Two group actions on $S_1,S_2$ are equivalent if there is a bijection $\varphi :S_1\to S_2$ such that $\varphi (gs)=g\varphi (s)$ for all $g,s$. Such a bijection is equivariant.

The orbit-stabilizer theorem in this form says that the action on $O_s$ is equivalent to the left coset action on $G/G_s$, by $g{G}_s\mapsto gs$. This map is well-defined and injective because $gs=hs$ exactly when $h^{-1}g\in G_s$, and it is onto by the definition of the orbit. It is equivariant by multiplication.

Thus every transitive action is a coset action. Two such actions $G/H$ and $G/K$ are equivalent exactly when $H,K$ are conjugate: stabilizers at different points of one orbit are conjugate, and a conjugacy explicitly supplies a coset-action equivalence. This last classification is what makes counting by orbit types possible.

On the set of left cosets $G/H$, define $g\cdot (xH)=(gx)H$. Given $xH$ and $yH$, the element $y{x}^{-1}$ sends the first to the second. Thus the coset action is transitive. In particular the orbit of $H$ is the entire coset space, and its stabilizer is $H$.

The orbit of $s$ is $O_s=\{gs:g\in G\}$, and its stabilizer is $G_s=\{g\in G:gs=s\}$. The identity fixes $s$; if $g,h$ fix it then $(gh)s=g(hs)=s$; and $gs=s$ implies $g^{-1}s=s$. Thus $G_s$ is a subgroup.

A group action is transitive when it has just one orbit: for every $s,t\in S$ there is $g\in G$ with $gs=t$.

Every column has $n$ entries of magnitude at most one, so $\Vert a_j{\Vert }^2\le n$. Substituting in part (d) givesThe bound comes from the maximum allowed column lengths, together with Hadamard's inequality.

Use a QR factorization, whose existence is supplied. Since $|\det Q|=1$ and $R$ is triangular,Each diagonal entry is one component of its column, so $|r_{jj}|\le \Vert r_j\Vert =\Vert a_j\Vert$. Multiplication yields Hadamard determinant inequality:The argument includes singular matrices, since a zero diagonal factor makes the determinant zero.

Comparing the $j$th columns of $A=QR$ gives $a_j=Q{r}_j$. ThusTaking nonnegative square roots proves $\boxed{\Vert a_j\Vert =\Vert r_j\Vert }$. This is simply norm preservation by an orthogonal matrix.

A QR decomposition is $\boxed{A=QR}$ with $Q$ orthogonal and $R$ upper triangular. One algorithm for computing it is Householder reflection reduction; a sequence of such reflections produces a QR factorization. The diagonal of $R$ may include zero when $A$ is singular, and no nonsingularity assumption is needed here.

A real square matrix is orthogonal when $Q^{T}Q=I$, equivalently when its columns form an orthonormal basis. Taking determinants gives $(\det Q{)}^2=1$, soIn particular it is invertible with inverse $Q^T$ and preserves the Euclidean norm.

The degree-zero polynomial is constant and therefore even. Assume inductively that $p_j(-x)=(-1{)}^j{p}_j(x)$ for $j<k$. Define $q_k(x)=(-1{)}^k{p}_k(-x)$, which has the same leading coefficient as $p_k$.

Changing variable $x\mapsto -x$ and using the even weight and inductive hypothesis gives $⟨q_k,p_j{⟩}_w=(-1{)}^{k+j}⟨p_k,p_j{⟩}_w=0$ for $j<k$. Hence $q_k-p_k$ has degree below $k$ and is orthogonal to every $p_0,\dots ,p_{k-1}$. Those polynomials form a basis for the polynomials of degree below $k$, so the difference is orthogonal to itself. Positivity of the weighted inner product forces it to be zero.

Thereforecompleting the induction and proving parity of orthogonal polynomials for an even weight. Thus even indices give even functions and odd indices give odd functions, regardless of the nonzero normalization chosen.

The weight is even, so opposite-parity polynomials are automatically orthogonal. Set $p_0=1$, $p_1=x$, then subtract lower-degree components within each parity class. The needed moments areThus $p_2=x^2-{\mu }_2/{\mu }_0=x^2-2/5$ is orthogonal to $p_0$, and $p_3=x^3-({\mu }_4/{\mu }_2)x=x^3-9x/14$ is orthogonal to $p_1$. Parity supplies every other pair. One valid answer isAll have the required degree and positive norm; arbitrary nonzero rescalings are equally valid.

With the weighted inner product $⟨p,q{⟩}_w={\int }_{-1}^{1}p(x)q(x)w(x)dx$, a family of orthogonal polynomials has $\mathrm{deg}{p}_j=j$, each $p_j$ nonzero, andContinuity and positivity of the weight make the integral finite and give $⟨p_j,p_j{⟩}_w>0$. Unit norm or monic leading coefficient is an additional optional normalization.

For a fully specified parameter value, do not maximize within the null. Substitute ${\theta }_0$ directly:Under the same regularity and large-count conditions, compare with ${\chi }_{k-1}^2$, since no null parameter has been estimated. Reject in its upper tail at the desired significance level. Fixing the parameter increases the reference degrees of freedom by one relative to part (b).

The unrestricted likelihood is maximized at the empirical proportions, while the null likelihood is maximized at ${\pi }_i(\hat{\theta })$. Twice the log of their ratio isLarge values mean the fitted null likelihood is much smaller than the unrestricted maximum, so rejection is in the upper tail.

For a regular identifiable scalar parameter lying in the interior, positive true cell probabilities, and sufficiently large expected counts, Wilks theorem gives an approximate chi-squared distribution with $(k-1)-1=k-2$ degrees of freedom under the null. The unrestricted probability vector has dimension $k-1$ and the fitted null has dimension one. At significance level $\alpha$, reject if $\Lambda (\hat{\theta })>F_{k-2}^{-1}(1-\alpha )$, or report the approximate p-value $1-F_{k-2}(\Lambda )$.

Cells with very small expected counts may require a prespecified pooling scheme or simulation calibration; the asymptotic reference is not an exact finite-sample law. For a regular null with $d$ fitted parameters, the corresponding degrees of freedom are $k-1-d$.

Ignoring a factor independent of the probabilities, the likelihood function is $L(\pi )={\prod }_i{\pi }_i^{n_i}$, so its log is $\ell (\pi )={\sum }_i{n}_i\log {\pi }_i$ subject to ${\sum }_i{\pi }_i=1$. A Lagrange multiplier gives $n_i/{\pi }_i=\lambda$. Summing ${\pi }_i=n_i/\lambda$ makes $\lambda =n$.

ThusSince all counts are positive, a zero fitted cell probability would have zero likelihood. On the interior the log-likelihood is strictly concave, so this stationary point is its unique global maximum.

At a true variance ${\sigma }^2>0$, $S/{\sigma }^2\sim {\chi }_n^2$. Therefore the power function isThe argument of $F_n$ decreases as ${\sigma }^2$ increases. Since the chi-squared density is positive on the positive half-line, the power is continuous and strictly increasing, withThese features determine the requested rough sketch. The curve passes through the size at the null value and approaches one for very large variance.