For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

We use the normalized Gowers U3 norm on an interval, for which a quadratic phase has norm one. Let ${\mathcal{C}}_N$ consist of integer tuples $(x,h_1,h_2,h_3)$ whose eight vertices $x+{\omega }_1{h}_1+{\omega }_2{h}_2+{\omega }_3{h}_3$, $\omega \in \{0,1{\}}^3$, all lie in $[1,N]$. If $\mathcal{C}$ denotes complex conjugation, defineEquivalently, choose a prime $8N<M<16N$, extend $f$ by zero to $G=\mathbb{Z}/M\mathbb{Z}$, call that extension $F$, and put $I=1_{[1,N]}$. ThenThe large ambient modulus prevents wraparound in cubes supported on the interval, so the ratio is independent of the chosen such $M$. This also proves nonnegativity and the norm properties by the Gowers uniformity norm on $G$. Some conventions omit the denominator; their interval norm differs by a fixed bounded factor, and the quadratic-phase norm is then $\Vert I{\Vert }_{U^3(G)}$.

For $f_0(x)=e(\alpha x^2)$, the exponent in every conjugated cube product is a third additive difference of a quadratic polynomial and is zero. Thus $\Vert f_0{\Vert }_{U^3(N)}=1$ in the normalized interval convention.

The generalized von Neumann inequality for four-term progressions connects this norm to counting arithmetic progressions. For functions bounded by one on a cyclic group of prime order greater than three,Three applications of Cauchy-Schwarz prove the bound. Zero extension and division by the number of interval progressions give the analogous interval estimate up to an absolute constant. In particular, if $g=1_A-\alpha 1_{[1,N]}$, where $\alpha =|A|/N$, has small Gowers U3 norm on an interval, expansion of the progression count shows that it differs from ${\alpha }^4$ times the count for $[1,N]$ by $O(\Vert g{\Vert }_{U^3(N)}{N}^2)$. A large Gowers U3 norm detects structure capable of changing four-term progression counts; quadratic phases are the basic example.

For the remaining proof use the ambient $G$ just specified, define ${\partial }_{h}F(x)=F(x+h)\overline{F(x)}$, and take averages uniformly on $G$. For frequencies not on the character grid, an exponential is evaluated at the chosen integer representatives; the general proof below selects actual characters $\theta =k/M$. The printed question does not define $G$ or the interval normalization, so these conventions make the assertion precise. If instead ${\partial }_{h}f(x)=f(x)\overline{f(x+h)}$ is used, conjugate the correlations and reverse every frequency sign; the quadratic example then has $\theta (h)=-2\alpha h$.

Lagrange's theorem states that for a subgroup $L$ of a finite group $G$, $|G|=[G:L]|L|$. In particular $|L|$ divides $|G|$; the equal-sized cosets partition $G$.

Apply the theorem to $H\cap K$ inside both subgroups. Its order divides both coprime orders, so $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $K$ by $K$ being a normal subgroup, and in $H$ because $k{h}^{-1}{k}^{-1}\in H$. It is therefore the identity. Thus every element of $H$ commutes with every element of $K$.

Define $\theta :H\times K\to G$ by $\theta (h,k)=hk$. The cross-commutation just proved gives $\theta (h,k)\theta (h^{\prime },k^{\prime })=h{h}^{\prime }k{k}^{\prime }=\theta (h{h}^{\prime },k{k}^{\prime })$, so this is a group homomorphism from the direct product of groups. The product assumption makes it surjective. If $hk=1$, then $h=k^{-1}\in H\cap K$, so both entries are the identity and the kernel of a group homomorphism is trivial. HenceThis proves the internal direct product theorem in the present coprime-order setting, rather than merely matching the numbers of elements.

By the partition and the law of total probability, $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\cap B_j)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Also $\mathbb{P}(B_i\mid A)=\mathbb{P}(A\cap B_i)/\mathbb{P}(A)$. Combining these identities proves Bayes' theorem:The assumed positive probabilities make the conditional expressions well defined.

Two nonzero vectors are linearly independent when $s\mathbf{x}+t\mathbf{y}=0$ forces $s=t=0$. Their linear span then has dimension two. If they are linearly dependent, one is a nonzero scalar multiple of the other, and their span has dimension one. Thus the two requested dimensions are $\boxed{2\text{ and }1}$.

The Euclidean scalar product and its norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_{j=1}^n{x}_j{y}_j$ and $\Vert \mathbf{x}\Vert =\sqrt{\mathbf{x}\cdot \mathbf{x}}$. To prove the Cauchy-Schwarz inequality, first handle $\mathbf{y}=0$ trivially. Otherwise the squared normgivesEquality holds exactly when the residual vector is zero, meaning the vectors are linearly dependent; this includes the zero-vector cases. Expanding $\Vert \mathbf{x}+\mathbf{y}{\Vert }^2$ and applying Cauchy-Schwarz then givesTaking nonnegative square roots proves the triangle inequality.

For the unit-vector optimization, let $\mathbf{w}=\mathbf{x}+\mathbf{y}$. Independence ensures $\mathbf{w}\ne 0$. The variable part of $S$ is $\mathbf{z}\cdot \mathbf{w}$, which Cauchy-Schwarz bounds below by $-\Vert \mathbf{w}\Vert$. Equality occurs only for the antiparallel unit vector. ThusThis is minimizing a linear functional on a sphere. The two consequences follow below.