infty$. When all are finite one can use $T_A(z)=(z-a_1)(a_2-a_3)/[(z-a_3)(a_2-a_1)]$; the formulas for infinity are its limits. Define $T_B$ in the same way. Then $T_B^{-1}\circ T_A$ sends $A$ to $B$ and restricts to a biholomorphism $X\setminus A\to X\setminus B$.

Let $P_m$ be the linear subspace of polynomials of degree at most $m$. It has finite dimension $m+1$, is proper in $P$, and $P={\bigcup }_{m\ge 0}{P}_m$. For any norm on $P$, the restricted norm on $P_m$ is equivalent to its coefficient Euclidean norm: on the coefficient unit sphere the continuous norm has a positive minimum and a finite maximum. Thus $P_m$ is complete. A complete subspace of a metric space is closed, since any sequence converging to an ambient limit is Cauchy and has the same limit in the subspace.

Part (b) makes each $P_m$ a closed set with empty interior, hence nowhere dense. If $P$ were a Banach space, this countable union would contradict the Baire category theorem. No norm makes the space of all polynomials complete.

If a linear subspace $V$ of a normed space contains an open ball $B(v,r)$, then $v\in V$ and subtraction gives $B(0,r)\subseteq V$. For any $x\in X$, choose a positive scalar $t$ with $\Vert tx\Vert <r$. Then $tx\in V$ and hence $x\in V$. Thus $V=X$. Every proper linear subspace has empty interior.