Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 37 / 1

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.

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.

Put $q=1-p$. The count has the zero-based geometric distribution, whose probability generating function is $G_N(z)=p/(1-qz)$. Hence the aggregate has a compound geometric distribution, withThis transform requires $q{M}_{X_1}(t)<1$; it holds for every $t\le 0$. To differentiate explicitly, let $g(t)=M_{X_1}(t)$. ThenandAt zero, $g=1$ and $g^{(j)}=m_j$, givingEvery term is positive for $0<p<1$ and strictly positive claim sizes. Therefore the unstandardized skewness is positive, even if the claim-size third central moment itself is negative. The zero-based convention matters: there is no factor $M_{X_1}(t)$ in the numerator of this aggregate transform.

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.