Compound Poisson Aggregate Probabilities by Recursion

Why Use a Recursion?

For an aggregate loss, the number of claims is random and each claim has a random severity. When claim counts follow a Poisson distribution and severities lie on a discrete grid, the aggregate probabilities can be calculated successively rather than by separately listing every possible claim count and combination of severities. This is useful when a calculation needs probabilities at many aggregate amounts: each new value uses probabilities already found at smaller amounts.

The recursion gives probabilities for the discrete model being used. If a continuous severity distribution was first rounded or discretized onto a grid, the recursion is exact for that grid model; the result may only approximate the original continuous model. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Model and Notation

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$. Given $N$, let $X_1,\ldots,X_N$ be independent, identically distributed claim severities, independent of $N$. Assume each severity is strictly positive and lies on a grid with spacing $h>0$: $P(X=jh)=p_j$ for integers $j\geq 1$, where $\sum_{j\geq 1}p_j=1$. The aggregate loss is $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. Define $f_k=P(S=kh)$ for integers $k\geq 0$.

The index $k$ counts grid units, not dollars unless $h=\$1$. For example, if $h=\$500$, then $f_3$ is the probability of an aggregate loss of $\$1{,}500$. The recursion requires severity probabilities $p_j$ on the same grid as the aggregate amounts. Strictly positive severity is an important assumption: it ensures that zero aggregate loss occurs exactly when there are no claims.

Theorem: Under the model above, the zero-loss probability and the positive-grid aggregate probabilities satisfy $$ f_0=e^{-\lambda}, \qquad f_k=\frac{\lambda}{k}\sum_{j=1}^{k}j p_j f_{k-j},\quad k\geq 1. $$ A severity probability $p_j$ is taken to be zero if that grid point has no probability mass.

Why the Recursion Holds

A short probability generating function argument explains both the formula and why the probabilities can be computed in order. Let $G_X(z)=\sum_{j\geq 1}p_jz^j$ be the probability generating function for severity measured in grid units, and let $G_S(z)=\sum_{k\geq 0}f_kz^k$. Conditioning on the Poisson claim count gives

G_S(z)=\exp\!\bigl(\lambda(G_X(z)-1)\bigr).

Indeed, given $N=n$, the generating function of the sum is $G_X(z)^n$. Averaging over $N$ gives $\sum_{n\geq 0}e^{-\lambda}\lambda^nG_X(z)^n/n!=\exp(\lambda(G_X(z)-1))$. Differentiating this identity gives $G'_S(z)=\lambda G'_X(z)G_S(z)$. Now $G'_S(z)=\sum_{k\geq 1}k f_kz^{k-1}$ and $G'_X(z)=\sum_{j\geq 1}j p_jz^{j-1}$. The coefficient of $z^{k-1}$ on the right is $\lambda\sum_{j=1}^k j p_j f_{k-j}$. Equating coefficients proves the recursion.

For $k=0$, the model assumption gives a direct argument: $S=0$ if and only if $N=0$, because every claim has positive severity. Thus $f_0=P(N=0)=e^{-\lambda}$. In the recursion for $f_k$, only $f_0,\ldots,f_{k-1}$ occur, so start with $f_0$ and proceed upward.

Using the Recursion in Order

Set the grid and identify the masses.
Confirm the grid spacing $h$, the Poisson parameter $\lambda$, and each $p_j$. Check that the severity probabilities sum to one.

Initialize at zero.
For strictly positive severities, calculate $f_0=e^{-\lambda}$.

Calculate each requested grid point.
For $k=1,2,\ldots$, use the recursion and previously calculated aggregate probabilities. Include only severity indices $j\leq k$.

Answer the event asked for.
An exact aggregate amount uses one $f_k$; a cumulative event such as $S\leq kh$ uses $\sum_{r=0}^k f_r$.

Worked Calculations

Worked Example: A Two-Point Severity and a Cumulative Probability

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims. Claim severities are independent and have $P(X=h)=0.5$ and $P(X=2h)=0.5$, where $h=\$1{,}000$. Calculate $P(S\leq \$2{,}000)$. State the method and avoid treating this cumulative probability as just the probability of exactly $\$2{,}000$.

Method. Use the compound Poisson recursion with $\lambda=1$, $p_1=p_2=0.5$, and $p_j=0$ for $j>2$. First, $f_0=e^{-1}=0.3678794$.

For $k=1$, only the $j=1$ term contributes: $f_1=(1/1)(1)(0.5)f_0=0.5f_0=0.1839397$.

For $k=2$, $f_2=(1/2)\{(1)(0.5)f_1+(2)(0.5)f_0\}$. The terms inside braces are $0.0919699$ and $0.3678794$, so $f_2=0.2299247$.

The requested event includes zero, one, or two grid units. Therefore $P(S\leq \$2{,}000)=f_0+f_1+f_2=0.3678794+0.1839397+0.2299247=0.7817438$. Rounded to four decimals, the answer is 0.7817.

Check and common mistake. The three probabilities are nonnegative and their sum is at most one. Using $f_2$ alone would answer $P(S=\$2{,}000)$, not the cumulative event.

Worked Example: A Missing Severity Grid Point

For a hypothetical coverage, $N\sim\operatorname{Poisson}(2)$. Each claim is $\$1{,}000$ with probability $0.25$ and $\$3{,}000$ with probability $0.75$; no claim has severity $\$2{,}000$. Calculate $P(S\leq \$3{,}000)$. The grid spacing is $h=\$1{,}000$.

Method. Use $\lambda=2$, $p_1=0.25$, $p_2=0$, $p_3=0.75$, and $f_0=e^{-2}=0.1353353$. Keeping the grid point $j=2$ in the model with probability zero prevents skipping a grid index.

For $k=1$, $f_1=(2/1)(1)(0.25)f_0=0.5f_0=0.0676676$.

For $k=2$, the $j=2$ term is zero, so $f_2=(2/2)\{(1)(0.25)f_1+(2)(0)f_0\}=0.25f_1=0.125f_0=0.0169169$.

For $k=3$, $f_3=(2/3)\{(1)(0.25)f_2+(2)(0)f_1+(3)(0.75)f_0\}$. Substituting $f_2=0.125f_0$, the expression in braces is $0.03125f_0+2.25f_0=2.28125f_0$. Hence $f_3=1.5208333f_0=0.2058224$.

The cumulative probability is $f_0+f_1+f_2+f_3=0.4257422$, or 0.4257 to four decimals.

Check and common mistake. There is no severity mass at grid point 2, but aggregate loss of two grid units is still possible from two claims of one unit; accordingly, $f_2$ is not zero. Setting $f_2=0$ merely because $p_2=0$ would omit a possible combination of claims.

Worked Example: Several Severity Masses and a Threshold

A hypothetical group has $N\sim\operatorname{Poisson}(1.5)$ claims. Each claim is $h=\$500$, $2h$, or $4h$, with probabilities $0.2$, $0.3$, and $0.5$, respectively. Calculate $P(S\leq \$2{,}000)$, and also report the probability of exactly $\$2{,}000$.

Method. Work in units of $h=\$500$, so the threshold $\$2{,}000$ is grid index $k=4$. Set $p_1=0.2$, $p_2=0.3$, $p_3=0$, and $p_4=0.5$. Begin with $f_0=e^{-1.5}=0.2231302$.

For clarity, write $f_k=c_ke^{-1.5}$. Then $c_0=1$. For $k=1$, $c_1=1.5(0.2)c_0=0.3$. For $k=2$, $c_2=(1.5/2)\{(1)(0.2)c_1+(2)(0.3)c_0\}=0.75(0.06+0.6)=0.495$.

For $k=3$, $c_3=(1.5/3)\{(1)(0.2)c_2+(2)(0.3)c_1+(3)(0)c_0\}$. Thus $c_3=0.5(0.099+0.18)=0.1395$.

For $k=4$, $c_4=(1.5/4)\{(1)(0.2)c_3+(2)(0.3)c_2+(3)(0)c_1+(4)(0.5)c_0\}$. The terms inside braces are $0.0279$, $0.297$, $0$, and $2$, giving $c_4=0.375(2.3249)=0.8718375$.

Therefore $P(S=\$2{,}000)=f_4=0.8718375e^{-1.5}=0.1945332$, or 0.1945 to four decimals. For the cumulative event, sum all five probabilities from zero through four grid units: $P(S\leq \$2{,}000)=(1+0.3+0.495+0.1395+0.8718375)e^{-1.5}=0.6261785$, or 0.6262 to four decimals.

Check and common mistake. The threshold includes every aggregate amount from $\$0$ to $\$2{,}000$, not only amounts that are individual claim severities. Also, the factor $j$ in the recursion is the grid index of a severity; omitting it would give incorrect probabilities.

Interpretation and Exam Checks

The recursion is particularly efficient for a requested range of aggregate amounts: to obtain $f_k$, retain the earlier $f_r$ values and use only severity masses through index $k$. A mass $p_j$ with $j>k$ cannot contribute to $S=kh$, because a single claim of that size already exceeds the target. This does not mean that $f_k$ is zero when $p_k=0$: several smaller claims may sum to $kh$.

Keep the probability meanings separate. $p_j$ describes one claim, while $f_k$ describes the sum of a random number of claims. The zero-loss probability is initialized from the frequency model, not from the severity distribution: under the positive-severity assumption, it is $e^{-\lambda}$. And a cumulative probability must include all grid points up to its threshold.

Check that $\lambda>0$ and that the severity masses are nonnegative and sum to one.
Keep the severity grid and aggregate grid in the same units.
Use $f_0=e^{-\lambda}$ before applying the recursion.
Include zero-probability severity grid points when they lie between positive-mass points; aggregate probabilities at those indices can still be positive.
For $P(S\leq kh)$, sum $f_0,\ldots,f_k$; for $P(S=kh)$, use $f_k$ alone.

Key takeaway: For Poisson frequency and strictly positive discrete severity, $f_0=e^{-\lambda}$, and each later aggregate probability is a weighted sum of earlier probabilities: $f_k=(\lambda/k)\sum_{j=1}^k j p_j f_{k-j}$. The grid units, positive-severity assumption, and distinction between exact and cumulative probabilities are essential.

Check Your Understanding

For each problem, use the compound Poisson recursion and retain full precision until the final probability, which should be rounded to four decimals.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.8)$, with each claim equal to $h=\$1{,}000$ or $2h$, each with probability $0.5$. Find $P(S=0)$ and $P(S=\$1{,}000)$. (Answer: $P(S=0)=0.4493$; $P(S=\$1{,}000)=0.1797$)
Let $N\sim\operatorname{Poisson}(1)$, and let claim severity be $h$ with probability $0.25$ and $2h$ with probability $0.75$, where $h=\$500$. Find $P(S\leq \$1{,}000)$. (Answer: 0.7473)
A hypothetical model has $N\sim\operatorname{Poisson}(1.2)$, with severity $h=\$1{,}000$ with probability $0.2$ and $3h$ with probability $0.8$. Find $P(S\leq \$3{,}000)$. (Answer: 0.6720)
For $N\sim\operatorname{Poisson}(2)$, suppose severities are $h=\$1{,}000$ with probability $0.4$ and $2h$ with probability $0.6$. Find $P(S\leq \$2{,}000)$, including the zero-loss probability. (Answer: 0.4493)
Claims follow $N\sim\operatorname{Poisson}(0.5)$, and each has severity $\$500$ with probability $0.6$ or $\$1{,}000$ with probability $0.4$. Find $P(S\leq \$500)$. (Answer: 0.7885)

What You'll Learn