Why Discretize a Continuous Severity?
Aggregate-loss calculations often need a discrete severity distribution, even when individual claim amounts are modeled continuously. Replacing a continuous severity with probabilities on a lattice makes methods such as discrete convolution and Panjer recursion applicable. The choice of discretization matters: it changes the severity distribution and can therefore change calculated aggregate probabilities.
Let \(X\) be a nonnegative continuous claim severity, let \(h>0\) be a lattice spacing, and use lattice points \(0,h,2h,\ldots\). Write \(S_X(x)=P(X>x)\). The interval endpoints are immaterial for a continuous distribution, but the convention below assigns each interval \([jh,(j+1)h)\) to its left endpoint for lower discretization and its right endpoint for upper discretization.
With \(p_j=P(jh\le X<(j+1)h)\), the lower lattice probability at \(jh\) is \(p_j\); the upper lattice probability at \((j+1)h\) is \(p_j\). For nonnegative continuous \(X\), the interval probabilities can be calculated as \(p_j=F_X((j+1)h)-F_X(jh)\). An unbounded severity needs no finite “last bin”: continue the lattice over all \(j\ge 0\), including the probability in the tail.
Three Ways to Place Severity on the Lattice
Lower and upper discretizations are deterministic mappings. Local-moment discretization is a randomized mapping: conditional on a particular severity \(x\in[jh,(j+1)h]\), it places probability \(((j+1)h-x)/h\) at \(jh\), and probability \((x-jh)/h\) at \((j+1)h\). These probabilities sum to one, and their weighted average is \(x\).
To obtain the unconditional lattice probabilities, integrate these conditional weights over the severity distribution in each interval, then add contributions to any lattice point from its two neighboring intervals. A point at the edge of the support or at zero simply receives the contribution from the interval that can reach it.
Worked Example: Discretizing a Uniform Severity
A hypothetical loss has \(X\sim\operatorname{Uniform}(0,3)\), measured in thousands of dollars. Use spacing \(h=1\) and find the lower, upper, and local-moment lattice distributions. Check the means.
Lower discretization. Each of the three intervals \([0,1)\), \([1,2)\), and \([2,3)\) has probability \(1/3\). Map them to 0, 1, and 2, respectively. Thus \(P(Y=0)=P(Y=1)=P(Y=2)=1/3\), and \(E[Y]=(0+1+2)/3=1\).
Upper discretization. Map those same intervals to 1, 2, and 3. Each point again has probability \(1/3\), so \(E[Y]=(1+2+3)/3=2\).
Local-moment discretization. Within any unit interval, a uniform severity has a uniform fractional part. Its average rounding weight is split equally between the two endpoints. The interval \([0,1)\) therefore contributes probability \(1/6\) at 0 and \(1/6\) at 1; \([1,2)\) contributes \(1/6\) at 1 and \(1/6\) at 2; and \([2,3)\) contributes \(1/6\) at 2 and \(1/6\) at 3. Combining contributions gives probabilities \(1/6,1/3,1/3,1/6\) at 0, 1, 2, and 3. Its mean is \(0(1/6)+1(1/3)+2(1/3)+3(1/6)=1.5\), equal to \(E[X]=1.5\). The lower and upper means differ from the continuous mean, as expected.
Why Local-Moment Discretization Preserves the Mean
The mean-preservation result follows directly from the endpoint weights. For fixed \(x\in[jh,(j+1)h]\), the conditional expectation of the rounded value \(Y\) is
Taking expectations over \(X\) gives \(E[Y]=E[E[Y\mid X]]=E[X]\). This argument requires a finite severity mean. It preserves the mean for one claim, but it does not claim that every feature of the severity distribution—or every aggregate probability—is preserved. In particular, local-moment rounding can move individual outcomes either upward or downward.
Worked Example: Calculating Local-Moment Probabilities for an Exponential Severity
A hypothetical claim severity \(X\), in thousands of dollars, is exponential with mean 1, so its density is \(f_X(x)=e^{-x}\) for \(x\ge0\). With \(h=1\), find the local-moment probabilities at lattice points 0 and 1. Show how the contributions arise.
Probability at 0. Only severities in \([0,1)\) can be rounded to 0. Given \(X=x\) there, the weight at 0 is \(1-x\). Therefore
Probability at 1. The point 1 receives contributions from the upper endpoint of \([0,1)\) and the lower endpoint of \([1,2)\). The corresponding weights are \(x\) and \(2-x\):
The probabilities at 0 and 1 sum to \(0.7675\), the lattice CDF at 1. The remaining probability lies at lattice points 2 or higher. As a check on the method, local-moment discretization preserves this exponential severity’s mean of 1; the first two probabilities alone do not need to sum to one.
Effects on Compound Aggregate CDFs
Let \(N\) be a claim count independent of the severities \(X_1,X_2,\ldots\), and define the aggregate \(S=\sum_{i=1}^{N}X_i\), with an empty sum equal to zero. Apply the same discretization independently to each severity and use the same claim count in all models. Lower and upper discretization have a useful CDF ordering; local-moment discretization does not have the same guarantee.
For the first two claims, couple each discretized severity to the same continuous severity. Claim by claim, \(X_i^{L}\le X_i\le X_i^{U}\), where \(X_i^{L}\) and \(X_i^{U}\) are the lower- and upper-rounded values. Summing these inequalities for the same realized \(N\) gives \(S^{L}\le S\le S^{U}\), including when \(N=0\). Consequently, \(\{S\le s\}\subseteq\{S^{L}\le s\}\) and \(\{S^{U}\le s\}\subseteq\{S\le s\}\). Taking probabilities proves the stated CDF bounds. Local-moment rounding is not ordered claim by claim, so this coupling argument cannot establish a one-sided bound for it.
Worked Example: Comparing Compound CDF Errors
A hypothetical portfolio has \(N\sim\operatorname{Binomial}(2,0.5)\) claims. The severities are independent of \(N\) and of one another, and each has an exponential distribution with mean 1 thousand dollars. Use \(h=1\) thousand dollars. Compare the continuous aggregate CDF and the three discretized aggregate CDFs at \(s=1\) thousand dollars. Give probabilities to four decimals.
Continuous model. The claim-count probabilities are \(P(N=0)=0.25\), \(P(N=1)=0.50\), and \(P(N=2)=0.25\). For one claim, \(P(X\le1)=1-e^{-1}=0.632121\). For two claims, the sum has density \(xe^{-x}\) for \(x\ge0\), so
Thus \(P(S\le1)=0.25+0.50(0.632121)+0.25(0.264241)=0.6321\).
Lower discretization. The probabilities at 0 and 1 are \(p_0=1-e^{-1}=0.632121\) and \(p_1=e^{-1}-e^{-2}=0.232544\). For one claim, \(P(Y\le1)=p_0+p_1=1-e^{-2}=0.864665\). For two claims, their sum is at most 1 only if both are 0 or one is 0 and the other is 1. The probability is \(p_0^2+2p_0p_1=0.693567\). Therefore
Upper discretization. The probability at 1 is \(P(0<X\le1)=1-e^{-1}=0.632121\); there is no mass at 0. For two claims the discretized sum is at least 2, so it cannot be at most 1. Hence \(P(S^U\le1)=0.25+0.50(0.632121)+0.25(0)=0.5661\).
Local-moment discretization. From the preceding example, \(p_0=e^{-1}=0.367879\) and \(p_1=(1-e^{-1})^2=0.399576\). Thus the one-claim CDF at 1 is \(p_0+p_1=0.767455\). For two claims, the probability of a sum no greater than 1 is \(p_0^2+2p_0p_1=0.429327\). Therefore \(P(S^{LM}\le1)=0.25+0.50(0.767455)+0.25(0.429327)=0.7411\).
The absolute errors against the continuous value \(0.6321\), calculated before rounding, are 0.2236 for lower, 0.0661 for upper, and 0.1089 for local-moment discretization. The lower result exceeds the continuous CDF, and the upper result is smaller, as the coupling argument requires. Local-moment rounding preserves the severity mean but does not force its aggregate CDF to equal the continuous CDF or to fall on a particular side.
Practical Checks and Exam Traps
A finer spacing reduces the width of each interval, but increases the number of lattice probabilities needed in the aggregate calculation. Always preserve the probability in the upper tail: dropping it changes the severity model rather than merely discretizing it. When the lattice is expressed in thousands of dollars, both the spacing and the threshold must use that same unit.
A common mistake is to reverse the CDF inequality because lower-rounded losses are smaller. The wrong argument is: “The lower-rounded aggregate is smaller, so \(P(S^L\le s)\le P(S\le s)\).” The inequality is reversed: a smaller realized aggregate is more likely to be at or below the same threshold. In the worked example, the wrong direction would claim \(0.8557\le0.6321\), which is false. The correct coupling gives \(P(S^L\le s)\ge P(S\le s)\); for upper rounding it gives \(P(S^U\le s)\le P(S\le s)\). These are CDF statements, not claims that every point probability or every other aggregate measure has the same ordering.
Before using a lattice distribution in a discrete aggregate calculation, check that its probabilities are nonnegative and sum to one, confirm the lattice spacing and loss units, and verify whether the selected method preserves the severity mean. Lower and upper discretizations are useful when a directional CDF bound matters; local-moment discretization is useful when preserving the mean is important. The resulting probabilities are approximations to the continuous aggregate model, and the spacing should be chosen with the calculation’s accuracy and computational cost in mind.
Check Your Understanding
For each problem, use the stated lattice convention and keep the claim-count and severity assumptions distinct.
- A hypothetical severity is uniform on \([0,4]\), and the lattice spacing is \(h=2\). Give the lower- and upper-discretized probability at each lattice point and their means, in the severity’s original units. (Answer: Lower: probability \(1/2\) at 0 and 2, mean 1; upper: probability \(1/2\) at 2 and 4, mean 3.)
- For a continuous severity and the same claim count, state the direction of the lower- and upper-discretized aggregate CDFs relative to the continuous aggregate CDF. (Answer: Lower CDF is at least the continuous CDF; upper CDF is at most the continuous CDF.)
- A severity of exactly \(1.4\) is locally moment-discretized on a unit lattice. Give the conditional probabilities at the two possible lattice points and check the conditional mean. (Answer: Probability \(0.6\) at 1 and \(0.4\) at 2; mean \(0.6(1)+0.4(2)=1.4\).)
- A portfolio has either zero or one claim, each with probability \(1/2\). A continuous claim has probability \(0.7\) of being at most 1. Its lower discretization places probability \(0.8\) at 0 and 0.2 at 1. Find the continuous and lower-discretized aggregate CDFs at 1. (Answer: Continuous \(0.5+0.5(0.7)=0.8500\); lower \(0.5+0.5(1)=1.0000\).)
- For the same zero-or-one claim count, suppose local-moment discretization preserves the severity mean. Does that alone establish whether its aggregate CDF at a specified threshold is above or below the continuous model’s CDF? (Answer: No. Mean preservation does not give a one-sided CDF bound; local-moment rounding can move severity values either upward or downward.)