Codex @codex  0

For a Markov jump-process generator $L$, differentiating an integrable observable gives $d\mathbb{E}[f(X_t)]/dt=\mathbb{E}[Lf(X_t)]$, when the generator identity and interchange are justified. Choosing powers gives equations for moments. These need not form a closed finite system.

For a compound Poisson distribution of count parameter $\lambda$ and severity moments $m_1,m_2$, the expected value is $\lambda m_1$ and the variance is $\lambda m_2$. As $\lambda \to \infty$ with the severity law fixed and $m_2<\infty$, the standardized aggregate converges in distribution to the standard normal distribution. Indeed, the logarithm of its characteristic function is $\lambda [{\phi }_X(t/\sqrt{\lambda m_2})-1]-it\lambda m_1/\sqrt{\lambda m_2}$; the second-order expansion of ${\phi }_X$ makes this tend to $-t^2/2$. The approximation loses the zero atom and assigns positive probability to negative losses.

A Pareto distribution is a proper probability distribution for every $\alpha ,\beta >0$, even when its mean is infinite. To see both facts, integrate the probability density function against $x^q$. For real $q\ge 0$,The improper integral converges exactly when $q-\alpha -1<-1$. At $q=\alpha$ it diverges logarithmically; above that it diverges as a positive power of the upper endpoint. Setting $q=0$ gives one, proving normalization regardless of whether higher moments exist.

Taking $q=1,2$ and using $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X]{)}^2$ when both terms are finite givesFor $0<\alpha \le 1$, the nonnegative random variable has expected value $+\infty$; a centered variance about a finite mean is unavailable. For $1<\alpha \le 2$, the mean is finite and the variance is infinite.^[ref] A finite total probability and a finite expected value answer different questions.

At the chosen minimum $\beta =10$, $\alpha =3$ gives $\mathbb{E}[X]=15$, $\mathbb{E}[X^2]=300$ and variance $300-15^2=75$, but an infinite third moment. At $\alpha =3/2$, the mean is 30 and the variance is infinite; at $\alpha =3/4$, even the mean is infinite, while the median remains the finite number $102^{4/3}\simeq 25.198$. At the boundary $\alpha =2$, the earlier example has mean 20 but infinite variance. A finite sample contains finite values almost surely; that fact cannot establish that the underlying mean or variance is finite.

For a nonnegative random variable with finite positive mean, the share of its total amount carried by the largest fraction $q$ of outcomes can be defined by dividing their contribution to the expected value by the whole expected value. For a Pareto distribution, let $0<q\le 1$ and choose $x_q=\beta q^{-1/\alpha }$, so that $\mathbb{P}(X\ge x_q)=q$. If $\alpha >1$, direct integration givesThis is a share of amount, distinct from the fraction of outcomes. It depends on $\alpha$, but not on the unit of measurement $\beta$.^[ref]

For the chosen $\alpha =2$, the top 20% starts at $x_{0.2}=10\sqrt{5}\simeq 22.361$ and carries $T(0.2)=\sqrt{0.2}\simeq 44.72\%$ of the amount. To make the top 20% carry exactly 80%, solveThe resulting Pareto distribution has a finite mean and infinite variance. The 80/20 rule selects one shape parameter; it is not a universal law, or even a property of every Pareto distribution. For $\alpha \le 1$, the population-share definition above is unavailable because its denominator is infinite; substituting such an $\alpha$ into the finite-mean formula is invalid.

The left panel compares three quantiles in units of the minimum. The right panel shows $T(0.2)$ only where the mean is finite, and marks the chosen 80/20 value. The thresholds $\alpha =1$ and $\alpha =2$ separate the moment regimes; they do not separate valid and invalid probability distributions.

A straight-looking empirical log-log plot does not by itself establish a Pareto distribution or a reliable tail exponent; estimating an exponent and assessing goodness of fit are separate tasks.^[ref] If a Pareto distribution is used only above a fitted lower threshold, $T(q)$ describes that conditional tail population, not automatically the whole population. A genuine finite upper cutoff gives a truncated distribution with finite nonnegative moments; its survival function, quantiles and top shares must be recalculated. Truncation is a modelling change, distinct from merely observing a finite sample. None of the chosen numbers here estimates an actual wealth distribution.

The parity construction of pairwise independent bits turns two independent fair bits into three pairwise independent random variables. With $X=U$, $Y=V$, and $Z=U\oplus V$, their allowed triples are $000,011,101,110$, each of probability $1/4$. Every coordinate has the Bernoulli distribution with parameter $1/2$. Every pair has the same joint probability mass function:All twelve pair cells equal the product of the corresponding probabilities in their marginal distributions. This establishes pairwise independent random variables, a requirement involving every pair of value-events, rather than merely one chosen event per coordinate. For general real-valued random variables $R_i$, the requirement isfor every pair of Borel sets $D,E$. For these binary random variables, checking all four cells of each pair suffices: any value-event is a union of the checked cells. In fact, for binary random variables, independence of their one-value events also gives the other cells by taking complements. Siegrist gives the all-set definition and its connection to indicator random variables in Random Services: independence of random variables.

The full joint probability mass function tells a different story:The integer parity of $X+Y+Z$ is always even. Any two bits determine the third: $Z=X\oplus Y$, $X=Y\oplus Z$, and $Y=X\oplus Z$. Each coordinate is independent of either other coordinate individually, but is dependent on the pair together. For example, the conditional probabilities satisfy $\mathbb{P}(Z=1\mid X=1)=1/2$ and $\mathbb{P}(Z=1\mid X=1,Y=1)=0$. See also pairwise independence does not imply joint independence for the general distinction.

The constraint changes the sum's probability distribution without changing its expected value or variance. Let $S=X+Y+Z$. Only $000$ gives $S=0$; each other allowed triple gives $S=2$. Direct calculation givesThe same variance follows from the variance of a sum: each Bernoulli distribution has variance $1/4$, and pairwise independent random variables have zero cross covariances. Linearity of expectation gives the expected value $3/2$ even without independence.

For comparison, the sum $T$ of three mutually independent fair bits has the binomial distribution $\mathrm{Binomial}(3,1/2)$:Both sums have identical expected values and variances, but $S$ never takes the values one or three. Even their third moments differ: $\mathbb{E}[S^3]=6$, whereas $\mathbb{E}[T^3]=27/4$. Pairwise independence preserves these first two sum summaries while allowing a different joint probability distribution and a different sum law. The calculation is an exact finite example.

For a moment of order $q$ with $\theta +q>-1$, integration of the probability density function givesThus the expected value and variance areThe cumulative distribution function on $(0,1)$ is $F(x)=x^{\theta +1}$. Its unique median $m$ satisfies $F(m)=1/2$, giving

Let ${\xi }_j$ be the displacement in step $j$, so the random walk position is $x(n)={\sum }_{j=1}^n{\xi }_j$. The single-step expected value and second moment areBy linearity of expected value, $\boxed{⟨x(n)⟩=0}$. The steps are independent random variables, so for $i\ne j$, $\mathbb{E}[{\xi }_i{\xi }_j]=\mathbb{E}[{\xi }_i]\mathbb{E}[{\xi }_j]=0$. Expanding the square gives the mean squared displacementThese are ensemble expected values; the empirical average over a finite number $N$ of independent cells fluctuates about them. The expected mean squared displacement itself does not depend on $N$.

A transformation model consists of a group $G$ acting on the sample space and on the set of statistical parameters, with the compatibility propertyIn words, transforming a sample from $P_{\theta }$ produces a sample from $P_{g\theta }$. A transitive transformation model is generated from one reference distribution by the group action. For densities, the compatibility property includes the Jacobian determinant of the sample transformation; it is not merely equality of density values at transformed points.

A maximal invariant $A$ is constant on orbits of a group action and separates them: $A(gx)=A(x)$, and $A(x)=A(y)$ implies $y=gx$ for some $g$. An equivariant estimator $T$ of the parameter satisfies $T(gx)=gT(x)$. Equivariance means the estimate follows the transformed parameter, whereas invariance means the reported value stays unchanged.

First suppose the set of statistical parameters can be identified with the group, as happens for a simply transitive action. An equivariant estimator then provides a group-valued fit $\hat{g}(x)$ with $\hat{g}(gx)=g\hat{g}(x)$. Normalize the sample by this fit:Indeed $A(gx)=\hat{g}(x{)}^{-1}{g}^{-1}gx=A(x)$. Conversely, if $A(x)=A(y)$, then $y=\hat{g}(y)\hat{g}(x{)}^{-1}x$, placing $x,y$ in the same orbit of a group action. This establishes maximality, rather than only invariance.

For a transitive action with a nontrivial stabilizer $H$ of a reference parameter ${\theta }_0$, a parameter-valued equivariant estimator does not specify a unique normalizing transformation. Choose a section $h_{\theta }$ with $h_{\theta }{\theta }_0=\theta$ and form $b(x)=h_{T(x)}^{-1}x$. Under $g$,Thus the $H$-orbit of $b(x)$ is invariant. It is maximal: if $b(y)=hb(x)$ with $h\in H$, then $y=h_{T(y)}h{h}_{T(x)}^{-1}x$. Appropriate measurable sections and orbit of a group action coordinates are assumed when treating this as a statistic. This qualification explains why an arbitrary equivariant statistic alone is not automatically a complete set of invariant coordinates.

For the location-scale family, write $X_i=\mu +\sigma Z_i$ with $\sigma >0$ and a fixed joint distribution of $Z$. The group consists of maps $x_i\mapsto a+b{x}_i$ with $b>0$, acting on parameters by $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. For a nonconstant sample of size $n\ge 2$, take the equivariant estimatorIt gives the maximal invariantIts coordinates satisfy ${\sum }_i{A}_i=0$ and ${\sum }_i{A}_i^2=n$. They stay unchanged under location and positive-scale transformations. If two samples have the same standardized vector, thenso a single group transformation maps the first sample onto the second. The vector retains the observation labels: sorting it would discard information not removed by this location-scale group. The construction is algebraic and does not require the underlying distribution to have finite moments. Constant samples can be assigned a separate invariant value; they form a single exceptional orbit of a group action and have probability zero under an independent continuous base distribution.

Finally, in a transitive transformation model every invariant statistic is ancillary. Choose $g$ taking ${\theta }_0$ to $\theta$ and use $X_{\theta }\stackrel{d}{=}g{X}_{{\theta }_0}$; invariance gives $A(X_{\theta })\stackrel{d}{=}A(X_{{\theta }_0})$. Thus the standardized residual configuration provides a natural conditioning variable, while the equivariant fit contains the location and scale information.

Partition a regular parameter as $(\psi ,\lambda )$, with interest $\psi$ and nuisance $\lambda$. Orthogonal statistical parameters have zero cross-block of expected Fisher information,This may hold at one parameter value or throughout a parametrization. It is expected-information orthogonality, and does not generally force the observed information cross-block to vanish in each sample.

First, the asymptotic covariance of the joint maximum-likelihood estimator is the inverse information. Orthogonality makes this matrix block diagonal, giving first-order asymptotic independence of interest and nuisance fits. The efficient information with unknown nuisance isFor orthogonal parameters it equals $I_{\psi \psi }$. Thus estimating the nuisance creates no first-order information loss relative to knowing it, at the orthogonal point. This is a local asymptotic statement, not a general finite-sample independence theorem.

Second, orthogonality stabilizes constrained nuisance estimates. Differentiating ${\ell }_{\lambda }(\psi ,{\tilde{\lambda }}_{\psi })=0$ givesWhen the expected cross-information vanishes, regular sampling fluctuations give $j_{\lambda \psi }=O_p(\sqrt{n})$ and $j_{\lambda \lambda }=O_p(n)$. The derivative is therefore $O_p(n^{-1/2})$ locally. Changing the interest parameter by $O_p(n^{-1/2})$ changes its constrained nuisance fit by only $O_p(n^{-1})$, smaller than the ordinary nuisance estimation error. This reduces sensitivity to how the nuisance is fitted and simplifies higher-order likelihood adjustments.

For scalar interest, local orthogonal nuisance reparametrization can often be constructed explicitly. Keep $\psi$ and write the original nuisance as $\lambda (\psi ,\eta )$. The transformed interest score is $U_{\psi }+{\lambda }_{\psi }^{\mathsf{T}}{U}_{\lambda }$. Its covariance with the new nuisance score vanishes whenWith smooth nonsingular nuisance information, this ordinary differential equation has a local solution for suitable initial nuisance coordinates. For several interest coordinates the corresponding equations need compatibility conditions; global orthogonality is not automatic.

The normal model gives a transparent example. For independent observations with mean $\mu$ and variance $\nu$, direct score expectations giveThe cross-information is zero because the centered normal third moment is zero. When fitting $\nu$ at fixed $\mu$,so the change is indeed quadratic in a local change of the mean.

Orthogonality is also useful for removing nuisance effects beyond first order. Take variance $\nu$ as interest and mean $\mu$ as nuisance. Put $Q={\sum }_i(X_i-\overline{X}{)}^2$. The profile log-likelihood is ${\ell }_p(\nu )=-\frac{n}{2}\log \nu -Q/(2\nu )$ up to a constant. The constrained mean is ${\tilde{\mu }}_{\nu }=\overline{X}$ for all $\nu$, so its sample-coordinate Jacobian determinant is one, and $j_{\mu \mu }=n/\nu$. The information adjustment givesThis agrees with the exact likelihood based on $Q/\nu \sim {\chi }_{n-1}^2$: estimating the mean has used one degree of freedom. In this setting the Cox-Reid adjusted profile likelihood has the same form. The exact distribution follows by projecting the centered normal vector onto the $(n-1)$-dimensional subspace orthogonal to the constant vector; an orthonormal basis there supplies $n-1$ independent standard normal coordinates whose squares sum to $Q/\nu$.

The example shows both the benefit and the limits. Orthogonality organizes first-order precision and nuisance stability; it does not eliminate all higher-order nuisance effects. Nor does it identify an ancillary statistic or justify dropping the sample-coordinate factor in every modified profile likelihood. In the exponential regression calculation that factor cancels the information adjustment, so an information-only correction would give a different answer.

A large molecular Schmidt number $\mathrm{Sc}=\nu /\kappa$ separates the velocity's smallest smooth length scale from the scalar's smaller diffusive length scale. Follow a Lagrangian trajectory $X(t)$ and measure displacement $x$ from it. Within a sufficiently small scalar packet, Taylor expansion of the smooth velocity field givesThe moving frame removes uniform sweeping, leaving a linear random velocity gradient tensor. Thus the local scalar equation is the advection-diffusion equation with velocity $Cx$, and averaging over realizations of $C$ gives the corresponding ensemble average. The approximation requires the packet to lie within the velocity's locally smooth region. Its thin, diffusive direction can satisfy this condition even when its long direction eventually becomes too large for one global linearization; applications to a spatially varying flow then use local packet descriptions. The statistical conclusions below refer to the linear-flow model and the usual mixing assumptions, not to every conceivable trace-free random matrix process.

Write $Q=\int \chi dA$ and assume sufficiently rapid decay to discard boundary terms. Integration by parts givesFor $J_{ij}=\int x_i{x}_j\chi dA$, integrate the advective term once and the diffusive term twice. Since ${\partial }_k(x_i{x}_j)={\delta }_{ki}{x}_j+{\delta }_{kj}{x}_i$ and $\Delta (x_i{x}_j)=2{\delta }_{ij}$, one obtainsThese are mass conservation and the evolution of the unnormalized second moment tensor. The first moment obeys $\dot{M}=CM$, so a centered packet stays centered.

For a positive Gaussian scalar packet in a linear incompressible flow, $B$ must be a positive-definite matrix. Diagonalizing its symmetric matrix and applying the Gaussian integral givesThe factor $Q^2$ in the last identity matters because $J$ has not been divided by the scalar mass. Substitution into the advection-diffusion equation also givesThese identities agree with the second moment equation and prove that the Gaussian family is preserved. For an initially negative packet the same analysis applies after replacing $\chi$ by $-\chi$ and taking its positive mass.

To see the roles of stretching and diffusion explicitly, let the deformation gradient $F$ solve $\dot{F}=CF$, $F(0)=I$. Incompressibility gives $\det F=1$. Differentiating $F^{-1}J{F}^{-T}$ integrates the moment equation:Without diffusion, $J=FJ(0)F^T$ and $\det J$ is constant. In a statistically stationary, mixing flow with positive Lyapunov exponent $\lambda$, the two stretching exponents are $\lambda$ and $-\lambda$. Writing the ordered eigenvalues of $J$ as $e^{2{\rho }_1},e^{2{\rho }_2}$, their typical initially nondiffusive evolution isThe fluctuation scale presumes the usual central limit theorem for accumulated strain; it is not implied by incompressibility alone. During this stage the packet becomes long and thin but keeps its area and peak concentration.

In the instantaneous principal-axis basis, write $\tilde{C}$ for the rotated velocity gradient tensor. Differentiation of an eigenvalue givesThe formulas hold through a repeated eigenvalue by continuity of their sum. The compressed physical variance is $e^{2{\rho }_2}/Q$. Its contraction is arrested when it is of order $\kappa /\lambda$, the square of the Batchelor scalar microscale ${\ell }_B\sim \sqrt{\kappa /\lambda }$. If its initial length is $L\gg {\ell }_B$, the crossover occurs atThereafter ${\rho }_2$ has stationary fluctuations associated with the strain and the diffusive floor, while ${\rho }_1$ continues to grow typically at rate $\lambda$. Under sufficiently fast mixing, the central part of the distribution of ${\rho }_1$ is approximately Gaussian with mean $\lambda t+O(1)$ and variance $D_{\rho }t+o(t)$ for a strain-dependent coefficient $D_{\rho }$. This central approximation does not determine far tails, which control high concentration moments; those require a full large deviation principle. The two logarithmic widths need not be independent. The area now increases and $f$ decreases typically as $e^{-\lambda t}$. This is diffusion arrest of Gaussian packet compression.

For example, a constant principal strain $C=\mathrm{diag}(\lambda ,-\lambda )$ gives the exact physical variancesThe thin variance approaches its diffusive floor, the long variance grows exponentially, and the mass-to-area relation gives the quoted typical amplitude decay. A random strain additionally makes rare histories important for concentration moments.

For a precise statistical formulation define the logarithmic area increaseThe Gaussian identities give the exact relation $f(t)=f(0)e^{-H_t}$; henceAssume the long-time strain statistics give a large deviation principle for $H_t/t$ on $s\ge 0$, with convex good rate function $I(s)$, $I(\lambda )=0$. At exponential accuracy the probability of $H_t/t$ near $s$ is $e^{-tI(s)}$. Combining that cost with the amplitude factor and applying the Laplace principle for probability measures gives the large-deviation decay rates of a Gaussian scalar packetThe initial amplitude only changes a time-independent prefactor. One can see the minimization directly by dividing the possible area-growth rates into small intervals: each contributes its probability cost times $e^{-\mu ts}$, and the interval with least total exponential cost dominates. An interval around a minimizing $s$ gives the matching lower bound. The nonnegative area increase bounds the exponential weight by one, so unusually large positive growth cannot produce a divergent weighted tail. Using the area rather than only ${\rho }_1$ also avoids discarding possibly important fluctuations of the compressed direction.

Suppose $I$ is strictly convex and differentiable, with its minimum at $\lambda >0$ and finite negative right derivative at zero. DefineFor $0<\mu <{\mu }_{\ast }$ the minimizing rate $s_{\mu }$ is positive and uniquely solvesFor $\mu \ge {\mu }_{\ast }$, convexity puts the minimum at the endpoint $s=0$, givingHigh moments are dominated by rare histories with little net area growth, in which a packet keeps a substantial concentration. Their probability cost is $I(0)$, independent of the moment order. A finite threshold is conditional: if $I^{\prime }(0+)=-\infty$, the minimizer can approach zero without reaching it at any finite $\mu$.

Below the threshold, differentiation of the minimizing expression cancels the terms involving $s_{\mu }^{\prime }$ because $I^{\prime }(s_{\mu })+\mu =0$. Thus ${\gamma }_{\mu }^{\prime }=s_{\mu }$, andStrict negativity uses a nondegenerate rate function with a unique zero at $\lambda$: for positive $\mu$, $s_{\mu }<\lambda$ and $I(s_{\mu })>0$. For deterministic strain the ratio can instead be constant. Near the typical rate, if $I(s)\simeq (s-\lambda {)}^2/(2D)$ with $D>0$, then ${\gamma }_{\mu }\simeq \lambda \mu -D{\mu }^2/2$ for small $\mu$. If this quadratic expression holds all the way to zero, it gives the illustrative full resultThe quadratic form is an example, not a universal law of turbulent strain.

A thin scalar packet is stretched until molecular diffusion broadens its compressed direction; conservation of its scalar mass then converts area growth into dilution. Ensemble averages sample different stretching histories, producing different decay rates for different moments and saturation when weakly stretched histories dominate. The present moment calculation is specifically for the centered Gaussian packet and averaging over the flow; signed random initial fields with overlapping packets can also have cancellations and require their initial-field statistics.

Finally, positive exponential decay is a statistical conclusion requiring the strain assumptions above, not a consequence of the local differential equation by itself. For $C\equiv 0$ and initially isotropic physical variance $L^2$, the exact solution has $J=Q(L^2+2\kappa t)I$ and $f=Q/[2\pi (L^2+2\kappa t)]$. Its moments decay algebraically as $t^{-\mu }$. Thus the general equations for the exponential rates are the rate-function minimization above; numerical rates and a finite saturation threshold require the long-time statistics of the flow.

The covariance is symmetric, because multiplication of real random variables is commutative. Expanding with the bilinearity just proved givesIdentical probability distributions imply equal second moments and equal means, hence equal variances. Therefore the assertion is true, withThe cancellation shows exactly why no assumption of independence is needed.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable under this hypothesis. More generally the definition works whenever the displayed expectations exist. Assertions about arbitrary random variables below are understood on this domain: without suitable moments a covariance need not be defined at all.

A probability density function must integrate to one. ThereforeThe first two moments areConsequently the mean and variance are

For fixed $\theta >0$, use the Lyapunov condition for the independent centered epoch counts $Z_i=Y_{i+1}-\theta /i$. Their total variance is $s_n^2=\theta H_{n-1}+{\theta }^2{H}_{n-1}^{(2)}\sim \theta \log n$.

Write ${\mu }_i=\theta /i$. Differentiating the geometric distribution probability generating function gives $\mathbb{E}[Y_{i+1}^3]={\mu }_i+6{\mu }_i^2+6{\mu }_i^3$. Since $|Y_{i+1}-{\mu }_i{|}^3\le 4Y_{i+1}^3+4{\mu }_i^3$, the sum of the centered absolute third moments is $O(\log n)$. ConsequentlyThe Lindeberg-Feller central limit theorem applies through the Lyapunov condition, giving $(S-\theta H_{n-1})/s_n\stackrel{d}{\to }N(0,1)$. Now $\sqrt{\log n}{s}_n/H_{n-1}\to \sqrt{\theta }$, so the Slutsky theorem proves the asymptotic normality of the Watterson estimator:The limiting variance is $\theta$. When $\theta =0$, every count vanishes and the limiting law is the point mass at zero.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

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 $\lambda \to \infty$.