A credibility estimate combines a risk's own observed experience with a collective or prior mean, in the form $Z{\hat{\mu }}_{\mathrm{experience}}+(1-Z){\mu }_{\mathrm{collective}}$. The credibility factor $Z\in [0,1]$ is the weight given to that risk's experience. Bayesian credibility starts with a prior distribution for a latent risk parameter, updates it using observations, and estimates the conditional risk mean by its Bayesian posterior expected value under squared-error loss. In conjugate models this posterior mean can have exactly the displayed credibility form.

For the unequal policy exposures, defineConditional independence of the binomial distribution observations gives the likelihood, up to a factor not depending on $\theta$,Multiplication by the beta distribution prior therefore gives Beta-binomial conjugacy:Both posterior parameters are positive, since $0\le s\le M$ and the prior parameters are positive.

The conditional target is $\mu (\theta )=m_{k+1}\theta$. To verify the quadratic-loss rule directly, for any proposed estimate $d$ and observed data $\mathcal{D}$,The variance term does not depend on $d$, so the unique Bayes estimator under squared error loss is the posterior mean. Applying the beta mean formula givesIt is also the posterior expected future claim count, by the tower property of conditional expectation.

Let $p_0=\alpha /(\alpha +\beta )$ be the collective prior claim probability and $\hat{p}=s/M$ the observed claims per policy. ThenTherefore the exact beta-binomial credibility with unequal exposures isThe individual experience must be weighted by exposure: $s/M$ is the policy-weighted frequency, not the unweighted average of $x_i/m_i$ when the $m_i$ differ. The factor multiplies next-year expected counts after scaling both component probabilities by $m_{k+1}$.

Let $P(z)={\sum }_{n\ge 0}{p}_n{z}^n$, $F(z)={\sum }_{j\ge 1}{f}_j{z}^j$, and $G(z)={\sum }_{k\ge 0}{g}_k{z}^k$ be the count, claim-size and aggregate probability generating functions. Independence in the aggregate claims model givesThe Panjer claim-count class recurrence can be written $n{p}_n=(an+b)p_{n-1}$. Multiply by $z^{n-1}$ and sum to obtainUsing the chain rule in $G=P\circ F$ then yieldsThe coefficient of $z^{k-1}$ on the left is $k{g}_k-a{\sum }_{j=1}^k{f}_j(k-j)g_{k-j}$, where the $j=k$ term is zero. On the right it is $(a+b){\sum }_{j=1}^{k}j{f}_j{g}_{k-j}$. Rearranging gives the Panjer recursionThe initial value is $p_0$ because all claim sizes are positive: zero total loss occurs exactly when $N=0$. Only already computed aggregate probabilities occur on the right. The count normalization determines $p_0$ from $a,b$ as well: with $w_0=1$ and $w_n={\prod }_{j=1}^n(a+b/j)$, we have $p_n=p_0{w}_n$ and $p_0=({\sum }_{n\ge 0}{w}_n{)}^{-1}$. The assumed valid count law makes this sum finite.

Finally $p_0$ must be positive. If it were zero, the recurrence would make every $p_n$ zero, contradicting normalization. At $n=1$ the recurrence therefore givesThis proves the sign constraint without assuming any particular count family. The three parts below also provide explicit initial probabilities for the aggregate recursion.

Let $M_n={\max }_{1\le j\le n}{W}_j$ for independent and identically distributed random variables with distribution function $F$. Then $P(M_n\le u)=F(u{)}^n$. Membership in the maximum domain of attraction of a nondegenerate distribution function $G$ means that some deterministic $a_n>0$ and $b_n\in \mathbb{R}$ satisfyor equivalently $F(a_{n}x+b_n{)}^n\to G(x)$ at every continuity point of $G$.

Under the stated convergence, put $Z_n=(M_n-b_n)/a_n$, $r_n={\alpha }_n/a_n\to a>0$, and $s_n=({\beta }_n-b_n)/a_n\to b$. The Slutsky theorem says that combining a sequence with convergence in distribution with sequences converging in probability to constants preserves the corresponding sum, product and ratio limits, provided a limiting denominator is nonzero. Applying it here givesThe limiting distribution function is $P((Z-b)/a\le x)=G(ax+b)$, continuous for every real $x$ because $G$ is continuous and $a>0$. Consequently,

Let $m={\overline{Y}}_{+++}$, $r_i={\overline{Y}}_{i++}$ and $c_j={\overline{Y}}_{+j+}$. The proposed least-squares estimators satisfy the sum-to-zero constraints because ${\sum }_i{r}_i=Im$ and ${\sum }_j{c}_j=Jm$. Define $e_{ijk}=Y_{ijk}-r_i-c_j+m$. Direct summation givesNow write any other admissible parameters as $\mu =m+u$, ${\alpha }_i=r_i-m+a_i$, ${\beta }_j=c_j-m+b_j$, where ${\sum }_i{a}_i={\sum }_j{b}_j=0$. Expanding the ordinary least squares criterion, the cross-products with $e$ vanish by the preceding identities. The cross-products between $u$, $a_i$ and $b_j$ vanish by their zero sums. ConsequentlyEvery added term is nonnegative and all vanish only at $u=0$, $a_i=b_j=0$. This proves the unique constrained minimum, withThe additive-model residual sum of squares and intercept-only residual sum of squares are thereforeThe same orthogonality gives the analysis of variance identityFitting factor B alone yields fitted values $c_j$ and residual sum ${\mathrm{RSS}}_0-S{S}_B$. Fitting A alone gives residual sum ${\mathrm{RSS}}_0-S{S}_A$. Adding B after A then reduces it to ${\mathrm{RSS}}_1$, a reduction of $S{S}_B$ again. The reduction due to B is $S{S}_B$ in either order. This is balanced factorial orthogonality; it generally fails for unequal cell replication.

There are $12$ observations. The additive model estimates $1+(3-1)+(2-1)=4$ independent parameters, so the missing degrees of freedom are A: 2, B: 1, residuals: 8. The individual requested residual sums are below.

To check for an interaction, extend the normal linear model to $\mu +{\alpha }_i+{\beta }_j+{\gamma }_{ij}$, with each row and column sum of $\gamma$ zero. Equivalently, fit the six unrestricted cell means. Its residual sum of squares is the within-cell sumThe interaction adds $(I-1)(J-1)=2$ parameters; the cell-means residual has $IJ(K-1)=6$ degrees of freedom. Under the no-interaction null hypothesis and the independent equal-variance normal error model,The numerator and denominator arise from orthogonal Gaussian projections, proving the exact F-test. Reject for a sufficiently large value. The printed additive analysis of variance table does not supply ${\mathrm{RSS}}_{\mathrm{cell}}$, so it cannot determine this test statistic numerically; the replicated cell data would do so.

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.