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For the Poisson process of claims and relative safety loading $\rho$, the premium rate in the classical risk model is $c=(1+\rho )\lambda \mu$. Dividing the adjustment coefficient equation $\lambda (M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_{\infty }$ as the upper endpoint of the finite moment-generating function domain, so $M(r)<\infty$ for $0\le r<r_{\infty }$ and its stated divergence occurs at that endpoint.

Set $F(r)=M(r)-1-(1+\rho )\mu r$. ThenDifferentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite moments of every order. Thus $F$ is strictly convex and negative just to the right of zero. If $r_{\infty }<\infty$, divergence of $M(r)$ makes $F(r)\to \infty$ at the endpoint. If $r_{\infty }=\infty$, choose $a>0$ with $\mathbb{P}(X_1\ge a)>0$; then $M(r)\ge \mathbb{P}(X_1\ge a)e^{ar}$, which eventually exceeds every linear function. Again $F(r)\to \infty$. The intermediate value theorem gives a positive zero.

There is at most one positive zero: strict convexity and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho \mu$ and crosses zero exactly once. HenceThis proves the secant-slope existence criterion for an adjustment coefficient in the present setting.

Put $m_j=\mathbb{E}{X}_1^j$. For $u>0$, the positive remainder of the exponential series gives $e^u>1+u+u^2/2$. Apply it to $u=R{X}_1$ and take expectations:Dividing by $R>0$ yieldsKeeping the cubic term similarly givesThe polynomial $g(r)=m_3{r}^2+3m_{2}r-6\rho \mu$ is strictly increasing for $r\ge 0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, soFinally, $3m_2{r}_1=6\rho \mu$, whence $g(r_1)=m_3{r}_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested upper bounds can in fact be sharpened to the strict ordering $0<R<r_2<r_1$. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and polynomial moments elsewhere. These are polynomial moment bounds for the adjustment coefficient.

For rate-$\alpha$ exponential distribution claims,The positive root satisfies $1/(\alpha -R)=(1+\rho )/\alpha$ after dividing out $R$, soThe cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho {\alpha }^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho >0$.

Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Condition on the claim count, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.

Let $X_i=Y_i-{\theta }_0$. Under the symmetric continuous null hypothesis, positive and negative signs have probability $1/2$ and zeros have probability zero. For the sign test, take $N_{+}={\sum }_i{1}_{\{X_i>0\}}$ and reject for small values: $N_{+}\sim \mathrm{Bin}(n,1/2)$ under the null. Choose $k$ with ${\mathbb{P}}_0(N_{+}\le k)\le \alpha$; randomization at the next value attains exact size if desired. The lower tail is appropriate for $\theta <{\theta }_0$.

For the Wilcoxon signed-rank test, rank $|X_i|$ increasingly and put $W_{+}={\sum }_i{R}_i{1}_{\{X_i>0\}}$. Continuous observations have no absolute-value ties almost surely. Conditional on the absolute values, the signs are independent fair signs because symmetry makes the sign independent of magnitude. Hence $W_{+}$ has the distribution of ${\sum }_{r=1}^{n}r{B}_r$ for independent Bernoulli random variables $B_r$ with parameter $1/2$. Again reject in the lower tail, using this exact subset-sum distribution or a normal approximation. Independence gives the requested null moments:For a symmetric kernel $h$ of fixed order $m$, a one-sample U-statistic is $U_n={\left(\genfrac{}{}{0ex}{}{n}{m}\right)}^{-1}{\sum }_{i_1<\cdots <i_m}h(X_{i_1},\dots ,X_{i_m})$. State the U-statistic central limit theorem: if $\mathbb{E}{h}^2<\infty$, $\eta =\mathbb{E}h$, and ${\zeta }_1=\mathrm{Var}(\mathbb{E}[h\mid X_1])>0$, then $\sqrt{n}(U_n-\eta )\Rightarrow N(0,m^2{\zeta }_1)$. The nondegeneracy condition is important.

For $h(x,y)=1_{\{x+y>0\}}$, its order-two U-statistic satisfies the exact Wilcoxon signed-rank statistic identityTo prove it, in an unordered pair the sign of the sum is the sign of whichever observation has larger absolute value. Counting the pairs for which a positive observation is larger contributes its rank minus one; its own positive-sign indicator supplies the remaining one.

Write $F_0$ for the distribution function of $X$ under the null. Symmetry gives $\eta =1/2$ andThe probability integral transform makes $F_0(X_1)$ uniform on $(0,1)$, so ${\zeta }_1=1/12$. Therefore $\sqrt{n}(U_n-1/2)\Rightarrow N(0,1/3)$. The centred $N_{+}$ is $O_p(\sqrt{n})$ and is negligible on the scale $n^{3/2}$ of $W_{+}$. Combining the exact decomposition and its mean givesThe lower-tail approximate level-$\alpha$ rejection rule compares this standardized statistic with ${\Phi }^{-1}(\alpha )$; a half-unit continuity correction can refine discrete-tail calculations.

Put $S_1={\sum }_i{Y}_i$, $S_2={\sum }_i{Y}_i^2$ and $R=\sqrt{S_2}$. Apart from a data-only constant, the log-likelihood isThe natural parameters in the ambient two-parameter exponential family of normal distributions are ${\eta }_1=1/\mu$ and ${\eta }_2=-1/(2{\mu }^2)$, restricted to the nonaffine curve ${\eta }_2=-{\eta }_1^2/2$, ${\eta }_1>0$. Its parametrization has rank one. Thus this is a one-dimensional curved exponential family inside a two-dimensional full family.

The Fisher-Neyman factorization theorem makes $(S_1,S_2)$ sufficient. For two samples $y,z$, the parameter-dependent part of their log density ratio is $\Delta S_1/\mu -\Delta S_2/(2{\mu }^2)$. A polynomial in $1/\mu$ is constant on $(0,\infty )$ only if both nonconstant coefficients vanish. The likelihood-ratio criterion for minimal sufficiency therefore proves $(S_1,S_2)$ is minimal sufficient.

Write $Y_i=\mu Z_i$ with independent $Z_i\sim N(1,1)$. Since $\mu >0$, $a=\sqrt{n}\sqrt{{\sum }_i{Z}_i^2}/{\sum }_i{Z}_i$ has a probability distribution independent of $\mu$: it is an ancillary statistic. The denominator is nonzero almost surely. The Cauchy-Schwarz inequality gives $|a|\ge 1$; on the requested positive branch, $S_1>0$ and $a\ge 1$.

The likelihood equation is $n{\mu }^2+S_1\mu -S_2=0$. Its positive root isThere is exactly one positive root. The sign of the score function changes from positive to negative there, and $\ell \to -\infty$ at both ends of $(0,\infty )$ when $S_2>0$, so it is the unique maximum-likelihood estimator. The useful identities areSubstitution yields, dropping the constant $-n/2$,Differentiating twice and using $q/a=q^2-1$ at the fitted value gives the observed informationThe scalar P-star approximation uses $\sqrt{j(v)}\exp \{\ell (\mu ;v,a)-\ell (v;v,a)\}$. Consequently, with its normalizing factor explicitly retained,Here $C_n(a)$ is chosen to make the density integrate to one, and is independent of $\mu$. To see this without assuming an approximate normalizer, defineThe integral is finite and positive, and normalized $p^{\ast }$ is exactlyIndeed $C_n(a)=\{\sqrt{n(1+q^2)/(2\pi )}\exp [n{q}^2/2-nq/a]J_n(a){\}}^{-1}$.

There is a direct check of this conditional scale density in a curved Gaussian family. In polar coordinates $y=Ru$, the sample density times volume element is proportional to ${\mu }^{-n}{R}^{n-1}\exp [-R^2/(2{\mu }^2)+R{\sum }_i{u}_i/\mu ]dRd\omega (u)$. Conditioning on $a$ fixes ${\sum }_i{u}_i=\sqrt{n}/a$. The remaining angular factor is independent of $R,\mu$, while $R=q\sqrt{n}v$. Thus the exact conditional radial density has the same normalized form. For $n=1$ the positive ancillary branch fixes the positive sign and the same one-dimensional radial calculation applies. In this scale model the normalized $p^{\ast }$ expression equals the exact conditional density, not merely its leading approximation.

For $t>0$, compute its conditional distribution by one-dimensional quadrature:It is zero for $t\le 0$; with the conditional-density calculation just given, this normalized formula is exact for almost every admissible $a$. Numerically, subtract the maximum log integrand before exponentiating to avoid overflow. The sampling probability still depends on the stipulated true $\mu$; replacing it by a fitted value would be a separate plug-in inference step.

A transformation model consists of a group $G$ acting measurably and invertibly on the sample space, an induced group action on the parameter space, and a family $\{P_{\theta }\}$ satisfying $Y\sim P_{\theta }\Rightarrow gY\sim P_{g\theta }$. In a generated, transitive model, every $P_{\theta }$ is the image of a reference distribution under some transformation. In the density formulation, the transformed density includes the appropriate sample-space Jacobian determinant; the model is a statement about whole probability laws, not merely about changing parameter names.

A maximal invariant $A$ has two properties: $A(gy)=A(y)$ for every $g$, and $A(y)=A(z)$ implies $z=gy$ for some $g$. It is therefore a coordinate for the group orbits, retaining exactly the information unaffected by the transformations. An equivariant estimator obeys $T(gy)=gT(y)$, where the right action is the induced parameter action.

For the normalization construction, suppose the estimator determines a group element $t(y)\in G$ with $t(gy)=gt(y)$. Define $A(y)=t(y{)}^{-1}y$. ThenIf $A(y)=A(z)$, rearranging gives $z=t(z)t(y{)}^{-1}y$. This is an actual transformation in $G$, so the normalized sample is a maximal invariant. This group-valued assumption matters: an estimator of a parameter with a nontrivial stabilizer does not necessarily determine a unique normalization. For a transitive parameter space $G/H$, choose a transformation carrying the estimated parameter to the reference value and take the resulting normalized sample modulo the remaining $H$-action. Different choices then have the same residual orbit. Under a free action $H$ is trivial and the simpler formula applies.

For a location-scale model, $g_{a,b}(y_1,\dots ,y_n)=(a+b{y}_1,\dots ,a+b{y}_n)$, $b>0$, with composition $(a,b)(c,d)=(a+bc,bd)$. Its parameter action is $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. Choose location and scale estimates satisfyingFor instance, on nonconstant samples take the sample mean and the positive square root of $n^{-1}{\sum }_i(y_i-\bar{y}{)}^2$. The standardized vectoris invariant. Conversely, equality of these vectors gives $z_i=c+d{y}_i$ for all $i$, with $d=\hat{s}(z)/\hat{s}(y)>0$ and $c=\hat{m}(z)-d\hat{m}(y)$. Hence it is maximal, not just invariant. Degenerate constant samples have zero fitted scale and form a separate orbit; define a separate invariant value there. The construction is on labelled sample vectors: one must not discard their order unless permutations are also included in the transformation group.

The profile likelihood for the parameter of interest replaces the nuisance parameter by its constrained best fit: $L_p(\psi )={\sup }_{\lambda }L(\psi ,\lambda )$, with ${\ell }_p(\psi )=\ell (\psi ,{\hat{\lambda }}_{\psi })$. It is not in general a normalized sampling density. A conditional likelihood instead uses the sampling law conditional on a statistic carrying the nuisance dependence. In the canonical exponential family here, conditioning on $S$ removes $\lambda$ because the factor involving $\lambda$ is constant over the conditioning fibre.

Let ${\bar{T}}_1=n^{-1}{\sum }_i{\tau }_1(y_i)$ and let the observed $S$ equal $s$. For the cumulant-generating function of ${\tau }_2(Y)$ at fixed $\psi$,The saddlepoint solves $K^{\prime }(\hat{t})=s$, equivalently $d_{\lambda }(\psi ,{\hat{\lambda }}_{\psi })=s$ with ${\hat{\lambda }}_{\psi }=\lambda +\hat{t}$. This is also the constrained likelihood equation, since ${\ell }_{\lambda }=n(s-d_{\lambda })$. Assume an interior saddle, $d_{\lambda \lambda }>0$, and the usual smooth nonlattice conditions under which $S$ has a density. Exponential tilting to ${\hat{\lambda }}_{\psi }$ makes $s$ the tilted mean. A local normal approximation gives tilted density $\{n/(2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })){\}}^{1/2}$ at that mean; undoing the tilt gives the saddlepoint density approximationThe factor $n^{1/2}$ is for the sample mean, rather than the sum. A transform-based derivation gives the same expression by expanding $K$ to second order at its stationary point.

The joint log-likelihood is $n\psi {\bar{T}}_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(y_i)$. Subtracting $\log {\hat{f}}_S(s;\psi ,\lambda )$ cancels every appearance of the original $\lambda$, leavingAny change from the ambient density to a conditional density on $S=s$ contributes only a data-dependent Jacobian determinant. Thus, up to a constant independent of $\psi$, the saddlepoint conditional likelihood adjustment isEquivalently $B=\frac{1}{2}\log j_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })$ up to the constant $-\frac{1}{2}\log n$, where $j_{\lambda \lambda }=n{d}_{\lambda \lambda }$ is the nuisance observed information. The adjustment has a positive sign. The printed exponential-family form alone does not guarantee a density for $S$: for example a continuous $Y$ can have ${\tau }_2(Y)=1_{\{Y>0\}}$. In a lattice case one must approximate the probability mass of the sum instead; the displayed density calculation uses the regular continuous interpretation.