= Solution
Let $M_X(t)=\mathbb Ee^{tX}$. Conditioning on $N\sim\operatorname{Pois}(\lambda)$ gives
$$
\mathbb Ee^{tS}=\mathbb E[M_X(t)^N]
=\exp\{\lambda(M_X(t)-1)\},\qquad
\boxed{K_S(t)=\lambda(M_X(t)-1).}
$$
Where the <moment-generating function> is finite around zero, differentiating the <cumulant-generating function> gives
$$
\boxed{\kappa_j(S)=K_S^{(j)}(0)=\lambda\mathbb E[X^j].}
$$
In particular $\mathbb ES=\lambda\mathbb EX$ and $\operatorname{Var}S=\lambda\mathbb EX^2$. These <compound Poisson cumulants> use raw claim <moments>. The first two formulas also follow by conditioning: $\mathbb E[S\mid N]=N\mathbb EX$ and $\operatorname{Var}(S\mid N)=N\operatorname{Var}X$, so
$$
\operatorname{Var}S=\mathbb E[N]\operatorname{Var}X+\operatorname{Var}N(\mathbb EX)^2
=\lambda\mathbb EX^2.
$$
This 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 $W$ gives
$$
\boxed{\begin{aligned}
\mathbb ES_I&=\lambda\int_0^M\overline F(x)\,dx,&
\operatorname{Var}S_I&=2\lambda\int_0^M x\overline F(x)\,dx,\\
\mathbb ES_R&=\lambda\int_M^\infty\overline F(x)\,dx,&
\operatorname{Var}S_R&=2\lambda\int_M^\infty(x-M)\overline F(x)\,dx.
\end{aligned}}
$$
For example, $\mathbb EY^2=\int_0^\infty\mathbb P(Y^2>t)dt=2\int_0^M x\overline F(x)dx$ 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 $d>0$,
$$
\overline F(x)=\begin{cases}1,&0\le x\le d,\\d^3/x^3,&x\ge d,\end{cases}
\qquad\mathbb EX=3d/2,\quad\mathbb EX^2=3d^2.
$$
If $0\le M\le d$, every claim exceeds the <reinsurance retention>. Therefore
$$
\begin{aligned}
\mathbb ES_I&=\lambda M,&\operatorname{Var}S_I&=\lambda M^2,\\
\mathbb ES_R&=\lambda(3d/2-M),&\operatorname{Var}S_R&=\lambda(3d^2-3dM+M^2).
\end{aligned}
$$
If $M\ge d$, the tail integrals instead give
$$
\begin{aligned}
\mathbb ES_I&=\lambda\left(\frac{3d}{2}-\frac{d^3}{2M^2}\right),&
\operatorname{Var}S_I&=\lambda\left(3d^2-\frac{2d^3}{M}\right),\\
\mathbb ES_R&=\frac{\lambda d^3}{2M^2},&
\operatorname{Var}S_R&=\frac{\lambda d^3}{M}.
\end{aligned}
$$
Consequently the requested sum of <variances> is
$$
V(M)=\lambda\begin{cases}
3d^2-3dM+2M^2,&0\le M\le d,\\
3d^2-d^3/M,&M\ge d.
\end{cases}
$$
On the first branch, $V'(M)=\lambda(4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V'(M)=\lambda d^3/M^2>0$. For $\lambda>0$ the minimum is unique; if $\lambda=0$, all <variances> are zero and every <reinsurance retention> minimizes them. Thus
$$
\boxed{M^*=\frac{3d}{4},\qquad V(M^*)=\frac{15}{8}\lambda d^2.}
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-41-retention.png]
{title=Sum of insurer and reinsurer aggregate variances for shape-three Pareto claims, minimized at retention three quarters of the minimum claim size}
The two aggregate payouts are dependent. Conditioning on the common count gives <covariance> $\lambda\operatorname{Cov}(Y,W)+\lambda\mathbb EY\mathbb EW=\lambda\mathbb E[YW]=\lambda M\mathbb EW$, so $V(M)+2\operatorname{Cov}(S_I,S_R)=3\lambda d^2$ 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 $t>0$, $\int_d^\infty e^{tx}3d^3x^{-4}dx=\infty$, and <moments> of order $j\ge3$ 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>.
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