Aggregate Loss On Lattice Supports

Why Aggregate Losses Have Restricted Supports

An aggregate loss is a sum of individual claim severities. Even when each severity can take several values, the sum cannot take every nonnegative value. If severities occur only in fixed increments, the aggregate must also obey an increment rule. Recognizing that rule can eliminate impossible outcomes before any detailed probability calculation.

Let $N$ be the number of claims, let $X_j$ be the severity of claim $j$, and define the aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Unless stated otherwise, assume $N$ is independent of the severities and that, conditional on $N=n$, the $n$ severities are independent and identically distributed as $X$. The lattice restrictions below follow from the severity support and do not require a particular distribution for $N$.

Definition: A set is contained in a lattice with spacing $h>0$ and offset $a$ if every value in the set has the form $a+kh$, where $k$ is an integer. The offset is understood modulo $h$: offsets that differ by an integer multiple of $h$ describe the same lattice. For example, the values $2,5,8,\ldots$ lie on the lattice with spacing $3$ and offset $2$.

For integer-valued severities, the useful spacing is often the greatest common divisor of differences between support values. For example, the support $\{2,5\}$ has spacing $3$, since the difference is $5-2=3$, and offset $2$. More generally, if all severity values belong to $a+h\mathbb{Z}$, their sum for a fixed claim count $n$ belongs to $na+h\mathbb{Z}$. This is a restriction on the support; it does not assert that every point on that lattice can actually occur.

From the Severity Lattice to the Aggregate Lattice

If every possible severity can be written as $a+kh$, then for $n$ claims the aggregate is a sum of $n$ such values. Its offset is therefore multiplied by $n$, while its spacing remains $h$. In particular, when severities are integer multiples of $h$, so that $a=0$, every fixed-count aggregate is also a multiple of $h$.

X_j\in a+h\mathbb{Z}\ \text{for every }j \quad\Longrightarrow\quad S\mid(N=n)\in na+h\mathbb{Z}.

If the claim count can take several values, the unconditional aggregate support is contained in the union of these count-specific lattices over counts with positive probability. The zero-claim outcome must also be included: $S=0$ when $N=0$. A convenient preliminary check is to compare a proposed aggregate value with the relevant lattice for each possible claim count. If it belongs to none of them, its probability is zero.

Important distinction: Passing a lattice test is necessary, but it may not be sufficient for an aggregate value to be possible. A severity distribution may occupy only some points on its lattice. The actual aggregate support consists of sums of values that the severity distribution permits, not every point on the containing lattice.

Why the Lattice Restriction Holds

The result follows directly from closure under addition. If each of $n$ severities is $a+k_jh$, then their sum has offset $na$ and an integer multiple of $h$. This gives the restriction for a fixed count without requiring a formula for the full aggregate distribution.

Write each severity on the lattice.
For each claim $j$, write $X_j=a+k_jh$, where $k_j$ is an integer.

Add the severities.
For $n$ claims, $S=\sum_{j=1}^{n}(a+k_jh)=na+h\sum_{j=1}^{n}k_j$.

Apply the integer restriction.
Since $\sum_{j=1}^{n}k_j$ is an integer, $S$ must belong to $na+h\mathbb{Z}$. The severity support may impose additional restrictions within that lattice.

For exact probabilities, the same support logic can be combined with conditioning on the count:

P(S=s)=\sum_{n\geq 0}P(N=n)\,P\!\left(\sum_{j=1}^{n}X_j=s\mid N=n\right).

Terms for counts whose aggregate lattice does not contain $s$ are zero. Even for a count whose lattice does contain $s$, the conditional probability may still be zero because of gaps in the severity support. The examples show both checks in practice.

Worked Example: A Count Mixture with Integer Severity Values

A hypothetical portfolio has $P(N=0)=0.20$, $P(N=1)=0.30$, and $P(N=2)=0.50$. Conditional on the count, claim severities are independent, with $P(X=2)=0.40$ and $P(X=5)=0.60$. Find $P(S=7)$ and $P(S=8)$.

Method. First list the severity lattice and the aggregate values for each possible count. Then calculate the probability of the target value only for counts that can produce it.

The severities lie on the lattice $2+3\mathbb{Z}$. For zero claims, the aggregate is $0$. For one claim, the only aggregate values are $2$ and $5$. For two claims, the possible sums are

2+2=4,\qquad 2+5=7,\qquad 5+2=7,\qquad 5+5=10.

Thus $S=7$ can occur only when $N=2$, with one severity of each size. Independence gives

P(S=7)=P(N=2)\left[P(X_1=2,X_2=5)+P(X_1=5,X_2=2)\right] $$ $$ =0.50\left[(0.40)(0.60)+(0.60)(0.40)\right] =0.50(0.48)=0.24.

The value $8$ is not among the aggregate values for $N=0,1,$ or $2$. Therefore $P(S=8)=0$. Check: The two-claim probabilities sum to $0.16+0.48+0.36=1$; multiplying by $P(N=2)=0.50$ gives $0.50$, as required for that count.

Common mistake avoided: A candidate might use only the fact that the severity spacing is $3$ and overlook the offset. With two claims, the aggregate is congruent to $4$ modulo $3$, not $0$ modulo $3$. Listing the count-specific lattice, or adding the possible severity values, prevents that error.

Worked Example: A Shifted Lattice in Dollar Amounts

A hypothetical coverage has claim severities $X=\$1.25,\ \$2.00,$ or $\$2.75$, with probabilities $0.20,\ 0.50,$ and $0.30$, respectively. The count probabilities are $P(N=1)=0.20$, $P(N=2)=0.50$, and $P(N=3)=0.30$. Find $P(S=\$4.75)$.

Method. Express the severity values in quarter-dollar units to identify their lattice, then check which claim counts can produce the target. In those units the severities are $5,8,11$. Their differences are $3$ and $6$, so the spacing is $3$ quarter-dollar units, or $\$0.75$. The offset is $5$ quarter-dollar units, or $\$1.25$.

For $n$ claims, the aggregate belongs to the lattice with offset $n(\$1.25)$ and spacing $\$0.75$. The target $\$4.75$ is not on the one-claim lattice: its difference from $\$1.25$ is $\$3.50$, which is not an integer multiple of $\$0.75$. It is also not on the three-claim lattice: its difference from $\$3.75$ is $\$1.00$, again not a multiple of $\$0.75$. It is on the two-claim lattice because

\$4.75-\$2.50=\$2.25=3(\$0.75).

For two claims, a total of $\$4.75$ requires $\$2.00+\$2.75$. There are two possible orders, so

P(S=\$4.75) =P(N=2)\left[2(0.50)(0.30)\right] =0.50(0.30)=0.15.

Check: The conditional probability of the required two-claim combination is $0.30$, and the count probability is $0.50$; their product is $0.15$. Common mistake avoided: Reporting $0.30$ would omit the probability that $N=2$. A conditional probability given the count is not yet an unconditional aggregate probability.

Worked Example: A Lattice-Compatible Value That Is Impossible

A hypothetical portfolio has $P(N=1)=0.25$, $P(N=2)=0.50$, and $P(N=3)=0.25$. Independent severities have probabilities $P(X=1)=0.50$, $P(X=4)=0.30$, and $P(X=10)=0.20$. Determine whether $S=17$ is possible, and calculate $P(S=11)$.

Method. Use the lattice to narrow down the claim count, then check the actual sums of permitted severity values. The severities lie on $1+3\mathbb{Z}$, so an $n$-claim aggregate lies on $n+3\mathbb{Z}$.

The target $17$ is not on the one-claim lattice, since $17-1=16$ is not divisible by $3$. It is not on the three-claim lattice, since $17-3=14$ is not divisible by $3$. It is on the two-claim lattice, since $17-2=15$ is divisible by $3$. The lattice test alone therefore leaves $N=2$ as a candidate. But the possible two-claim sums are

1+1=2,\quad 1+4=5,\quad 1+10=11,\quad 4+4=8,\quad 4+10=14,\quad 10+10=20,

including the corresponding reversed orders for unequal severities. None equals $17$, so $P(S=17)=0$. This is a lattice-compatible value that the severity support does not permit.

For $S=11$, only $N=2$ can produce the total: the required severities are $1$ and $10$, in either order. Hence

P(S=11)=P(N=2)\left[2P(X=1)P(X=10)\right] =0.50\left[2(0.50)(0.20)\right] =0.10.

Check: The conditional probability of one severity of each size is $0.20$; multiplying by $P(N=2)=0.50$ gives $0.10$. Common mistake avoided: Concluding that $17$ is possible just because it belongs to the two-claim lattice. A lattice identifies a containing set, not necessarily the exact support.

Using Lattices Efficiently

Lattice reasoning is most useful as a first step in a probability calculation. It can identify which claim counts deserve further attention, show immediately that a proposed value has probability zero, and flag a unit-conversion error. For example, if severities are measured in cents, a support on multiples of $25$ cents gives aggregate totals in multiples of $25$ cents when the offset is zero. A proposed total of $\$12.13$ is then impossible, regardless of the count probabilities.

A reliable calculation separates three questions:

Does the target belong to the count-specific lattice? If not, that count contributes zero probability.
Can the permitted severity values actually sum to the target? A lattice match can still leave a gap in the support.
What is the probability of the required count and severity combination? Include the count probability when calculating an unconditional probability.

Keep the offset as well as the spacing. Saying that an aggregate is a multiple of the spacing is correct only when the lattice has zero offset. And when amounts are rescaled, rescale both the spacing and the offset: converting dollars to cents changes their numerical values, but not which outcomes are possible.

Key takeaway: If severities lie on $a+h\mathbb{Z}$, then for $N=n$ the aggregate lies on $na+h\mathbb{Z}$. Across a random count, take the union of those count-specific restrictions, including $S=0$ for no claims. Then check the actual severity support: lattice membership is necessary, but not always sufficient, for a positive aggregate probability.

Check Your Understanding

For each problem, use the support restriction first, then use the stated probabilities if an exact probability is requested.

Two claims are certain. Each severity is either $3$ or $7$. What are the possible aggregate values, and what spacing and offset describe the severity lattice? (Answer: Aggregate values $6,10,14$; severity spacing $4$, offset $3$.)
Severity values are $\$0.50,\ \$1.50,$ and $\$2.50$, and the claim count can be $0,1,$ or $2$. State the aggregate lattice for each count. (Answer: $N=0:\ S=0$; $N=1:\ \$0.50+\$1.00\mathbb{Z}$; $N=2:\ \$1.00+\$1.00\mathbb{Z}$.)
Let $P(N=0)=0.10$, $P(N=1)=0.40$, and $P(N=2)=0.50$. Independent severities are $2$ with probability $0.25$ and $6$ with probability $0.75$. Find $P(S=8)$. (Answer: $0.1875$.)
Severities can be $1,4,$ or $10$, and $N$ can be $1,2,$ or $3$. Does the lattice test alone rule out $S=17$, and is $17$ actually possible? (Answer: The lattice test does not rule it out because it is on the two-claim lattice; it is actually impossible because no two permitted severities sum to $17$.)
Let $P(N=1)=0.20$, $P(N=2)=0.50$, and $P(N=3)=0.30$. Independent severities are $1,4,$ or $10$, with probabilities $0.50,0.30,$ and $0.20$, respectively. Find $P(S=8)$. (Answer: $0.0450$.)

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