Panjer Recursion

Study guide: Exam ASTAM

Panjer Recursion computes the distribution of aggregate claims $S = X_1 + \cdots + X_N$ one value at a time, when the claim count $N$ is in the (a,b,0) or (a,b,1) class of frequency distributions and the severity $X$ is arithmetic, on the integers $0, 1, 2, \ldots$ in some monetary unit. Each probability $f_S(x)$ is a short weighted sum of the ones before it, so the recursion replaces the many convolutions of the collective risk model with a single pass.

$f_S(0) = P_N\big(f_X(0)\big)$

$f_S(x) = \frac{1}{1 - a f_X(0)} \sum_{y=1}^{x} \left(a + \frac{b\,y}{x}\right) f_X(y)\, f_S(x-y)$

$\begin{aligned} f_S(x) = \frac{1}{1 - a f_X(0)} \Big\{ & \big[p_1 - (a+b)p_0\big] f_X(x) \\ & + \sum_{y=1}^{x} \left(a + \tfrac{b\,y}{x}\right) f_X(y)\, f_S(x-y) \Big\} \end{aligned}$

Symbols: Following Loss Models (chapter 9), $f_X$ and $f_S$ are the probability functions of the severity and of aggregate claims, $p_k = \Pr(N = k)$ , and $a$ and $b$ are the constants of the frequency distribution, so that $p_k = (a + b/k)\,p_{k-1}$ . $P_N(z) = E[z^N]$ is the probability generating function of $N$ . The second block is the (a,b,0) recursion and the third is the (a,b,1) recursion.
Starting value: The recursion must start from $f_S(0)$ , the probability that every claim is zero, which is $\sum_k p_k\,f_X(0)^k = P_N(f_X(0))$ . When $f_X(0) = 0$ it is simply $p_0$ , and for a Poisson $N$ it is $e^{-\lambda(1 - f_X(0))}$ .
The (a,b,1) version: A zero-truncated or zero-modified frequency follows the relation $p_k = (a + b/k)\,p_{k-1}$ only from $k = 2$ , so its $p_1$ is not $(a+b)p_0$ . The extra term $[p_1 - (a+b)p_0]\,f_X(x)$ corrects for this, and it vanishes for an (a,b,0) distribution, which turns the third block back into the second.
Compound Poisson: With $a = 0$ and $b = \lambda$ , the recursion is $f_S(x) = \frac{\lambda}{x}\sum_{y} y\,f_X(y)\,f_S(x-y)$ , with no denominator. A sum of independent compound Poisson variables is again compound Poisson, with $\lambda = \sum_j \lambda_j$ and a severity that mixes the $f_j$ with weights $\lambda_j/\lambda$ , so a whole book can be run through one recursion. A continuous severity must first be put on a lattice by discretization.

Compound Poisson Aggregate Claims The number of claims is Poisson with $\lambda = 2$ . Each claim is 1, 2 or 3 (in $1,000s) with probabilities 0.5, 0.3 and 0.2. Find $\Pr(S \le 3)$ .
Answer For the Poisson, $a = 0$ and $b = 2$ , and $f_X(0) = 0$ , so $f_S(0) = e^{-2}$ and each step is $f_S(x) = \frac{2}{x}\sum_y y\,f_X(y)\,f_S(x-y)$ .
 $\begin{align*} f_S(0) &= e^{-2} \\ &= 0.135335 \\ f_S(1) &= 2(1)(0.5)(0.135335) \\ &= 0.135335 \\ f_S(2) &= \tfrac{2}{2}\big[0.5(0.135335) + 2(0.3)(0.135335)\big] \\ &= 0.148869 \\ f_S(3) &= \tfrac{2}{3}\big[0.5(0.148869) + 0.6(0.135335) + 0.6(0.135335)\big] \\ &= 0.157891 \end{align*}$ 
Adding the four probabilities gives $\Pr(S \le 3) = 0.577430$ . Aggregate claims stay at or below $3,000 in about 58% of years, and convolving the severity up to three times gives the same answer.

Zero-Modified Frequency With a Deductible The number of losses is zero-modified geometric with $\beta = 1$ and $p_0^M = 0.4$ . After a per-loss deductible, the payment on a loss is 0, 1 or 2 with probabilities 0.2, 0.5 and 0.3. Find the probability that aggregate payments are at most 2.
Answer For the geometric, $a = \beta/(1+\beta) = 0.5$ and $b = 0$ . The zero-modified probabilities are $p_k^M = 0.6\,p_k^T$ , so $p_1^M = 0.6(0.5) = 0.3$ , and the (a,b,1) term is $p_1 - (a+b)p_0 = 0.3 - 0.5(0.4) = 0.1$ . The denominator is $1 - 0.5(0.2) = 0.9$ . The starting value uses the zero-truncated generating function at $f_X(0) = 0.2$ , which is $P^T(0.2) = (1/1.8 - 0.5)/0.5 = 0.111111$ .
 $\begin{align*} f_S(0) &= 0.4 + 0.6(0.111111) \\ &= 0.466667 \\ f_S(1) &= \frac{0.1(0.5) + 0.5(0.5)(0.466667)}{0.9} \\ &= 0.185185 \\ f_S(2) &= \frac{0.1(0.3) + 0.5\big[0.5(0.185185) + 0.3(0.466667)\big]}{0.9} \\ &= 0.162551 \end{align*}$ 
Adding them gives $\Pr(S \le 2) = 0.814403$ . Starting from $p_0^M = 0.4$ would leave out the years in which every loss falls below the deductible, and dropping the term $0.1\,f_X(x)$ would treat the frequency as an ordinary geometric.

How it appears on the exam

ASTAM asks candidates to use convolution and recursive formulas to derive probability and distribution functions for aggregate claims with (a,b,0) or (a,b,1) frequency and a discrete severity. A question gives the frequency's parameters and a severity on a few points, often after a per-loss deductible, and asks for $f_S(x)$ or $F_S(x)$ for small $x$ .

Stop-loss premiums: The first few probabilities often feed a stop-loss premium, through $E[(S-d)_+] = E[S] - d + \sum_{x<d}(d-x)\,f_S(x)$ .

Continuous severity: When the question gives a continuous severity and a span, the severity must be discretized first, and the answer is then in units of the span.

Exam traps

The denominator: When $f_X(0) > 0$ , every step divides by $1 - a f_X(0)$ . The Poisson has $a = 0$ , which hides the factor, so it is easy to forget for the negative binomial and the binomial. For the binomial, $a = -q/(1-q)$ is negative, and the denominator is greater than 1.

The starting value: With a mass at zero in the severity, $f_S(0) = P_N(f_X(0))$ , not $p_0$ . Starting from $p_0$ leaves out every year in which claims occur but none of them pays.

The (a,b,1) term: For a zero-truncated or zero-modified frequency, leaving out $[p_1 - (a+b)p_0]\,f_X(x)$ gives the answer for the wrong frequency. The term uses the modified $p_0^M$ and $p_1^M$ , not the probabilities of the corresponding (a,b,0) distribution.

The weights: The weight is $a + by/x$ , where $y$ is the claim amount and $x$ is the total. Swapping them, or writing $b/x$ without the factor $y$ , gives wrong probabilities from $f_S(2)$ onward.