Why Aggregate Tails and Quantiles Need Care
An aggregate loss model describes the total loss from a collection of claims. Two questions often matter: how likely is the total to exceed a specified amount, and what amount is exceeded with a specified probability? These are related questions, but exact equality at the threshold can change the answer when aggregate losses have positive probability at individual values.
The first task is to identify the event precisely. “Exceeds 100” means strictly greater than 100; “at least 100” includes equality. The distinction is especially important for discrete aggregate distributions, which can have substantial probability at a threshold. The calculations below use exact distributions when available and a clearly specified approximation when one is used.
Strict and Non-Strict Exceedance Probabilities
For any real-valued random variable, the strict exceedance probability follows directly from the complement of the event \(S_N\le x\). The non-strict probability requires the probability strictly below \(x\), which is the left-hand limit of the cumulative distribution function.
If there is no probability mass at \(x\), then \(F_{S_N}(x)=F_{S_N}(x-)\), and the two exceedance probabilities agree. If there is an atom at \(x\), their difference is exactly \(P(S_N=x)\). This is why it is not safe to substitute a threshold into a tail formula before deciding whether the problem asks for \(>\) or \(\ge\).
Worked Example: Exact Tails From a Compound Binomial Model
A hypothetical portfolio has two independent exposure units. Each unit produces a claim with probability \(0.5\), so \(N\sim\operatorname{Binomial}(2,0.5)\). Given a claim, its amount is 1 or 2 units with probability \(0.5\) each, independently of the other claim and of \(N\). Find \(P(S_N>2)\), \(P(S_N\ge2)\), and the 0.80 quantile.
Method: construct the exact aggregate probability function. The count probabilities are \(P(N=0)=0.25\), \(P(N=1)=0.50\), and \(P(N=2)=0.25\).
If \(N=0\), the aggregate is 0, contributing probability 0.25. If \(N=1\), the aggregate is 1 or 2, each with conditional probability 0.5. Their unconditional contributions are \(0.50(0.5)=0.25\) each. If \(N=2\), the possible sums are 2, 3, and 4, with conditional probabilities 0.25, 0.50, and 0.25. Their unconditional contributions are 0.0625, 0.125, and 0.0625.
Combining contributions at the same aggregate amount gives:
The probabilities sum to \(0.25+0.25+0.3125+0.125+0.0625=1\), a check that the distribution is normalized. Strictly exceeding 2 includes only amounts 3 and 4:
Meeting or exceeding 2 also includes the probability at 2:
The cumulative probabilities at 0, 1, and 2 are 0.25, 0.50, and 0.8125. Therefore the smallest amount at which the cumulative probability reaches 0.80 is 2, so the 0.80 quantile is 2. Common mistake avoided: reporting 0.50 for \(P(S_N>2)\) by including the mass at 2. That is the answer to \(P(S_N\ge2)\), not the strict exceedance question.
Quantiles: The General Definition
A quantile gives a threshold associated with a cumulative probability. For \(0<p<1\), use the lower quantile definition below. It works for continuous and discrete distributions, including distributions for which a quantile is not itself a value with positive probability.
For a finite ordered support, the practical procedure is to accumulate the probabilities from the smallest value upward and stop at the first value whose cumulative probability is at least \(p\). The qualification matters: for a general discrete distribution, there need not be a first support value where the cumulative probability reaches \(p\). The infimum definition remains the general definition.
A quantile should also be checked at its boundary. For a finite-support distribution, if \(q_p\) is selected as the first support value crossing \(p\), then \(F_{S_N}(q_p)\ge p\); at the preceding support value, the cumulative probability is less than \(p\). This check catches errors caused by rounding or by selecting the wrong side of a jump.
Worked Example: A Quantile That Is Not a Support Value
Consider a hypothetical aggregate loss whose distribution assigns probability \(2^{-n}\) to \(1-1/n\) for each integer \(n\ge2\), and probability 0.5 to 2. Find its 0.50 quantile and compare the strict and non-strict exceedance probabilities at that amount.
Method: apply the infimum definition, not an unqualified first-support-value rule. The probabilities below 1 sum to
Every one of the values \(1-1/n\) is below 1, and these values approach 1 as \(n\) increases. For any \(x<1\), at least one of these support values is greater than \(x\), with positive probability. Consequently \(F_{S_N}(x)<0.5\). At \(x=1\), all the mass below 1 has been included, so \(F_{S_N}(1)=0.5\). Thus the set \(\{x:F_{S_N}(x)\ge0.5\}\) begins at 1 and
There is no probability mass at 1, so 1 is not a support value. At this threshold, the mass strictly above 1 is the 0.5 at 2; the mass at all the points \(1-1/n\) is below 1. Therefore \(P(S_N>1)=P(S_N\ge1)=0.5\). Common mistake avoided: claiming that a discrete quantile must be a support value. The infimum definition correctly gives 1 even though no support point equals 1.
Quantiles and Tails Under a Specified Approximation
An approximate distribution must be stated before using it to calculate an aggregate tail or quantile. For an integer-valued aggregate, a normal approximation is often used with a continuity correction: an event at integer thresholds is matched to the interval halfway between adjacent integers. The correction reflects the fact that a continuous normal variable assigns zero probability to any single point, whereas an integer-valued aggregate can have positive mass at each integer.
Worked Example: Approximate Tails and an Integer Quantile
A hypothetical count of small, unit-sized claims has a Poisson distribution with mean 100. Its aggregate amount in units is therefore integer-valued, with mean 100 and variance 100. Use the specified normal approximation \(Y\sim N(100,100)\), with continuity correction, to approximate \(P(S_N>110)\), \(P(S_N\ge110)\), and the 0.95 quantile.
Method: standardize the corrected normal thresholds. The standard deviation is \(\sqrt{100}=10\). For strict exceedance, the corrected boundary is 110.5, giving \(z=(110.5-100)/10=1.05\). Hence
where \(Z\sim N(0,1)\) and \(\Phi\) is its cumulative distribution function. For the non-strict event, the corrected boundary is 109.5, so \(z=(109.5-100)/10=0.95\). Thus
For the 0.95 quantile, seek the smallest integer \(k\) for which the approximate cumulative probability through \(k\) is at least 0.95. With the correction, this is \(\Phi((k+0.5-100)/10)\ge0.95\). Since the 0.95 standard normal quantile is approximately 1.6449, the condition becomes \(k+0.5\ge100+10(1.6449)=116.449\). The smallest integer satisfying it is 116. The adjacent-threshold check confirms the choice: at \(k=115\), the standardized corrected boundary is 1.55 and \(\Phi(1.55)\approx0.9394<0.95\); at \(k=116\), it is 1.65 and \(\Phi(1.65)\approx0.9505\ge0.95\). The approximate 0.95 quantile is therefore 116 units. Probabilities are rounded to four decimal places.
Common mistake avoided: using 110 as the normal boundary for both discrete tail events. The correction is 110.5 for \(>110\) and 109.5 for \(\ge110\). The approximate difference, 0.1711 minus 0.1469, is 0.0242, representing the estimated probability at the integer threshold.
Checks That Prevent Threshold Errors
Before finalizing a result, check the wording, the distribution and the boundary. For an exact distribution, add the relevant probability masses explicitly or use the appropriate CDF value. For an approximation, state the approximating distribution and whether a continuity correction was applied. For a quantile, verify that the cumulative probability reaches \(p\) at the reported threshold and, where the support is ordered with a first crossing, remains below \(p\) at the previous support value.
- For “exceeds \(x\),” exclude the mass at \(x\); for “at least \(x\),” include it.
- For a discrete quantile, use the infimum definition. Use the first-support-value procedure only when a first crossing exists.
- For an integer aggregate approximated by a continuous normal distribution, apply the half-unit correction consistently to tails and quantiles.
- Do not report an approximate normal result as an exact aggregate probability.
These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. The central skill is not merely evaluating a distribution function; it is translating the wording into the correct event and then applying the matching definition.
Check Your Understanding
For each problem, identify whether the requested event is strict or non-strict before calculating.
- An exact aggregate distribution has probability 0.20 at 0, 0.30 at 2, and 0.50 at 5. Find \(P(S_N>2)\) and \(P(S_N\ge2)\). (Answer: 0.5000 and 0.8000, respectively.)
- For the distribution in question 1, find the 0.49 quantile and verify it using the cumulative probabilities. (Answer: \(q_{0.49}=2\), since \(F(0)=0.20<0.49\) and \(F(2)=0.50\ge0.49\).)
- Suppose a specified continuous approximation is \(Y\sim N(200,20^2)\). Approximate the probability that the aggregate exceeds 220, without a continuity correction because this specified approximation treats the aggregate as continuous. (Answer: \(1-\Phi(1)=0.1587\).)
- An integer-valued aggregate is approximated by \(N(50,5^2)\). Using continuity correction, approximate the probability that it is at least 56. (Answer: \(1-\Phi(1.1)=0.1357\).)
- A discrete distribution assigns probability \(2^{-n}\) to \(1-1/n\) for every integer \(n\ge2\), and probability 0.5 to 2. Find its 0.50 quantile and state whether that amount is a support value. (Answer: \(q_{0.50}=1\); 1 is not a support value.)