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Write the one-column design as $x\in {\mathbb{R}}^n\setminus \{0\}$ and $S=x^{T}x$. Maximizing the normal linear model likelihood first gives $\hat{\beta }=x^{T}Y/S$. Its minimized residual sum of squares is $R=\Vert Y-x\hat{\beta }{\Vert }^2$. For variance $v>0$, the remaining log likelihood is $-\frac{n}{2}\log v-R/(2v)$ plus a constant, whose maximum is atHere $R>0$ almost surely because $n>1$ and the true variance is positive. In the exceptional data case $R=0$, the likelihood is unbounded as $v\downarrow 0$, so there is no positive-variance maximum.

Let $P=x{x}^T/S$, the orthogonal projection matrix onto the design column. Then $\hat{\beta }-\beta =x^T\epsilon /S$ and $Y-x\hat{\beta }=(I-P)\epsilon$. These are orthogonal linear projections of an isotropic multivariate normal distribution, hence independent. An orthonormal coordinate system adapted to $x$ makes the residual squared length a sum of $n-1$ independent squared standard normals. Thus the joint distribution of least-squares and variance estimators is the product lawIn particular the maximum likelihood variance estimator is biased: $\mathbb{E}{\hat{\sigma }}^2=(n-1){\sigma }^2/n$; the unbiased residual estimator uses denominator $n-1$.

The Gauss-Markov theorem states that with full-column-rank design, mean-zero errors and covariance ${\sigma }^{2}I$, the ordinary least squares estimator is the best linear unbiased estimator: its covariance is no larger than any other such estimator's in the Loewner order. Normality is unnecessary for that theorem.

In the given normal linear model, maximizing the likelihood function is equivalent to minimizing $\Vert Y-X\beta {\Vert }^2$. The normal equations and full rank giveFor a direct variance comparison, write ${\beta }^{\ast }=AY$. Unbiasedness for every $\beta$ requires $AX=I_p$. Set $B=(X^{T}X{)}^{-1}{X}^T$ and $D=A-B$, so $DX=0$. The cross covariance terms vanish because $B{D}^T=(X^{T}X{)}^{-1}(DX{)}^T=0$. ThereforeThis proves the requested comparison for every contrast $t$ without relying only on the theorem's name.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.

Let the smaller charge be $q$ at $\mathbf{a}$ and the other charge be $-Q$ at $\mathbf{b}$, where $0<q<Q$; an overall sign reversal changes no zero-potential locus. Put $\lambda =q/Q\in (0,1)$. The electrostatic potential vanishes exactly when $|\mathbf{x}-\mathbf{a}|=\lambda |\mathbf{x}-\mathbf{b}|$. Completing the square gives the Apollonius sphereThe distances from its center to the charge positions are $|\mathbf{c}-\mathbf{a}|=\lambda R<R$ and $|\mathbf{c}-\mathbf{b}|=R/\lambda >R$. Thus it encloses exactly the charge of smaller magnitude. Unequal magnitudes are essential: the equal-magnitude case instead gives the perpendicular bisector plane.

Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital angular momentum operator annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The angular momentum of a radial function times a linear polynomial is determined by its degree-one angular factor: direct application of the rotation derivatives givesEquivalently its restriction to the sphere is a degree-one spherical harmonic. ThusFurthermore ${\hat{L}}_3\psi =-i\hslash K(x_1-x_2)e^{-\alpha r}$. Its expected value is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. HenceThe state need not be an eigenstate of ${\hat{L}}_3$. If normalization is desired, $\int P^2{e}^{-2\alpha r}{d}^{3}x=6\pi /{\alpha }^5$, so $K={\alpha }^{5/2}/\sqrt{6\pi }$; the zero expectation holds after normalization for any nonzero initial $K$.

Take a normalized state for which the indicated variances and operator pairings exist. Set $A={\hat{L}}_3-⟨{\hat{L}}_3⟩$ and $B={\hat{x}}_1-⟨{\hat{x}}_1⟩$. For real $\lambda$, positivity of the squared norm givesThe canonical commutation relations yield $[{\hat{L}}_3,{\hat{x}}_1]=i\hslash {\hat{x}}_2$. Thus the quadratic polynomial in $\lambda$ is nonnegative everywhere only if its discriminant is nonpositive. This gives the Robertson uncertainty principleIf either variance is zero, the same polynomial argument forces the commutator expectation to vanish, so that edge case is included. Centering the operators is essential to obtain variances rather than uncentered second moments.

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf{v}={\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime }$ and $W=|\mathbf{v}\times \mathbf{b}|>0$. Since ${\sigma }_x=\mathbf{v}$ and ${\sigma }_y=\mathbf{b}$, its first fundamental form isFor the unit normal $\mathbf{n}=(\mathbf{v}\times \mathbf{b})/W$, the mixed coefficient of the second fundamental form iswhere the term containing ${\mathbf{b}}^{\prime }$ twice vanishes. Also $g=\mathbf{n}\cdot {\sigma }_{yy}=0$, while $EG-F^2=W^2$. ThusThe Gaussian curvature vanishes exactly when the displayed scalar triple product is zero. This is the Gaussian curvature of a ruled surface criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.

At a regular point, choose the unit normal $\mathbf{n}=({\sigma }_x\times {\sigma }_y)/|{\sigma }_x\times {\sigma }_y|$. The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, $K=(eg-f^2)/(EG-F^2)$, independent of the normal orientation.

For the given graph, ${\sigma }_x=(1,0,y)$ and ${\sigma }_y=(0,1,x)$, so with $S=1+x^2+y^2$,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

Write the Gaussian integer $\alpha =a+bi$, with $a,b\in \mathbb{Z}$, and set $R=\mathbb{Z}[\alpha ]$. Complex conjugation preserves $R$, because $\bar{\alpha }=2a-\alpha$. Choose a nonzero $\beta \in P$. Then $N=\beta \bar{\beta }$ is a positive integer in $P$. It cannot be one because $P$ is a proper prime ideal. Factoring $N$ into rational primes and applying primality repeatedly gives a prime $p\in P$.

The element $\alpha$ satisfies the monic polynomial $F(t)=t^2-2at+a^2+b^2$. Evaluation induces a ring homomorphism that is a surjectionThe source has $p^2$ elements by part (b), soIf $b=0$, use $F(t)=t-a$ instead and get at most $p$ elements. Moreover $R/P$ is an integral domain, so its finiteness makes it a field. This proves the finite residue-field theorem for Gaussian integer orders without incorrectly assuming that every Gaussian integer order is a principal ideal domain.

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that for each $a\in R$ and $b\ne 0$ there are $q,r\in R$ with $a=qb+r$ and either $r=0$ or $\nu (r)<\nu (b)$. Some conventions add a multiplicative monotonicity condition; it is unnecessary for this implication.

Let $I$ be a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing $\nu (d)$, possible because the values lie in the nonnegative integers. For $a\in I$, division by $d$ gives $a=qd+r$, with $r=a-qd\in I$. A nonzero remainder would contradict minimality, so $r=0$ and $a\in (d)$. Since $d\in I$, the reverse inclusion also holds. The zero ideal is $(0)$. Thus

For the same multiplier convention, the Lagrangian dual problem isFor every feasible $x$ and every $\lambda$, the equality constraint implies $q(\lambda )\le L(x,\lambda )=f(x)$. Taking first the infimum over feasible $x$ and then the supremum over $\lambda$ proves weak duality:The inequalities remain meaningful with the usual extended-value conventions, including $q(\lambda )=-\infty$ when the Lagrangian is unbounded below. Equality of optimum values, or their attainment, does not follow from weak duality. If the sufficient pair in part (a) exists, it attains equality, since $q({\lambda }^{\ast })=L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$.

For a vector equality constraint, take the Lagrangian function in constrained optimizationThe sign of the multiplier is a convention; equality multipliers are unrestricted. The Lagrange sufficiency theorem says that if $x^{\ast }\in X$ is feasible and, for some ${\lambda }^{\ast }$, globally minimizes $L(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes the constrained objective.

Indeed for any feasible $x$, $f(x)=L(x,{\lambda }^{\ast })\ge L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$. Thus $\boxed{x^{\ast }\text{ is a global primal minimizer}}$. The theorem requires a global Lagrangian minimum, not merely a stationary point. It needs neither differentiability nor convexity when stated in this form; those hypotheses can be used separately to establish the required minimum.

Let $U_{(1)}=\min (U_1,U_2)$ and $U_{(2)}=\max (U_1,U_2)$ be the order statistics. For the symmetric uniform distribution, each observation is above or below $\theta$ with probability $1/2$, independently. ThusSo $\boxed{[U_{(1)},U_{(2)}]}$ is an exact 50% confidence interval.

The observed interval $[10,11.5]$ contains values impossible under the sampling model. Both observations must lie within one unit of the parameter, so the interval of all compatible values is $[U_{(2)}-1,U_{(1)}+1]=[10.5,11]$. It contains the true parameter almost surely. A support-restricted confidence interval for a uniform location parameter can therefore reportMore precisely, the procedure $[U_{(1)},U_{(2)}]\cap [U_{(2)}-1,U_{(1)}+1]$ retains exact 50% unconditional coverage, since the second interval contains $\theta$ almost surely. The intersection is a nonempty interval for every sample in the model and equals $[10.5,11]$ for these data. Reporting the full compatible interval alone is also valid and has coverage one. On this sample's large-spread event it lies wholly between the observations; this does not turn the unconditional 50% procedure into a generally 100% procedure.

The sample mean $\overline{X}$ has normal distribution $N(\mu ,1/n)$ because the observations are independent with known variance 1. The pivotal quantity $Z=\sqrt{n}(\overline{X}-\mu )$ has the standard normal distribution. Consequentlyand inversion gives the confidence intervalReplacing 1.96 by the exact $0.975$ normal quantile gives exact 95% coverage; the printed numerical value gives the usual rounded interval. No variance estimation or Student distribution is needed.

A $100\gamma \%$ confidence interval is a pair of statistics $L(X)\le U(X)$ whose random interval satisfiesAn exact confidence interval has equality throughout. The probability refers to repeated sampling at a fixed unknown parameter; the endpoints are random before observation. It is not a posterior probability assigned to the parameter after seeing the data. Here $0<\gamma <1$.

Let $X=\mathbb{R}$ with the usual metric and let $Y=\mathbb{R}$ with the discrete metric, and take $f(x)=x$. Every subset of $Y$ is an open set, so the image of every open subset of $X$ is open: $f$ is an open map.

It is not continuous at any point. For $ϵ=1/2$, every $\delta >0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between metric spaces.

The continuity condition at $x\in X$ between metric spaces isThe function is continuous when this holds at every $x$.

If $f$ is continuous and $U\subseteq Y$ is an open set, take $x\in f^{-1}(U)$. Some $ϵ$-ball around $f(x)$ lies in $U$. Continuity gives a $\delta$-ball around $x$ mapped into that ball and hence contained in $f^{-1}(U)$. Thus $f^{-1}(U)$ is open, including the empty set, in the notation for the image and preimage of a function.

Conversely, suppose every open set has open preimage. For fixed $x$ and $ϵ>0$, the preimage of the open ball $B_e(f(x),ϵ)$ is open and contains $x$, so contains some $B_d(x,\delta )$. This gives the displayed metric definition. Therefore $\boxed{f\text{ is continuous}⟺f^{-1}(U)\text{ is open for every open }U}$.

No: local uniform convergence need not be global uniform convergence. For a bounded counterexample use the continuous functions $f_n(x)=\min (1,|x|/n)$ and $f(x)=0$. On any fixed interval $[-M,M]$, for $n>M$,Thus there is locally uniform convergence to zero. On all of $\mathbb{R}$, however, ${\sup }_x|f_n(x)-f(x)|=1$ for every $n$, so uniform convergence fails. The cutoff needed to control the tails cannot be chosen independently of the interval.