Actuarial Notes

Individual Risk Model

Study guide: Exam FAM

The Individual Risk Model represents the aggregate losses of a portfolio as the sum of the losses on a fixed number n of independent policies, S = X_1 + \cdots + X_n, where most of the X_j are zero. Each loss is written as X_j = I_j B_j, where the indicator I_j shows whether policy j has a claim and B_j is the claim amount if it does. It contrasts with the collective risk model of the aggregate loss model, which sums a random number of identically distributed claims, and it is unrelated to Individual Risk Rating, an Exam 8 pricing topic.

S = X_1 + X_2 + \cdots + X_n, \quad X_j = I_j B_j

E[S] = \sum_{j=1}^{n} q_j\,\mu_j

\text{Var}(S) = \sum_{j=1}^{n} \left[q_j\,\sigma_j^2 + q_j(1 - q_j)\,\mu_j^2\right]

  • Symbols: Following Loss Models (chapter 9), q_j = \Pr(I_j = 1) is the probability of a claim on policy j, and \mu_j and \sigma_j^2 are the mean and variance of its claim amount B_j. The policies are independent but not identically distributed, so each can have its own claim probability and benefit. With a fixed benefit b_j, as in group term life, \sigma_j^2 = 0, and X_j is b_j times a Bernoulli variable.
  • Where the variance comes from: Conditioning on the indicator, the law of total variance gives E[\text{Var}(X_j \mid I_j)] = q_j\sigma_j^2 and \text{Var}(E[X_j \mid I_j]) = q_j(1 - q_j)\mu_j^2. Independence then lets the means and the variances add across policies.
  • Individual versus collective: In the collective model, a random claim count N is independent of identically distributed claim amounts. Modelling the same book collectively, with N Poisson with mean \sum q_j and a severity that mixes the B_j with weights proportional to q_j, keeps the mean but adds \sum q_j^2\mu_j^2 to the variance, because a Poisson count lets a policy claim more than once.
  • Approximating the distribution: The exact distribution of S needs a convolution of n different distributions, so for a large portfolio S is approximated by a normal or a lognormal distribution with the same mean and variance.

Group Life With Two Classes A group life policy covers 400 employees with a death probability of 0.002 and a benefit of $100,000 each, and 200 employees with a death probability of 0.005 and a benefit of $50,000 each. Deaths are independent. Find the mean and variance of total claims, and the premium that covers claims with probability 0.95 under the normal approximation.

Answer The benefits are fixed, so each variance term is q_j(1 - q_j)\,b_j^2.

\begin{align*}
E[S] &= 400(0.002)(100{,}000) + 200(0.005)(50{,}000) \\
&= 80{,}000 + 50{,}000 \\
&= 130{,}000 \\
\text{Var}(S) &= 400(0.002)(0.998)(100{,}000)^2 \\
&\quad + 200(0.005)(0.995)(50{,}000)^2 \\
&= 7.984 \times 10^{9} + 2.4875 \times 10^{9} \\
&= 1.04715 \times 10^{10}
\end{align*}

The standard deviation is 102{,}330, and the premium is 130{,}000 + 1.645(102{,}330) = 298{,}333. The premium is more than twice the expected claims, because a single death in the first class costs $100,000 against a mean of $130,000. With so few expected deaths, the distribution of S is very skewed and the normal approximation is rough.

Random Claim Amounts, Individually and Collectively Each of 1,000 independent health policies has a probability of 0.1 of a claim in the year. Given a claim, the amount has mean 2,000 and standard deviation 1,500. Find E[S] and \text{Var}(S) in the individual risk model, and compare the variance with a collective model that has a Poisson number of claims with the same mean.

Answer Each policy has the same q, \mu and \sigma, so the sums are 1,000 times one policy's values.

\begin{align*}
E[S] &= 1{,}000(0.1)(2{,}000) \\
&= 200{,}000 \\
\text{Var}(X_j) &= 0.1(1{,}500)^2 + 0.1(0.9)(2{,}000)^2 \\
&= 225{,}000 + 360{,}000 \\
&= 585{,}000 \\
\text{Var}(S) &= 1{,}000(585{,}000) \\
&= 5.85 \times 10^{8}
\end{align*}

The collective model has \lambda = 100, and its variance is \lambda\,E[B^2] = 100(1{,}500^2 + 2{,}000^2) = 6.25 \times 10^{8}. The difference, 4 \times 10^{7}, is exactly \sum q_j^2\mu_j^2 = 1{,}000(0.01)(4{,}000{,}000). The standard deviations are 24,187 and 25,000, so the collective model is slightly conservative.

How it appears on the exam

  • FAM asks candidates to define the collective and individual risk models and to calculate their means and variances. A question describes a group of policies in classes, each with its own claim probability and benefit, and asks for E[S] and \text{Var}(S).
  • Approximations: FAM also asks candidates to use the normal or lognormal approximation for the aggregate distribution. The lognormal matches the moments through \sigma^2 = \ln\big(1 + \text{Var}(S)/E[S]^2\big) and \mu = \ln E[S] - \sigma^2/2.
  • Telling the models apart: A question that lists policies that can each claim at most once is set in the individual model, while one that gives a distribution for the number of claims is set in the collective model.

Exam traps

  • The missing term: The variance of X_j is q_j\sigma_j^2 + q_j(1 - q_j)\mu_j^2. Using q_j\sigma_j^2 alone ignores the uncertainty over whether a claim happens, and using q_j E[B_j^2] gives the larger compound Poisson variance.
  • Fixed benefits still vary: A fixed benefit has \sigma_j^2 = 0, but X_j still has variance q_j(1 - q_j)b_j^2, because the policy pays either nothing or the full benefit.
  • A fixed number of policies: In the individual model, n is the fixed number of policies, and the number of claims, \sum I_j, is random. Treating n as the number of claims overstates E[S] by a factor of 1/q.
  • The namesake page: Individual Risk Rating, an Exam 8 topic, adjusts one insured's premium for its own characteristics or loss experience through experience, schedule or retrospective rating. It has nothing to do with this model.