Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.

Use the Sylow theorems. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.

If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\ge p$ and $n_r\ge q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at leastplaces, since the excess is $(p-1)(q-1)>0$. Hence some Sylow subgroup of order $s\in \{q,r\}$ is normal.

In the quotient by this normal subgroup, the largest prime $p$ has a normal Sylow subgroup: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a normal subgroup of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that normal subgroup and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.

Let $P$ be this normal Sylow subgroup. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a normal Hall $\{p,q\}$-subgroup. The serieshas factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.

A Hall subgroup for a prime set $\pi$ is a subgroup $H\le G$ whose order has only prime divisors in $\pi$ and whose index has no prime divisor in $\pi$:Here one is allowed in either class, and ${\pi }^{\prime }$ is the complementary set of primes. Equivalently $|H|$ contains the complete prime-power contribution to $|G|$ for each prime in $\pi$.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.

If $H\le G$, induction gives $H^{(i)}\le G^{(i)}$, because commutators of elements of a subgroup are also commutators in the larger group. Thus termination of the derived series of $G$ forces termination for $H$.

For a normal subgroup $N$, the quotient map sends $[x,y]$ to the commutator of their images. Surjectivity then givesConsequently subgroups and quotient groups of a soluble group are soluble. The assertion about a quotient uses a normal subgroup; it is not a quotient by an arbitrary subgroup.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the commutator subgroup. The group is soluble ifEquivalently it has a finite series with abelian factors. The trivial group is included.

An invertible time-series representation recovers the driving white noise from current and past observations. In the inverse series the support condition is thereforeAgain the series must converge. Stable invertibility uses ${\sum }_{r\ge 0}|b_r|<\infty$, which ensures mean-square convergence when $X_t$ has finite variance. Merely writing a bilateral inverse is not invertibility in this one-sided sense: it may require future observations. For a general correlated input $X$, square summability of $b_r$ alone is not the same sufficient condition as it is for a white noise input.

A causal time-series representation uses only the present and past driving white noise. Thus the coefficient condition isThe series must have its stated convergence meaning. For centered white noise of positive finite variance, ${\sum }_{r\ge 0}|a_r{|}^2<\infty$ is sufficient and necessary for mean-square convergence. In the usual stable-filter convention one imposes the stronger ${\sum }_{r\ge 0}|a_r|<\infty$. A bilateral stationary linear process need not be causal: terms with $r<0$ involve future driving values.

Write $m(\lambda )=\mathbb{E}(X_i\mid \lambda )$ and $v(\lambda )=\mathrm{Var}(X_i\mid \lambda )$. The Bühlmann model uses the finite structural parametersHere $v$ is the expected process variance and $a$ is the variance of hypothetical means. The target is the latent conditional mean $m(\lambda )$, rather than the realized next count. The Bühlmann credibility estimate is the best affine predictor of that target under mean squared error.

The law of total expectation and law of total variance give $\mathbb{E}{X}_i=m_0$ and $\mathrm{Var}(X_i)=v+a$. Conditional independence gives $\mathrm{Cov}(X_i,X_j\mid \lambda )=0$ for $i\ne j$, and the law of total covariance therefore yieldsTo derive the optimal predictor, consider $d=A+{\sum }_{i=1}^n{b}_i{X}_i$. Minimizing with respect to the intercept gives $A=m_0(1-{\sum }_i{b}_i)$. Thus $d=m_0+{\sum }_i{b}_i(X_i-m_0)$. The linear least-squares projection normal equations areFor $v>0$, subtraction of any two equations forces all $b_i$ equal. Substituting a common coefficient then gives $b_i=a/(v+na)$. Equivalently, the mean squared error of the centered predictor is $a-2a{\sum }_i{b}_i+v{\sum }_i{b}_i^2+a({\sum }_i{b}_i{)}^2$, a convex quadratic with precisely these normal equations. HenceThe ratio notation assumes $a>0$; the credibility factor formula $na/(na+v)$ also handles $a=0,v>0$. If $a=0$, the target is the constant $m_0$ almost surely. If $v=0,a>0$, one observation already equals $m(\lambda )$ almost surely, and the average gives it exactly. If both vanish, the target and observations are constant.

The result optimizes over affine functions of the observations. It need not equal the unrestricted posterior mean; exact Bayesian inference generally depends on the whole prior and likelihood, whereas the Bühlmann credibility estimate uses these second-moment structural parameters.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

For quota share reinsurance the insurer retains the same fraction of each claim, soThe annual retained loss is consequently $T_I=\alpha T$. Each transformed risk severity has probability density function $f_i(y/\alpha )/\alpha$ for $y>0$. The mixed transformed severity, with the same weights ${\lambda }_i/({\lambda }_1+{\lambda }_2)$, still gives a compound Poisson distribution. Scaling the expected value and variance givesThis agrees with the general retained-claim formulas because $\mathbb{E}g(X_i)=\alpha {\mu }_i$ and $\mathbb{E}[g(X_i{)}^2]={\alpha }^2({\sigma }_i^2+{\mu }_i^2)$.

Let $G(z)={\sum }_{n\ge 0}{p}_n{z}^n$ be the probability generating function. For $|z|<1$ the series can be differentiated term by term: $\sum n{p}_n|z{|}^{n-1}$ is finite, even without a finite claim-count expected value. Multiply the Panjer claim-count class recurrence by $n$ and put $m=n-1$. ThenHenceThis argument is valid throughout the open unit disk. Values at boundary points may be obtained by a limit when the required derivatives exist; one need not assume $\mathbb{E}N<\infty$ to establish the identity.

Take increasing to mean strict growth on every nonempty time interval; otherwise a constant family has no uniquely determined driving point. The half-plane-capacity parameterization imposed below guarantees strict growth. For $s<t$, define the increment hull in the mapped domain byThis definition includes filling and boundary conventions automatically. Its mapping-out function is $g_{s,t}=g_t\circ g_s^{-1}$. The Loewner local growth property meansfor every finite $T$ within the parameter range. Equivalently, the mapped increments have uniformly vanishing diameter on compact time intervals.

For fixed $t$, the nonempty compact Euclidean closures ${\overline{K}}_{t,t+h}$ are nested as $h$ decreases. Their diameters tend to zero, so their intersection is a singleton. Its point lies on the real axis: the imaginary part of any point in a hull is at most its enclosing radius. Define the Loewner transform bySince ${\xi }_t$ lies in each closure, every point of $K_{t,t+h}$ is at distance at most $2\mathrm{rad}(K_{t,t+h})$ from it.

To prove continuity, choose $z\in K_{t+2h}\setminus K_{t+h}$ and write $w=g_t(z)$, $w^{\prime }=g_{t+h}(z)$. Then $w\in K_{t,t+2h}$, $w^{\prime }\in K_{t+h,t+2h}$ and $w^{\prime }=g_{t,t+h}(w)$. The continuity estimate from part (i) givesThe right-hand side tends to zero uniformly on compact time intervals. Applying the same inequality with the earlier time as the base proves left continuity as well. Thus the Loewner transform is continuous.

Now impose $\mathrm{hcap}(K_t)=2t$. The half-plane-capacity composition rule follows by composing Laurent expansions at infinity and givesFor a point not yet swallowed, put $z_s=g_s(z)$ and $r_{s,t}=2\mathrm{rad}(K_{s,t})$. The increment is contained in the half-disc of radius $r_{s,t}$ centred at ${\xi }_s$. The continuity estimate first proves continuity of $s\mapsto g_s(z)$: its increment is bounded by $3\mathrm{rad}(K_{s,t})$. The differentiability estimate for a mapping-out function then gives, when $t-s$ is small enough,On a compact interval before swallowing the denominator stays away from zero. Divide by $t-s$ and let $t\downarrow s$. The local-growth property makes the error tend to zero. The analogous backward quotient has the same limit, using continuity of $g_s(z)$ and ${\xi }_s$. Thus the Chordal Loewner equation follows:The initial value follows from $\mathrm{hcap}(K_0)=0$, hence $K_0=\varnothing$. Without the capacity parameterization, the same argument gives $d{g}_t(z)=d\mathrm{hcap}(K_t)/(g_t(z)-{\xi }_t)$, interpreted with the capacity clock.