2.1. The pgf and moments of the GP distribution

Consider a rv $N$ follow the distribution of GP $(\lambda, \theta )$ . Ambagaspitiya & Balakrishnan (Reference Ambagaspitiya and Balakrishnan1994) showed that when $\theta \gt 0$ , its probability generating function (pgf) could be written as

(1) \begin{equation} G_{N}(z)=\mathbb{E}[z^{N}]=\exp \!\left \{-\frac{\lambda }{\theta } \left [W({-}\theta e^{-\theta } z) +\theta \right ]\right \}, \end{equation}

where $W$ is the Lambert’s $W$ function defined as

\begin{equation*} W(x)\exp\! (W(x))=x. \end{equation*}

For more details about the properties of $W$ function, see Corless et al. (Reference Corless, Gonnet, Hare, Jeffrey and Knuth1993).

Let $\mu _{N}^{(\alpha )}\,:\!=\,\mathbb{E}[N^{(\alpha )}]$ be the $\alpha$ th factorial moment of $N$ (see Definition 2). Because $\mathbb{E}[N^{(\alpha )}]=\frac{d^{\alpha } }{dz^{\alpha }}G_{N}(z) \big |_{z=1}$ , it is easy to see by differentiating that

\begin{align*} &\mu _{N}^{(1)}=\mathbb{E}[N]=\frac{\lambda }{1-\theta },\\ &\mu _{N}^{(2)}=\mathbb{E}[N\cdot (N-1)]=\frac{\lambda }{(1-\theta )^3}+\left(\frac{\lambda }{1-\theta }\right)^2-\frac{\lambda }{1-\theta }, \end{align*}

and thus the variance of $N$ is

\begin{equation*} \mathbb {V}\mathrm {ar}(N)=\frac {\lambda }{(1-\theta )^3}. \end{equation*}

2.2. Recursive formula for GP and a family of $\mathbf{\mathcal{H}}$ distributions

Definition 1 (Ambagaspitiya, Reference Ambagaspitiya1995). A rv $N$ is said to belong to the $\mathcal{H}$ family characterized by $(h_1(a,b), h_2(a,b))$ , i.e., $X\in \mathcal{H}(h_1(a,b), h_2(a,b))$ , if its pmf satisfies the recursive formula

(2) \begin{equation} p_n(a,b)=\left ( h_1(a,b)+\frac{h_2(a,b)}{n}\right ) p_{n-1}(a+b,b), \qquad n=1,2,\ldots . \end{equation}

It is easy to check that both GP and GNB belong to the $\mathcal{H}$ family. For GP $(a,b)$ , $h_1(a,b)=\frac{ab}{a+b}$ and $h_2(a,b)=\frac{a^2}{a+b}$ , while for GNB $(a,b, \alpha )$ , $h_1(a,b)=\frac{a(b-1)\alpha }{(1-\alpha )(a+b)}$ and $h_2(a,b)=\frac{a(a+1)\alpha }{(1-\alpha )(a+b)}$ . More specifically, the pmf of GP $(a,b)$ satisfies the recursive relation

(3) \begin{equation} p_n(a,b)=\frac{a}{a+b}\left (b+\frac{a}{n}\right )p_{n-1}(a+b,b), \qquad n=1,2,\ldots, \end{equation}

and the pmf of GNB $(a,b, \alpha )$ satisfies the recursive relation

\begin{equation*} p_n(a,b,\alpha )=\frac {a\alpha }{(1-\alpha )(a+b)}\left (b-1+\frac {a+1}{n}\right )p_{n-1}(a+b,b,\alpha ), \qquad n=1,2,\ldots . \end{equation*}

2.3. The GP distribution as a compound Poisson distribution itself

The GP distribution is closely related to the Galton-Watson branching process (Goovaerts & Kaas, Reference Goovaerts and Kaas1991), which models the spreading of certain objects (e.g., family names, viruses) as follows. For example, suppose that there are $M$ individuals originally. Each of these gives rise to $L_i$ other individuals, $i=1,2,\ldots, M$ . These, in turn, give rise to $L_{ij}$ new victims, $j=1,2,\ldots, L_i$ , and so on. Now if $M$ is a Poisson $(\lambda )$ distributed rv, and $L_i, L_{ij},\ldots$ are independent Poisson $(\theta )$ ( $\theta \lt 1$ ) rv’s, the total number $N$ of people infected before extinction has a GP $(\lambda,\theta )$ distribution.

If the original number of infected is fixed at one, then the total number of infected, $B$ , has a Borel ( $\theta$ ) distribution, with pmf

\begin{equation*} \mathbb {P}(B=n)=\frac {(n\theta )^{n-1}e^{-n\theta }}{n!}, \qquad n=1,2,\ldots, \end{equation*}

mean $ \mathbb{E}(B)=\frac{1}{1-\theta }$ and variance $\mathbb{V}\mathrm{ar}(B)=\frac{\theta }{(1-\theta )^3}.$

Therefore, a GP $(\lambda,\theta )$ is a compound Poisson sum of independent Borel $(\theta )$ rv’s, i.e.,

(4) \begin{equation} N=\sum _{i=1}^M B_i, \end{equation}

where the primary rv $M$ is Poisson distributed with parameter $\lambda$ , and the secondary rv is independent and identically distributed (iid) as Borel( $\theta$ ).

Note that one relationship between the Borel and GP distributions is

\begin{equation*} B\stackrel {d}{=} 1+N^*, \end{equation*}

where $N^*\sim$ GP $(\theta,\theta )$ . Therefore, the pgf of $B\sim$ Borel( $\theta$ ) is

(5) \begin{equation} G_B(z)=zG_{N^*}(z)=ze^{\theta (G_B(z)-1)}, \end{equation}

where the compound form of $N^*$ using Equation (4) is applied here.

2.4. The compound GP random variable

Denote a compound GP rv $X$ as

\begin{equation*} X=\sum _{i=1}^N C_i, \end{equation*}

where $N$ follows GP $(\lambda,\theta )$ and $C_i$ are iid rv’s independent of $N$ .

Goovaerts & Kaas (Reference Goovaerts and Kaas1991) derived a recursive algorithm for computing the distribution of the compound sum when $C_i$ ’s take positive integer-valued using the representation in Equation (4). Their derivation starts by rewriting the compound GP rv as

\begin{equation*} X=\sum _{i=1}^M Y_i, \qquad Y_i=\sum _{j=1}^{B_i} C_{ij}, \end{equation*}

where $M$ is a Poisson( $\lambda$ ) rv, $B_i$ is a Borel( $\theta$ ) rv, and $C_{ij}$ is an iid sequence of claim amounts having the same distributions as $C_i$ . Each $Y_i$ has a compound Borel distribution. Hence, the pgf of $X$ becomes (see Equation (12) of Goovaerts & Kaas, Reference Goovaerts and Kaas1991)

\begin{align*} G_X(u)&=\exp \!\left (\lambda \!\left [G_Y(u)-1\right ]\right )\\ & =\exp \!\left (\lambda \!\left [G_B(G_C(u))-1\right ]\right )\\ & =\exp (\lambda (t-1)), \end{align*}

with $t=t(u)$ such that $G_C(u)=t e^{-\theta (t-1)}$ via Equation (5). This gives a two-step algorithm for calculating the distribution of $X$ . The first step is to compute the coefficients of $G_B(G_C(u))$ (which amounts to the pdf of $Y_i$ ); and the second step is to evaluate the coefficients of $G_X(u)$ by Panjer’s recursion formula.

Ambagaspitiya & Balakrishnan (Reference Ambagaspitiya and Balakrishnan1994) made use of the recursion formula for GP in Equation (3) and showed that, when claim sizes are discrete on positive integers (or have absolutely continuous pdf), the pmf (or pdf) of the compound GP rv also satisfies a recursive formula. In particular, their Theorem 5.1 stated that for the discrete claim size with pmf $f(x)=\mathbb{P}(C=x)$ for $x=1,2,\ldots$ , the pmf of $X$ can be calculated as

(6) \begin{equation} \mathbb{P}(X=x;\,\lambda, \theta ) =\frac{\lambda }{\lambda +\theta }\sum _{j=1}^x \left (\theta +\lambda \frac{j}{x}\right ) \mathbb{P}(X=x-j;\,\lambda +\theta, \theta )f(j), \end{equation}

with the initial values $\mathbb{P}(X=0;\,\lambda +j\theta,\theta )=\exp\! ({-}\lambda -j\theta )$ for $j=0,1,2,\ldots$ . Later on, Ambagaspitiya (Reference Ambagaspitiya1995) extended the recursive formula for the pmf of compound distribution when $N$ is in $\mathcal{H}$ family and $C$ has support on positive integers,

(7) \begin{equation} \mathbb{P}(X=x;\,a, b) =\sum _{j=1}^x \left (h_1(a,b)+h_2(a,b)\frac{j}{x}\right ) \mathbb{P}(X=x-j;\,a+b, b)f(j). \end{equation}

3.1. Recursive formula for the pmf of the moment transforms of GP

We have the following theorem regarding the size-biased transform of a general distribution in the $\mathcal{H}$ family.

Theorem 1. For $N\sim \mathcal{H}(h_1(a,b), h_2(a,b))$ , let $\tilde N$ be its size-biased transform and $M=\tilde N-1$ . Then $M\sim \mathcal{H}(h_1^*(a,b),h_2^*(a,b))$ with

\begin{equation*} h_1^*(a,b)=1-\frac {h_1(a,b)+h_2(a,b)}{\mu _N(a,b)}, \end{equation*}

and

\begin{equation*} h_2^*(a,b) =\frac {h_1(a,b)+h_2(a,b)}{h_1(a,b)}h_1^*(a,b). \end{equation*}

Proof. Equation (2) can be written as, for $k=1,2,\ldots$ ,

(8) \begin{align} k p_k (a,b) &= (h_1(a,b) k+h_2(a,b))p_{k-1}(a+b,b) \notag \\ &=\left [h_1(a,b) (k-1)+(h_1(a,b)+h_2(a,b))\right ]p_{k-1}(a+b,b). \end{align}

Dividing both sides of Equation (8) by $\mu _N(a,b)$ , we have, for $k=1,2,\ldots$

\begin{align*} g_k (a,b) = &\ h_1(a,b)\frac{\mu _N(a+b,b)}{\mu _N(a,b)} g_{k-1}(a+b,b) \\ & + \left [h_1(a,b)+h_2(a,b)\right ] \frac{\mu _N(a+b,b)}{\mu _N(a,b)}\frac{p_{k-1}(a+b,b)}{\mu _N(a+b,b)}. \end{align*}

For $k=2,3,\ldots$ , multiplying both side by $(k-1)$ yields

\begin{align*} (k-1)g_k (a,b) =&\ h_1(a,b)\frac{\mu _N(a+b,b)}{\mu _N(a,b)} (k-1) g_{k-1}(a+b,b) \\ &+ \left [h_1(a,b)+h_2(a,b)\right ] \frac{\mu _N(a+b,b)}{\mu _N(a,b)} g_{k-1} (a+b,b), \end{align*}

which can be written as

\begin{equation*} k g_{k+1}(a,b)= h_1(a,b)\frac {\mu _N(a+b,b)}{\mu _N(a,b)} k g_{k} (a+b,b) + \left [h_1(a,b)+h_2(a,b)\right ] \frac {\mu _N(a+b,b)}{\mu _N(a,b)} g_{k}(a+b,b), \end{equation*}

for $k=1, 2,3,\ldots .$

Now, since $M\,:\!=\,\tilde N -1$ , we have its pmf denoted as $f_{k-1}(a,b)\,:\!=\,\mathbb{P}({M}=k-1)=\mathbb{P}(\tilde N=k)=g_k(a,b)$ for $k=1,2,3,\ldots$ , thus

\begin{equation*} k f_k (a,b)= h_1(a,b)\frac {\mu _N(a+b,b)}{\mu _N(a,b)} k f_{k-1}(a+b,b)+\left [h_1(a,b)+h_2(a,b)\right ] \frac {\mu _N(a+b,b)}{\mu _N(a,b)} f_{k-1}(a+b,b), \end{equation*}

which is equivalent to

\begin{align*} f_k (a,b)&= \left ( h_1(a,b)\frac{\mu _N(a+b,b)}{\mu _N(a,b)} +\left [h_1(a,b)+h_2(a,b)\right ] \frac{\mu _N(a+b,b)}{\mu _N(a,b)}\frac{1}{k}\right ) f_{k-1}(a+b,b)\\ &\,=\!:\,\left ( h_1^*(a,b) + \frac{h_2^*(a,b)}{k} \right ) f_{k-1}(a+b,b), \qquad k=1, 2,3,\ldots . \end{align*}

Hence, the distribution of $M$ still satisfies the general recursive formula (2) as desired.

To further simplify the representation of $h_1^*(a,b)$ , we write Equation (2) as

\begin{equation*} k p_k (a,b) =h_1(a,b) (k-1) p_{k-1}(a+b,b)+(h_1(a,b)+h_2(a,b))p_{k-1}(a+b,b), \qquad k=1,2,\ldots . \end{equation*}

Summing up both sides from $k=1$ to $\infty$ yields

\begin{equation*} \mu _N(a,b)=h_1(a,b)\mu _N(a+b,b)+\left [h_1(a,b)+h_2(a,b)\right ], \end{equation*}

or equivalently

\begin{equation*}\mu _N(a+b,b)=\frac {\mu _N(a,b)-\left [h_1(a,b)+h_2(a,b)\right ]}{h_1(a,b)}. \end{equation*}

Therefore,

\begin{equation*} h_1^*(a,b)=h_1(a,b)\frac {\mu _N(a+b,b)}{\mu _N(a,b)}= 1-\frac {h_1(a,b)+h_2(a,b)}{\mu _N(a,b)}, \end{equation*}

which completes the proof.

Since both GP and GNB distributions are in the family $\mathcal{H}$ , we have the following corollary as the direct application of the above theorem.

Actually, Theorem 1 could be generalized to the $\alpha$ th factorial moment transform in a similar manner. The result is stated in the following Theorem. Its proof is provided in Appendix A.

Theorem 2. Let $N\sim \mathcal{H}(h_1(a,b), h_2(a,b))$ . For an integer $\alpha \ge 1$ , let $\tilde N_{\alpha }$ be its $\alpha$ th factorial moment transform and $M_{\alpha }=\tilde N_{\alpha }-\alpha$ . Then $M_{\alpha }\sim \mathcal{H}(h_1^*(a,b), h_2^*(a,b))$ with

\begin{equation*} h_1^*(a,b)=h_1(a,b)\frac {\mu _N^{(\alpha )}(a+b,b)}{\mu _N^{(\alpha )}(a,b)} \qquad \text {and} \qquad h_2^*(a,b) = \frac {\alpha \cdot h_1(a,b)+ h_2(a,b)}{h_1(a,b)}h_1^*(a,b). \end{equation*}

3.2. Generating function for the moment transforms of GP

As illustrated in Remark 2.5 of Blier-Wong et al. (Reference Blier-Wong, Cossette and Marceau2022), another way to study the moment transform of distributions is through the generating function. This is due to the fact that the pgf of the factorial moment transform of a discrete rv $N$ is given by

\begin{equation*}G_{\tilde N_{\alpha }}(z)=\frac {z^{\alpha } }{\mathbb {E}[N^{(\alpha )}]}\cdot \frac {d^{\alpha } }{dz^{\alpha }}G_N(z).\end{equation*}

In particular, the pgf for a first-moment transformed GP distribution, $\tilde N$ , is given by

\begin{align*} G_{\tilde N}(z)&=\frac{z G^{\prime }_N(z)}{\mathbb{E}[N]} = \frac{z}{\mathbb{E}[N]} \left \{ \frac{ \lambda }{\theta } \theta e^{-\theta } W^{\prime } ({-}\theta e^{-\theta } z) G_N(z) \right \} \\ &=(1-\theta ) e^{-\theta } W^{\prime } ({-}\theta e^{-\theta } z)z G_N(z)\\ &=-\frac{1-\theta }{\theta } \frac{W({-}\theta e^{-\theta }z)}{1+W({-}\theta e^{-\theta }z)} G_N(z), \end{align*}

where we have made use of the identity for Lambert’s $W$ function that

\begin{equation*} W^{\prime }(x)=\frac {W(x)}{x(1+W(x))}. \end{equation*}

Then the pgf for $M=\tilde N-1$ is

(9) \begin{align} G_{M}(z)&=-\frac{1-\theta }{\theta } \frac{W({-}\theta e^{-\theta }z)}{1+W({-}\theta e^{-\theta }z)} \frac{G_N(z)}{z}. \end{align}

5.1. Application I: risk measures on compound GP distribution

Consider the aggregate loss model

\begin{equation*} X=\sum _{i=1}^{N} C_i, \end{equation*}

where $N$ is the counting rv and $C_i$ follows some discrete distribution with $\mathbb{P}(C_i=k)=f_k$ . In this section, we provide methods for computing the tail probability $\mathbb{P}(X\gt x)$ and the CTE defined as $\mathbb{E}[X|X\gt x].$

To calculate the CTE, we use the result from Denuit (Reference Denuit2020), which states that, for any rv $X$ and a measurable function $g$ ,

(17) \begin{equation} \mathbb{E}[Xg(X)]=\mathbb{E}[X]\mathbb{E}[g(\tilde{X})], \end{equation}

where $\tilde{X}$ is the size-biased transform of $X$ . When $X$ follows a compound distribution, the distribution $\tilde{X}$ is given by Theorem 3 in Equation (10). Hence, to evaluate the CTE, one could apply Equation (17) with $g(x) = \mathbb{I} (X \gt x)$ being an indicate function and obtain

\begin{equation*} \mathbb {E}[X|X\gt x] = \mathbb {E}[X] \frac {\mathbb {P}(\tilde {X}\gt x)}{\mathbb {P}(X\gt x)}. \end{equation*}

Thus, the calculation boils down to the two tail probabilities $\mathbb{P}(\tilde{X}\gt x)$ and $\mathbb{P}(X\gt x)$ . In what follows, we provide three methods to carry out the calculation, namely, the recursive method, the transformation method via FFT, and the simulation method. Note that the recursive method has the restriction that the claim size distribution has positive support. The other two methods do not have such a restriction.

5.1.1. Recursive method

The recursive formula for computing the distribution of a compound GP $X$ with positive discrete severity is given in Equation (6).

To compute the distribution of $\tilde{X}$ , we need to use our main result in Theorem 3. Recall from Equation (10) that

\begin{equation*} \tilde X \stackrel {d}{=} \tilde C + Y, \end{equation*}

where $Y=\sum _{i=1}^M C_i$ with $M=\tilde N-1$ and $\tilde C$ being the size-biased version of $C$ . When $N$ is GP distributed, we showed in Corollary 1 that the distribution of $M$ still in $\mathcal{H}$ family, therefore, the distribution of $Y=\sum _{i=1}^{M} C_i$ can be evaluated recursively using Equation (7). Then, the distribution of $\tilde{X}$ can be computed through a direct convolution of $\tilde{C}$ and $Y$ .

5.1.2. Transformation method via FFT

The mgf of a compound GP rv $X$ is given by

\begin{equation*}M_X(t) =\mathbb {E}[e^{tX}] = G_N(M_C(t)).\end{equation*}

Applying the pgf of the GP distribution in Equation (1), we have

(18) \begin{equation} M_X(t)=\exp \!\left \{-\frac{\lambda }{\theta } \left [W({-}\theta e^{-\theta } M_C(t)) +\theta \right ]\right \}. \end{equation}

Then by Equation (14), the mgf of $\tilde{X}$ becomes

(19) \begin{align} M_{\tilde{X}}(t)&= G_{M}(M_C(t)) \cdot M_{\tilde{C}}(t)\nonumber \\[4pt] &= -\frac{1-\theta }{\theta } \frac{W({-}\theta e^{-\theta }M_C(t))}{1+W({-}\theta e^{-\theta }M_C(t))} \frac{G_N(M_C(t))}{M_C(t)} \frac{M_C'(t)}{\mathbb{E}[C]}, \end{align}

where we applied the pgf of $M$ given in Equation (9).

With the mgf of $X$ and $\tilde{X}$ given by Equations (18) and (19), respectively, both the tail probability and the CTE can be computed through FFT. An algorithm is provided in the following; see Wang (Reference Wang1998) and Embrechts & Frei (Reference Embrechts and Frei2009).

1. 1. Parameter Definition: Define the parameters for the GP claim frequency and the claim severity distributions.

2. 2. Fourier Transform of Severity Distribution: Compute the FFT of the claim severity distribution. If the severity distribution is continuous, it needs to be discretized.

3. 3. Fourier Transform of Compound Distribution: The FFT of compound distribution $X$ (or $\tilde{X}$ ) can be computed according to Equation (18) (or Equation (19)). The Lambert $W$ function can be evaluated using, for example, the emdbook package in R programming language.

4. 4. Inverse Fourier Transform: In this step, we recover the aggregate loss distribution by applying inverse FFT.

These steps together provide a practical and computationally efficient way of calculating the compound GP distribution.

5.2. Application II: capital allocations for a portfolio of risks

In this section, we consider a capital allocation problem of a portfolio of $n$ independent losses $(X_1, X_2,\ldots, X_n)$ , each of which is a compound distribution. Specifically,

\begin{equation*} X_i=\sum _{k=1}^{N_i}C_{ik}, \qquad i=1,\ldots, n, \end{equation*}

where $N_i\sim$ GP $(\lambda _i,\theta _i)$ and $C_{ik}$ , $k=1,2,\ldots$ , are iid with the common distribution $C_i$ . All rv’s are independent. For $i=1,\ldots,n$ , let $\tilde X_i$ and $\tilde C_i$ be the size-biased version of $X_i$ and $C_i$ , respectively, and let

\begin{equation*} Y_i=\sum _{j=1}^{M_i} C_{ij}, \end{equation*}

where $M_i=\tilde N_i -1$ . Then as shown earlier, we have

\begin{equation*} \tilde {X}_i\stackrel {d}{=} Y_i +\tilde C_i. \end{equation*}

Now consider the aggregate risk of the portfolio, denoted as

\begin{equation*} S=\sum _{i=1}^n X_i. \end{equation*}

We look into details of the CTE and Euler allocation rules in the following.

CTE allocation rule

Under the CTE risk measure, the total capital requirement for the portfolio is $ \mathbb{E}[S|S\gt s]$ , which can be calculated using

\begin{equation*} \mathbb {E}[S|S\gt s]=\mathbb {E}[S]\frac {\mathbb {P}(\tilde {S}\gt s)}{\mathbb {P}(S\gt s)}, \end{equation*}

where $\tilde{S}$ denotes the size-biased transform of $S$ .

Furthermore, from Property 1 in Section 3.1 of Denuit (Reference Denuit2020), we have

\begin{equation*} \tilde {S} \stackrel {d}{=} S-X_K +\tilde {X}_K=\sum _{j\not =K}^n X_j+\tilde X_K, \end{equation*}

where $K$ is a discrete rv, independent of $\{X_i\}$ and $\{\tilde X_i\}$ , with pmf

\begin{equation*} \mathbb {P}(K=i)=\frac {\mathbb {E}[X_i]}{\mathbb {E}[S]}, \qquad i=1,2,\ldots, n. \end{equation*}

Putting these together, the capital requirement for the portfolio becomes

(20) \begin{align} \mathbb{E}[S|S\gt s]&= \mathbb{E}[S]\frac{\mathbb{P}(\tilde{S}\gt s)}{\mathbb{P}(S\gt s)} \nonumber \\ &=\sum _{i=1}^n \frac{\mathbb{E}(X_i)}{\mathbb{P}(S\gt s)}\mathbb{P}\!\left (\sum _{j\not =i}^n X_j+\tilde X_i\gt s\right ). \end{align}

In addition, according to the CTE capital allocation rule, the capital allocated to risk $X_i$ is

(21) \begin{equation} \mathbb{E}[X_i|{S\gt s}]= \frac{\mathbb{E}(X_i)}{\mathbb{P}(S\gt s)}\mathbb{P}(S-X_i+\tilde X_i \gt s)=\frac{\mathbb{E}(X_i)}{\mathbb{P}(S\gt s)}\mathbb{P}\!\left (\sum _{j=1,j\not =i}^n X_j+\tilde X_i \gt s\right ). \end{equation}

Euler allocation rule

When the distribution of $S$ is discrete, we may study the conditional mean allocation $ \mathbb{E}[X_i|{S=s}]$ . This quantity is based on the Euler allocation rule and plays a unique role in peer-to-peer insurance prices. It has been studied in the literature. For example, Denuit & Robert Reference Denuit and Robert2022) derived a list of desirable properties of this allocation principle. Gribkova et al. (Reference Gribkova, Su and Zitikis2023) proposed an empirical estimator of $ \mathbb{E}[X_i|{S=s}]$ and established its asymptotic normality under minimal conditions. Blier-Wong et al. (Reference Blier-Wong, Cossette and Marceau2022) presented an ordinary generating function of $\mathbb{E}[X_i \times I_{\{S=s\}})]$ , which leads a method to compute it using FFT.

Here, because the distribution of $X_i$ and its moment transform $\tilde{X}_i$ can be evaluated using recursive or FFT methods as shown in Section 4, we took a different approach from Blier-Wong et al. (Reference Blier-Wong, Cossette and Marceau2022) and propose to compute conditional mean allocation $ \mathbb{E}[X_i|{S=s}]$ by

(22) \begin{align} \mathbb{E}[X_i|S=s] &= \frac{\mathbb{E}(X_i)}{\mathbb{P}(S = s)} \mathbb{P}\!\left (\sum _{j\not =i}^n X_j+\tilde X_i = s\right ). \end{align}

Numerically, in order to apply the recursive methods to evaluate Equations (21) and (22), all the claim size $C_{ij}$ has to have positive support, which is an inconvenient constraint. Therefore, we recommend using FFT method because it can be flexibly used to compute the distribution function of $X_i$ , its moment transform, as well as the required convolution.

6.1. Example 1

Consider a compound GP distribution

\begin{equation*}X=\sum _{i=1}^{N} C_i,\end{equation*}

where $N$ follows GP $(\lambda =5, \theta =0.5)$ and $C_i$ ’s are iid following zero truncated Poisson distribution with parameter $\mu =5$ , that is

\begin{equation*}p_C^T(k)=\frac {1}{1-e^{-\mu }}\frac {\mu ^k e^{-\mu }}{k!}, \qquad k=1,2,3,\ldots . \end{equation*}

This assumption of a zero-truncated claim size distribution allows the application of the recursive method for computing the compound GP distribution.

In this example, we calculate the tail probability $\mathbb{P}(X\gt x)$ and CTE $\mathbb{E}[X|X\gt x]$ for different risk levels $x$ .

The computation is carried out using the recursive and FFT methods. The results are compared with those obtained from the simulation methods with a sample size of $n = 10^6$ . The simulations were performed using the inversion method and the branching algorithm. The results are shown in Table 1, from which we observe that all methods result in nearly identical results.

Table 1. Computation of tail probabilities and CTE of compound GP distribution using different methods

$^{\rm a}$ FFT method.

$^{\rm b}$ $N$ Simulated by Branching process based simulation.

$^{\rm c}$ $N$ Simulated by direct inversion simulation method.

$^{\rm d}$ Recursive method.

In addition, in Tables 2 and 3, we report the computation time of the different methods and the standard error of the simulation methods (estimated by repeating the simulation 100 times). All computations are performed on a personal laptop with Intel Core i7 CPU and 16 GB RAM. It is seen that the FFT method is by far the fastest, whereas the simulation methods are the slowest. Based on these results, we conclude that the FFT method is reliable and efficient for evaluating the risk measures of the compound GP distributions.

Table 2. Comparison of tail probability results

Table 3. Comparison of CTE results

Furthermore, we compare our results with the compound Poisson case with a Poisson parameter $10$ such that the mean of the frequency distribution equals. As seen in Table 4, the tail of compound GP distribution is heavier, and its CTE risk measures are larger than the compound Poisson case.

Table 4. Tail probabilities and CTE of compound Poisson distribution

6.2. Example 2

Consider the capital allocation example in Section 4.2.4 in Denuit (Reference Denuit2020). Let $(X_1, X_2, X_3)$ be three independent risks follow the compound GP’s whose parameters are given in Table 5. The severity parameters are the same as those adopted in Denuit (Reference Denuit2020), whereas the claim frequencies are assumed to be GP distributed instead of Poisson distributed while keeping the same mean. The significance of using GP is that it has over-dispersion: the means of the frequencies are $(10, 15, 20)$ , whereas their variances are $(40, 60, 142)$ . Consequently, compared with the case of Poisson frequency, $X_3$ is much more risky than $X_2$ , which in turn is more risky than $X_1$ .

Table 5. The portfolio of three compound GP risks

The portfolio aggregate risk is thus given by $S=\sum _{i=1}^3 X_i$ .

CTE allocation rule

The CTE-based total capital requirement is given as $\mathbb{E}(S|S\gt s)$ , and the proportions allocated to individual risk are

\begin{equation*} \alpha _i\,:\!=\,\frac {\mathbb {E}(X_i|S\gt s)}{\mathbb {E}(S|S\gt s)}, \quad i=1,2,3. \end{equation*}

We implemented the capital allocation computation described in Section 5.2 using the FFT method. The results are shown in Table 6, from which we conclude that:

• • When the threshold $s$ increases, the proportions of risk capital allocated to $X_1$ and $X_2$ slightly decrease, whereas those allocated to $X_3$ increase.

• • The risks $X_2$ and $X_3$ have the same loss severity but the frequency component in $X_3$ is more volatile. As a result, their proportions of risk capital allocated change with the risk levels $s$ ; more precisely, a greater proportion is assigned to $X_3$ as $s$ increases. The ratio $\alpha _2/\alpha _3$ decreases from 75% to around 60% as $s$ increases from 0 to 100. Note that $\alpha _2/\alpha _3 = \mathbb{E}(N_2)/\mathbb{E}(N_3)=0.75$ when $s=0$ .

Table 7 shows the CTE capital allocation and tail behavior for the compound Poisson sums under the parameter setup in Table 5 (except changing the primary distributions to Poisson with the same means), which is also calculated in Table 2 of Ren (Reference Ren2022). Comparing Tables 6 and 7, we have the following observations:

• • Overall, the tail of the compound GP distribution is heavier than compound Poisson, and so is the CTE risk measure.

• • When the threshold $s$ increases, the proportion of risk capital allocated to $X_2$ decreases under the compound GP assumption, while it increases slightly under the compound Poisson case.

• • At the same risk level $s$ , more capital is allocated to the risk $X_3$ under the compound GP case.

These differences show the model risk for the frequency assumption, which is one of the motivations for future analysis on statistical modeling of the loss frequency distribution based on real data. This will be a future research topic.

Table 6. CTE capital allocation for the portfolio of three compound GP risks

Table 7. CTE capital allocation and tail risk for the portfolio of three compound Poisson risks

Euler allocation rule

We re-do the above calculation with Euler allocation rule, where the proportions allocated to individual risks are

\begin{equation*} \beta _i\,:\!=\,\frac {\mathbb {E}(X_i|S=s)}{\mathbb {E}(S|S=s)}=\frac {\mathbb {E}(X_i|S=s)}{s}, \quad i=1,2,3, \end{equation*}

where the numerator is given in Equation (22). The results for the GP case are shown in Table 8 and those for the Poisson case are shown in Table 9. We can observe that the capital allocation patterns for the three risks are similar under the Euler and CTE capital allocation rules.

Table 8. Euler capital allocation for the portfolio of three compound GP risks

Table 9. Euler capital allocation and tail risk for the portfolio of three compound Poisson risks