= Solution
Under <quota share reinsurance>, the direct insurer retains a fixed fraction $\alpha$ of each claim and the reinsurer pays the other fraction. Thus their annual claim totals are $\alpha S$ and $(1-\alpha)S$. Under <excess of loss reinsurance> with retention $d$, the direct insurer pays $\min(X_j,d)$ on each individual claim and the reinsurer pays $(X_j-d)_+$. Under <stop loss reinsurance> with annual retention $D$, the direct insurer pays $\min(S,D)$ and the reinsurer pays $(S-D)_+$. The last two contracts differ because one caps each claim and the other caps the aggregate.
Conditioning on the annual <Poisson distribution> count gives the <compound Poisson distribution> transform:
$$
M_S(t)=\sum_{n\ge0}e^{-\lambda}\frac{\lambda^n}{n!}M_X(t)^n
=\boxed{\exp\{\lambda(M_X(t)-1)\}.}
$$
This equality is understood on the domain where the <moment-generating function> is finite. By the <expected value premium principle>, annual premium income is $c=(1+\theta)\lambda\mu$. Without <reinsurance>, final capital is $W_0=u+c-S$, and therefore
$$
\boxed{M_{W_0}(t)=\exp\{t(u+(1+\theta)\lambda\mu)+\lambda(M_X(-t)-1)\}.}
$$
For the quota-share contract, the reinsurer's expected claim cost is $(1-\alpha)\lambda\mu$, so its annual premium is
$$
\boxed{c_R=(1+\theta_R)(1-\alpha)\lambda\mu.}
$$
The direct insurer keeps its original premium income, pays this reinsurance premium and retains only $\alpha S$. Thus
$$
W=u+c-c_R-\alpha S
=u+\lambda\mu\{\theta-\theta_R+(1+\theta_R)\alpha\}-\alpha S,
$$
and
$$
\boxed{M_W(t)=\exp\left[t\left(u+\lambda\mu\{\theta-\theta_R+(1+\theta_R)\alpha\}\right)
+\lambda\{M_X(-\alpha t)-1\}\right].}
$$
The reinsurer's loading is applied to the ceded expected loss, not to the direct insurer's original premium.
Maximizing expected <exponential utility> $\mathbb E[-e^{-\beta W}]$ is equivalent to minimizing the exponent $H(\alpha)=\log\mathbb E[e^{-\beta W}]$. For <exponential distribution> claim sizes of mean $\mu$,
$$
M_X(r)=\frac1{1-\mu r},\qquad r<1/\mu,
$$
so on the finite-utility domain $\beta\mu\alpha<1$,
$$
H(\alpha)=-\beta\left[u+\lambda\mu\{\theta-\theta_R+(1+\theta_R)\alpha\}\right]
+\lambda\left(\frac1{1-\beta\mu\alpha}-1\right).
$$
For a nontrivial positive arrival rate, differentiation gives
$$
H'(\alpha)=\lambda\beta\mu\left[(1-\beta\mu\alpha)^{-2}-(1+\theta_R)\right],
\qquad
H''(\alpha)=\frac{2\lambda(\beta\mu)^2}{(1-\beta\mu\alpha)^3}>0.
$$
The unique unconstrained stationary retention is therefore
$$
\boxed{\alpha_*=
\frac{1-(1+\theta_R)^{-1/2}}{\beta\mu}.}
$$
It is positive and lies below $1/(\beta\mu)$. If $\alpha_*<1$, it is the unique optimal retention in the printed range $0<\alpha<1$. If $\alpha_*\ge1$, expected utility increases throughout that open interval and its supremum is approached as $\alpha\uparrow1$; \b[there is no attained maximum in the strictly open range]. Allowing the usual no-reinsurance endpoint gives the compact formula
$$
\boxed{\alpha_{\mathrm{opt}}=\min\{1,\alpha_*\}\quad\text{when }0\le\alpha\le1.}
$$
Retentions with $\beta\mu\alpha\ge1$ have infinite exponential loss moment and expected utility $-\infty$, so they cannot displace the finite stationary solution. The stationary point and full-retention endpoint just described stay within the finite-transform domain whenever they are selected.
For the requested comparisons,
$$
\frac{\partial\alpha_*}{\partial\theta_R}
=\frac{1}{2\beta\mu(1+\theta_R)^{3/2}}>0.
$$
\b[Higher reinsurer loading increases optimal retention], up to full retention: reinsurance is more expensive. \b[The direct insurer's loading $\theta$ does not affect the optimizer]. It adds only the constant $-\beta\lambda\mu\theta$ to $H$, reflecting the translation property of constant absolute risk aversion. These observations give the <quota-share retention under exponential utility> rule; $u$ and the positive arrival rate also drop out of the stationary equation. In the degenerate case $\lambda=0$, no claims or premiums occur, $W=u$, and every retained fraction has the same expected utility.
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