Why Combine Portfolio Losses?
An insurer may want to assess the total loss from two blocks of business, such as two hypothetical groups of policies. If the aggregate losses from the blocks are independent, the distribution of their total can be calculated from the two separate aggregate distributions. This calculation is a discrete convolution: each possible pair of portfolio outcomes contributes to one possible combined outcome.
The key is to combine loss amounts and probabilities, not just expected values. Adding the expected losses gives the expected total, but it does not determine the probability of exceeding a threshold. That probability depends on how the possible outcomes from the two portfolios combine.
Independence concerns the aggregate losses themselves. It is not enough that the portfolios have different names or contain different policies: a shared event or common exposure could make their losses dependent. The convolution formula below requires independence, and the variance-addition formula also requires it.
Convolution for Discrete Aggregate Losses
Suppose \(S_1\) and \(S_2\) take values on discrete supports. To find the probability that the combined loss equals \(s\), list all pairs of outcomes \(x\) and \(y\) for which \(x+y=s\). The events corresponding to those pairs are mutually exclusive, so their probabilities can be added.
The sum includes only values in the supports of the two portfolio distributions. Equivalently, each pair \((x,y)\) contributes \(P(S_1=x)P(S_2=y)\) to the combined outcome \(s=x+y\). If multiple pairs give the same total, all their contributions must be combined.
Why the Means and Variances Add
Provided the expectations exist, linearity of expectation gives the mean of the combined loss whether or not the portfolios are independent:
For the variance, expand the square in the definition of variance. The resulting covariance term explains why independence matters.
If \(S_1\) and \(S_2\) are independent and have finite variances, then \(E[S_1S_2]=E[S_1]E[S_2]\), so \(\operatorname{Cov}(S_1,S_2)=0\). Thus:
Independence is essential for this variance result, but not for adding means. If the portfolio losses are dependent, the covariance term must be included; it may be positive or negative.
Worked Example: Combining Two Single-Exposure Portfolios
Two hypothetical portfolios each contain one exposure, and their losses are independent. In Portfolio A, a claim occurs with probability \(0.20\) and has a fixed loss of $100; otherwise the loss is zero. In Portfolio B, a claim occurs with probability \(0.30\) and has a fixed loss of $50; otherwise the loss is zero. Find the combined loss distribution, its mean and variance, and the probability that the combined loss is at least $100.
Method and assumptions. Each portfolio aggregate is a Bernoulli loss with the stated fixed claim amount. The two aggregate losses are independent, so multiply the probabilities of each pair of outcomes and combine pairs with the same total.
For A, the probabilities of losses $0 and $100 are \(0.80\) and \(0.20\). For B, the probabilities of losses $0 and $50 are \(0.70\) and \(0.30\). The four outcome pairs give:
- Both losses are zero: total $0, with probability \(0.80(0.70)=0.56\).
- A is zero and B loses $50: total $50, with probability \(0.80(0.30)=0.24\).
- A loses $100 and B is zero: total $100, with probability \(0.20(0.70)=0.14\).
- Both have claims: total $150, with probability \(0.20(0.30)=0.06\).
The probabilities sum to \(0.56+0.24+0.14+0.06=1.00\), as required. Therefore, the probability of a combined loss of at least $100 is \(0.14+0.06=0.20\).
For a Bernoulli loss that is \(b\) with probability \(p\) and zero otherwise, the mean is \(bp\) and the variance is \(b^2p(1-p)\). Thus:
Independence gives \(E[S]=20+15=\$35\) and \(\operatorname{Var}(S)=1{,}600+525=2{,}125\) dollars squared. As a check on the mean, the combined distribution gives \(50(0.24)+100(0.14)+150(0.06)=12+14+9=\$35\). The common mistake is to use the probability of a claim in either portfolio as though it were the probability of a loss of at least $100; the $50-only outcome does not meet that threshold.
Combining Distributions With More Than Two Outcomes
When either aggregate can take several values, a product table helps ensure that every pair is considered. Multiply across each pair of independent outcomes, assign the product to their sum, and then add products that correspond to the same total. Moments can be calculated separately and added as a check.
Worked Example: Convolving Two Binomial-Frequency Portfolios
Two independent hypothetical portfolios each have a binomial claim count. Portfolio A has \(N_1\sim\operatorname{Binomial}(2,0.25)\), and each claim has a fixed loss of $40. Portfolio B has \(N_2\sim\operatorname{Binomial}(2,0.50)\), and each claim has a fixed loss of $30. Within each portfolio, the claim indicators are independent, and the two portfolios are independent. Find the combined distribution, \(P(S\leq \$70)\), and the mean and variance.
Method and assumptions. First calculate each binomial count distribution, translate counts into aggregate losses, and then convolve the two loss distributions. The fixed severities mean that a portfolio's loss is its claim count multiplied by its claim amount.
For A, the probabilities for counts \(0,1,2\) are \(0.75^2=0.5625\), \(2(0.25)(0.75)=0.3750\), and \(0.25^2=0.0625\). Its aggregate losses are therefore $0, $40, and $80 with those probabilities. For B, the count probabilities are \(0.25,0.50,0.25\), giving aggregate losses $0, $30, and $60.
Each entry in the following table is the product of its row and column probabilities. The row identifies A's loss and the column identifies B's loss.
| A loss | B: $0 (0.25) | B: $30 (0.50) | B: $60 (0.25) |
|---|---|---|---|
| $0 (0.5625) | $0: 0.140625 | $30: 0.281250 | $60: 0.140625 |
| $40 (0.3750) | $40: 0.093750 | $70: 0.187500 | $100: 0.093750 |
| $80 (0.0625) | $80: 0.015625 | $110: 0.031250 | $140: 0.015625 |
There are no duplicate totals in this table, so the combined distribution is: $0 with probability 0.140625; $30 with 0.281250; $40 with 0.093750; $60 with 0.140625; $70 with 0.187500; $80 with 0.015625; $100 with 0.093750; $110 with 0.031250; and $140 with 0.015625. These probabilities sum to 1.000000.
The outcomes no greater than $70 are $0, $30, $40, $60, and $70. Hence \(P(S\leq \$70)=0.140625+0.281250+0.093750+0.140625+0.187500=0.843750\), or 0.8438 to four decimal places.
For a binomial count with parameters \(n,p\) and fixed loss \(b\) per claim, the aggregate mean is \(bnp\), and the variance is \(b^2np(1-p)\). For A these are \(40(2)(0.25)=\$20\) and \(40^2(2)(0.25)(0.75)=\$600\) squared. For B they are \(30(2)(0.50)=\$30\) and \(30^2(2)(0.50)(0.50)=\$450\) squared. Therefore \(E[S]=\$50\) and \(\operatorname{Var}(S)=\$1{,}050\) squared. As a distribution check, summing each combined loss times its probability also gives $50. A common error is to multiply probabilities for one pair and stop there; totals such as $70 must include every pair that produces $70, if more than one does.
Worked Example: Combining Distributions With Overlapping Totals
Two independent hypothetical portfolios have the following aggregate loss distributions. Portfolio A loses $0, $25, or $50 with probabilities 0.50, 0.40, and 0.10. Portfolio B loses $0, $25, or $75 with probabilities 0.60, 0.30, and 0.10. Find the combined distribution, the probability that the total exceeds $50, and its mean and variance.
Method and assumptions. The aggregate distributions are given directly, and the two portfolio losses are independent. Multiply the probabilities for each pair of losses; when pairs yield the same total, add their contributions.
The nine outcome pairs give the following contributions:
- A $0 with B $0, $25, or $75 gives totals $0, $25, or $75 with probabilities \(0.50(0.60)=0.30\), \(0.50(0.30)=0.15\), and \(0.50(0.10)=0.05\).
- A $25 with B $0, $25, or $75 gives totals $25, $50, or $100 with probabilities \(0.40(0.60)=0.24\), \(0.40(0.30)=0.12\), and \(0.40(0.10)=0.04\).
- A $50 with B $0, $25, or $75 gives totals $50, $75, or $125 with probabilities \(0.10(0.60)=0.06\), \(0.10(0.30)=0.03\), and \(0.10(0.10)=0.01\).
The repeated totals are $25, $50, and $75. Adding their contributions produces the combined distribution:
| Combined loss | Probability |
|---|---|
| $0 | 0.30 |
| $25 | 0.39 |
| $50 | 0.18 |
| $75 | 0.08 |
| $100 | 0.04 |
| $125 | 0.01 |
The probabilities sum to \(0.30+0.39+0.18+0.08+0.04+0.01=1.00\). Thus \(P(S>\$50)=0.08+0.04+0.01=0.13\).
For A, \(E[S_1]=25(0.40)+50(0.10)=\$15\), and \(E[S_1^2]=25^2(0.40)+50^2(0.10)=250+250=500\). Therefore \(\operatorname{Var}(S_1)=500-15^2=275\) dollars squared. For B, \(E[S_2]=25(0.30)+75(0.10)=\$15\), and \(E[S_2^2]=25^2(0.30)+75^2(0.10)=187.5+562.5=750\). Therefore \(\operatorname{Var}(S_2)=750-15^2=525\) dollars squared. Independence gives \(E[S]=\$30\) and \(\operatorname{Var}(S)=275+525=800\) dollars squared. The mean from the combined distribution is \(25(0.39)+50(0.18)+75(0.08)+100(0.04)+125(0.01)=\$30\), which checks the calculation. The main trap is to report, for example, only the contribution \(0.15\) to a $25 total and overlook the other pair contributing \(0.24\).
Practical Checks and Common Traps
A reliable calculation separates the construction of the distribution from the calculation of moments and threshold probabilities. Once the distribution is assembled, it should satisfy basic checks: probabilities are nonnegative, their sum is one, and every combined loss is the sum of a possible outcome from each portfolio.
Identify each portfolio's aggregate loss distribution and verify that independence is part of the assumptions. Do not infer independence merely because the portfolios are distinct.
For every possible pair \((x,y)\), calculate the combined loss \(x+y\) and probability \(P(S_1=x)P(S_2=y)\).
Add the probabilities of all pairs with the same total before using the resulting distribution.
Check that probabilities sum to one. Add means and, under independence, variances. Calculate a threshold probability from the combined distribution using the inequality exactly as stated.
- Adding variances without independence: In general, include \(2\operatorname{Cov}(S_1,S_2)\). Adding variances alone is justified when the covariance is zero; independence is a sufficient condition.
- Forgetting duplicate totals: Different pairs can lead to one combined loss. Their probabilities belong together in the probability mass at that total.
- Misreading a threshold: “At least” includes equality, whereas “exceeds” does not. For a discrete distribution, that difference can change the answer substantially.
- Using only the means for a probability question: The combined mean does not determine the probability of a specified loss threshold. Use the convolution when the combined distribution is needed.
These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. They apply whenever two independent portfolio aggregates are available as discrete distributions; if dependence is present, account for it rather than applying independence formulas automatically.
Check Your Understanding
For each problem, treat the stated aggregate losses as independent unless the problem explicitly says otherwise.
- Portfolio A has a 0.10 probability of a fixed $200 loss and otherwise loses zero. Portfolio B has a 0.20 probability of a fixed $100 loss and otherwise loses zero. Find the combined mean and the probability of a positive combined loss. (Answer: mean $40; probability 0.2800.)
- Portfolio A loses $0 or $50 with probabilities 0.70 and 0.30. Portfolio B loses $0 or $50 with probabilities 0.60 and 0.40. Find the probability of a combined loss of $100, the combined mean, and its variance. (Answer: probability 0.1200; mean $35; variance 1,125 dollars squared.)
- Portfolio A has loss $0 or $20 with probabilities 0.50 and 0.50. Portfolio B has loss $0 or $40 with probabilities 0.25 and 0.75. Find the probability the combined loss is at least $40, the mean, and the variance. (Answer: probability 0.7500; mean $40; variance 400 dollars squared.)
- Portfolio A has loss $0 or $10 with probabilities 0.40 and 0.60. Portfolio B has loss $0 or $20 with probabilities 0.50 and 0.50. Find the combined distribution and the probability of a combined loss of at least $20. (Answer: $0: 0.20, $10: 0.30, $20: 0.20, $30: 0.30; probability 0.5000.)
- Two portfolio losses are identical: each equals $100 with probability 0.20 and zero otherwise. They are therefore dependent, not independent. Find the mean and variance of their combined loss. (Answer: mean $40; variance 6,400 dollars squared.)