On a Markovian Game Model for Competitive Insurance Pricing

Abstract

In this paper, we extend the non-cooperative one-period game of Dutang et al. (Journal of Operational Research 231(3):702–711, 2013) to model a non-life insurance market over several periods by considering the repeated (one-period) game. Using Markov chain methodology, we derive general properties of insurer portfolio sizes given a price vector. In the case of a regulated market (identical premium), we are able to obtain convergence measures of long run market shares. We also investigate the consequences of the deviation of one player from this regulated market. Finally, we provide some insights of long-term patterns of the repeated game as well as numerical illustrations of leadership and ruin probabilities.

Appendices

Appendix 1

1.1 Proofs of Section 3

We prove below Propositions 1, 2, 3, 5 with standard probabilistic arguments. We recall that bolded notation are reserved for vectors, $G^P$ and $G^M$ stand for the probability and the moment generating functions, $\otimes$ the Kronecker product.

1.1.1 Properties of the Lapse Model

Proof (Proof of Prop. 1)

By A3, $(C_{i,t})_t$ is a Markov chain with transition matrix $P_\rightarrow (\varvec{x}_t)$ defined as

$$\begin{aligned} P_\rightarrow (\varvec{x}_t)= \left( \begin{matrix} p_{1 \rightarrow 1}(\varvec{x}_t) &{} \dots &{} p_{1 \rightarrow J}(\varvec{x}_t) \\ &{} \dots &{} \\ p_{J \rightarrow 1}(\varvec{x}_t) &{} \dots &{} p_{J \rightarrow J}(\varvec{x}_t) \end{matrix}\right) . \end{aligned}$$

In fact, $P_\rightarrow$ is a matrix function. By Proposition 14, the transition matrix has no null terms. It is immediate that the transition from $C_{i,0}$ to $C_{i,t}$ is the multiplication of the t matrices $P_\rightarrow (\varvec{x}_1)$,..., $P_\rightarrow (\varvec{x}_t)$. On the finite state space $\{1,\dots , J\}$, the Markov chain is both irreducible and aperiodic using Proposition 14 in Appendix 2. By (Norris 1997, p. 41), the process $(C_{1,t},C_{2,t})_t$ is still a Markov chain on the space $E^2=\{1,\dots ,J\}^2$ with transition matrix $P_\rightarrow (\varvec{x}_t)\otimes P_\rightarrow (\varvec{x}_t)$. Iterating $N-1$ more times leads to the result.

Proof (Proof of Prop. 2)

Let $\mathcal N_j$ be the set of customers of Insurer j at time 0. That is $\forall i\in \mathcal N_j, C_{i,0}=j$. As $(C_{i,t})_t$ is a Markov chain, the transition from Insurer j to Insurer k is governed by the jth row $\tilde{p}_j$ of the matrix $P_\rightarrow ^{(t)}=P_\rightarrow (\varvec{x}_1)\times \dots \times P_\rightarrow (\varvec{x}_t),~ \tilde{p}_j = \left( P_{\rightarrow ,j,1}^{(t)},\dots , P_{\rightarrow ,j,J}^{(t)}\right) .$ Thus $\forall i\in \mathcal N_j, C_{i,t}~\vert ~ C_{i,0}=j\sim \mathcal M_J(1, \tilde{\varvec{p}}_{j})$. By A1 and A2, those policyholders of Insurer j will choose insurers according to a multinomial distribution $\overline{\varvec{C }}_{j,t}\sim \mathcal M_J(n_{j,t-1}, \tilde{\varvec{p}}_{j})$ given $N_{j,0}=n_{j,0}$. From period $t-1$ to period t, the transition matrix simplifies to $P_\rightarrow (\varvec{x}_t)$ and $\tilde{\varvec{p}}_j = \varvec{p}_{j\rightarrow }(\varvec{x}_t)$.

Proof (Proof of Prop. 3)

The portfolio sizes vector is the sum of choice vectors $\varvec{N}_{t} =\overline{\varvec{C }}_{1,t}+\dots + \overline{\varvec{C }}_{J,t}.$ By A2, Proposition 2 and given $\varvec{N}_{t-1}=\varvec{n}$, $(\overline{\varvec{C }}_{j,t})_j$ are independent multinomial vectors with parameters $\mathcal M_J(n_{j}, \varvec{p}_{j\rightarrow }(\varvec{x}_t))$ for $j=1,\dots ,J$. Therefore, $\varvec{N}_t$ (obtained by summing over j) has a known distribution given $\varvec{N}_{t-1}=\varvec{n}$. Since $(\varvec{N}_t)_t$ is a discrete-time process taking values in $\mathcal S_{ms}$, $(\varvec{N}_t)_t$ is a Markov chain. By recurrence, the number of elements of $\mathcal S_{ms}$ is $\mathrm{Card}(\mathcal S_{ms})=\left( {\begin{array}{c}N+J-1\\ N\end{array}}\right)$. The transition matrix of size $\mathrm{Card}(\mathcal S_{ms})\times \mathrm{Card}(\mathcal S_{ms})$ has a complex expression $\overline{P}_t = \left( P(\varvec{N}_t=\varvec{m}\vert \varvec{N}_{t-1}=\varvec{n}) \right) _{\varvec{n},\varvec{m}}$ where $\varvec{n},\varvec{m}\in \mathcal S_{ms}$ and

$$\begin{aligned}P(\varvec{N}_t=\varvec{m}\vert \varvec{N}_{t-1}=\varvec{n}) = \sum _{ \underset{ \text {s.t. } \sum \limits _l c_{1l} = n_{1} }{ 0\le c_{11}, \dots , c_{1J} \le N, } } \dots \sum _{ \underset{ \text {s.t. } \sum \limits _l c_{Jl} = n_{J} }{ 0\le c_{J1}, \dots , c_{JJ} \le N, } } \prod _{\underset{\sum \limits _{k} c_{kj} = m_{j}}{j=1,}}^J \frac{n_j!}{c_{j1}!\dots c_{jJ}!} \left( p_{j \rightarrow 1}(\mathbf{x}_t)\right) ^{c_{j1}}\dots \left( p_{j \rightarrow i}(\mathbf{x}_t)\right) ^{c_{jJ}}. \end{aligned}$$

The probability $P(\overline{\varvec{C }}_{j,t}=\varvec{c}_{j} \vert N_{j,t-1}=n_j)$ depends on the price vector $\varvec{x}_t$, and therefore is time dependent. By A2, the probability generating function of $\varvec{N}_t\vert \varvec{N}_{t-1}=\varvec{n}$ is in constrast simpler

$$\begin{aligned} G^P_{\varvec{N}_t\vert \varvec{N}_{t-1}=\varvec{n}}(\varvec{z})= & {} G^P_{\overline{\mathbf{C }}_{1,t}}(\varvec{z})\times \dots \times G^P_{\overline{\mathbf{C }}_{J,t}}(\varvec{z}) = \left( \varvec{z}^T p_{1\rightarrow }(\varvec{x}_t)\right) ^{n_{1}}\times \dots \times \left( \varvec{z}^T p_{J\rightarrow }(\varvec{x}_t)\right) ^{n_{J}}, \end{aligned}$$

where $z\in \mathbb R^J$, ${}^{T}$ denotes the matrix transpose and $G^P(.)$ denotes the probability generating function. Using Proposition 2, we have $\overline{\varvec{C }}_{j,t}~\vert ~ \varvec{N}_{j,0}=\varvec{n}$ follows a multinomial distribution with parameters $\mathcal M_J(n_{j}, \tilde{\varvec{p}}_{j})$. By similar arguments, $G^P_{\varvec{N}_t\vert \varvec{N}_{0}=\varvec{n}}(\varvec{z}) =(P_\rightarrow ^{(t)}\times \varvec{z})^{\varvec{n}}$. If $\mu$ is the invariant measure of $(C_{i,t})_t$, then

$$\begin{aligned} P_\rightarrow ^{(t)} \underset{t\rightarrow +\infty }{\longrightarrow } \left( \begin{matrix}\mu \\ \dots \\ \mu \end{matrix}\right) \Rightarrow P_\rightarrow ^{(t)}\times \varvec{z}= \left( \begin{matrix}\mu ^T \varvec{z}\\ \dots \\ \mu ^T \varvec{z}\end{matrix}\right) \Rightarrow G^P_{\varvec{N}_t\vert \varvec{N}_{t-1}=\varvec{n}}(\varvec{z}) = (\mu ^T \varvec{z})^{\sum _i n_i}. \end{aligned}$$

In other words, the probability generating function of $\varvec{N}_t~\vert ~\varvec{N}_{0}=\varvec{n}$ is the p.g.f. of a multinomial distribution. Since we obtain a limiting distribution for the Markov chain $(\varvec{N}_t)_t$ is also its invariant measure, see, e.g., (Norris 1997, p. 33).

Proof (Proof of Prop. 4)

By A4, the probability generating function of the sum constituting $N_{j,t}$ is the product of generating function of each binomially distributed random variables

$$\begin{aligned} G^P_{N_{j,t}\vert \varvec{N}_{t-1}=\varvec{n}}(z) = \prod _{k=1}^J (1-p_{k \rightarrow j}(\varvec{x}_t) + p_{k \rightarrow j}(\varvec{x}_t) z)^{n_{k}}. \end{aligned}$$

Differentiating with respect to z, we get

$$\begin{aligned}G^P_{N_{j,t}\vert \varvec{N}_{t-1}=\varvec{n}}{}^\prime(z) = \sum _{k=1}^J n_kp_{k \rightarrow j}(\varvec{x}_t) (1-p_{k \rightarrow j}(\varvec{x}_t) + p_{k \rightarrow j}(\varvec{x}_t) z)^{n_{k}-1} \prod _{l\ne k} (1-p_{l \rightarrow j}(\varvec{x}_t) + p_{l \rightarrow j}(\varvec{x}_t) z)^{n_{l}}. \end{aligned}$$

Taking $z=1$ leads to the result. The mass probability function of the portfolio size $N_{j,t}$is given by

$$\begin{aligned} P(N_{j,t} = m_{j}\vert \varvec{N}_{t-1}=\varvec{n}) = \sum _{ \underset{ \text { s.t. } \sum \limits _k c_k = m_{j} }{ 0 \le c_1, \dots , c_{J} \le n } } \prod _{l=1}^J \left( {\begin{array}{c}n_{l}\\ c_j\end{array}}\right) (p_{l\rightarrow j}(\mathbf{x}_t))^{c_j} \left( 1- p_{l\rightarrow j}(\mathbf{x}_t) \right) ^{ n_{l} - c_j}. \end{aligned}$$

1.1.2 Properties of the Loss Model

Proof (Proof of Prop. 5)

Using assumptions A5, A6, A7, the moment generating function of $S_{j,t}$ given that $N_{j,t}=n_{j}$ using (5) is

$$\begin{aligned} \forall u, G^M_{S_{j,t}\vert N_{j,t}=n_{j}}(u) = E\left( e^{uS_{j,t}} \vert N_{j,t}=n_{j}\right) = E\left( \prod _{i=1}^{n_j} e^{uY_{i,t}}\right) = \left( G^M_{Y_{i,t}}(u)\right) ^{n_j}. \end{aligned}$$

Since $Y_{i,t}$ is a compound distribution by (4), we get $G^M_{Y_{i,t}}(u)= G^P_{M_{i,t}}( G^M_Z(u)).$ As the claim frequency belongs to the (a, b, 0) class, the resulting distribution for total claim number can be easily derived, see Table 1. Again using assumptions A5, A6, A7, we have

$$\begin{aligned} \forall u, G^M_{S_{j,t} }(u) = E\left( E\left( e^{uS_{j,t}}\vert N_{j,t}\right) \right) = E\left( \left( G^M_{Y_{i,t}}(u)\right) ^{N_{j,t}} \right) =G^P_{N_{i,t}}\left( G^P_{M_{i,t}}\left( G^M_Z(u)\right)\right). \end{aligned}$$

Using Lemma 1, with $f=G^P_{N_{i,t}}$ $g=G^P_{M_{i,t}}$ and $h=G^M_Z$, we set $x=0$ in order to compute moments so that $h^{(j)}(0)= (G^M_Z)^{(j)}(0) = E(Z^j)$, $h(0)=1$, $g^{(l)}(1)=E(M_{i,t}\dots (M_{i,t}-l+1))$, $f^m(1)= E(N_{i,t}\dots (N_{i,t}-m+1))$. So we have

$$\begin{aligned} \frac{d^n G^M_{S_{j,t} }(0) }{dx^n}= & {} \sum _{m_j\in \mathbb N} \frac{n! E(N_{i,t}\dots (N_{i,t}-m_.+1)) }{ m_1!\,\cdots \,m_n!} \prod _{j=1}^n\left( \sum _{l_j\in \mathbb N} \frac{E(M_{i,t}\dots (M_{i,t}-l_.+1)) }{ l_1!\,\cdots \,l_j!} \right) ^{m_j} \\&\prod _{k=1}^j\left( \frac{E(Z^j) }{ k!}\right) ^{l_k m_j}, \end{aligned}$$

with $m_.=m_1+\cdots +m_n$ and $l_.=l_1+\cdots +l_j$ where the multiple summation is a finite set of integers, see Lemma 1. In particular, $n=1$ leads to $m_1=1$, $l_1=1$. Hence $E(S_{j,t})= \frac{E(N_{i,t}) }{ 1!} \left( \frac{E(M_{i,t}) }{ 1!} \right) ^{1} \left( \frac{E(Z) }{ 1!}\right) ^{1} = E(N_{i,t}) E(M_{i,t}) E(Z).$

Using recursively the Faà di Bruno formula, we obtain the following lemma.

Lemma 1

Assuming f, g and h are nth-time differentiable, we have

$$\begin{aligned} \frac{d^n f(g(h(x)))}{dx^n} = \sum _{m_j\in \mathbb N} \frac{n! f^{(m_1+\cdots +m_n)}(g(h(x)))}{m_1!\,m_2!\,\cdots \,m_n!} \prod _{j=1}^n\left( \sum _{l_j\in \mathbb N} \frac{g^{(l_1+\cdots +l_j)}(h(x))}{l_1!\,\cdots \,l_j!} \right) ^{m_j} \prod _{k=1}^j\left( \frac{h^{(j)}(x)}{k!}\right) ^{l_k m_j}, \end{aligned}$$

where the multiple summation is over integers

$$\begin{aligned} \left\{ m_j\in \mathbb N, \sum _{j=1}^n j m_j=n \right\} , \left\{ l_k\in \mathbb N, \sum _{k=1}^j k l_k=j \right\} . \end{aligned}$$

Appendix 2

1.1 Proofs of Section 4

1.1.1 Properties of the Transition Probability

Proposition 14

Transition probability $p_{l \rightarrow j}(\varvec{x})$ is a strictly decreasing function of $x_j$ given $\varvec{x}_{-j}$ and verifies $0 < p_{l \rightarrow j}(\varvec{x}) < 1$.

Proof

The expression of $p_{j \rightarrow k}$ can be rewritten as

$$\begin{aligned} p_{j \rightarrow k}(\varvec{x}) = p_{j \rightarrow j}(\varvec{x}) \left( \delta _{jk} + (1-\delta _{jk} ) e^{\kern 0.14emf_j(x_{j},x_{k}) } \right) , p_{j \rightarrow j}(x) = \frac{ 1 }{ 1 + \sum \limits _{l \ne j} e^{\kern 0.14emf_j(x_j,x_l) } }, \end{aligned}$$

with $\delta _{ij}$ denoting the Kronecker delta where the summation is over $l \in \{1, \dots , J\} \setminus \{ j \}$ and $f_j$ is the price function. The price function $f_j$ goes from $(t,u) \in \mathbb R^2 \mapsto f_j(t,u) \in \mathbb R$. Partial derivatives are denoted by $\frac{ \partial f_j (t,u) }{ \partial t } = f^\prime_{j1}(t,u)$ and $\frac{ \partial f_j (t,u) }{ \partial u } = f^\prime_{j2}(t,u)$. Derivatives of higher order use the same notation principle. The $p_{j \rightarrow k}(\varvec{x})$ function has the good property to be infinitely differentiable. Since we have

$$\begin{aligned} \frac{ \partial }{ \partial x_i } \sum \limits _{l \ne j} e^{\kern 0.14em f_j(x_j,x_l) } = \delta _{ji} \sum \limits _{l \ne j} f^\prime_{j1}(x_j,x_l) e^{\kern 0.14em f_j(x_j,x_l) } + (1-\delta _{ji}) f^\prime_{j2}(x_j,x_l) e^{\kern 0.14em f_j(x_j,x_i) } , \end{aligned}$$

we deduce

$$\begin{aligned} \frac{ \partial p_{j \rightarrow j}(x) }{ \partial x_i }= & {} - \left( \sum _{l \ne j} f^\prime_{j1}(x_j,x_l) \lg _j^l(x) \right) p_{i\rightarrow j}(x) \delta _{ij} - f^\prime_{j2}(x_j,x_l) p_{j\rightarrow i}(x) p_{j \rightarrow j}(x) (1-\delta _{ij}) . \end{aligned}$$

Furthermore,

$$\begin{aligned} \frac{ \partial }{ \partial x_i } \left( \delta _{jk} + (1-\delta _{jk} ) e^{ \kern 0.14emf_j(x_j,x_k) } \right) = (1-\delta _{jk} ) \left( \delta _{ik} f^\prime_{j2}(x_j,x_k) e^{\kern 0.14em f_j(x_j,x_k) } + \delta _{ij} f^\prime_{j1}(x_j,x_k) e^{\kern 0.14em f_j(x_j,x_k) } \right) . \end{aligned}$$

Hence, we get

$$\begin{aligned} \frac{ \partial p_{j \rightarrow k}(x) }{ \partial x_i } =& - \delta _{ij} \left( \sum _{l \ne j} f^\prime_{j1}(x_j,x_l) p_{j \rightarrow l}(x) \right) p_{j \rightarrow k}(x) - (1-\delta _{ij}) f^\prime_{j2}(x_j,x_i) p_{j \rightarrow i}(x) p_{j \rightarrow k}(x) \\ &+ (1-\delta _{jk}) \left[ \delta _{ij} f^\prime_{j1}(x_j,x_k) p_{j \rightarrow k}(x) + \delta _{ik} f'_{j2}(x_j,x_k) p_{j \rightarrow k}(x) \right] . \end{aligned}$$

Let $\phi _l$ be the family function $x_j \mapsto p_{l \rightarrow j}(\varvec{x})$ for $l=1,\dots ,J$. $\phi _j$ has the following derivative

$$\begin{aligned} \phi _j^\prime(x_j) = - \left( \sum _{l \ne j} f^\prime_{j1}(x_j,x_l) p_{j \rightarrow l}(\varvec{x}) \right) p_{j \rightarrow j}(\varvec{x}). \end{aligned}$$

Since for the two considered price function, we have $\bar{f}{^\prime_{j1}}(x_j,x_l) = \alpha _j /x_l>0$ and $\tilde{f}{^\prime_{j1}}(x_j,x_l) = \tilde{\alpha }_j>0$, then the function $\phi _j$ is strictly decreasing. For $l\ne j$, the function $\phi _l$ has the following derivative $\phi _l^\prime(x_j) = f^\prime_{j2}(x_l, x_j) p_{l \rightarrow j}(\varvec{x}) ( 1- p_{l \rightarrow j}(\varvec{x})).$ Again, for the two considered price function, we have $f^\prime_{j2}(x_j,x_l) = - \alpha _j x_j/x_l^2<0$ and $\tilde{f}{^\prime_{j2}}(x_j,x_l) = - \tilde{\alpha }_j<0$. So, the function $\phi _l$ is strictly decreasing.

Futhermore, the function $\phi _l$ decreases from 1 to 0 such that $\phi _l(x_j)\rightarrow 1$ (resp. $\phi _l(x_j)\rightarrow 0$) when $x_j\rightarrow -\infty$ (resp. $x_j\rightarrow -\infty$). When $i\ne j$, functions $x_i \mapsto p_{i \rightarrow j}(x)$ are also strictly increasing. Let $\overline{\underline{\varvec{x}}^{j}} =(\underline{x}, \dots , \underline{x}, \overline{x}, \underline{x}, \dots , \underline{x})$ and $\underline{\overline{\varvec{x}}^{j}}=(\overline{x}, \dots , \overline{x}, \underline{x}, \overline{x}, \dots , \overline{x})$. We have

$$\begin{aligned} \forall \varvec{x}\in [\underline{x}, \overline{x}]^J, 0< p_{i \rightarrow j}(\underline{\overline{\varvec{x}}^{j}})< p_{i \rightarrow j}(\varvec{x})< p_{i \rightarrow j}(\overline{\underline{\varvec{x}}^{j}}) < 1. \end{aligned}$$

1.1.2 Properties of a Constant Regulated Price Vector

Proof (Proof of Prop. 6)

Using Proposition 3 and when $\mathbf{x}=(x,\dots ,x)$ using (13), we have

$$\begin{aligned} P_\rightarrow =\left( \begin{matrix} p_{1 \rightarrow 1}&{} \frac{1-p_{1 \rightarrow 1}}{J-1} &{} \dots &{} \dots &{} \frac{1-p_{1 \rightarrow 1}}{J-1} \\ &{} \ddots \\ \dots &{} \frac{1-p_{j \rightarrow j}}{J-1} &{} p_{j \rightarrow j}&{} \frac{1-p_{j \rightarrow j}}{J-1} &{} \dots \\ &{}&{}&{} \ddots \\ \frac{1-p_{J \rightarrow J}}{J-1} &{} \dots &{} \dots &{} \frac{1-p_{J \rightarrow J}}{J-1} &{} p_{J \rightarrow J}\end{matrix}\right) \in \mathbb R^{J\times J}. \end{aligned}$$

(21)

We deduce that the probabilities appearing in the proof of Proposition 3 simplify to

$$\begin{aligned} \frac{n_j!}{c_{j1}!\dots c_{jJ}!} \left( p_{j \rightarrow j}\right) ^{c_{jj}} (p_{j \ne })^{c_{j1}} \dots \left( p_{j \ne }\right) ^{c_{jJ}}= & {} \frac{n_j!}{c_{j1}!\dots c_{jJ}!} \left( p_{j \rightarrow j}\right) ^{c_{jj}} \left( p_{j \ne }\right) ^{n_j - c_{jj}}. \end{aligned}$$

We get the desired result by summing over appropriate indexes.

Proof (Proof of Prop. 7)

For a constant price vector, identical players ($p_{j \rightarrow j}=p_=$ and $p_{j \ne }=p_{\ne }$) and Proposition 4, the probability generating function is

$$\begin{aligned} G^P_{N_{j,t}\vert \varvec{N}_{t-1}=\varvec{n}}(z)= & {} (1-p_{j \rightarrow j}+ p_{j \rightarrow j}z)^{n_{j}} \prod _{l\ne j}(1-p_{\ne } + p_{\ne } z)^{n_{l}} \\= & {} (1-p_{j \rightarrow j}+ p_{j \rightarrow j}z)^{n_{j}} (1-p_{\ne } + p_{\ne } z)^{N-n_j}. \end{aligned}$$

$N_{j,t}\vert \varvec{N}_{t-1}=\varvec{n}$ is a sum of two binomially distributed random variables $\mathcal B(n_j, p_{j \rightarrow j})$ and $\mathcal B(n-n_j, p_{j \ne })$.

Proof (Proof of Th. 1)

When $\varvec{x}_t=\varvec{x}$, $(C_{i,t})_t$ is a Markov chain with a transition matrix $P_\rightarrow (\varvec{x}_t)=P_\rightarrow$. When $\varvec{x}=(x,\dots ,x)$ (13) leads to (21). Since the number of state is finite (J) and the Markov chain is irreducible by Prop. 14, there exists a unique invariant measure $\mu$, see e.g. Norris (1997).

Let us consider the general matrix $M\in \mathbb R^{J\times J}$ with general term $M_{i,j} = a_i (1-\delta _{ij}) + (1-(J-1)a_i) \delta _{ij}$. Note that the rows of M equal 1 and M has only two different terms by row. The reversibility conditions for a measure $\mu$ are

$$\begin{aligned} \left\{ \begin{array}{ll}\mu _1 M_{1,2} = \mu _2 M_{2,1} \\ \dots \\ \mu _1 M_{1,J} = \mu _J M_{J,1} \\ \end{array}\right. \Leftrightarrow \left\{ \begin{array}{ll}\mu _1 a_1/a_2 = \mu _2 \\ \dots \\ \mu _1 a_1/a_J = \mu _J. \\ \end{array}\right. \end{aligned}$$

Let $a_{-i}^\Pi = \prod \limits _{j=1, j\ne i}^J a_j$. Using $\mu _1+\dots +\mu _J=1$, we get by multiplying both sides by $a_{-i}^\Pi$

$$\begin{aligned} \mu _1 + \sum _{i>2} \mu _1 \frac{a_1}{a_i} =1 \Leftrightarrow \mu _1 = \frac{a_{-1}^\Pi }{\sum \limits _{i=1}^J a_{-i}^\Pi } \Leftrightarrow \mu = \left( \frac{a_{-1}^\Pi }{\sum \limits _{i=1}^J a_{-i}^\Pi }, \dots , \frac{a_{-J}^\Pi }{\sum \limits _{i=1}^J a_{-i}^\Pi }\right) . \end{aligned}$$

The measure $\mu$ above is in detailed balance with M and also an invariant measure for M. Setting $a_j = p_{j \ne }$ leads to the desired result. In the special case where $p_{j \rightarrow j}$ are identical across insurers, $p_{j \ne }= p_{\ne }$ is constant. Hence for all $j=1,\dots ,J$

$$\begin{aligned} a_{-i}^\Pi = \prod _{j\ne i} p_{j \ne }= (p_{\ne })^{J-1} \Rightarrow \mu _i = \frac{(p_{\ne })^{J-1}}{\sum \limits _{l=1}^J (p_{\ne })^{J-1}} = \frac{(p_{\ne })^{J-1}}{J \times (p_{\ne })^{J-1}} = \frac{1}{J}. \end{aligned}$$

Proof (Proof of Prop. 8)

Using (4) and Proposition 1, for large t, we have

$$\begin{aligned} P(\varvec{N}_t = \varvec{n}) = \frac{N!}{(c_1^\Pi +\dots +c_J^\Pi )^N} \prod _{j=1}^J \frac{(c_j^\Pi )^{n_j}}{n_j!}. \end{aligned}$$

The asymptotic marginal distribution is binomial $\mathcal B(N, \mu _j)$ with $P(N_{j,t} = m) = \left( {\begin{array}{c}N\\ m\end{array}}\right) (\mu _j)^m (1-\mu _j)^{N-m} .$ This leads to the desired result for the aggregate claim amount.

Appendix 3

1.1 Proofs of Section 5

Proof (Proof of Th. 2)

Since $\sum _k p_{j \rightarrow k}= 1$, we get $p_{j\ne } = (1-p_{j \rightarrow j}-p_{j\rightarrow 1})/(J-2)$, and $p_{1\ne }=\frac{1-p_{1\rightarrow 1}}{J-1}$. We use the following notation

$$\begin{aligned} P_\rightarrow =\left( \begin{matrix} p_{1 \rightarrow 1}&{} p_{1\ne } &{} \dots \\ p_{2\rightarrow 1} &{} p_{2\rightarrow 2} &{} p_{2\ne } &{} \dots \\ p_{3\rightarrow 1} &{} p_{3\ne } &{} p_{3\rightarrow 3} &{} p_{3\ne } &{} \dots \\ &{}&{}&{} \ddots &{} \\ p_{J\rightarrow 1} &{} p_{J\ne } &{} \dots &{} p_{J\ne } &{} p_{J\rightarrow J} \\ \end{matrix}\right) = \left( \begin{matrix} a_1 &{} b_1 &{} \dots \\ b_2 &{} a_2 &{} c_2 &{} \dots \\ b_3 &{} c_3 &{} a_3 &{} c_3 &{} \dots \\ &{}&{}&{} \ddots &{} \\ b_J &{} c_J &{} \dots &{} c_{J} &{} a_J \\ \end{matrix}\right) . \end{aligned}$$

(22)

A first series of equation for the invariant measure is obtained from $\mu ^\prime=\mu ^\prime P_\rightarrow$. Ignoring the first equation and subtracting the second equation from all others, we get

$$\begin{aligned} \left\{ \begin{array}{ll}\mu _2 = b_1 \mu _1 +a_2\mu _2 +\dots + c_J \mu _J \\ \mu _3 -\mu _2= (c_2-a_2)\mu _2+ (a_3-c_3)\mu _3 \\ \vdots \\ \mu _J -\mu _2= (c_2-a_2)\mu _2 + (a_J-c_J)\mu _J \end{array}\right. \Leftrightarrow \left\{ \begin{array}{ll}-b_1 \mu _1 =-\mu _2 +a_2\mu _2 + c_3\mu _3+\dots + c_J \mu _J \\ \mu _3(1-a_3+c_3) = (c_2-a_2+1)\mu _2 \\ \vdots \\ \mu _J(1-a_J+c_J) = (c_2-a_2+1)\mu _2. \end{array}\right. \end{aligned}$$

The $J-2$ equations give $\mu _j = \mu _2 \frac{c_2-a_2+1}{c_j-a_j+1}, j>2.$ Recalling that any row of M sums up to 1, $a_i+b_i+c_i(J-2)=1$ for $i\ne 1$, we have

$$\begin{aligned} \forall j=3,\dots , J, a_j = 1- c_j(J-2) - b_j \Rightarrow c_j-a_j+1 = (J-1)c_j +b_j =:d_j. \end{aligned}$$

For $j=3,\dots , J$, $\mu _j = \mu _2 \frac{d_2}{d_j}$. The first equation for $\mu _1$ becomes

$$\begin{aligned} b_1 \mu _1 =\mu _2 -a_2\mu _2 - \sum _{j=3}^J c_j\mu _2 \frac{d_2}{d_j} \Leftrightarrow \mu _1 =\mu _2 \frac{b_2}{b_1} + \mu _2 \frac{d_2}{b_1} \sum _{j=3}^J\left(\frac{c_2}{d_2} - \frac{c_j}{d_j} \right). \end{aligned}$$

Using the condition $\sum _i \mu _i=1$ yields to $\mu _2 = \left( 1+ \frac{b_2}{b_1} + \frac{d_2}{b_1} \sum _{j=3}^J\left(\frac{c_2}{d_2} - \frac{c_j}{d_j} +\frac{b_1}{d_j}\right) \right) ^{-1}.$ Reintroducing the product notation $d^\Pi _{-1,-j}= \prod _{l\ne 1, j} d_l$ yields the following reformulation

$$\begin{aligned} \mu _j = \frac{d^\Pi _{-1,-j}b_1}{ d^\Pi _{-1} + \sum \limits _{j=2}^Jd^\Pi _{-1,-j}(b_1-c_j) }, j> 2, \mu _1 =\frac{d^\Pi _{-1} + \sum \limits _{j=2}^Jd^\Pi _{-1,-j}(-c_j) }{ d^\Pi _{-1} + \sum \limits _{j=2}^Jd^\Pi _{-1,-j}(b_1-c_j) }. \end{aligned}$$

Let us go back to the original transition matrix (22) with $a_1 = p_{1 \rightarrow 1}$, $b_1 = p_{1\ne } = \frac{1-p_{1 \rightarrow 1}}{J-1}$, $\forall j>1, a_j = p_{j\rightarrow j}$, $b_j = p_{j\rightarrow 1}$, $c_j = p_{j\ne }$. With $d_j = (J-1)c_j+b_j = (J-1) p_{j\ne } +p_{j\rightarrow 1}$, we obtain the desired result. In the special case of identical insurers, we have $\forall j \ne 1, p_{j\ne }=p_{2\ne }$ and $p_{j\rightarrow 1}=p_{2\rightarrow 1}$. With $d_j = d_2\Rightarrow d_{-1,-j}^\Pi = d_2^{J-2} \Rightarrow d_{-1}^\Pi = d_2^{J-1}$, we obtain the desired result.

Proof (Proof of Prop. 9)

Consider Insurer j is the cheapest, i.e. $x_j < x_k$ for all $k \ne j$. $p_{k \rightarrow j}(x) > p_{k \rightarrow l}(x)$ for $l\ne j$ given the initial portfolio sizes $n_j$’s are constant, since the change probability $p_{k \rightarrow j}$ (for $k\ne j$) is a decreasing function (see Appendix 2). Using the stochastic order ($\le _{\text {st}}$), the convex order ($\le _{\text {cx}}$), the majorization order ($\le _{\text {m}}$), see (Shaked and Shanthikumar 2007, resp. Chap. 1, Chap. 3, p. 2), we can show a stochastic order of the portfolio size by applying the convolution property of the stochastic order J times: $N_k(\varvec{x}) \le _{\text {st}}N_j(\varvec{x}) , \forall k \ne j.$  

Let us consider the empirical loss average of an insurer with portfolio size n $\overline{A}(n) = \frac{1}{n} \sum _{i=1}^{n} Y_i ,$ where $Y_i$ denotes the total claim amount for Policy i. For $n < \tilde{n}$, we define two policy numbers $\varvec{a}_{n}, \varvec{b}_{\tilde{n}} \in \mathbb R^{ \tilde{n}}$ as

$$\begin{aligned} \varvec{b}_{\tilde{n}} = \left( \frac{1}{\tilde{n}}, \dots , \frac{1}{\tilde{n}} \right) \text { and } \varvec{a}_{ n} = \left( \underbrace{ \frac{1}{ n}, \dots , \frac{1}{ n} }_{\text {size } n} , \underbrace{ 0, \dots , 0 }_{\text {size } \tilde{n} - n} \right) . \end{aligned}$$

Since $\varvec{b}_{\tilde{n}} \le _{\text {m}}\varvec{a}_{ n}$ and $(Y_i)_i$’s are i.i.d. random variables, we have

$$\begin{aligned} \sum _i b_{\tilde{n}, i } Y_i \le _{\text {cx}}\sum _i a_{ n, i } Y_i \Leftrightarrow \sum _{i=1}^{\tilde{n}} \frac{1}{\tilde{n}} Y_i \le _{\text {cx}}\sum _{i=1}^{n} \frac{1}{n} Y_i \Leftrightarrow \overline{A}(\tilde{n}) \le _{\text {cx}}\overline{A}(n). \end{aligned}$$

Using Theorem 3.A.23 of Shaked and Shanthikumar (2007), except that for all $\phi$ convex, $E(\phi (\overline{A}(n)))$ is a decreasing function (rather an increasing function) of n and $N_k(\varvec{x}) \le _{\text {st}}N_j( \varvec{x})$, we can show $\overline{A}(N_j( \varvec{x})) \le _{\text {cx}}\overline{A}(N_k(\varvec{x})).$  

Proof (Proof of Prop. 10)

Using Prop. 9, we have $\sum _{i=1}^{\tilde{n}} \frac{1}{\tilde{n}} Y_i \le _{\text {cx}}\sum _{i=1}^{n} \frac{1}{n} Y_i.$ For all increasing convex functions $\phi$, the function $x\mapsto \phi (x+a)$ is still increasing and convex. Thus for all random variables X, Y such that $X\le _{\text {icx}}Y$ and real numbers a, b, $a\le b$, we have

$$\begin{aligned} E(\phi (X+a)) \le E(\phi (X+b)) \le E(\phi (Y+b)) \Leftrightarrow a+X \le _{\text {icx}}b+Y. \end{aligned}$$

Consider Insurer j is the cheapest and $x_j(1-e_j) \le x_k (1-e_k)$ and using the fact that $X \le _{\text {cx}}Y$ is equivalent to $-X \le _{\text {cx}}-Y$, we have $uw_j(\varvec{x},\tilde{n}) \le _{\text {icx}}uw_k(\varvec{x},n) , \forall k \ne j.$ Using Theorem 3.A.23 of Shaked and Shanthikumar (2007), except that for all $\phi$ convex, $E(\phi (uw_j( \varvec{x}, n)))$ is a decreasing function of n and $N_k(\varvec{x}) \le _{\text {st}}N_j( \varvec{x})$, we can show $UW_j = uw_j(\varvec{x},N_j( \varvec{x})) \le _{\text {icx}}uw_k(\varvec{x}, N_k(\varvec{x})) = UW_k.$  

Appendix 4

1.1 Proofs of Section 6

Proof (Proof of Prop. 11)

Assuming $\# J_t \ge 2$, since the strategy set is $[\underline{x},\overline{x}]^{J_t}$, it guarantees the market proxy $m_j= m_j(\varvec{x})$ or $m_j= m_j(\varvec{x},\varvec{n})$ to be positive. Given $\varvec{x}_{-j,t}$, the function $x_{j,t}\mapsto O_j(\varvec{x}_t)$ is a quadratic (hence concave) function. Given that the constraint functions (11) are linear, by Theorem 1 of Rosen (1965), the existence of a premium equilibrium at time t is guaranteed. The proof of uniqueness is exactly the same as in Dutang et al. (2013). Omitting the time index t, consider a generic objective function

$$O_j(\varvec{x}) = \left( a_j - b_j \frac{x_j}{m_j(\varvec{x})}\right) (x_j-c_j), \text {with } m_j(\varvec{x}) = \sum _{i\ne j} w_i x_i,$$

with known weights $w_i>0$ and positive constant $a_j,b_j,c_j>0$. In the case of no active constraint functions. Similarly to Dutang et al. (2013), if $\varvec{x}^\star$ is a Nash equilibrium, $\varvec{x}^\star$ must verify

$$\forall j, \nabla _{x_j} O_j(\varvec{x}^\star ) = 0 \Leftrightarrow M_1M_2 \varvec{x}= v,$$

with

$$M_1 =\left( \begin{matrix}2b_1/w_1 &{} -a_1 &{} \dots \\ -a_2 &{} 2b_2/w_2 &{} -a_2 &{} \dots \\ &{}&{} \ddots &{} \\ &{}\dots &{} -a_J &{} 2b_J/w_J \end{matrix}\right) , M_2= \left( \begin{matrix}w_1 &{} 0 &{} \dots \\ &{} \ddots \\ \dots &{} 0 &{} w_J \end{matrix}\right) , v = \left( \begin{matrix} b_1 c_1\\ \vdots \\ b_J c_J \end{matrix}\right) .$$

NB: When the market proxy is the arithmetic mean (8), we set $w_j = 1$ and $b_j=\beta _j(J-1)$. When the market proxy is the weighted mean (9), we set $w_j = n_j$ and $b_j=\beta _j(N-n_j)$.

Remark 14

Getting a linear system for the premium equilibrium, we are looking for a necessary and sufficient condition. The linear system for the premium equilibrium has a solution when the determinant of M is non null. Since $\det {M}=\det {M_1}\det {M_2}$, $M_2$ being diagonal yields a positive determinant $\det {M_2}=\prod _i w_i >0$ for the weights considered. So $\det {M_0}=0$ is equivalent to $\det {M_1}=0$.

Using Lemma 2, we have with $a_j=1+\beta _j$ and $b_j=\beta _j$

$$\begin{aligned} \det {M_1}= & {} \frac{2b_1}{w_1} \prod _{k\ne 1} (a_k+2b_k/w_k) - \sum _{j=2}^J a_j \prod _{k\ne j} (a_k+2b_k/w_k) . \end{aligned}$$

Using $\tilde{\beta }_j= 1+\beta _j+2\beta _jw_{-j}^\Sigma /w_j$, $w_{-j}^\Sigma =\sum _{k\ne j} w_k$, $\tilde{\beta }_{-j}^\Pi =\prod _{k\ne j} \tilde{\beta }_k$,

$$\begin{aligned} \det {M}=0 \Leftrightarrow 2\beta _1 w_{-1}^\Sigma \tilde{\beta }_{1} = w_1\sum _{j=2}^J (1+\beta _j) \tilde{\beta }_{j} . \end{aligned}$$

There are many solutions to this equation, but there is a unique solution when $w_j=w$ is constant and $\beta _j$’s are all identical. Say $\beta _j=\beta$ leading to $\tilde{\beta }_j = 1+\beta +2\beta (J-1)$. Hence $\det {M}=0$ yields to $\beta =1$

A sufficient condition for the linear system to have a solution is M to be diagonally dominant. That is $\forall j=1,\dots , J$,

$$\begin{aligned} |2b_j|> \sum _{k\ne j} | -a_jw_k| \Leftrightarrow 2b_j> a_j \sum _{k\ne j} w_k \Leftrightarrow 2b_j > a_j w_{-j}^\Sigma \end{aligned}$$

In the case of objective function (10), we choose $a_j=1+\beta _j$, $b_j=\beta _j w_{-j}^\Sigma$. So the sufficient condition is $2\beta _j w_{-j}^\Sigma> (1+\beta _j)w_{-j}^\Sigma \Leftrightarrow \beta _j >1.$ In that case, one can check that the determinant $\det {M_1}$ is strictly positive. This fact was also seen in Dutang et al. (2013).

Lemma 2

Consider the following multi-diagonal matrix for $n\ge 2$

$$\begin{aligned} M_n= \left( \begin{matrix}u_1 &{} v_1 &{} \dots \\ v_2 &{} u_2 &{} v_2 &{} \dots \\ &{} &{} \ddots \\ &{}\dots &{}v_{n-1} &{} u_{n-1} &{}v_{n-1} \\ &{}&{}\dots &{}v_n &{} u_n \end{matrix}\right) . \end{aligned}$$

With $w_1=u_1$ and $w_j=v_j, \forall j=2,\dots ,n$, the determinant is given by $\det {M_n} =(-1)^{n+1} \sum _{j=1}^n w_j \prod _{k\ne j} (v_k-u_k).$

Proof

If $\det {M_n}=(-1)^{n+1} \sum _{j=1}^n w_j \prod _{k\ne j} (v_k-u_k)$, then

$$\begin{aligned} \det {M_{n+1}}= & {} (u_{n+1}-v_{n+1}) \det {M_{n}} +(-1)^{n+1} (v_1-u_1) \dots (v_{n}-u_{n})v_{n+1} \\= & {} (u_{n+1}-v_{n+1}) (-1)^{n+1} \sum _{j=1}^n w_j \prod _{k\ne j} (v_k-u_k) +(-1)^{n+1} (v_1-u_1) \dots (v_{n}-u_{n})v_{n+1} \\= & {} (-1)^{n+2} \sum _{j=1}^{n+1} w_j \prod _{k\ne j} (v_k-u_k). \end{aligned}$$

For $n=2$, $\det {M_2} =u_1u_2 - v_1v_2 = (u_2-v_2) u_1 +(-1)^{2+1} (v_1-u_1)v_2.$

Proof (Proof of Prop. 13)

Using Proposition 14, we have

$$\begin{aligned} 0< p_{i \rightarrow j}(\underline{\varvec{x}}^{j-})< p_{i \rightarrow j}(\varvec{x})< p_{i \rightarrow j}(\overline{\varvec{x}}^{j}_-) < 1, \end{aligned}$$

for all $\mathbf{x}\in [\underline{x}, \overline{x}]^J$. Taking supremum and infimum on player j, we get $0 < \underline{p}_l= \underset{j}{\inf } p_{i \rightarrow j}(\underline{\mathbf{x}}^{j-})$ and $\underset{j}{\sup }~p_{i \rightarrow j}(\overline{\mathbf{x}}^{j}_-) = \overline{p}_l < 1.$ Using the proof of Th. 4 in Appendix 1, we have

$$\begin{aligned}P\left( N_{j,t}(\mathbf{x}) = m_j | N_{j,t-1}> 0 , Card(J_{t-1})> 1 \right) &= \sum _{ \underset{ \text { s.t. } \sum _l \tilde{m}_l = m_j }{ \tilde{m}_1, \dots , \tilde{m}_{J_{t-1}}\ge 0 } } \prod _{ l \in J_{t-1} } \left( {\begin{array}{c}n_{l,t-1}\\ \tilde{m}_l\end{array}}\right) p_{i \rightarrow j}(\mathbf{x})^{\tilde{m}_l} \left( 1- p_{i \rightarrow j}(\mathbf{x}) \right) ^{ n_{l,t-1} - \tilde{m}_j} \\& > \sum _{ \underset{ \text { s.t. } \sum _l \tilde{m}_l = m_j }{ \tilde{m}_1, \dots , \tilde{m}_{J_{t-1}}\ge 0 } } \prod _{ l \in J_{t-1} } \left( {\begin{array}{c}n_{l,t-1}\\ \tilde{m}_l\end{array}}\right) \underline{p}_l^{\tilde{m}_l} (1- \overline{p}_l)^{ n_{l,t-1} - \tilde{m}_j} = \xi > 0. \end{aligned}$$

Therefore,

$$\begin{aligned}P(Card(J_t) = 0 | Card(J_{t-1})> 1) &\ge P\left( \forall j \in J_{t-1}, N_{j,t}(\varvec{x})> 0, K_{j,t-1} + N_{j,t}(\varvec{x}) x^\star _{j,t}(1-e_j)< \sum _{i=1}^{N_{j,t}(\varvec{x}) } Y_i | Card(J_{t-1})> 1 \right) \\& \ge \sum _{ m_j =1 }^{N} P_t\left( N_{j,t}(\varvec{x}) = m_j | Card(J_{t-1})> 1 \right) P\left( K_{j,t-1} + m_j x^\star _{j,t}(1-e_j)< \sum _{i=1}^{m_j } Y_i \right) \\& > \sum _{ m_j =1 }^{N} \xi P\left( K_{j,t-1} + m_j x^\star _{j,t}(1-e_j) < \sum _{i=1}^{m_j } Y_i \right) =\bar{\xi }> 0. \end{aligned}$$

Thus, we have

$$\begin{aligned}P(Card(J_t)> 1 | Card(J_{t-1})> 1) = & \ 1- P(Card(J_t) = 0 | Card(J_{t-1})> 1) \\& -P(Card(J_t)= 1 | Card(J_{t-1})> 1) \\ \le& \ 1- P(Card(J_t) = 0 | Card(J_{t-1}) > 1) \\ < &\ 1- \bar{\xi }< 1. \end{aligned}$$

By successive conditioning, we get

$$\begin{aligned} P(Card(J_t)> 1 ) = P(Card(J_0)> 1) \prod _{s=1}^t P(Card(J_s)> 1 | Card(J_{s-1}) > 1) < \left( 1- \bar{\xi }\right) ^t. \end{aligned}$$

So, the probability $P(Card(J_t) > 1 )$ decreases geometrically as t increases.

Proof (Proof of Prop. 12)

In a regulated market ($x_{j,t}^\star =x_j$), using Theorems 1 and 2, the asymptotic distribution of $\varvec{N}_t$ is a multinomial distribution. Hence the leadership probability is independent of initial condition $\phi _j(k_j,n_j)=\phi _j$. Using the probability mass function leads to the desired result.

See Table 7, 8 and 9

Table 7 Parameters of the lapse model

Table 8 Parameters of the loss model PLN

Table 9 Parameters of insurers