= Expected value premium principle
{title2=$\Pi(L)=(1+\theta)\mathbb E L$}
= Expected value principle
{synonym}
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 $\theta\ge0$, it sets
$$
\Pi(L)=(1+\theta)\mathbb E[L]
=\underbrace{\mathbb E[L]}_{\text{expected claim cost}}
+\underbrace{\theta\mathbb E[L]}_{\text{loading amount}}.
$$
Here $\theta=0$ 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 $\theta$. https://www.stats.ox.ac.uk/~winkel/bs4b11.pdf[Matthias Winkel, Actuarial Science II, §16.1, Definition 68 and Example 70]{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 $1000(0.1)=£100$. 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.
\Image[/expected-premium-bernoulli-loading.png]
{title=Toy Bernoulli policy: two possible losses and an expected cost plus loading}
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 https://www.stats.ox.ac.uk/~winkel/bs4b11.pdf[Actuarial Science II]{ref}.
To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually <independent> <Bernoulli random variables> $B_1,\ldots,B_n$, each of parameter $0.1$. Every <claim size> is £1,000, with at most one claim per policy. The <aggregate claims model> is
$$
S_n=1000\sum_{i=1}^{n}B_i,
\qquad \sum_{i=1}^{n}B_i\sim\operatorname{Binomial}(n,0.1).
$$
Using <linearity of expectation> and addition of <variances> for <independent> <random variables>, the <expected value> and <standard deviation> of this <binomial distribution> aggregate are
$$
\mathbb E[S_n]=100n,\qquad
\operatorname{Var}(S_n)=1000^2n(0.1)(0.9)=90000n,\qquad
\operatorname{SD}(S_n)=300\sqrt n.
$$
Thus 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> $£300/\sqrt n$. Equivalently, the aggregate <coefficient of variation>, its <standard deviation> divided by its positive <expected value>, is
$$
\frac{\operatorname{SD}(S_n)}{\mathbb E[S_n]}
=\sqrt{\frac{1-p}{np}}=\frac{3}{\sqrt n}.
$$
It 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>.
\Image[/expected-premium-pooling-scaling.png]
{title=Toy claim pooling: relative variability falls with independence but survives a common shock}
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 exactly
$$
\Pr(S_{100}>12000)
=\sum_{k=13}^{100}\binom{100}{k}(0.1)^k(0.9)^{100-k}
\approx0.19818.
$$
A 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>:
$$
\operatorname{Var}\left(\sum_{i=1}^n L_i\right)
=\sum_{i=1}^n\operatorname{Var}(L_i)
+2\sum_{i<j}\operatorname{Cov}(L_i,L_j).
$$
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 https://www.stats.ox.ac.uk/~winkel/bs4b11.pdf[§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. https://www1feb-uva.nl/ke/act/people/kaas/Recap123456.pdf[Rob Kaas's Modern Actuarial Risk Theory formula sheet, page 2, Ruin theory]{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.
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