Define the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. It is symmetric and idempotent, the orthogonal projection onto the column space of $X$. Since $HX=X$,The covariance is ${\sigma }^{2}H{H}^T={\sigma }^{2}H$. This normal distribution can be singular when $p<n$, since all fitted vectors lie in the $p$-dimensional model subspace.

The least-squares estimator minimizes $\Vert Y-Xb{\Vert }^2$. Differentiating in $b$ gives $-2X^T(Y-Xb)=0$, hence $\boxed{X^{T}X\hat{\beta }=X^{T}Y}$. Full column rank makes $X^{T}X$ positive definite, soThis is a linear transformation of a multivariate normal distribution. Its mean is $\beta$ and its covariance is ${\sigma }^2(X^{T}X{)}^{-1}$. Therefore $\boxed{\hat{\beta }\sim N_p(\beta ,{\sigma }^2(X^{T}X{)}^{-1})}$.

For the linear test equation $f(y)=\lambda y$ put $z=h\lambda$. The second stage solves $(1-az)k_2=\lambda y[1+(1-a)z]$, and substitution into the update gives the stability functionA-stability requires no poles and $|R(z)|\le 1$ throughout the closed left half-plane. If $a\ne 1/2$, the quadratic numerator gives unbounded magnitude as $z\to -\infty$; when $a=0$ the denominator is constant and the same conclusion holds. Thus $a=1/2$ is necessary. At this value $R(z)=(1+z/2)/(1-z/2)$, whose only pole is at positive $z=2$. Moreoveron the left half-plane. Hence the method is A-stable exactly for $\boxed{a=1/2}$. Its limit at negative infinity is $-1$, so it is not L-stable.

Use the stage solution continuous from $k_2=f(y)$ as $h\to 0$. Its displacement is $\delta =h[(1-a)f+a{k}_2]=hf+a{h}^2{f}^{\prime }f+O(h^3)$. Taylor expansion of $f(y+\delta )$ givesConsequently the Runge-Kutta method advances byThe exact solution has third-order term $h^3(f^{\prime 2}f+f^{\prime \prime }{f}^2)/6$. The first two orders agree for every real $a$, but the $f^{\prime \prime }{f}^2$ coefficient is always $1/4$ instead of $1/6$. Therefore the general nonlinear method has $\boxed{\text{order }2}$ for every fixed real $a$. The special value $a=1/3$ matches the linear test equation at third order but does not raise the nonlinear order. This is the two-stage Runge-Kutta family with an implicit second stage.

Put $\Psi =\mu e^{-i(kx+2k^{2}t)\sigma }$. The Lax pair becomes ${\Psi }_x=U\Psi$, ${\Psi }_t=V\Psi$, withEquality of mixed derivatives, using invertibility, gives the zero-curvature condition $U_t-V_x+[U,V]=0$. The terms quadratic in $k$ cancel. The terms linear in $k$ cancel because $\sigma Q=-Q\sigma$, $\sigma Q_x=-Q_x\sigma$ and $Q^2$ commutes with $\sigma$. For the constant term, the identitiescancel the differentiated $Q^2$ term, leaving $Q_t+i{Q}_{xx}\sigma -2i{Q}^3\sigma =0$. Multiplying by $i$ provesOnce $S$ is known, its time dependence enters through the explicit exponential conjugation. The boundary relation and normalization form a linear problem for $\mu$, and $Q=i[\sigma ,{\mu }_1]$ reconstructs the nonlinear solution. This is the requested linearization. Its disadvantage as an initial-value method is that it does not itself determine $S$ from a prescribed $Q(x,0)$. The inverse scattering transform supplies that direct-scattering step, including any discrete spectral data and their pole conditions, before reconstruction. Solvability cannot be inferred merely from writing an arbitrary jump.

The $k^{-1}$ coefficient of the first equation is ${\mu }_{1x}+i[\sigma ,{\mu }_2]=Q{\mu }_1$. Substitution in the expression for $B$ gives $B=-2{\mu }_{1x}$. Decompose ${\mu }_1$ into diagonal and off-diagonal parts. Since $Q=i[\sigma ,{\mu }_1]$, its off-diagonal part is $iQ\sigma /2$. ThusThe commutator in the $k^{-1}$ equation has zero diagonal; multiplication of two off-diagonal two-by-two matrices is diagonal. Combining the two parts proves

There is a literal inconsistency in the stated Riemann-Hilbert problem: a jump matrix $S(k)$ tending to zero is incompatible with ${\mu }_{\pm }\to I$. Along the real axis the conjugating exponentials are unitary, so the right side of the jump tends to zero while the left side tends to $I$. No normalized solution satisfies the printed decay condition. The intended condition is $S(k)\to I$, or replacement of $S$ in the jump by $I+S$ when $S$ denotes its decaying part.

For this corrected problem assume a regular, invertible, pole-free solution, differentiable in $x,t$, with the required expansion uniformly at infinity. These conditions matter: arbitrary jump data need not be solvable, and poles would contribute extra terms. Write $\sigma ={\sigma }_3$ and $\mu =I+{\mu }_1/k+{\mu }_2/k^2+\cdots$. Matching the nondecaying coefficients in the two differential expressions yieldsThe first expression makes $Q$ off-diagonal.

These matches really establish the differential equations, rather than just formal identities. Let $G=e^{-i(kx+2k^{2}t)\sigma }S(k)e^{i(kx+2k^{2}t)\sigma }$. It satisfies $G_x=-ik[\sigma ,G]$ and $G_t=-2i{k}^2[\sigma ,G]$. Differentiating ${\mu }_{+}={\mu }_{-}G$ consequently shows thathave matching boundary values. They therefore extend holomorphically across the contour. The chosen coefficients make both residuals $O(k^{-1})$ at infinity. Their absence of poles and Liouville theorem imply $R_x=R_t=0$, proving the required Lax pair under the corrected hypotheses. This is matrix NLS reconstruction from a normalized Riemann-Hilbert problem.

Put $P(t)={\prod }_i(t-x_i)$. For every polynomial $q$ of degree at most $n-1$, $Pq$ has degree at most $2n-1$ and vanishes at all nodes; thus ${\int }_{-1}^{1}Pq=0$. The monic degree-$n$ orthogonal polynomial for this integral is unique: the difference of two such monic polynomials has degree below $n$ and is orthogonal to itself, hence is zero. Consequently $P$ is the monic normalization of the degree-$n$ Legendre polynomial, proving the given nodes are exactly its roots. Together the three results characterize this Gaussian quadrature rule.

Apply exactness to the constant polynomial one. Its integral is two and its value at every node is one, so $\boxed{\sum _{i=1}^n{a}_i=2}$.

For each node define its Lagrange interpolation cardinal polynomial ${\ell }_i(t)={\prod }_{j\ne i}(t-x_j)/(x_i-x_j)$. Its square has degree $2n-2<2n$ and nodal values ${\ell }_i(x_j{)}^2={\delta }_{ij}$. Exactness therefore gives $a_i={\int }_{-1}^1{\ell }_i(t{)}^{2}dt$. The polynomial is nonzero, so continuity makes this integral strictly positive. Thus $\boxed{a_i>0}$ for every node.

If the first move is bad but the buyer misses exercise, the remaining European claim is worth $2/3$, whereas the existing hedge is worth three. At that node rehedge the continuation claim: its stock holding is $(0-5)/(16-10)=-5/6$ and its bank holding is $2/3+(5/6)12=32/3$.

Starting from the original short $1/6$ and bank balance five, short another $2/3$ share for eight, giving bank cash thirteen. Retain $32/3$ for the new hedge and withdraw $\boxed{7/3}$ as guaranteed time-one profit. At time two the retained bank grows to $40/3$. If price is sixteen, covering the short costs $40/3$ and the put expires worthless. If price is ten, covering costs $25/3$ and the remaining five pays the put. Thus the profit does not depend on the second move. Investing the withdrawn profit at the riskless rate makes it $35/12$ at time two.

The terminal European put option payoffs are zero at forty-five, thirty-six and sixteen, and five at ten. At the good first node its value is zero. At the bad node its continuation value is $(4/5)[(5/6)0+(1/6)5]=2/3$, below the exercise payoff three, so optimal exercise occurs there. At time zero the continuation value is $(4/5)[(3/8)0+(5/8)3]=\boxed{3/2}$, exceeding the immediate payoff zero.

To hedge one sold American put option, short $1/6$ of a share and put four in the bank at time zero. The portfolio costs $-15/6+4=3/2$, exactly the premium. At time one the bank holds five. After a good move the short position costs five to close, exhausting the portfolio and liability. After a bad move it costs two to close, leaving three for the exercised put. Thus after two bad periods the rational buyer already exercised at time one, and the hedge pays three and closes then; there is no remaining time-two obligation. Physical delivery gives the same accounting by immediately selling the stock received on exercise.

At gross rate $R=5/4$, the conditional up probabilities of the equivalent martingale measure areAll lie strictly between zero and one. The four terminal path probabilities are $1/16,5/16,25/48,5/48$ in the order good-good, good-bad, bad-good, bad-bad.

At the node with stock price thirty, both possible next prices exceed thirty-three, the value of thirty invested at gross rate $11/10$. Take no position at time zero. If the first move is good, borrow thirty at time one and buy one share; if it is bad, do nothing. At time two sell the share for forty-five or thirty-six and repay thirty-three, leaving twelve or three. The strategy has zero initial cost, no possible loss, and positive profit on the good first move, so it is an arbitrage.

Working with measurable functions modulo equality almost everywhere, the sequence is Cauchy in Lp space if for every $ϵ>0$ there exists $N$ such that $\Vert f_n-f_m{\Vert }_p<ϵ$ for all $m,n\ge N$. This is a statement about integral norm distance, not pointwise convergence. Without quotienting null functions the displayed quantity is a seminorm, so the conventional $L^p$ normed space uses these equivalence classes.

Because $k>0$ and the original smooth closed curve has periodic Frenet data, $\arctan (\tau /k)$ is a continuous real-valued periodic function. Therefore $\int k_{g}d\bar{s}=-\oint d\arctan (\tau /k)=0$. The simple closed normal curve divides the sphere into two disks by the Jordan theorem. Apply Gauss-Bonnet theorem to either disk, choosing its consistent boundary orientation. Its Gaussian curvature is one and its total boundary curvature remains zero. Thus each region has area $2\pi$, proving the two regions have equal area.

Let $T,n,B$ be the Frenet frame of the original space curve, with $T^{\prime }=kn$, $n^{\prime }=-kT+\tau B$, $B^{\prime }=-\tau n$. Put $w=\sqrt{k^2+{\tau }^2}$. The normal-indicatrix arc-length satisfies $d\bar{s}/ds=w>0$, and its unit tangent is $\overline{T}=(-kT+\tau B)/w$. On the unit sphere its outward normal is $n$, and $\overline{T}\times n=(-kB-\tau T)/w$.

Differentiate $\overline{T}$ in $s$. The derivative of $1/w$ is parallel to $\overline{T}$ and contributes nothing to its projection onto $\overline{T}\times n$. The remaining derivative is $(-k^{\prime }T+{\tau }^{\prime }B-(k^2+{\tau }^2)n)/w$, givingWith the alternative left-normal signed-curvature convention the displayed scalar changes sign, while the area conclusion is unchanged.

For an oriented surface normal $N$ and a unit-speed curve tangent $T$, take signed geodesic curvature $k_g=(dT/ds)\cdot (T\times N)$. This convention uses boundary orientation with the region on the right; reversing to the usual left-hand boundary convention reverses both orientation and the corresponding curvature convention. With positive orientation consistent this way, the Gauss-Bonnet theorem for a region $D$ with piecewise smooth boundary isFor a disk with smooth boundary there are no corner terms and $\chi (D)=1$. Stating the signed convention is necessary to reproduce the sign in part (ii).

Put $\zeta =e^{2\pi i/7}$ and induce the characters ${\chi }_r(a)={\zeta }^r$ of $H=⟨a⟩$. Since $H$ is normal, the induced value outside $H$ is zero, while at $a^m$ it is ${\zeta }^{rm}+{\zeta }^{2rm}+{\zeta }^{4rm}$. The nonzero $r$ split into the two doubling orbits, so the two distinct induced characters have class valueswhere $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6=\overline{A}$. Summing seventh roots gives $A+B=-1$, and multiplication gives $AB=2$, hence $|A{|}^2=|B{|}^2=2$. Their character norms are $(9+3\cdot 2+3\cdot 2)/21=1$, proving irreducibility. Together with the three degree-one characters, the sum of squared degrees is $3+2\cdot 9=21$. Thus these are all the remaining irreducible characters.

For a character $\chi$ of a representation of $H\le K$, the induced character isThe sum is zero when no conjugate of $g$ meets $H$.