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For every nonzero-risk portfolio,Equality holds precisely forThe maximum is therefore $\sqrt{q^T{V}^{-1}q}$. For each resulting mean, part (a) shows that this portfolio has the least possible variance, so every maximizer lies on the mean-variance efficient ray.

PutFor a risky holding vector $\theta$ and the remaining wealth in the bank,soWhen $V$ is nonsingular it is positive definite. Cauchy--Schwarz in the
$V$ inner product giveswith equality exactly when $\theta =\lambda V^{-1}q$. Under
${\theta }^{T}q=h$,

For the constraint $\mathbb{E}{X}_1\ge m$, if $h\le 0$ the zero risky portfolio is feasible and the minimum variance is zero. If $h>0$, the inequality binds and the preceding optimizer and minimum apply.

False. The coordinate axesare both closed nowhere dense subsets of ${\mathbb{R}}^2$, but every $(x,y)$ is $(x,0)+(0,y)$, sowhich is not meagre by the Baire category theorem.

True. Write $C={\bigcup }_{n\ge 1}{C}_n$ with every $C_n$ nowhere dense in $\mathbb{R}$. ThenThe closure of $C_n\times \mathbb{R}$ is
$\overline{C_n}\times \mathbb{R}$, which has empty interior because $\overline{C_n}$ has empty interior. Each product is therefore nowhere dense, so $A$ is meagre in ${\mathbb{R}}^2$.

False. Fix $y_0\in \mathbb{R}$ and takeThis closed set has empty interior in ${\mathbb{R}}^2$, so it is nowhere dense and hence meagre. Its section at $y_0$ is $C_{y_0}=\mathbb{R}$, which is not meagre in itself by the Baire category theorem.

A set is of first Baire category, or meagre, when it is a countable union of nowhere dense sets. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant Baire category theorem says that a nonempty complete metric space is not meagre in itself.

No. Every estimator in parts (b) and (c) has $a\ne 1$, so its risk contains the nonconstant quadratic term
$((a-1)\theta +b{)}^2$ and has infinite supremum over $\theta \in \mathbb{R}$. The estimator ${\overline{X}}_n$ has constant risk $1/n$, so those estimators cannot be minimax.

The risk of ${\overline{X}}_n$ is constantly $1/n$. Suppose an estimator $\delta$ dominates it, and put
$m(\theta )={\mathbb{E}}_{\theta }\delta$ and $b(\theta )=m(\theta )-\theta$. The biased Cramér-Rao bound, using total information $n$, givesDomination would implyfor every real $\theta$. The standard differential-inequality argument shows that the only globally defined differentiable solution is $b\equiv 0$: a nonzero value forces $b^{\prime }$ to retain a sign and magnitude that makes $b$ leave the permitted bounded interval in one time direction.

Thus $m(\theta )=\theta$. The Cramer--Rao inequality now gives
${\mathrm{Var}}_{\theta }\delta \ge 1/n$, so domination forces equality everywhere. Equality in Cramer--Rao makes the centred estimator proportional to the Gaussian score:and hence $\delta ={\overline{X}}_n$ almost surely. No strict improvement exists, so ${\overline{X}}_n$ is admissible.

Put ${\theta }_0=-b/(a-1)$ and compare with the constant estimator
$\delta ={\theta }_0$. Its risk is $(\theta -{\theta }_0{)}^2$. For $a<0$,for every $\theta$, since $|a-1|>1$. Hence the original estimator is inadmissible.

If $a>1$, compare withIts bias is the negative of the original bias, so the squared-bias terms agree, whileIts risk is therefore strictly smaller for every $\theta$. Thus
${\hat{\theta }}_{a,b}$ is not an admissible estimator.

Since ${\overline{X}}_n\sim N(\theta ,1/n)$, the estimator has variance
$a^2/n$ and bias $(a-1)\theta +b$. Its quadratic risk function is

Write $d_G(x)$ for the degree of $x$ in $G$ and $d_A(x)$ for the number of its neighbours lying in $A$. Each trial succeeds with probability
$d_A(x)/d_G(x)$. The stated geometric-sum fact shows that the total holding time at $x$ is exponential with rate $d_A(x)/d_G(x)$. Conditional on success, each neighbour in $A$ is chosen uniformly. Hence

LetFor every edge $xy$ of $A$,Thus detailed balance holds, so $\pi$ is invariant. Connectedness of $A$ makes the chain irreducible and this invariant distribution unique.

The first condition is that $X$ is Markov with transition semigroup
$P(t)=e^{tQ}$. The second is that, for every function
$f:S\to \mathbb{R}$,is a martingale.

The first condition implies the second by conditioning over a short time interval, using
$P(h)f=f+hQf+o(h)$, summing increments, and passing to the limit. This is Dynkin's formula.

Conversely assume the martingale problem for a continuous-time Markov chain. Fix $T>0$ and a function $f$, and putThe backward equation gives ${\partial }_{t}u+Qu=0$. Applying the martingale identity to this time-dependent function shows that $u(t,X_t)$ is a martingale. Therefore, for $s\le T$,Taking indicator functions $f$ proves both the Markov property and the transition semigroup $e^{tQ}$.

Since $f_n\uparrow f$, the functionsare nonnegative and increase to $g=f-f_1$. The monotone convergence theorem gives
$\mu (g_n)\uparrow \mu (g)$. Because $f_1$ is integrable, adding the finite number
$\mu (f_1)$ yields

Integrability of $f_1$ is essential. On $\mathbb{R}$ with Lebesgue measure, letThen $f_n\uparrow 0$ pointwise, but every integral is $-\infty$, so
$\mu (f_n)$ does not increase to $\mu (0)=0$.

The monotone convergence theorem states that if
$0\le f_n\uparrow f$ pointwise, thenallowing the value $+\infty$.

Monotonicity of the integral shows that the increasing limit $L$ of the left side is at most $\int f$. Conversely, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. The setsincrease to the support of $s$. HenceLet $c\uparrow 1$ and take the supremum over all simple $s\le f$. This gives
$L\ge \int f$, proving equality.

Lift $f$ to a doubly periodic meromorphic function on $\mathbb{C}$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the residue theorem gives

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
${\sum }_i{m}_i-1$, and the kernel of constants has dimension one. For $n\ge 1$,Equivalently, this is the genus-one case of the Riemann-Roch theorem for a positive divisor.

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the maximum modulus principle, every such function is constant, so the kernel has dimension one. Rank--nullity givesFor $n=0$, this simply says $\dim V=1$.

The sum of two meromorphic functions has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.