Convolution and recursive formulas for aggregate claims

Exam ASTAM – Advanced Short-Term Actuarial Mathematics · SOA (Society of Actuaries) · 6 min read · Updated: 8 Nov 2026

Welcome to the World of Aggregate Claims!

Hi there! If you've made it to Exam ASTAM, you already know that insurance companies don't just care about one single claim—they care about the total amount they have to pay out over a period. This total amount is called the Aggregate Claim, usually denoted by $S$.

In this chapter, we are going to learn how to calculate the probability distribution of $S$. Why does this matter? Because if an actuary can't predict the likelihood of total claims exceeding a certain amount, the company might not have enough money to pay them! We will focus on two main toolkits: Convolution (for when the number of claims is fixed) and Recursive Formulas (Panjer's recursion and its extension to the $(a, b, 0)$ and $(a, b, 1)$ classes, for when the number of claims is a random variable).

Don't worry if this seems heavy on the math at first. We’ll break it down into bite-sized pieces!

1. The Building Blocks: What is $S$?

Before we dive into the formulas, let's remember our core collective risk model:

$S = X_1 + X_2 + \dots + X_N$

Where:
• $N$ is the Frequency (the number of claims).
• $X_i$ is the Severity (the cost of the $i$-th claim).
• $S$ is the Aggregate Loss (the total pile of money paid out, with $S = 0$ when $N = 0$).

In this chapter, we assume that all $X_i$ are independent and identically distributed (i.i.d.) non-negative random variables and that they are also independent of $N$.

2. The Convolution Method

Imagine you know for a fact that exactly 2 claims occurred ($N=2$). To find the probability that the total cost is \$100, you have to look at all the ways two non-negative numbers can add up to 100 (e.g., 20+80, 50+50, 0+100).

Convolution is the formal mathematical method of summing over all combinations of independent random variables.

How it works:

If we define $f_X(x)$ as the probability mass function (pmf) of a discrete severity $X$, then the pmf of the sum of $n$ claims, denoted by the $n$-fold convolution $f_X^{*n}(s)$, is calculated as:

$f_X^{*n}(s) = \sum_{x=0}^{s} f_X(x) \cdot f_X^{*(n-1)}(s-x)$

Step-by-Step Breakdown:

1. Start with $n=0$ and $n=1$: By definition, $f_X^{*0}(0) = 1$ (and 0 otherwise), while $f_X^{*1}(s) = f_X(s)$.
2. Move to $n=2$: Calculate the 2-fold convolution: $f_X^{*2}(s) = \sum_{x=0}^{s} f_X(x) f_X(s-x)$.
3. Keep going: For $n=3$, convolve $f_X(x)$ with the result of $f_X^{*2}$.
4. Aggregate distribution: If $N$ is random, the overall pmf is $f_S(s) = \sum_{n=0}^{\infty} p_n f_X^{*n}(s)$, and the cumulative distribution function is $F_S(s) = \sum_{n=0}^{\infty} p_n F_X^{*n}(s)$.

Key Takeaway: Convolution is conceptually straightforward, but evaluating it for compound models with unbounded $N$ requires summing infinite convolutions—which is computationally impractical by hand.

3. Panjer's Recursive Formula: The $(a, b, 0)$ Class

To bypass computing endless convolutions, Harry Panjer introduced an exact recursive relationship when the frequency $N$ belongs to the $(a, b, 0)$ class.

The $(a, b, 0)$ Relationship

Recall that for distributions in the $(a, b, 0)$ family, probabilities satisfy:
$p_k = \left(a + \frac{b}{k}\right) p_{k-1}, \quad k = 1, 2, 3, \dots$

The three member distributions are:
• Poisson($\lambda$): $a = 0$, $b = \lambda$, $p_0 = e^{-\lambda}$
• Binomial($m, q$): $a = -\frac{q}{1-q} < 0$, $b = \frac{(m+1)q}{1-q}$, $p_0 = (1-q)^m$
• Negative Binomial($r, \beta$): $a = \frac{\beta}{1+\beta} > 0$, $b = \frac{(r-1)\beta}{1+\beta}$, $p_0 = (1+\beta)^{-r}$

The $(a, b, 0)$ Recursive Formula

For a discrete severity distribution with pmf $f_X(x)$ defined on non-negative integers $x = 0, 1, 2, \dots$, the aggregate claims pmf $f_S(s)$ satisfies:

$f_S(s) = \frac{1}{1 - a f_X(0)} \sum_{x=1}^{s} \left(a + \frac{b x}{s}\right) f_X(x) f_S(s-x), \quad s = 1, 2, 3, \dots$

Starting Value: $f_S(0)$

To initialize the recursion, we need $f_S(0) = P(S = 0)$. Total claims equal zero if no claims occur, or if all claims that do occur have severity zero:
$f_S(0) = P_N(f_X(0))$

Where $P_N(z) = E[z^N]$ is the probability generating function (pgf) of $N$.
If $f_X(0) = 0$ (claims are strictly positive), this simplifies nicely to $f_S(0) = P_N(0) = p_0$, and the factor $\frac{1}{1 - a f_X(0)}$ becomes 1.

4. Extending the Recursion: The $(a, b, 1)$ Class

In Exam ASTAM, aggregate models frequently feature zero-truncated or zero-modified frequency distributions. These form the $(a, b, 1)$ class, where the recursion starts at $k = 2$:

$p_k = \left(a + \frac{b}{k}\right) p_{k-1}, \quad k = 2, 3, 4, \dots$

Because the relationship does not hold from $k=0$ to $k=1$, an adjustment term appears at step $x=s$.

The $(a, b, 1)$ Recursive Formula

For frequency $N$ in the $(a, b, 1)$ class with discrete severity on $x = 0, 1, 2, \dots$, the recursion for $s = 1, 2, 3, \dots$ is:

$f_S(s) = \frac{\left[p_1 - (a + b) p_0\right] f_X(s) + \sum_{x=1}^{s} \left(a + \frac{b x}{s}\right) f_X(x) f_S(s-x)}{1 - a f_X(0)}$

Notice the connection: When $N$ is in the $(a, b, 0)$ class, $p_1 = (a + b) p_0$, so the bracketed term $\left[p_1 - (a+b)p_0\right]$ drops to zero, perfectly matching the original Panjer formula!

Starting Value for $(a, b, 1)$:

As before, $f_S(0) = P_N(f_X(0))$. If $f_X(0) = 0$, then $f_S(0) = p_0$. For a zero-truncated distribution, $p_0 = 0$, so $f_S(0) = 0$.

Deriving the Distribution Function $F_S(s)$

Once you calculate $f_S(0), f_S(1), \dots, f_S(s)$, the cumulative distribution function is readily obtained by accumulating the probabilities:
$F_S(s) = \sum_{y=0}^{s} f_S(y)$

5. Step-by-Step Recursive Process & Continuous Severities

How to Apply the Recursion on the Exam:

1. Classify the frequency: Determine whether $N$ is $(a, b, 0)$ or $(a, b, 1)$ and extract $a$, $b$, $p_0$, and $p_1$.
2. Find $f_S(0)$: Calculate $P_N(f_X(0))$.
3. Compute $f_S(1)$: Apply the formula with $s=1$ (uses $f_S(0)$).
4. Compute $f_S(2), f_S(3), \dots$: Each step uses the previously computed $f_S(s-x)$ values.
5. Accumulate $F_S(s)$: Add the calculated values if the question asks for a cumulative probability $P(S \le s)$ or a stop-loss premium.

What if Severity is Continuous?

Recursive formulas require a discrete severity distribution. When the true severity is continuous, we first discretize it into a grid with step size $h$ using either:
• Method of Rounding: Assign probability mass to grid point $jh$ based on the interval around it: $f_j = F_X(jh + h/2) - F_X(jh - h/2)$ for $j \ge 1$, and $f_0 = F_X(h/2)$.
• Method of Local Moment Matching: Match moments over sub-intervals so that local mean and/or variance are preserved.

6. Quick Review & Exam Traps

Watch the Index: In both Panjer formulas, the summation is from $x=1$ to $s$, never from $x=0$. The $x=0$ piece is accounted for by the denominator $1 - a f_X(0)$.
Weighting Term: The weight inside the summation is always $\left(a + \frac{b x}{s}\right)$. Notice that $x$ is the current severity index and $s$ is the aggregate loss index.
Zero-Modified vs. Zero-Truncated: For $(a, b, 1)$ models, do not forget the extra term $\left[p_1 - (a+b)p_0\right] f_X(s)$. If the frequency is zero-truncated, $p_0 = 0$, and the term simplifies to $p_1 f_X(s)$.

References

SOA Exam ASTAM Syllabus, Spring 2026 — Society of Actuaries

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the collective risk model representation for aggregate claims $S$?

$S = X_1 + X_2 + \dots + X_N$ (with $S = 0$ when $N = 0$), where $N$ is claim frequency and $X_i$ are i.i.d. non-negative severities independent of $N$.

How is the $n$-fold convolution $f_X^{*n}(s)$ defined recursively for discrete severity $X$?

$f_X^{*n}(s) = \sum_{x=0}^{s} f_X(x) f_X^{*(n-1)}(s-x)$, initialized with $f_X^{*0}(0) = 1$ and $f_X^{*1}(s) = f_X(s)$.

What defining recurrence relation characterizes the $(a, b, 0)$ class of frequency distributions?

$p_k = \left(a + \frac{b}{k}\right) p_{k-1}$ for all $k = 1, 2, 3, \dots$

What is Panjer's recursive formula for aggregate claims when $N$ belongs to the $(a, b, 0)$ class?

$f_S(s) = \frac{1}{1 - a f_X(0)} \sum_{x=1}^{s} \left(a + \frac{b x}{s}\right) f_X(x) f_S(s-x)$ for $s = 1, 2, 3, \dots$

How is the starting value $f_S(0)$ initialized in Panjer's recursion?

$f_S(0) = P_N(f_X(0))$, which simplifies to $f_S(0) = p_0$ when $f_X(0) = 0$.

What recursive formula computes $f_S(s)$ when frequency $N$ belongs to the $(a, b, 1)$ class?

$f_S(s) = \frac{[p_1 - (a+b)p_0] f_X(s) + \sum_{x=1}^{s} \left(a + \frac{b x}{s}\right) f_X(x) f_S(s-x)}{1 - a f_X(0)}$ for $s \ge 1$.

How is a continuous severity distribution discretized using the Method of Rounding for use in recursive formulas?

Over grid step $h$, set $f_0 = F_X(h/2)$ and $f_j = F_X(jh + h/2) - F_X(jh - h/2)$ for $j \ge 1$.

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Convolution and recursive formulas for aggregate claims

Welcome to the World of Aggregate Claims!

1. The Building Blocks: What is \(S\)?

2. The Convolution Method

How it works:

Step-by-Step Breakdown:

3. Panjer's Recursive Formula: The \((a, b, 0)\) Class

The \((a, b, 0)\) Relationship

The \((a, b, 0)\) Recursive Formula

Starting Value: \(f_S(0)\)

4. Extending the Recursion: The \((a, b, 1)\) Class

The \((a, b, 1)\) Recursive Formula

Starting Value for \((a, b, 1)\):

Deriving the Distribution Function \(F_S(s)\)