= Solution
The <expected value> and <variance> under <quota share reinsurance> follow by scaling the <exponential distribution>:
$$
\boxed{\mathbb E S_I^*=\alpha\mu_S,\qquad
\operatorname{Var}(S_I^*)=\alpha^2\mu_S^2.}
$$
The <retained stop loss moments for an exponential aggregate> follow from the payout $Y=\min(S,M)$, its <survival function> equals $e^{-x/\mu_S}$ for $0\leq x<M$ and zero for $x\geq M$. The <tail integral formula for moments> gives, with $m=M/\mu_S$ and $r=e^{-m}$,
$$
\mathbb EY=\int_0^M e^{-x/\mu_S}\,dx=\mu_S(1-r),
\qquad
\mathbb EY^2=2\int_0^M xe^{-x/\mu_S}\,dx
=2\mu_S^2[1-(1+m)r].
$$
Consequently \b[the retained moments are]
$$
\boxed{\mathbb E\widetilde S_I=\mu_S(1-e^{-M/\mu_S}),\qquad
\operatorname{Var}(\widetilde S_I)
=\mu_S^2[1-2(M/\mu_S)e^{-M/\mu_S}-e^{-2M/\mu_S}].}
$$
Matching the two <expected values> forces $\alpha=1-r$, which lies strictly between zero and one. The difference of the <variances> simplifies to
$$
\boxed{\operatorname{Var}(S_I^*)-\operatorname{Var}(\widetilde S_I)
=2\mu_S^2e^{-M/\mu_S}
\left(\frac M{\mu_S}-1+e^{-M/\mu_S}\right)\geq0.}
$$
Indeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h'(m)=1-e^{-m}\geq0$ for $m\geq0$. Because $M>0$, the difference is actually positive. At equal retained <expected value>, <aggregate stop loss reinsurance> reduces the <variance> more than <quota share reinsurance>.
Back to article page
