Panjer Recursion

As I mentioned in my post on the Cramer-Lundberg model, I’ve been trying to teach myself some insurance maths. One of the books I’ve been consulting is David Dickson’s Insurance Risk and Ruin. One of the sections was on computationally useful formulae for distribution of the aggregate claims process

\displaystyle S(t)=\sum_{j=1}^{N(t)}X_{j},\indent\forall t\geq 0

where X_{j} are the i.i.d. individual claims. The formula I want to share with you today is the Panjer recursion formula due to Harry Panjer. Given a “suitable” claims number process N(t) and nonnegative-integer-distributed claims X_{j}, the probability mass function of the compound process S(t) can be computed recursively.

We begin by specifying what “suitable” above mathematically means. A discrete random variable N is said to belong to belong to the (a,b,0) class if its probability mass function \left\{p_{n}\right\}_{n=0}^{\infty} can be calculated recursively by

\displaystyle p_{n}=\left(a+\dfrac{b}{n}\right)p_{n-1},\indent\forall n\geq 1

where a,b are fixed constants. We will show that there are precisely three distributions in the (a,b,0) class: the Poisson, the binomial, and the negative binomial (where we allow noninteger parameters) distributions.

We assume that p_{0} > 0, which is indicated by the 0 in (a,b,0). Observe that p_{1}=(a+b)p_{0}, so it is necessary that a+b\geq 0. We proceed by examining the cases for a+b. If a+b=0, then p_{1}=0 and hence p_{n}=0 for all n\geq 1 and has a degenerate distribution at 0. If a=0, then p_{n}=\frac{b}{n}p_{n-1}, so that

\displaystyle p_{n}=\dfrac{b^{n}}{n!}p_{0},\indent\forall n\geq 0

Since \sum_{n=0}^{\infty}p_{n}=1, we obtain that

\displaystyle1=\sum_{n=0}^{\infty}p_{n}=p_{0}\sum_{n=0}^{\infty}\dfrac{b^{n}}{n!}=p_{0}e^{b}\Longrightarrow p_{0}=e^{-b}

We conclude that, if a = 0, then \left\{p_{n}\right\}_{n=0}^{\infty} is a Poisson distribution with parameter b.

Lastly, consider the case a>0 and a+b>0. Repeatedly applying the recursion formula, we obtain the identity

\begin{array}{lcl}\displaystyle p_{n}&=&\left(a+\dfrac{b}{n}\right)\left(a+\dfrac{b}{n-1}\right)\cdots\left(a+\dfrac{b}{2}\right)\left(a+b\right)p_{0}\\[.7 em]&=&\displaystyle\dfrac{a^{n}}{n!}\left(n+\dfrac{b}{a}\right)\left(n-1+\dfrac{b}{a}\right)\cdots\left(2+\dfrac{b}{a}\right)\left(1+\dfrac{b}{a}\right)p_{0}\end{array}

Set \alpha := 1+\frac{b}{a}>0, so that

\displaystyle p_{n}={{\alpha+n-1}\choose n}a^{n}p_{0}

I claim that a < 1. By the ratio test, it suffices to show a \neq 1.

\displaystyle 1 > \sum_{n=1}^{\infty}p_{n}>\sum_{n=1}^{\infty}\dfrac{a^{n}(n-1)!}{n!}\alpha p_{0}=\alpha\sum_{n=1}^{\infty}\dfrac{a^{n}}{n}p_{0}=\alpha p_{0}\sum_{n=1}^{\infty}\dfrac{1}{n},

which is a contradiction since the harmonic series diverges. Recall that a negative binomial distribution \text{NBin}(k,p) satisfies

\displaystyle p^{k}\sum_{n=1}^{\infty}{{k+n-1}\choose n}q^{n}=1-p^{k},\indent p+q=1

where k \in (0,\infty). Since 1+\sum_{n=1}^{\infty}{{\alpha+n+1}\choose n}a^{n}=(1-a)^{-\alpha}, for all \left|a\right|<1, we see that p_{0}=(1-a)^{\alpha}, so that N\sim\text{NBin}(\alpha,1-a).

Lastly, consider the case where a+b>0 and a<0. I claim that there exists a positive integer \kappa such that a+\frac{b}{\kappa+1}=0, so that p_{n}=0 for n\geq \kappa+1. Indeed, since \frac{b}{n} \rightarrow 0 and b > 0, a+\frac{b}{n}\leq 0 for all sufficiently large n. If such a \kappa does not exist, then choosing n to be minimal such that a+\frac{b}{n}<0, we obtain p_{n}<0, which is a contradiction. We can write


so that

\begin{array}{lcl}\displaystyle p_{n}&=&\dfrac{a^{n}}{n!}\left(n+\dfrac{b}{a}\right)\left(n-1+\dfrac{n}{a}\right)\cdots\left(2+\dfrac{b}{a}\right)\left(1+\dfrac{b}{a}\right)p_{0}\\[.7 em]&=&\displaystyle\dfrac{a^{n}}{n!}\left(-\kappa+n-1\right)\left(-\kappa+n-2\right)\cdots\left(-\kappa+1\right)(-\kappa)p_{0}\\[.7 em]&=&\displaystyle (-1)^{n}\dfrac{a^{n}}{n!}\left(n-1-\kappa\right)\left(n-2-\kappa\right)\cdots\left(1-\kappa\right)\kappa p_{0}\\[.7 em]&=&\displaystyle{\kappa\choose{n}}(-a)^{n}p_{0}\end{array}

Set A:=-a>0. Since \sum_{n=0}^{\infty}p_{n}=1 and p_{n}=0 for n\geq \kappa+1, we see that

\displaystyle p_{0}\sum_{n=0}^{\kappa}{\kappa\choose n}A^{n}=1

By the binomial formula, \sum_{n=0}^{\kappa}{\kappa\choose n}A^{n}=(1+A)^{\kappa}, so that p_{0}=(1+A)^{-\kappa}. Since any positive number can be written as A=\frac{p}{1-p}, where p \in (0,1), we have that

\displaystyle p_{0}=\left(1+\dfrac{p}{1-p}\right)^{-\kappa}=(1-p)^{\kappa} \Longrightarrow p_{n}={\kappa \choose n}p^{n}(1-p)^{\kappa-n},

which shows that N \sim \text{Bin}(\kappa,\frac{a}{a-1}).

If N has distribution of (a,b,0) class, then there is a closed-form expression for its generating function P_{N}(r). Denote the generating function of N by P_{N}(r)=\sum_{n=0}^{\infty}r^{n}p_{n}. Since generating functions convergence uniformly for \left|r\right| < 1, we can differentiate term-by-term to obtain

\begin{array}{lcl}\displaystyle P_{N}'(r)=\sum_{n=1}^{\infty}nr^{n-1}p_{n}&=&\sum_{n=1}^{\infty}nr^{n-1}\left(a+\dfrac{b}{n}\right)p_{n-1}\\[.7 em]&=&\displaystyle a\sum_{n=1}^{\infty}nr^{n-1}p_{n-1}+b\sum_{n=1}^{\infty}r^{n-1}p_{n-1}\\[.7 em]&=&\displaystyle a\sum_{n=1}^{\infty}(n-1)r^{n-1}p_{n-1}+a\sum_{n=1}^{\infty}r^{n-1}p_{n-1}+bP_{N}(r)\\[.7 em]&=&\displaystyle arP_{N}'(r)+(a+b)P_{N}(r)\end{array}

Rearranging, we can write (1-ar)P_{N}'(r)=(a+b)P_{N}(r). For sufficiently small r > 0, we can divide both sides by 1-ar and solve the differential equation to obtain

\begin{array}{lcl}\displaystyle\dfrac{P_{N}'(r)}=\dfrac{a+b}{1-ar}P_{N}(r)&\Longrightarrow&\displaystyle\frac{d}{dr}\log\left(P_{N}(r)\right)=\dfrac{a+b}{1-ar}\\[.7 em]&\Longrightarrow&\displaystyle\log(P_{N}(r))=-\dfrac{a+b}{a}\log\left(1-ar\right)+C'\\[.7 em]&\Longrightarrow&\displaystyle P_{N}(r)=C\left(1-ar\right)^{\frac{a+b}{a}}\end{array}

Since P_{N}(0)=p_{0}, we conclude that C=p_{0}.

Panjer recursion gives us a formula for computing the probability distribution function of the aggregate claims S=\sum_{j=1}^{N}X_{j} when the counting random variable N has distribution of (a,b,0) class and the individual claim X_{j} has discrete distribution with probability function \left\{f_{j}\right\}_{j=0}^{\infty}. We restrict our attention to the case the individual claims are distributed on the nonnegative integers, so that S is also distributed on the nonnegative integers. Let g denote the probability distribution function of S. Since N is independent of the X_{j},

\begin{array}{lcl}\displaystyle g_{0}=\mathbb{P}\left(\sum_{j=1}^{N}X_{j}=0\right)&=&\displaystyle\mathbb{P}(N=0)+\sum_{n=1}^{\infty}\mathbb{P}\left(\sum_{j=1}^{n}X_{j}=0\mid N=n\right)\mathbb{P}(N=n)\\&=&\displaystyle p_{0}+\sum_{n=1}^{\infty}\mathbb{P}(X_{j} = 0)^{n}p_{n}\\&=&\displaystyle p_{0}+\sum_{n=1}^{\infty}f_{0}^{n}p_{n}\\&=&\displaystyle P_{N}(f_{0})\end{array}

We know that the probability generating function of S is given by

\displaystyle P_{S}(r)=P_{N}\left(P_{X}(r)\right)

Differentiating with respect to r, we obtain by the chain rule that

\displaystyle P_{S}'(r)=P_{N}'\left(P_{X}(r)\right)P_{X}'(r)

Substituting in the identity P_{N}'(r)=arP_{N}'(r)+(a+b)P_{N}(r) with r=P_{X}(r), we obtain

\begin{array}{lcl}\displaystyle P_{S}'(r)&=&\displaystyle\left[aP_{X}(r)P_{N}'\left(P_{X}(r)\right)+(a+b)P_{N}\left(P_{X}(r)\right)\right]P_{X}'(r)\\&=&\displaystyle aP_{X}(r)P_{S}'(r)+(a+b)P_{S}(r)P_{X}'(r)\end{array}

Substituting in the series expressions for the generating functions, we obtain


Multiplying both sides by r, we have

\begin{array}{lcl}\displaystyle\sum_{j=0}^{\infty}jg_{j}r^{j}&=&\displaystyle a\left(\sum_{j=0}^{\infty}f_{j}r^{j}\right)\left(\sum_{j=0}^{\infty}jg_{j}r^{j}\right)+(a+b)\left(\sum_{j=0}^{\infty}g_{j}r^{j}\right)\left(\sum_{j=0}^{\infty}jf_{j}r^{j}\right)\\&=&\displaystyle a\sum_{j=0}^{\infty}\left(\sum_{k=0}^{j}(j-k)f_{k}g_{j-k}\right)r^{j}+(a+b)\sum_{j=0}^{\infty}\left(\sum_{k=0}^{j}kf_{k}g_{j-k}\right)r^{j}\end{array}

where we use the fact that the product of convergent power series is the Cauchy product. Since a power series is uniquely determined by its coefficients, we can compare coefficients on both sides to obtain the formula

\displaystyle jg_{j}=\sum_{k=0}^{j}a(j-k)f_{k}g_{j-k}+(a+b)kf_{k}g_{j-k}=\sum_{k=0}^{j}\left(aj+bk\right)f_{k}g_{j-k}, \indent \forall j \geq 1

Rearranging, we obtain

\displaystyle (j-af_{0}j)g_{j}=\sum_{k=1}^{j}(aj+bk)f_{k}g_{j-k} \Longrightarrow g_{j}=\dfrac{1}{1-af_{0}}\sum_{k=1}^{j}\left(a+\dfrac{bk}{j}\right)f_{k}g_{j-k}

for all j\geq 1. Note that the RHS of the second identity only contains terms with g_{0},\cdots,g_{j-1}, hence defines a recursive formula for g_{j}. The preceding formula is called the Panjer recursion formula first introduced by Harry Panjer in 1981. One might ask how we a priori know that af_{0}\neq 1, but this follows from our examination of (a,b,0) class distributions above and our hypothesis that N belongs to the (a,b,0) class.

We have established a recursive formula for the probability mass function. Does an analogous formula exist for the cumulative distribution function (cdf)? According to one of the books I’m consulting, a recursion formula for the cdf does not exist, in general. However, an exception to this statement is when N has a geometric distribution p_{n}=pq^{n}. As the geometric distribution is just a \text{NBin}(1,p) distribution, we have that a=1-p, b=0, and g_{k}=\frac{q}{1-qf_{0}}\sum_{j=1}^{k}f_{j}g_{k-j} for k\geq 1. Hence,

\begin{array}{lcl}\displaystyle G(n)=\sum_{k=0}^{n}g_{k}=g_{0}+\sum_{k=1}^{n}\dfrac{q}{1-qf_{0}}\sum_{j=1}^{k}f_{j}g_{k-j}&=&\displaystyle g_{0}+\dfrac{q}{1-qf_{0}}\sum_{j=1}^{n}f_{j}\sum_{k=j}^{n}g_{k-j}\\&=&\displaystyle G(0)+\dfrac{q}{1-qf_{0}}\sum_{j=1}^{n}f_{j}G(n-j)\end{array}

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6 Responses to Panjer Recursion

  1. Mitja says:

    excelent article

  2. Mitja says:

    Dear Matt,

    maybe you have some article or lessons in pdf about De prill’s recursion? or extension of Panjer recursion?

    Thank in advance
    Mitja Benčina

  3. irma says:

    thank you for posting this article. It really helps me

  4. Nilson Lima says:

    great explanation. every step is very clear

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