Why Discretize Claim Severity?
A claim severity model may be continuous, while a calculation or data display calls for a finite table of possible amounts. Discretizing severity replaces each range of losses by one representative amount. The table is an approximation: its probabilities describe how much probability lies in each range, while its representative amounts stand in for all losses in that range.
This tutorial focuses on constructing that table consistently and using it to approximate the mean. The result depends on two choices that must be stated clearly: the intervals and the representative amount assigned to each interval. A correct probability total does not, by itself, guarantee a good mean approximation.
The intervals must form a partition: every possible loss must belong to exactly one interval. For example, one may use intervals closed on the left and open on the right, with the final interval closed at its upper endpoint if the support is bounded. If \(X\) is continuous, a single boundary point has probability zero, so assigning that point to either neighboring interval does not change the probabilities. With a distribution that has point masses, boundary assignment must be specified because a boundary may carry positive probability.
The representative amount can be a lower endpoint, an upper endpoint, a midpoint, or another stated value. The choice should reflect the purpose of the approximation. Do not assume that a table’s representative amounts are automatically midpoints, or that the instruction to round always means rounding to the nearest dollar.
Constructing the Discrete Table
For each interval, calculate its probability using the distribution of \(X\). If \(F_X(x)=P(X\le x)\), then for an interval \([a,b)\), \(P(a\le X<b)=F_X(b^-)-F_X(a^-)\), where the left limits handle possible point masses at the boundaries. For a continuous severity, this reduces to \(F_X(b)-F_X(a)\). Equivalently, using the survival function \(S_X(x)=P(X>x)\), a continuous model gives \(P(a\le X<b)=S_X(a)-S_X(b)\).
Once the probabilities are found, the discretized mean is the probability-weighted average of the assigned representatives. Each interval contributes its representative multiplied by the probability of that interval. This interval-based expression is useful even when two or more intervals have the same representative value.
Why the Approximate Mean Has This Form
The mean of a discrete random variable is the sum of each possible value times the probability of that value. If all the \(c_j\) are distinct, the probability of \(X^*=c_j\) is just \(p_j\). But the representatives need not be distinct: if two intervals use the same representative, their probabilities combine when describing the distribution of \(X^*\).
Therefore, summing over distinct representative values gives the usual discrete mean. Expanding each combined probability recovers the interval-based formula:
This derivation also explains why the interval formula is safe when representatives repeat: it counts each interval’s contribution once. If \(X\) has a finite mean, the difference between the original and approximate means can be written as \(E[X^*]-E[X]=E[X^*-X]\). Thus, if every loss in every interval is within \(\varepsilon\) of its representative, then \(|E[X^*]-E[X]|\le \varepsilon\). The bound follows from \(|X^*-X|\le\varepsilon\) for every outcome and the inequality \(|E[Z]|\le E[|Z|]\). It does not apply with a finite \(\varepsilon\) when an unbounded interval has a fixed finite representative.
Worked Example: Rounding Down a Uniform Severity
A hypothetical warranty portfolio is modeled with claim severity \(X\sim\operatorname{Uniform}(0,500)\), with amounts in dollars. Round each claim down to the lower $100 interval representative: 0 for losses in \([0,100)\), 100 for \([100,200)\), and so on, with 400 for \([400,500]\). There is no interest rate or mortality assumption; this is a claim severity calculation. Find the discrete table and approximate mean, then compare it with \(E[X]\).
Method. The uniform distribution on \([0,500]\) has density \(1/500\), so the probability of an interval is its width divided by 500. Each interval has width 100, giving probability \(100/500=0.2\).
| Loss interval | Representative \(c_j\) | Probability \(p_j\) | Contribution \(c_jp_j\) |
|---|---|---|---|
| \([0,100)\) | $0 | 0.2 | $0 |
| \([100,200)\) | $100 | 0.2 | $20 |
| \([200,300)\) | $200 | 0.2 | $40 |
| \([300,400)\) | $300 | 0.2 | $60 |
| \([400,500]\) | $400 | 0.2 | $80 |
The probabilities sum to \(5(0.2)=1.0\). The discretized mean is \(0+20+40+60+80=\$200\). The continuous uniform mean is \(E[X]=(0+500)/2=\$250\), so this lower-endpoint discretization underestimates the mean by $50. The difference is consistent with rounding every loss down by an amount between $0 and $100.
Check. The table assigns one representative to every interval, and the five probabilities cover the full uniform support. The approximate mean is below the continuous mean, as expected when every loss is rounded down.
Worked Example: Rounding to the Nearest $250
For a hypothetical equipment policy, let \(X\sim\operatorname{Uniform}(0,1000)\) dollars. Round to the nearest multiple of $250, using representatives 0, 250, 500, 750, and 1,000. The boundary ties may be assigned to either neighboring amount because \(X\) is continuous. There is no interest rate or mortality assumption. Construct the table and calculate \(E[X^*]\).
Method. The rounding intervals are \([0,125)\), \([125,375)\), \([375,625)\), \([625,875)\), and \([875,1000]\). Their widths are 125, 250, 250, 250, and 125 dollars. Since the uniform density is \(1/1000\), the probabilities are 0.125, 0.25, 0.25, 0.25, and 0.125.
| Loss interval | Representative | Probability | Contribution to mean |
|---|---|---|---|
| \([0,125)\) | $0 | 0.125 | $0 |
| \([125,375)\) | $250 | 0.25 | $62.50 |
| \([375,625)\) | $500 | 0.25 | $125.00 |
| \([625,875)\) | $750 | 0.25 | $187.50 |
| \([875,1000]\) | $1,000 | 0.125 | $125.00 |
Adding the contributions gives \(E[X^*]=0+62.50+125.00+187.50+125.00=\$500.00\). The probabilities sum to \(0.125+0.25+0.25+0.25+0.125=1\). Also, the true mean of the uniform distribution is \((0+1000)/2=\$500\). The approximate mean happens to equal the true mean because this distribution and these rounding intervals are symmetric about $500; equality here does not mean that rounding leaves every loss unchanged.
Check. The end intervals are only half as wide as the interior intervals, so they each receive half the probability of an interior interval. Treating all five intervals as equally likely would produce an incorrect table.
Worked Example: A Tail Interval and a Chosen Representative
A hypothetical severity model has \(X\sim\operatorname{Exponential}(\text{mean } \$1{,}000)\), so \(S_X(x)=e^{-x/1000}\) for \(x\ge0\). Group losses into \([0,500)\), \([500,1000)\), \([1000,1500)\), and \([1500,\infty)\). Assign representatives $250, $750, $1,250, and $1,750, respectively. There is no interest rate or mortality assumption. Calculate the table and approximate mean.
Method. For each finite interval \([a,b)\), use \(P(a\le X<b)=S_X(a)-S_X(b)\). For the final interval use \(P(X\ge1500)=S_X(1500)\). The four probabilities are \(1-e^{-0.5}=0.393469\), \(e^{-0.5}-e^{-1}=0.238651\), \(e^{-1}-e^{-1.5}=0.144749\), and \(e^{-1.5}=0.223130\), to six decimal places.
| Loss interval | Representative | Probability (6 decimals) | Contribution (dollars) |
|---|---|---|---|
| \([0,500)\) | $250 | 0.393469 | $98.3673 |
| \([500,1000)\) | $750 | 0.238651 | $178.9883 |
| \([1000,1500)\) | $1,250 | 0.144749 | $180.9363 |
| \([1500,\infty)\) | $1,750 | 0.223130 | $390.4775 |
Using the unrounded probabilities, the approximate mean is \(250(1-e^{-0.5})+750(e^{-0.5}-e^{-1})+1250(e^{-1}-e^{-1.5})+1750e^{-1.5}=\$848.77\), rounded to the nearest cent. The displayed probabilities sum to 0.999999 only because each has been rounded to six decimals; the exact probabilities sum to 1. The true exponential mean is $1,000. This approximation is lower because the chosen $1,750 representative is below many of the losses in the open-ended tail interval.
Check. The tail probability must be the survival probability at $1,500, not the probability of the preceding finite interval. The tail representative is a chosen approximation; it is not implied by the exponential model or by the interval itself.
Common Mistakes and Exam Traps
The most important bookkeeping trap occurs when different intervals share a representative. For example, suppose three intervals have probabilities 0.2, 0.3, and 0.5, with representatives 5, 5, and 25. The probability that \(X^*=5\) is \(0.2+0.3=0.5\), and the probability that \(X^*=25\) is 0.5. Therefore \(E[X^*]=5(0.5)+25(0.5)=15\).
A wrong calculation would write \(5P(X^*=5)+5P(X^*=5)+25P(X^*=25)\), counting the same probability for representative 5 twice. The correction is to sum over distinct representative values, or, more directly, to sum \(c_jP(X\in I_j)\) over the original intervals. The latter gives \(5(0.2)+5(0.3)+25(0.5)=15\).
- Changing the rounding rule. Rounding down, rounding up, and rounding to the nearest grid point produce different intervals and potentially different means. Write the intervals before calculating.
- Using equal probabilities for unequal intervals. Under a uniform model, probability is proportional to interval width. Under other models, it is determined by the CDF or survival function, not width alone.
- Ignoring the support or tail. The intervals must cover the full possible range. An unbounded final interval still needs a stated representative if a finite discrete approximation is required.
- Rounding intermediate probabilities too early. Keep sufficient precision in calculations and round the final monetary result to cents unless the question specifies another precision.
Check Your Understanding
For each problem, identify the intervals and representatives before calculating. Monetary answers are rounded to the nearest cent where needed.
- A hypothetical loss has \(X\sim\operatorname{Uniform}(0,600)\) dollars and is rounded down to the lower $200 representative, using intervals \([0,200)\), \([200,400)\), and \([400,600]\). Find the discrete probabilities and approximate mean. (Answer: probabilities \(1/3,1/3,1/3\); mean $200.)
- A hypothetical severity has \(X\sim\operatorname{Uniform}(0,800)\) dollars and is rounded to the nearest $200, with representatives 0, 200, 400, 600, and 800. Find the approximate mean. (Answer: $400.)
- A hypothetical claim has an exponential severity with mean $500. Use intervals \([0,250)\), \([250,500)\), \([500,750)\), and \([750,\infty)\), with representatives $125, $375, $625, and $875. Find the approximate mean. (Answer: $424.39.)
- Three intervals have probabilities 0.2, 0.3, and 0.5 and representatives 5, 5, and 25. Find the probability distribution of \(X^*\) over distinct values and its mean. (Answer: \(P(X^*=5)=0.5\), \(P(X^*=25)=0.5\); mean 15.)
- A hypothetical severity has \(X\sim\operatorname{Uniform}(0,1000)\) dollars and is rounded down using four intervals of width $250, represented by their lower endpoints. Find the approximate mean and its difference from the true mean. (Answer: approximate mean $375; true mean $500; the approximation is $125 lower.)