Let $B$ be a nondegenerate alternating bilinear form on $V$ of dimension $2m$ over ${\mathbb{F}}_q$. A symplectic basis has $B(e_i,f_j)={\delta }_{ij}$ and all $e_i,e_j$ and $f_i,f_j$ pairings zero. Such bases are built by choosing a nonzero $e_1$, finding $f_1$ with $B(e_1,f_1)=1$, and continuing in their nonsingular orthogonal complement for a sesquilinear form. In these coordinates the group preserving $B$ isIts action on symplectic bases is regular. There are $q^{2m}-1$ choices for $e_1$ and $q^{2m-1}$ for $f_1$, so induction gives the finite symplectic group orderThe center of a group consists of $aI$ with $a^2=1$, and has order $\gcd (2,q-1)$. To see there are no other central elements, a central element commutes with every symplectic transvection and hence fixes every symplectic transvection center line; a linear map fixing every line is scalar. Consequently the projective symplectic group over a finite field has this order divided by $\gcd (2,q-1)$.

Here is a proof of the main ingredients in simplicity. The symplectic transvectionpreserves $B$, has inverse $T_{v,-c}$, and fixes $v^{\perp }$ pointwise. These transformations generate $S{p}_{2m}(q)$. Indeed if $B(u,w)\ne 0$, then $T_{w-u,1/B(u,w)}$ sends $u$ to $w$. If the pairing is zero, choose an intermediate vector having nonzero pairing with both, so two symplectic transvections suffice. This aligns the first vector of any symplectic basis. For partners $f,f^{\prime }$ of the same $e$, their difference lies in $e^{\perp }$. If $B(f,f^{\prime })\ne 0$ a symplectic transvection centered at $f^{\prime }-f$ aligns them and fixes $e$. If it is zero and the partners differ, $h=f+e$ pairs nontrivially with both and still pairs to one with $e$; two such symplectic transvections work. After fixing the pair, induction in its orthogonal complement for a sesquilinear form aligns the rest of the basis.

The projective action on one-spaces is faithful. For $m\ge 2$, a line stabilizer has three orbits: the line itself, the other lines perpendicular to it, and the lines not perpendicular to it. Witt's lemma proves transitivity within the latter two orbits by matching the two-dimensional restricted form. This action is primitive. A block containing distinct perpendicular lines $L,M$ is invariant under both stabilizers and therefore contains a line perpendicular to $M$ but not to $L$. A block containing distinct nonperpendicular lines likewise contains a line perpendicular to $L$ but not to $M$, by using the $M$ stabilizer. Such lines exist by nonsingularity in dimension at least four. Either possibility makes the block contain both nontrivial stabilizer orbits, hence every line. For $m=1$ the action on the projective line is two-transitive and thus primitive.

Symplectic transvections with one fixed center line form an Abelian normal subgroup of its stabilizer, and their conjugates generate the whole group. The Iwasawa simplicity lemma now reduces simplicity to being perfect. Its short proof is useful: any nontrivial normal subgroup of a faithful primitive group is transitive, since its orbits are blocks. Writing $G=NH$, where $H$ is a point stabilizer, shows that all conjugates of the specified Abelian subgroup have the same image modulo $N$. Thus $G/N$ is Abelian. If $G$ is perfect, $N=G$.

The groups can also be shown to be perfect concretely. For $q>3$ choose $a$ with $a^2\ne 1$ and let $D$ scale $e$ by $a$ and $f$ by $a^{-1}$. Thenso every symplectic transvection is a group commutator. For $q=3,m\ge 2$, symplectic transvections of coefficient one are conjugate, have order three, and the four with centers $e_1,e_2,e_1+e_2,e_1-e_2$ in a two-dimensional totally isotropic subspace multiply to the identity. They commute, and the sum of their rank-one terms is zero modulo three. Their common class in the abelianization is therefore annihilated by both three and four, hence is zero. For $q=2,m\ge 3$, use the seven nonzero vectors of a three-dimensional totally isotropic subspace. The symplectic transvections commute and their product is the identity, because each diagonal coefficient occurs four times and each cross coefficient twice. Their common abelianized class is annihilated by two and seven, hence is zero. Generation now proves these groups are perfect.

It follows thatThe exceptions are $S_3$, $A_4$ and $S_6$ respectively. For odd $q$ the full symplectic group over a finite field has nontrivial center of a group, so simplicity refers to its quotient group; for even $q$ the center of a group is trivial.

A symplectic parabolic subgroup stabilizes a flag of totally isotropic subspaces. The maximal such parabolic subgroups $P_r$ stabilize a single $r$-space, $1\le r\le m$. In an adapted basis their Levi subgroup is $G{L}_r(q)\times S{p}_{2m-2r}(q)$, with an arbitrary block of $r(2m-2r)$ parameters and a remaining symmetric block of $r(r+1)/2$ parameters. ThusThese are the single-member flag stabilizers, equivalently the standard maximal parabolic subgroups obtained by omitting one node of the symplectic Dynkin diagram. Two particular cases have transparent descriptions.

For the symplectic point stabilizer, write $V=⟨e,f⟩\perp W$. Its unipotent radical consists ofAll pairings are preserved by direct substitution. The product law is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, so this radical has order $q^{2m-1}$. With group commutator convention $uv{u}^{-1}{v}^{-1}$, its commutator has central parameter $2B(b,b^{\prime })$: it is Heisenberg-type in odd characteristic, but Abelian in characteristic two. A Levi subgroup scales $e,f$ inversely and acts as $Sp(W)$, givingThis index counts the one-spaces, all of which are isotropic for an alternating bilinear form.

For a Lagrangian subspace $E=⟨e_1,\dots ,e_m⟩$, the symplectic Lagrangian stabilizer has block formSubstituting into $g^{\mathsf{T}}Jg=J$ proves both the inverse-transpose lower block and symmetry of $S$. The unipotent radical is the additive group of symmetric matrices, and $A$ acts by $S\mapsto AS{A}^{\mathsf{T}}$. ThereforeThe last identity follows by canceling $q$-powers and $(q^j-1)$ factors in the two order formulas; it counts maximal totally isotropic subspaces.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

Here is Witt's lemma with the hypotheses appropriate to the classical groups. Let $V$ be finite-dimensional with a nonsingular alternating bilinear form, a nonsingular symmetric bilinear form in characteristic different from two, or a nonsingular Hermitian form with nontrivial involution. Every isometry of a space with a form $\varphi :U\to U^{\prime }$ between vector subspaces extends to a form-preserving automorphism of $V$. For quadratic forms, the same statement holds in every characteristic when the polar bilinear form of a quadratic form is nonsingular and the isometry of a space with a form preserves the quadratic form itself. The restricted forms on $U,U^{\prime }$ may be degenerate. We prove these versions, including the characteristic-two quadratic case needed later.

For an alternating bilinear form $B$, split the restricted space as $U=H\oplus R$, where $R=U\cap U^{\perp }$ and $B{|}_H$ is nonsingular. Choose a symplectic basis for $H$ and a basis $r_1,\dots ,r_t$ of $R$. The nondegenerate form on $V$ supplies vectors $s_j$ in $H^{\perp }$ with $B(r_i,s_j)={\delta }_{ij}$. Adding linear combinations of the $r_i$ to the $s_j$ makes $B(s_i,s_j)=0$: for each pair $i<j$ one may cancel its pairing by adding an appropriate multiple of $r_i$ to $s_j$, without disturbing the dual pairings or previously treated pairs. This uses no division by two. Thus $H$ together with the pairs $(r_i,s_i)$ is nonsingular and its orthogonal complement for a sesquilinear form has a symplectic basis.

Apply the same construction to $\varphi (H),\varphi (r_i)$, choosing partners $s_i^{\prime }$ there. The two completed bases have identical alternating Gram matrices and equal remaining dimensions. The linear map matching these bases preserves $B$ and agrees with $\varphi$ on $H\oplus R=U$. This proves the alternating version over an arbitrary field, including characteristic two.

For a quadratic form $Q$, define its polar bilinear form of a quadratic form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. Its symmetrypreserves $Q$. The zero-dimensional case is immediate. Induct on $\dim U$: choose $Q(t)\ne 0$ and a hyperplane $K\subseteq U\cap t^{\perp }$. Extending $\varphi {|}_K$ first reduces to $\varphi {|}_K=I$. Put $U=K\oplus Fu$, $v=\varphi (u)$, $d=u-v\in S=K^{\perp }$. If $Q(d)\ne 0$, $s_d$ completes the extension. Otherwise $B(u,d)=B(v,d)=0$. The hyperplanes $H=S\cap u^{\perp }$, $H^{\prime }=S\cap v^{\perp }$ are proper because $S^{\perp }=K$.

For $a\in S\setminus (H\cup H^{\prime })$ with $Q(a)\ne 0$, set $c=u-s_a(v)$. Thenand $s_a{s}_c$ fixes $K$ and sends $u$ to $v$. If no such $a$ exists and $|F|>2$, the identityforces $Q(z)=B(a,z)=0$. The complement of two hyperplanes spans $S$: adding a multiple of any outside vector avoids at most two forbidden scalars. Hence $d\perp S$, so $H=H^{\prime }$, forcing $Q{|}_S=0$, contrary to $t\in S$.

Over ${\mathbb{F}}_2$, choose $a$ outside both hyperplanes. In the remaining case $Q(a)=Q(d)=B(a,d)=0$, and $B(a,u)=B(a,v)=1$. The explicit maphas $g^2=I$, preserves $Q$ by cancellation of its two cross terms, fixes $K$ and sends $u$ to $v$. This closes the induction. The characteristic-independent symmetry argument and its binary exceptional step are also treated in Casselman's quadratic-form notes and Elman, Karpenko and Merkurjev, section 8.

In odd characteristic a symmetric bilinear form is recovered by the polarization identity for $Q(x)=B(x,x)/2$, so the quadratic proof also proves its extension theorem. For completeness, the Hermitian form case has a similarly explicit induction. Take $h$ linear in the first variable, with conjugation $a\mapsto \bar{a}$. For anisotropic vectors for a form $u,v$ with equal value $A=h(u,u)=h(v,v)\ne 0$, let $d=u-v$ and $c=h(u,d)$. When $c\ne 0$, the rank-one mapsends $u$ to $v$ and preserves $h$, since $h(d,d)=c+\bar{c}$ and the extra coefficient in $h(gx,gy)$ isWhen $c=0$, insert $\lambda u$ with $\lambda \bar{\lambda }=1$, $\lambda \ne 1$, and use two such maps. Such a scalar exists for a nontrivial involution: $\lambda =a/\bar{a}$ with $a\ne \bar{a}$ works. If the source vector subspace contains an anisotropic vector for a form, align it this way and induct in its nonsingular orthogonal complement for a sesquilinear form.

If instead every source vector is isotropic, the Hermitian polarization identity makes its whole restricted form zero. Choose $e$ in its basis and a partner $f$ pairing to one with $e$ and to zero with the remaining source basis. Make $h(f,f)=0$ by replacing $f$ by $f-te$, where $t+\bar{t}=h(f,f)$. The trace map is onto the fixed field: in odd characteristic divide by two; in characteristic two scale a nonzero trace. Choose $a+\bar{a}\ne 0$. The orthogonal vectors $e+af$ and $e-\bar{a}f$ have nonzero values $a+\bar{a}$ and its negative, so the preceding rank-one construction aligns this entire hyperbolic pair with its target pair. The remaining source basis lies in the pair's orthogonal complement for a sesquilinear form; induction finishes the Hermitian proof.

The hypotheses matter. For example, the identity-matrix symmetric bilinear form on ${\mathbb{F}}_2^3$ has canonical vector $c=(1,1,1)$ characterized by $B(x,c)=B(x,x)$, so every isometry of a space with a form fixes $c$. The one-dimensional map $e_1\mapsto c$ preserves the restricted bilinear form but cannot extend. Thus “all symmetric forms in characteristic two” is not a correct replacement for the quadratic or alternating versions proved above.

A credibility premium estimates a risk's expected future claims by combining its own observed experience with a collective or prior mean. In its basic form $Z\overline{X}+(1-Z)\mu$, the credibility factor $Z\in [0,1]$ is the weight assigned to individual experience. Larger credibility places less weight on pooling with the population. The amounts below are pure expected-claim premiums; no additional expense or profit loading is specified.

Under squared-error loss, the posterior risk for estimating the unknown conditional mean $\theta$ by $a$ isHence the Bayes estimator under squared error loss is the posterior mean. In the equal-year model, conditional independence makes the likelihood proportional to $\exp [-{\sum }_{j=1}^n(X_j-\theta {)}^2/(2{\sigma }_1^2)]$. Combining it with the normal prior and completing the square givesBecause $\mathbb{E}[X_{n+1}\mid \theta ]=\theta$, the required estimator is $m_n$; it also equals the predictive mean $\mathbb{E}[X_{n+1}\mid \mathbf{X}]$. The normal-normal credibility expression isThis is exact Bayesian credibility, not merely the best linear approximation to a possibly nonlinear posterior mean.

For unequal life counts use the usual exposure model: individual-life losses are conditionally independent with variance $v$, and the annual totals are conditionally independent given the same $\theta$. ThenThe dot denotes the stipulated normal approximation. The positive exposures $m_j$ are known. Define total past exposure and its experience mean byThe normal log-likelihood has quadratic term $-{\sum }_j{m}_j(X_j-\theta {)}^2/(2v)$. With the prior $N(\mu ,{\sigma }^2)$, normal-normal conjugacy with unequal exposures gives posterior variance and meanTherefore the estimated pure premium per life for year $n+1$ has the normal-normal credibility with unequal exposures formIt is also an exact normal-model Bühlmann–Straub credibility estimate. The individual experience is exposure-weighted: the unweighted mean of the yearly per-life observations would generally discard the different precisions.

For positive $W,v,{\sigma }^2$, differentiation givesThus increasing the prior variance increases credibility, because greater between-risk heterogeneity makes the collective mean less representative of this risk. Increasing the individual process variance decreases credibility, because a fixed amount of observed experience becomes noisier. The future exposure supplies no new observation, so it does not enter $Z_W$. It only scales the total premium:The independent-life variance scaling is part of the usual exposure interpretation. Without it, a known variance $v$ for one life does not determine the variance of a total: a common pairwise conditional covariance $c$ would add $m_j(m_j-1)c$ to $m_{j}v$. Similarly, cross-year conditional dependence would change the joint likelihood. Those assumptions are implicit in applying the standard credibility model to the final paragraph.

Let $Y=(\log X-\mu )/\sigma$ and write $\varphi$ for the standard normal density, with $\sigma >0$. The change of variables converts the density integral intoComplete the square using $e^{k\sigma y}\varphi (y)=e^{k^2{\sigma }^2/2}\varphi (y-k\sigma )$. Integration gives the truncated lognormal momentHere $\Phi$ is the standard normal distribution function; take $\log 0=-\infty$ and $\log \infty =\infty$. The identity holds for every real $k$, including the full-interval limits, because the shifted Gaussian integral remains finite.

The limited excess of loss reinsurance payment is $L(X)=\min \{(X-M{)}_{+},A\}$. It is zero below $M$, equals $X-M$ for $M<X<M+A$, and equals $A$ above $M+A$. Since the lognormal distribution has a continuous density,Put $d_j(t)=(\log t-\mu -j{\sigma }^2)/\sigma$, $j=0,1$, and $\overline{\Phi }=1-\Phi$. Applying the moment identity with $k=1$ and $k=0$ gives the conditional lognormal layer recoveryThe denominator is $\mathbb{P}(X>M)$, and payments outside that event are zero. The three numerator terms are respectively the truncated first moment, the attachment subtracted from partial payments, and the capped payments on claims above the layer. This proves the conditional expectation including the limit on each reinsurer payment.

For claims inflation by $h=1.1$, set $X_h=hX$. Its log-mean is ${\mu }_h=\mu +\log h$ and its log-standard-deviation stays $\sigma$. The attachment and payment limit stay fixed. Therefore the requested new conditional mean isExplicitly, in the preceding fraction replace $d_j(t)$ by $d_j^{(h)}(t)=(\log t-\mu -\log 1.1-j{\sigma }^2)/\sigma$ and replace $e^{\mu +{\sigma }^2/2}$ by $1.1e^{\mu +{\sigma }^2/2}$. Equivalently, claims inflation of a fixed reinsurance layer givesbecause $L_{M,A}(hX)=h{L}_{M/h,A/h}(X)$ and the new conditioning event is $X>M/h$. Merely multiplying the original conditional mean by $1.1$ would omit both of these threshold changes.

Use $\alpha$ for the intercept, to avoid confusing it with the observation means. Let $e_i=\exp (\hat{\alpha }+{\hat{\beta }}^T{x}_i)$ be the fitted Poisson regression mean. Apart from constants, its log-likelihood isAt a finite interior maximum-likelihood estimate, the intercept score equation gives the Poisson regression margin-matching score equations:The statistical saturated model fits each mean to $y_i$, taking the boundary mean zero if $y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson devianceThe intercept balance cancels the sum of linear terms. ConsequentlyThe unsimplified formula makes nonnegativity transparent; individual terms in the simplified formula need not each be positive.

For the deviance goodness-of-fit test, if the fitted design has rank $r$, compare a suitably calibrated $D$ with ${\chi }_{n-r}^2$, rejecting for large values. This approximation applies in a regular large-count regime, such as a fixed set of cells with increasing exposure and positive fitted proportions. It is not automatically accurate for many sparse Poisson observations. Check residual patterns and overdispersion; where the approximation is doubtful, simulate independent Poisson random variables with fitted means, refit each sample, and use a parametric bootstrap distribution of the same statistic. A large p-value means no detected lack of fit, not proof of the model. If a finite interior estimate fails to exist, the score proof needs a boundary-limit interpretation; for example all-zero data give a supremum with every fitted mean tending to zero and limiting deviance zero.

The prose names a second year that differs from the table label. The calculations use the four supplied counts, interpreting the first and second rows as the two displayed periods; no numerical conclusion depends on resolving that label discrepancy.

The first generalized linear model treats the four counts as independent Poisson random variables, with a log link and both row and column regression factors. The row-by-column interaction makes the log-linear model saturated. With $r,c\in \{0,1\}$, writeIts four maximum-likelihood fitted values equal the observed counts. Thus $\hat{\alpha }=\log 20$, $\hat{\beta }=\log (12/20)$, $\hat{\gamma }=\log (9/20)$, andThe interaction is the log odds ratio measuring association of row and column. Under independent Poisson sampling, the delta-method variances are $1/20$ for the intercept, $1/12+1/20$ for the row contrast, $1/9+1/20$ for the column contrast, and $1/20+1/9+1/12+1/11$ for the interaction. Their square roots reproduce the printed standard errors. Although the software labels the standardized ratios as t values, these are asymptotic normal Wald statistics, with known Poisson dispersion one, not exact finite-df Student tests.

The statistical saturated model has zero residual deviance and zero residual degrees of freedom; this is a perfect interpolation, not evidence of predictive adequacy. The null deviance $5.016056$ compares the four observed counts with an intercept-only model, whose four fitted means are $52/4=13$. Fisher scoring is the numerical score/information iteration used to fit the model; the iteration count is a convergence report, not a test statistic.

Dropping the interaction gives the independence log-linear model for a two-way contingency table, $\log {\mu }_{rc}=\alpha +\beta r+\gamma c$. Its Poisson regression margin-matching score equations match the row and column totals, soHere $r_r$ and $c_c$ denote the observed row and column totals. In particular the fitted row multiplier is $23/29$, the column multiplier is $20/32$, and the intercept is $\log (29\cdot 32/52)=2.881788$. These produce the second coefficient table. Their standard errors follow from the inverse Fisher information matrix; for the row and column contrasts their squared values are $1/29+1/23$ and $1/32+1/20$, while the intercept variance is $1/29+1/32-1/52$.

The Poisson deviance relative to the saturated fit isThe linear terms cancel because fitted and observed totals agree. The additive model has three coefficients, leaving one degree of freedom. Wilks theorem calibrates this likelihood-ratio test approximately by ${\chi }_1^2$, giving $p\simeq 0.2164$. The interaction Wald test instead gives $p\simeq 0.2192$; the two are only asymptotically equivalent.

Fisher's exact test conditions on both margins under independence. If $X$ is the upper-left count, thenIts usual two-sided p-value sums the probabilities of all tables no more probable than the observed $X=20$. That sum is $0.259724$, reproducing the output without an asymptotic approximation. It tests the same absence of row-column association, but conditioning and discreteness explain the numerical difference from the likelihood-ratio test.

In ordinary language, the male proportion among the cases is $20/29\simeq 69.0\%$ in the first row and $12/23\simeq 52.2\%$ in the second. The estimated case-sex odds ratio is $2.037$, but its approximate 95-percent confidence interval is wide, $(0.655,6.338)$. These small samples do not give convincing evidence that the sex composition of cases changed between the two periods. Failure to reject is not evidence that the compositions are exactly equal. Nor do these counts establish a male-to-female disease-incidence ratio or an incidence trend: population denominators and comparable ascertainment would be needed for those interpretations.

The S-Plus input and the data table differ in one entry: the former uses $33$ for the first year in the second male age group, whereas the table has $22$. The displayed coefficients are reproduced by the software vector with $33$, so that is the input analysed here. Using $22$ instead would give a different fit, including an intercept of $36.175$ rather than $37.275$.

The factor grid has $5\times 4\times 2=40$ rows. Its first factor varies fastest, so the five years run within each age group, and the age groups run within each sex. All three predictors are regression factors: in particular the year coefficients are category contrasts, not a fitted linear slope in calendar time. Under reference-level coding, take the first year, youngest age group and men as the reference levels in a regression factor. Write $s=1$ for women and $0$ for men, and $I_{yk},I_{aj}$ for the other year and age indicators. The normal linear model isThe star in the formula includes both main effects and their statistical interaction. There are twelve coefficients: an intercept, four year contrasts, one sex contrast, three age contrasts and three sex-by-age contrasts. Year effects are assumed common across all sex-age combinations; there is no year interaction term. Independent errors with common variance are additional assumptions of the reported exact tests.

The least-squares estimator is $\hat{b}=(X^{T}X{)}^{-1}{X}^{T}P$, where $X$ has rank twelve. Under the normal linear model,and these quantities are independent. This standard normal-projection theorem gives $s^2=\mathrm{RSS}/28$ and estimated coefficient standard errors $s\sqrt{(X^{T}X{)}_{jj}^{-1}}$. A coefficient estimate divided by its standard error has a Student t-distribution with 28 degrees of freedom under its zero-coefficient null. These are the printed t statistics and two-sided p-values; a printed zero is rounding, not a probability of exactly zero. The option suppresses the coefficient-correlation display.

The fitted intercept $37.275$ is the expected percentage for the reference sex, age and year. Relative to that year, the year changes are $0.375,0.500,1.750,3.000$ percentage points. The last two are individually significant at the conventional five-percent level; the first two are not. The sex coefficient $-18.8$ is the female-minus-male difference in the youngest group. The male age changes are $-7,-12.8,-23$ points. Adding the three interaction term coefficients gives the female age changes $-5.2,-8.2,-13.8$. Thus the sex gap shrinks with age. An individual insignificant interaction term coefficient does not justify deleting the whole interaction: the joint partial F-test for the three interaction terms gives $F=18.6822$ on $(3,28)$ degrees of freedom, with $p\simeq 7.46\times 10^{-7}$.

The regression residuals are observed minus fitted percentages; their printed five-number summary describes their spread. Direct calculation gives $\mathrm{RSS}=60.2$, so the residual standard error isThe coefficient of determination is $1-\mathrm{RSS}/\mathrm{TSS}=0.985827$. The overall F-test compares all eleven non-intercept coefficients with zero:Under the null it has an F-distribution on $(11,28)$ degrees of freedom; its very small p-value shows substantial explained variation. This does not independently validate the Gaussian error assumptions. Survey percentages can have different sampling variances and correlations, and the survey denominators and design are not supplied. The output's exact inferential interpretation is conditional on the stated normal linear model.

The final command makes an interaction plot, not another regression fit. Its points are arithmetic averages over the five years for each sex-age combination. It draws two decreasing, nonparallel lines, with coordinatesTheir gaps are $18.8,17.0,14.2,9.6$ percentage points, illustrating the sex-by-age statistical interaction. The average uses the software vector's disputed entry, consistently with the displayed fit. These are means of percentages, not denominator-weighted pooled prevalence estimates.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.