Portfolio Percentile Principle

Study guide: Exam FAM

The Portfolio Percentile Principle sets the premium for a portfolio of $N$ independent and identical policies so that the probability that the portfolio's aggregate future loss is negative, meaning that the portfolio makes a profit, equals a chosen level $\alpha$ such as $95\%$ . For large $N$ the aggregate loss is approximately normal by the central limit theorem, so the premium solves a single equation in $P$ (Dickson, Hardy & Waters, chapter 6).

$L = \sum_{i=1}^{N} L_{0,i}$

$\Pr(L < 0) = \alpha$

$N\,E[L_0] + z_{\alpha}\sqrt{N}\,\text{SD}(L_0) = 0$

Symbols: $L_{0,i}$ is the future loss at issue on policy $i$ , net or gross, and the losses are independent and identically distributed, so $E[L] = N\,E[L_0]$ and $\text{SD}(L) = \sqrt{N}\,\text{SD}(L_0)$ . $z_\alpha$ is the $\alpha$ -percentile of the standard normal distribution, 1.645 for $95\%$ . The third block follows from $\Pr(L < 0) = \Phi\left(-E[L]/\text{SD}(L)\right) = \alpha$ .
Solving for $P$ : For a fully discrete whole life insurance, $E[L_0] = S A_x - P\ddot{a}_x$ and $\text{SD}(L_0) = (S + P/d)\,\sigma$ , with $\sigma = \sqrt{{}^2A_x - A_x^2}$ . Both are linear in $P$ , so dividing the third block by $N$ gives $P = S\left(A_x + z_\alpha\sigma/\sqrt{N}\right)/\left(\ddot{a}_x - z_\alpha\sigma/(d\sqrt{N})\right)$ .
Size of the margin: The condition requires $E[L_0] = -z_\alpha\,\text{SD}(L_0)/\sqrt{N}$ , so the margin per policy shrinks like $1/\sqrt{N}$ , and the premium falls towards the equivalence principle premium as the portfolio grows. The principle allows only for the diversifiable risk of individual deaths. Risks that hit every policy together, such as a fall in interest rates or a general change in mortality, break the independence assumption and are not covered by it.

Whole Life Premium for 1,000 Policies An insurer sells fully discrete whole life insurances of $100,000 to $N = 1{,}000$ independent lives aged 60. On the SULT at $5\%$ , $A_{60} = 0.29028$ , ${}^2A_{60} = 0.10834$ and $\ddot{a}_{60} = 14.9041$ . Find the net premium for which the probability of an aggregate profit is $95\%$ , and compare it with the equivalence premium.
Answer Here $\sigma = \sqrt{0.10834 - 0.29028^2} = 0.155169$ and $z_{0.95}\,\sigma/\sqrt{1{,}000} = 1.645(0.155169)/31.6228 = 0.0080718$ . Dividing the condition by $N$ gives $29{,}028 - 14.9041P + 0.0080718(100{,}000 + 21P) = 0$ .
 $\begin{align*} P\left(14.9041 - 0.169508\right) &= 29{,}028 + 807.18 \\ P &= \frac{29{,}835.18}{14.734592} \\ &= 2{,}024.84 \\ P_{\text{EP}} &= \frac{29{,}028}{14.9041} \\ &= 1{,}947.65 \end{align*}$ 
The percentile premium is $4.0\%$ above the equivalence premium. With the same method, 100 policies need $2,197.97, a margin of $12.9\%$ , and 10,000 policies need $1,971.87, a margin of $1.2\%$ .

Portfolio Size for a Given Premium For the policy in the first example, the insurer charges a net premium of $2,000. How many policies must it sell for the probability of an aggregate profit to be at least $95\%$ ?
Answer
 $\begin{align*} E[L_0] &= 29{,}028 - 2{,}000(14.9041) \\ &= -780.20 \\ \text{SD}(L_0) &= \left(100{,}000 + 21(2{,}000)\right)(0.155169) \\ &= 22{,}034.00 \\ \sqrt{N} &\ge \frac{1.645(22{,}034.00)}{780.20} \\ &= 46.457 \\ N &\ge 2{,}158.3 \end{align*}$ 
The insurer needs at least 2,159 policies. With fewer, the expected profit of $780 a policy is too small, relative to the spread of the aggregate loss, to give a $95\%$ chance of a profit.

How it appears on the exam

FAM asks for a premium by the portfolio percentile principle, alongside the equivalence principle and the premium for a given EPV of profit, for whole life, term and endowment insurance and for annuities. The question gives $N$ , $\alpha$ and a table or survival model.

Reverse questions: A question can instead give the premium and ask for the smallest $N$ , as in the second example, or give $P$ and $N$ and ask for the probability of a profit.

The loss moments: Every such question first needs $E[L_0]$ and $\text{SD}(L_0)$ as functions of $P$ . For whole life and endowment insurance the $(S + P/d)$ shortcut gives the standard deviation, while for term insurance it must come from the loss distribution.

Exam traps

The sign of $z$ : The condition is $E[L] + z_\alpha\,\text{SD}(L) = 0$ with $z_\alpha = 1.645$ . Writing $E[L] - z_\alpha\,\text{SD}(L) = 0$ gives $1,872.20 in the first example, a premium below the equivalence premium that would make a loss likely.

$\sqrt{N}$ , not $N$ : The standard deviation of the aggregate loss is $\sqrt{N}\,\text{SD}(L_0)$ for independent policies. Using $N\,\text{SD}(L_0)$ treats the policies as perfectly dependent and gives a premium that does not fall as the portfolio grows.

$P$ inside the standard deviation: $\text{SD}(L_0)$ depends on $P$ through $S + P/d$ . Fixing it at the equivalence premium's value gives $2,023.96 instead of $2,024.84 in the first example, a small gap that still changes the answer at the precision a question asks for.

Rounding $N$ : The number of policies must be rounded up, because any smaller portfolio leaves the probability of a profit below $\alpha$ .