Use the right-action convention printed in the paper: $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. The orbit of a group action and stabilizer areThe identity belongs to $H$, products of elements of $H$ fix $\omega$, and $\omega g=\omega$ implies $\omega g^{-1}=\omega$. Thus $H$ is a subgroup. For $k\in G$,proving $\mathrm{Stab}(\omega g)=g^{-1}Hg$.

The rotational action on the eight vertices of a cube is transitive. Model the vertices as $(\pm 1,\pm 1,\pm 1)$. For any desired sign triple, choose a permutation whose parity makes the corresponding signed permutation matrix have determinant one; it maps $(1,1,1)$ to that vertex and preserves the cube. A rotation fixing a vertex fixes the axis through that vertex, the centre and the opposite vertex. It cyclically permutes the three incident edges, giving precisely rotations by $0,2\pi /3,4\pi /3$. Thus the stabilizer is cyclic of order three.

For any $q$ in the polynomial trial space, $f-q=(f-p)+(p-q)$ and $p-q$ is in that same space. The assumed orthogonality therefore gives the Pythagorean identityThus $p$ is the best approximation; equality is possible only when $q=p$ because the inner product is positive-definite.

Write the linear polynomial as $p=A+Bx$. Orthogonality of $x^2-p$ to $1$ and $x$ in the weighted inner product givesSolving yieldsThe weighted squared error is ${\int }_0^{1}x(x^2-6x/5+3/10{)}^{2}dx=1/600$, providing an additional exact check. The weight's zero value at one endpoint does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval of positive weight.

The first-column largest pivot element is four, so eliminate with multipliers $1/2,0,1/4$ below it. The augmented matrix becomesPartial pivoting in the second column swaps rows two and three because two is largest in modulus. The next elimination multipliers are $1/4$ and $1/2$. This givesThe third pivot element is already largest and the final multiplier below it is zero. Back substitution gives $z=1$, $y=3$, $x=-6$, $w=-1/2$. ThereforeThe row swap must also reorder the previously stored first-column multipliers in $L$. The factors areMultiplying these factors verifies $PA=LU$. This is the required elimination with partial pivoting, not an alternative solution method.

Condition on the state after $m$ steps and use the Markov property and time-homogeneous Markov property:This is the Chapman-Kolmogorov equation, or $P^{m+n}=P^m{P}^n$. A stationary distribution is a probability row vector with nonnegative entries summing to one and $\boxed{\pi P=\pi }$. Inductively $\pi P^n=\pi$, so a chain started with that distribution retains it at every time.

For the given matrix, its first stationary equation is ${\pi }_1={\pi }_1/2+({\pi }_2+{\pi }_3)/3$. Normalization gives ${\pi }_1=2/5$. The third equation gives ${\pi }_3=2{\pi }_2/3+{\pi }_3/3$, hence ${\pi }_3={\pi }_2$. ThereforeDirect multiplication verifies the second equation as well. The finite chain is irreducible, so this stationary probability distribution is unique.

A marginal density is obtained by integrating out the other variable: $\boxed{f_X(x)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dy}$. Here independence gives $f_{X,Y}=e^{-(x^2+y^2)/2}/(2\pi )$.

The polar transformation has absolute Jacobian determinant $r$. Consequentlywith density zero outside this range. Integrating over the angle gives $\boxed{f_R(r)=r{e}^{-r^2/2}}$, the Rayleigh distribution density. Integrating over radius, using ${\int }_0^{\infty }r{e}^{-r^2/2}dr=1$, gives $\boxed{f_{\Theta }(\theta )=1/(2\pi )}$ on the angle interval. The joint density factors as the product of these marginals, so $\boxed{R\text{ and }\Theta \text{ are independent}}$. The arbitrary angle assigned at $R=0$ is irrelevant because that event has probability zero.