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The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

The claim law is an Erlang distribution of shape two and rate $2/\beta$. Direct integration, using ${\int }_0^{\infty }x{e}^{-ax}dx=a^{-2}$ for $a>0$, gives its moment-generating functionDifferentiating at zero gives $\mathbb{E}X=\beta$. The positive relative safety loading condition for the classical risk model is thereforeThe adjustment coefficient is the positive root in the finite-transform domain ofSet $z=\beta R/2$ and $k=2c/(\lambda \beta )>2$. For $R>0$, division by $z$ reduces the equation toThe two roots are $z=(2k-1\pm \sqrt{1+4k})/(2k)$. The minus root lies strictly between zero and one: positivity follows from $(2k-1{)}^2-(1+4k)=4k(k-2)>0$, and the numerator is smaller than $2k$. The plus root exceeds one and is outside the moment-generating function's domain. Consequently the adjustment coefficient for shape-two Erlang claims isThe zero root of the original equation is not an adjustment coefficient. The increasing secant slope $(M_X(r)-1)/r$ on $(0,2/\beta )$ also establishes uniqueness of the positive root.

Now set $\beta =1$. Under quota share reinsurance, the retained payment is $Y_{\alpha }=\alpha X$. For $0<\alpha \le 1$, the change of variables $x=y/\alpha$ yieldsHere the second gamma parameter is the rate. In particular, $\mathbb{E}{Y}_{\alpha }=\alpha$ and $M_{Y_{\alpha }}(r)=(1-\alpha r/2{)}^{-2}$. At full cession $\alpha =0$ the retained payment instead has a point mass at zero.

The reinsurer's expected annual payments are $\lambda (1-\alpha )$, so its loaded premium is $(1+{\theta }_R)\lambda (1-\alpha )$. Subtract it from the original premium income to obtain the direct insurer's net premium rateIts expected retained claim outflow is $\lambda \alpha$. Thusby the stipulated lower bound on $\alpha$. In particular $c_{\alpha }>0$, so the retained portfolio satisfies the net profitability condition. Its relative safety loading is ${\theta }_R-({\theta }_R-\theta )/\alpha$.

Apply the same adjustment coefficient for shape-two Erlang claims formula with scale $\alpha$ and net premium $c_{\alpha }$. WritinggivesEquivalently it is the positive solution, with $R_{\alpha }<2/\alpha$, of

For fixed positive initial capital, a larger adjustment coefficient reduces the exponential upper bound ${\psi }_{\alpha }(u)\le e^{-R_{\alpha }u}$ from the Lundberg inequality. It also makes the large-capital decay exponent in the Cramér–Lundberg ruin asymptotic larger. These are the reasons for maximizing $R_{\alpha }$ as a ruin-risk criterion; the coefficient alone need not order exact finite-capital ruin probabilities for arbitrary different models.

If ${\theta }_R=\theta$, then $c_{\alpha }=(1+\theta )\lambda \alpha$ and $k_{\alpha }=2(1+\theta )$ is constant. The equal-loading quota share and vanishing retention relation becomesThe numerator is positive for $\theta >0$, so $R_{\alpha }$ decreases strictly with $\alpha$ and tends to infinity as $\alpha \downarrow 0$. There is no maximizing retention in the stipulated interval $0<\alpha \le 1$; the best limit is full cession. If the endpoint $\alpha =0$ is admitted as an idealized contract, all claims are transferred and the net premium income is zero, leaving the direct insurer's capital constant and its ultimate ruin probability zero. At that endpoint the coefficient equation is identically zero, so it no longer defines a unique finite adjustment coefficient. Calling $\alpha =0$ optimal requires this explicit extension of the original admissible set.

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: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

For the quota-share contract, the reinsurer's expected claim cost is $(1-\alpha )\lambda \mu$, so its annual premium isThe direct insurer keeps its original premium income, pays this reinsurance premium and retains only $\alpha S$. ThusandThe 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 $\mu$,so on the finite-utility domain $\beta \mu \alpha <1$,For a nontrivial positive arrival rate, differentiation givesThe unique unconstrained stationary retention is thereforeIt is positive and lies below $1/(\beta \mu )$. If ${\alpha }_{\ast }<1$, it is the unique optimal retention in the printed range $0<\alpha <1$. If ${\alpha }_{\ast }\ge 1$, expected utility increases throughout that open interval and its supremum is approached as $\alpha \uparrow 1$; there is no attained maximum in the strictly open range. Allowing the usual no-reinsurance endpoint gives the compact formulaRetentions with $\beta \mu \alpha \ge 1$ 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,Higher reinsurer loading increases optimal retention, up to full retention: reinsurance is more expensive. 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.

Under quota share reinsurance, $S_I=\alpha S$ and $S_R=(1-\alpha )S$. Write $v=\mathrm{Var}(S)=\lambda \mathbb{E}[X^2]$. ThenThe variance-minimizing quota share split is thereforeFor positive claims and $\lambda >0$ this minimizer is unique. If the aggregate has zero variance, every retained fraction gives the same zero objective.

Let $v=\mathrm{Var}(S)>0$ and $v_I=\mathrm{Var}(S_I)$. The variance-matching quota share reinsurance in this part has the nonnegative retained fraction ${\alpha }^{\ast }=\sqrt{v_I/v}$. For claimwise retentions $0\le h(x)\le x$, its existence within $[0,1]$ follows from $v_I=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]=v$. The comparison below only needs the matching contract specified in the question.

Because $S_R=S-S_I$, the variance and covariance identities giveThe Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le \sqrt{v{v}_I}={\alpha }^{\ast }v$. Therefore the quota share comparison at matched retained variance isThe last expression is the sum of variances under the matching quota share reinsurance. HenceIf $v=0$, the given matching contract forces $v_I=0$, so all relevant totals are constant and equality holds. For $v>0$, equality in the bound is equivalent to $S_I-\mathbb{E}[S_I]={\alpha }^{\ast }(S-\mathbb{E}[S])$ almost surely.

For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

For quota share reinsurance the insurer retains the same fraction of each claim, soThe annual retained loss is consequently $T_I=\alpha T$. Each transformed risk severity has probability density function $f_i(y/\alpha )/\alpha$ for $y>0$. The mixed transformed severity, with the same weights ${\lambda }_i/({\lambda }_1+{\lambda }_2)$, still gives a compound Poisson distribution. Scaling the expected value and variance givesThis agrees with the general retained-claim formulas because $\mathbb{E}g(X_i)=\alpha {\mu }_i$ and $\mathbb{E}[g(X_i{)}^2]={\alpha }^2({\sigma }_i^2+{\mu }_i^2)$.

The expected value and variance under quota share reinsurance follow by scaling the exponential distribution: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 $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. 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.

For quota share reinsurance, each claim and therefore its aggregate are retained in the same proportion. For aggregate stop loss reinsurance, the insurer pays the aggregate up to the retention, and the reinsurer pays the excess. Thus the insurer's payouts areThe subscript $+$ denotes the positive part. The stop loss contract here applies to the annual aggregate, rather than separately to each claim.

Suppose $S=S_I+S_R$ has finite positive variance $v$, and a matching quota share reinsurance contract exists with fraction ${\alpha }^{\ast }=\sqrt{\mathrm{Var}(S_I)/v}\in [0,1]$. The Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le {\alpha }^{\ast }v$. Expanding $\mathrm{Var}(S-S_I)$ then proves that the matching quota share minimizes the other party's variance, and hence the sum of party variances. Claimwise retentions $0\le h(x)\le x$ in a compound Poisson distribution automatically satisfy the required variance range, because $\mathrm{Var}(S_I)=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]$.

quota share reinsurance retains a fixed proportion $\alpha \in [0,1]$ of each claim. The annual retained aggregate is $\alpha S$, whose expected value and variance are $\alpha \mathbb{E}S$ and ${\alpha }^2\mathrm{Var}(S)$.

Reinsurance transfers part of an insurer’s claim liability to another insurer. If aggregate claims are $S$, a retained payout $g(S)$ with $0\le g(S)\le S$ leaves the reinsurer with $S-g(S)$. Quota share reinsurance retains a fixed fraction, whereas aggregate stop loss reinsurance retains losses only up to a fixed aggregate threshold.

If $S$ has exponential distribution with expected value $\mu$, the tail integral formula for moments gives $\mathbb{E}\min (S,M)=\mu (1-e^{-m})$ and $\mathrm{Var}(\min (S,M))={\mu }^2(1-2m{e}^{-m}-e^{-2m})$, where $m=M/\mu$. At matching retained expected value, the excess variance under quota share reinsurance is $2{\mu }^2{e}^{-m}(m-1+e^{-m})\ge 0$.

For any aggregate with positive finite variance $v$, quota share reinsurance with retained fraction $\alpha$ has total party variance $v({\alpha }^2+(1-\alpha {)}^2)=v/2+2v(\alpha -1/2{)}^2$. Equal sharing uniquely minimizes this objective. The two party totals remain dependent; adding their individual variances is a different objective from the fixed variance of their sum.