Matrix representations of life insurance payments

Abstract

A multi-state life insurance model is described naturally in terms of the intensity matrix of an underlying (time-inhomogeneous) Markov process which specifies the dynamics for the states of an insured person. Between and at transitions, benefits and premiums are paid, defining a payment process, and the technical reserve is defined as the present value of all future payments of the contract. Classical methods for finding the reserve and higher order moments involve the solution of certain differential equations (Thiele and Hattendorff, respectively). In this paper we present an alternative matrix-oriented approach based on general reward considerations for Markov jump processes. The matrix approach provides a general framework for effortlessly setting up general and even complex multi-state models, where moments of all orders are then expressed explicitly in terms of so-called product integrals of certain matrices. Thiele and Hattendorff type of theorems may be retrieved immediately from the matrix formulae. As a main application, methods for obtaining distributions and related properties of interest (e.g. quantiles or survival functions) of the future payments are presented from both a theoretical and practical point of view, employing Laplace transforms and methods involving orthogonal polynomials.

Appendix: On the \(L_2\) condition without piecewise constancy

Remark 7

Condition (iii) in Theorem 7 is not far from being necessary. Consider as an example the disability model in the time interval [0, 1] with states 0: active, 1: disabled, 2: dead and allowing recovery, with the same intensity $\lambda >0$ for transitions from 0 to 1 as from 1 to 0 and (for simplicity) mortality rate 0 and discounting rate $r=0$. The benefits are a lump sum $b^{01}=1$ upon transition from 0 to 1 and the contributions a constant payment rate $b^0<0$ when active.

When $Z(0)=0$, the total number N of transitions in [0, 1] is Poisson$(\lambda )$, the total benefits $U_1(0,1)$ are $M=\lceil N/2\rceil$, and the total contributions $U_0(0,1)$ equal $|b^1|$ times the total time $T_0$ spent in state 0. Thus $U(0,1)=$$U_1(0,1)-U_0(0,1)$$=M-U_0(0,1)$ Obviously, $U_0(0,1)$ is concentrated on the interval $[0,|b^1|]$ with a density g(x) which is bounded away from 0, say $h(x)>c_1>0$.

Assuming, again for simplicity, that $b^1>1$. Then the intervals $[m- b^1,m]$ overlap and so for a given x, at least one of them contribute to f(x). One candidate is the one with $m=\lceil x\rceil$. This gives

$$\begin{aligned} f(x)&\ge \ \mathbb {P}(M=\lceil x\rceil )h(\lceil x\rceil -x)\ \ge \ \mathbb {P}(N=2\lceil x\rceil )c_1\,. \end{aligned}$$

But using Stirling’s approximation to estimate the Poisson probability, it follows after a little calculus that $\mathbb {P}(N=2\lceil x\rceil )\ge$$c\mathrm {e}^{-4x\log x}$ for all large x, say $x\ge x_0$, and some $c>0$ (in fact the 4 can be replaced by any $c_3>2$). This gives for a normal$(\mu ,\sigma ^2)$$f_0$ that

$$\begin{aligned} \int \frac{f^2}{f_0}\ \ge \ \int _{x_0}^\infty \frac{f^2(x)}{f_0(x)}\,\mathrm {d}x \ \ge \ c_4\int _{x_0}^\infty \exp \{(x-\mu )/\sigma ^2-8x\log x\}\,\mathrm {d}x\ =\ \infty. \end{aligned}$$

That is, (8.1) fails and a similar calculation gives that (8.5) does so too.

The obvious way out is of course to take $f_0$ with a heavier tail than the normal, say doubly exponential (Laplacian) with density $\mathrm {e}^{-|x|}$ for $-\infty<x<\infty$. Given that this example and other cases where condition (iii) is violated do not seem very realistic, we have not exploited this further.

Fig. 5

The sample space partition. State 0= active = green, State 1 = disabled = blue, State 2 = dead = red

Example 4

Consider again the disability model with states 0: active, 1: disabled, 2: dead and no recovery. As in [11], we assume that the payment stream has the form $b^0(t)=-b_-^0$ for $t\le S$ and $b^0(t)=b_+^0$ for $t>S$,¹$b^1(t)\equiv b^1$ for all t. For example S could be the retirement age, say 65. Without recovery, the only non-zero transition rates are the $\mu _{01}(t)$, $\mu _{02}(t)$, $\mu _{12}(t)$. With the values used in [11], these are bounded away from 0 and $\infty$ on [0, T] for any T (say 75 or 80), and this innocent assumption is all that matters for the following.

Define the stopping times

$$\begin{aligned} \tau _0^1&=\ \inf \{t>S:\, Z(t)=1,\,Z(S)=0\text { for }S<t\}\\ \tau _1^2&=\ \inf \{t>\tau _0^1:\, Z(t)=2\}\\ \tau _0^2&=\ \inf \{t>S:\, Z(t)=2,\,Z(S)=0\text { for }S<t\} \end{aligned}$$

with the usual convention that the stopping time is $\infty$ if there is no t meeting the requirement in the definition. One then easily checks that the sets $F_0,\ldots ,F_{1_+2_+}$ defined in Fig. 5 defines a partition of the sample space. Here $F_0$ (corresponding to zero transitions in [S, T]) contributes with an atom at

$$\begin{aligned} a= & \, -b_-^0(1-\mathrm {e}^{-rS })/r+b_+^0(\mathrm {e}^{- rS}-\mathrm {e}^{- rT})/r\quad \text { with probability }\\ q= & \, \exp \left\{ \int _0^T\mu _{00}(u)\,\mathrm {d}u\right\} \end{aligned}$$

and the remaining 7 events with absolutely continuous parts, say with (defective) densities $g_{2_-},\ldots ,g_{1_+2_+}$. It is therefore sufficient to show that each of these is bounded. In obvious notation, the contribution to U(S, T) of the first 4 of these 7 events (corresponding to precisely one transition in [S, T]) are

$$\begin{aligned} A_{2-}= & \, \left[ -b^0_-(1-\mathrm {e}^{-r\tau _0^2})/r\right] \cdot 1_{\tau _0^2\le S}\,,\\ A_{2-}= & \, \left[ -b^0_-(1-\mathrm {e}^{-rS})/r+b_+^0(\mathrm {e}^{-rS}-\mathrm {e}^{-r\tau _0^2})/r\right] \cdot 1_{S<\tau _0^2\le T}\,,\\ A_{1-}= & \, \left[ -b^0_-(1-\mathrm {e}^{-r\tau _0^1})/r+b^1(\mathrm {e}^{-rT}-\mathrm {e}^{-r\tau _0^1})/r\right] \cdot 1_{\tau _0^1\le S}\,,\\ A_{1_+}= & \, \left[ -b^0_-(1-\mathrm {e}^{-rS})/r+b_+^0(\mathrm {e}^{-rS} -\mathrm {e}^{-r\tau _0^1})/r+b^1(\mathrm {e}^{-rT}-\mathrm {e}^{-r\tau _0^1})/r\right] \cdot 1_{S<\tau _0^1\le T}\,. \end{aligned}$$

The desired boundedness of $g_{2-},g_{2+},g_{1-},g_{1_+}$ therefore follows from the following lemma, where $\tau$ may be improper so that $\int h<1$:

Lemma 3

If a r.v. $\tau$ has a bounded density h and the function $\varphi$ is monotone and differentiable with $\varphi '$ bounded away from 0, then the density of $\varphi (\tau )$ is bounded as well.

Proof

The assumptions imply the existence of $\psi =\varphi ^{-1}$. Now just note that the density of $\varphi (\tau )$ is $h(\psi (x))/\varphi '(\psi (x))$ if $\varphi$ is increasing and $h(\psi (x))/\left| \varphi '(\psi (x))\right|$ if it is decreasing.$\square$

The cases of $g_{1_-2_-},g_{1_-2_+},g_{1_+2_+}$ (corresponding to precisely two transitions in (s, T]) is slightly more intricate. Consider first $g_{1_-2_-}$. The contribution to U(0, T) is here

$$\begin{aligned} A_{1_-2_-}\ =\ -b_-^0\int _0^{\tau _0^1}\mathrm {e}^{-ru}\,\mathrm {d}u\,+\, b^1\int _{\tau _0^1}^{\tau _1^2}\mathrm {e}^{-ru}\,\mathrm {d}u\ =\ c_0+c_1\mathrm {e}^{-r\tau _0^1}-c_2\mathrm {e}^{-r\tau _1^2} \end{aligned}$$

Now the joint density $h(t_1,t_2)$ of $(\tau _0^1,\tau _0^2)$ at a point $(t_1,t_2)$ with $0<t_1<t_2\le R$ is

$$\begin{aligned} \exp \left\{ \int _0^{t_1}\mu _{00}(u)\,\mathrm {d}u\right\} \mu _{01}(t_1) \cdot \exp \left\{ \int _{t_1}^{t_2}\mu _{11}(v)\,\mathrm {d}v\right\} \mu _{12}(t_2) \end{aligned}$$

so that $h(t_1,t_2)$ is bounded. Consider now the transformation taking $(\tau _0^1,\tau _0^2)$ into $(\tau _0^1,A_{12_-})$. The inverse of the Jacobian J is

$$\begin{aligned} \begin{vmatrix} 1&0 \\ -rc_1\mathrm {e}^{-r\tau _0^1}&rc_2\mathrm {e}^{-r\tau _1^2} \end{vmatrix}\ =\ rc_2\mathrm {e}^{-r\tau _1^2} \end{aligned}$$

which is uniformly bounded away from 0 when $\tau _1^2\le S$. Therefore also the joint density $k(z_1,z_2)$ of $(\tau _0^1,A_{12-})$ is bounded. Integrating $k(z_1,z_2)$w.r.t. $z_1$ over the finite region $0\le z_1\le S$ finally gives that $g_{1_-2_-}$ is bounded.

The argument did not use that $\tau _1^2\le S$ and hence also applies to $g_{1_-2_+}$. Finally note that the contribution to $g_{1_+2_+}$ from [0, S] is just a constant, whereas the one from (S, T] has the same structure as used for $g_{1_-2_-}$. Hence also $g_{1_+2_+}$ is bounded.

Remark 8

The calculations show that $F_{1+}$ also is an atom if $b_+^0=b^1$. However, since the contribution rate $b_-^0$ is calculated via the equivalence principle given the benefits $b^0_+,b^1$ and the transition intensities, this would be a very special case. The somewhat less special situation $b^0_+=b^1$ (same annuity to an disabled as to someone retired as active) also gives an atom, now at $F_{1+}$. In fact, $b_-^0=b^1$ is assumed in [11].

Remark 9

For a counterexample to (8.1), consider again the disability model with the only benefit being a lump sum of size $b^{01}(t)=$$\mathrm {e}^{rt}\varphi (t)$ being paid out at $\tau _0^1$ where $\varphi (t)=$$(t-a)^21_{t\le a}+b$, cf. Fig. 6.

Fig. 6

$\varphi (t)$

Then $U(s,T]=\varphi (\tau _0^1)$ has an atom at b (corresponding to $\tau _0^1>a$) and an absolutely continuous part on $(b,b+a^2]$. Letting $\psi :\,[b,b+a^2]\mapsto [0,a]$ be the inverse of $\varphi$, $t=\psi (y)$ satisfies

$$\begin{aligned} y=(t-a)^2+b\ \Rightarrow (y-b)^{1/2}=-(t-a)\ \Rightarrow \ t=\psi (y)=a-(y-b)^{1/2}\,, \end{aligned}$$

where the minus sign after the first $\Rightarrow$ follows since $\psi$ is decreasing. With h the density of $\tau _0^1$, we get as in Lemma 3 that the conditional density f of the absolutely continuous part is given by

$$\begin{aligned} f(y)\ =\ \frac{1}{\mathbb {P}(\tau _0^1\le a)}\frac{h(\psi (y))}{\left| \varphi '(\psi (y))\right| }\ =\ \frac{1}{\mathbb {P}(\tau _0^1\le a)}\frac{h(a-(y-b)^{1/2})}{(y-b)^{1/2}} \end{aligned}$$

But there are $c_1,c_2>0$ such that $f_0(y)\le c_1$ on $[b,b+a^2]$ and $h(t)\ge c_2$ on $[b,b+a^2]$, and so we get

$$\begin{aligned} \int \frac{f^2}{f_0}\ = \int _{b}^{b+a^2}\frac{f^2(y)}{f_0(y)}\,\mathrm {d}y\ \ge \ \int _{b}^{b+a^2}\frac{c_2^2}{c_1(y-b)}\,\mathrm {d}y\ =\ \infty \,, \end{aligned}$$

meaning that (8.1) does not hold.