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In the canonical ensemble, put $\beta =(k_{B}T{)}^{-1}$ and $Z={\sum }_j{e}^{-\beta E_j}$. For a temperature-independent observable $A$, differentiating its normalized expectation gives ${\partial }_{\beta }⟨A⟩=-⟨AE⟩+⟨A⟩⟨E⟩$. Taking $A=E$ yields ${\partial }_{\beta }⟨E⟩=-\mathrm{Var}E$. Since ${\partial }_{\beta }=-k_B{T}^2{\partial }_T$ at fixed volume,Differentiating $⟨E^2⟩-⟨E{⟩}^2$ with the same rule gives minus $⟨E^3⟩-3⟨E⟩⟨E^2⟩+2⟨E{⟩}^3$, namely minus the third central moment. ThusThese identities exhibit the second and third energy cumulants as successive derivatives of $\log Z$.

For the monatomic ideal gas, $⟨E⟩=3N{k}_{B}T/2$, $C_V=3N{k}_B/2$ and ${\partial }_T{C}_V=0$. The variance is $3N(k_{B}T{)}^2/2$ and the third central moment is $3N(k_{B}T{)}^3$. Division by the corresponding mean-energy powers gives

For the Poisson distribution, $G_N(z)=e^{\lambda (z-1)}$, so the aggregate has a compound Poisson distribution andThree differentiations immediately giveThe first two cumulants are $\lambda m_1$ and $\lambda m_2$. If $\lambda >0$ and claim sizes are strictly positive, $m_3>0$, so the paper's unstandardized skewness is strictly positive. For the degenerate count $\lambda =0$, the aggregate is zero and the third central moment is zero; strict positivity presumes a positive claim intensity.

For the binomial distribution, $G_N(z)=(1-p+pz{)}^n$. Thus the compound binomial distribution hasSet $h(t)=1-p+p{M}_{X_1}(t)$. Since $h(0)=1$ and $h^{(j)}(0)=p{m}_j$, differentiating $n\log h$ three times givesThe first two cumulants, for comparison, are $np{m}_1$ and $n(p{m}_2-p^2{m}_1^2)$.

Choose deterministic claim sizes $X_i=d>0$. Then $S=dN$, so $m_j=d^j$ andThis is strictly negative for $p>1/2$ and $n\ge 1$. A concrete negative-skewness example is $n=1$, $p=3/4$, $d=1$, for whichThere is no contradiction with positive claims: bounded Bernoulli-type totals can have more mass near their upper value and a longer left tail.

Write $m_j=\mathbb{E}{Y}^j$ and $M=M_Y$. The derivatives of the cumulant-generating function areSince $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three cumulants areThe last equality follows by expanding the cube and taking expectations. We use the paper's unstandardized third central moment throughout. The usual skewness coefficient divides this quantity by $\mathrm{Var}(Y{)}^{3/2}$; its sign is the same whenever the variance is positive.

There is a minor domain qualification: finite ordinary moments do not guarantee a finite moment-generating function for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite moments, all formulas also follow from left derivatives at zero, because $Y^j{e}^{tY}\le Y^j$ for $t\le 0$ and dominated convergence applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.

For the random sum of independent claims, condition on $N=k$. Independence of the claims and independence from the count giveTaking expectations and then logarithms yields the random-sum transform identity and the cumulant composition for a random sum:Here $G_N$ is the probability generating function. These identities hold wherever the transforms are finite, in particular for $t\le 0$ with positive claims.

For reuse in the three cases, write $m_j=\mathbb{E}{X}_1^j$ and $a_j={\kappa }_{N,j}$. Applying the chain rule three times givesor equivalentlyThis expresses the requested third central moment in raw claim moments, while keeping the contribution of claim-count variability explicit.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.

A gamma distribution translated by a real constant $k$. With shape $\alpha >0$ and rate $\nu >0$, its support is $(k,\infty )$ and its first three cumulants are $k+\alpha /\nu$, $\alpha /{\nu }^2$, and $2\alpha /{\nu }^3$. Its skewness is $2/\sqrt{\alpha }$. A negative shift permits negative values; introducing a shift does not automatically respect the nonnegative support of an insurance loss.

Approximate a compound Poisson distribution of positive claims by a shifted gamma distribution whose first three cumulants agree. Here $m_j=\mathbb{E}{X}^j$ are raw severity moments, while the compound cumulants are $\lambda m_j$. Solve $k+\alpha /\nu =\lambda m_1$, $\alpha /{\nu }^2=\lambda m_2$, and $2\alpha /{\nu }^3=\lambda m_3$ to obtain the displayed parameters. The approximation captures positive skewness, but not the atom $e^{-\lambda }$ at zero. Its shift can be negative; with exponential claims of mean $\mu$ it is $-\lambda \mu /3$. Matching three cumulants alone supplies no general tail-error bound.

The trispectrum is the Fourier transform of a fourth-order cumulant, a fourth-order counterpart of the bispectrum. It describes connected four-point correlations; the primordial trispectrum applies it to cosmological random fields.