Hattendorff’s theorem for non-smooth continuous-time
Markov models II: Application
Hartmut Milbrodt
Mathematisches Institut, Universität zu Köln, Weyertal 86-90, D-50931 Köln, Germany
Received September 1998; received in revised form June 1999
Abstract
The examples in this second part of our paper illustrate the broad scope of the generalized Hattendorff theorem exposed in Part I as well as some limitations concerning the interpretability of numerical results derived from Hattendorff type theorems. In particular, they show that “mixed” situations in which some transitions of the underlying Markov jump process are governed by smooth cumulative transition intensities, whereas others can only take place at discrete times come up quite naturally. Contrary to previous versions of Hattendorff’s theorem, our result applies to such examples as well as to fully discrete and to fully smooth situations. ©2000 Elsevier Science B.V. All rights reserved.
Keywords: Hattendorff’s theorem; Loss in a given state; Variance of the loss MSC: 62P05
1. Examples
The following examples serve to indicate how Hattendorff’s theorem works in its general form (see Part I; Milbrodt, 1999). On the one hand, they show that both, the variance decomposition with respect to time as well as the decomposition with respect to states, lead to interesting and well-interpretable results: they clearly indicate that the classical mean value calculation of premiums and reserves in life insurance disregards important aspects of insurance products such as the risk associated with them. On the other hand, they also exhibit certain limitations of the variance calculus of Hattendorff’s theorem. These are mainly due to the fact that distributions of losses in life insurance are, as a rule, extremely skewed and that, intuitively speaking, the variance of the loss as a measure of risk does not distinguish between the risk of the insurance company and the risk borne by the insured. Additionally, the examples illustrate that even simple “real-life” insurance products may lead to findings which are surprising at first sight (e.g. the phenomenon that additional built-in options of the insured may result in a reduced overall variance of the loss). Due to limitations of space, we have to be very brief with the examples. For details of calculations the reader is referred to Milbrodt and Helbig (1999). Without further mention, notations and the terminology in this Part II of our paper are taken from Part I (Milbrodt, 1999).
E-mail address: [email protected] (H. Milbrodt).
Fig. 1. Variances of annual losses: term life insurance (bullets), pure endowment (triangles) and endowment (crosses).
In the numerical examples below, the calculation bases for the biometric risks are chosen as follows. For the general mortality we use the German life tables DAV 1994 R and DAV 1994 T, both provided by the German Actuarial Association (Deutsche Aktuarvereinigung, DAV), see Schmithals and Schütz (1995) and Loebus (1994). The first of these is a generation life table with suitably prudent margins for life annuities, the latter is to be used with endowments and other forms of life insurance. Where necessary, withdrawal probabilities are taken from a table published by the DAV (1995). Of course, the qualitative essentials of the examples discussed in the following will not be affected by using comparable tables from other countries with a similar socio-demographic structure. We use compound interest with discount factor v := 1.04−1 (which corresponds to the actual first order interest rate in Germany) and force of interestδ := −logv. All numerical examples refer to a male person with entry-age 30 and entry-year 1997. Let us begin with very simple examples concerning the one-life-one-risk case.
Example 1. We consider a 30-year term life insurance with benefit 1 payable at the end of the year of death, a 30-year pure endowment with benefit 1 payable upon survival and a 30-year endowment with benefit 1 payable upon survival or at the end of the year of death. In all three cases we assume constant annual premiums payable at the beginning of each policy year. Fig. 1 shows the variances of annual losses for these insurances which can easily be obtained from (18) of Part I. To make the results comparable, in all three cases the life table DAV 1994 T has been used – despite the fact that in practice DAV 1994 R would be used for the pure endowment.
Fig. 2. Distributions of the total loss of a term life insurance (top left), a pure endowment (top right) and an endowment (lower plot).
However, as indicated already at the beginning of this section, there are also some problems connected with the interpretation of these findings. Fig. 2 exhibits the discrete distributions of the total losses, represented by their probability functions in a semi-logarithmic scale. On the abscissae possible valuesℓof the losses are marked, the ordinates are in a logarithmic scale and represent probabilitiespwith which these values are attained. Assuming the stationarity condition (see (17) of Part I), these distributions can be calculated from (16) of Part I. (In more complex examples the results of Hesselager and Norberg (1996) may be used). Taking into consideration the semi-logarithmic scale, one readily sees that these distributions are extremely skewed to the right for the term insurance and the endowment, whereas the distribution for the pure endowment is extremely skewed to the left. Now, observe that the economic interpretation of a positive deviation of the loss from its expectation(=0 because of the equivalence principle) is totally different from the interpretation of a negative deviation: the first means an economic gain of the insurer, whereas the latter means a true economic loss. In situations like these, the variance is an inappropriate measure of dispersion. Alternatively, the positive semi-variance Var+, i.e. the expectation of the
Table 1
Comparison of variances and positive semi-variances of losses.
Var Var+ Var+/Var(%)
Term insurance 0.0332 0.0288 86.50
Pure endowment 0.0089 0.0013 14.61
Endowment 0.0143 0.0133 93.16
In Table 1, the variances and the positive semi-variances of the total losses are compared for our examples. Its last column shows that in case of the term insurance and the endowment the insurer bears approximately 90% of the overall risk associated with the insurance contract, whereas in case of a pure endowment his share is less than 15%. In this sense, the latter is much less dangerous than the former two, a well-known fact, which can of course also be read off from Fig. 2.
To deal with examples which are a little bit more involved, we require extensions of the stationarity condition (see (17) of Part I) and of the well-known assumption of a uniform distribution of deaths between integer years. Let
S6= ∅be a finite state space andJ := {(y, z)∈S2|y6=z}. To formulate the general stationarity condition, assume that for every agex ≥ 0 we are given a Markov process(Ω,A, P , (Xt(x))t≥0)with state spaceS modelling the policy development for an insured(x). LetT(x)(s)denote the time of the first jump of(X(x)t )t≥0afters≥0(x≥0). The following definition is a natural generalization of (17) of Part I.
Definition 1. The processesX(x),x ≥ 0, satisfy the stationarity condition iff there are conditional distributions P (T(x)(s), X(x) time interval [s, t], provided it was there at its beginning, 0≤ s ≤t, y ∈S. As in the one-life-one-risk case the main significance of the stationarity condition is that these probabilities of persistence split multiplicatively if the time interval is partitioned into disjoint subintervals:
Lemma 1. Assume thatX(x), x ≥ 0, are Markov jump processes with state spaceS satisfying the stationarity condition. Let{k, t, x} ⊂[0,∞)andy∈S. Then
The proof of (1) follows easily from the stationarity condition and the Markov property ofX(x), (2) is an immediate consequence of (1).
Contrary to the stationarity condition, the assumption of a uniform distribution of exit times between integer years does not necessarily refer to a family of processesX(x), x≥0; it also makes sense for a single Markov jump process(Ω,A, P , (Xt)t≥0)with state spaceS. Lets≥0 andy∈S. As in Milbrodt and Stracke (1997), Definition
3.4, letT (s)denote the time of the first jump ofXafters,
and
Ey(s, t ):= X
z6=y
Eyz(s, t )=P (T (s)≤t|Xs =y), t≥s. (4)
Recall that the cumulative transition intensity matrixqis given by
qyz(s, t )=
Now, we follow Stracke (1997), Satz 2.15 and Bemerkung 2.16.
Lemma 2. LetXbe a Markov jump process with state spaceS, y∈Sandn∈N0be such thatP (Xn =y) >0,
The cumulative intensity matrixq is absolutely continuous on(n, n+1] with Lebesgue densities (“intensities”) given by
Proof. We restrict ourselves to the proof of (8). Relations (9)–(11) follow from that by summation overz6=y, by (5) and differentiating with respect totand again by summing overz.
Letz6=yandn≤s≤t≤n+1. The Markov property ofXthen yields
Eyz(n, t )−Eyz(n, s)=P (Xr =y, r ∈[n, s];T (s)≤t, XT (s)=z|Xn=y)
=P (T (s)≤t, XT (s)=z|Xs =y) P (T (n) > s|Xn=y)
=Eyz(s, t )(1−Ey(n, s)).
Fig. 3. Life insurance combined with deferred life annuity: States and transitions.
Remark 1.
1. Lety∈Sandn∈N0be such thatP (Xn=y) >0. IfEyz(n, n+1) >0, z6=y, then relation (7) is equivalent to
P (T (n)≤t|T (n)≤n+1, XT (n)=z, Xn=y)=t−n, z6=y, t ∈(n, n+1],
showing that (7) indeed amounts to the assumption that the exit timeT (n)is conditionally uniformly distributed on(n, n+1] givenT (n)≤n+1, XT (n)=zandXn=y.
2. In the one-life-one-risk case of Section 1 of Part I let 0 denote the state “alive” and 1 denote the state “dead” and putXt :=1[Tx,∞)(t ), t ≥0. In this situation, requiring (7) fory=0 and everyn∈N0such thatP (Tx> n) >0 is equivalent to the assumption thatKxandTx−Kxare stochastically independent and thatTx−Kxis uniformly distributed on(0,1].
3. As the following example illustrates, in the insurance of persons we often have “mixed” situations in which some transitions are governed by smooth transition probabilities and others can only take place at discrete times. (Typically, the first are of a “biological nature”, e.g. death or disablement, whereas the latter are typically ruled administratively such as withdrawal or the option to retire at certain ages.) Lemma 2 can easily be adapted to such situations. Details are left to the reader.
The following example is an extension of Example 4.17 of Milbrodt and Stracke (1997) by additional features.
Example 2. We consider a combination of a life insurance and a deferred life annuity for a male person(x), x <60, which is designed such that the insured can choose to retire with exact ages 60,61, . . . ,65. Moreover, he is allowed to withdraw from the contract immediately upon reaching integral agesx+1, . . . ,59. The state space is
S:= {a, w,0, . . . ,5, d}.
astanding for the initial state “active”,wdenoting “withdrawn”,m∈ {0, . . . ,5}meaning “retired with age 60+m” andd standing for the absorbing state “dead”. The subspaceP J ⊂J of possible jumps then is
P J := {(a, w), (a,0), . . . , (a,5), (a, d), (0, d), . . . , (5, d)}.
Fig. 3 illustrates the situation.
The actuarial payments are assumed to be as follows:
• A continuous payment of premiums with constant densityπas long as the insured is active, but at most until he reaches a limiting age 60+M, M∈ {0, . . . ,5}being fixed in advance. Of course,πequals the annual premium amount and is to be determined from the equivalence principle.
• A continuous payment of a life annuity with constant densityRmdepending on the retiring age 60+mchosen by the insured. The annual amountR5, say, is fixed in advance andR0, . . . ,R4are determined by a “fairness (no arbitrage) condition” asserting that at every potential retiring age 60+mthe actuarial value of future payments is independent of the decision of the insured:
• Immediately payable death benefits of amountD >0.
• Withdrawal benefits of amountαVa(k)+payable immediately upon withdrawal at timek, k =1, . . . ,29, i.e. upon withdrawing with agex+ka fractionα∈[0,1] of the non-negative prospective reserve is repaid. Thus, employing the notation from Section 2 of Part I, the actuarial payment function is given by
DT =Id, Dyz =
We assume that the policy development is described by a Markov jump process(Ω,A, P , (Xt)t≥0)with regular
transition matrixq, and that the stationarity condition is satisfied. For simplicity, we also assume that the cumulative force of mortality does not depend on the state immediately before death (active or age of retirement). (Obviously, due to self-selection, this will only hold approximately.) Since later on we will choose the parameters such that the annuity benefits will be in the fore, the one-year probabilities of death(qx+j)will be taken from the life table DAV 1994 R (see Schmithals and Schütz (1995)). To illustrate a common practice in Germany, we will modify assumption (7) slightly, requiring that(x)dies immediately upon reaching the ageω0=111 (which is the first age with entry 1 in the life table). This amounts to requiring thatKxandTx−Kxare conditionally independent given Tx< ω0−xand thatTx−Kxis conditionally uniformly distributed on(0,1] givenTx< ω0−x(compare Remark 1 2). Of course, this is only a minor technical detail without impact on reserves and losses for ages smaller thanω0, its advantage being that it makes the cumulative transition intensity matrix regular (see (18)). Employing Lemma 2 (respectively its modification according to Remark 1 3), we obtain the force of mortality for an active person,
dqad
dλ1 =1[0,65−x)λx, (16)
λ1denoting the one-dimensional Lebesgue measure and λx:τ 7−→
qx+[τ] 1−(τ −[τ]) qx+[τ]
1[0,ωx−x)(τ ), (17)
respectively, the cumulative force of mortality for pensioners,
qmd(s, u)= andr5=1. Obviously,rmstrongly depends on the socio-economic conditions at time 60+m−x, when the insured decides on the retirement. Without modelling this dependence, there is no meaningful way to estimatermfrom observations prior to that time. Hence, meaningful estimates will not be available, in general. The cumulative intensities of retirement are given by
In our numerical example below, we will use conditional one-year probabilities of withdrawal
sj :=P (Xj+1=w|Xj =a), j =0, . . . ,58−x,
which solely depends on the time passed since the beginning of the contract but neither on the age of entry nor on the sex of the insured. Since we have assumed that withdrawal is only possible when reaching integral ages x+1, . . . ,59, the cumulative intensity of withdrawal is given by
p∗as usual denoting the one-year probabilities of survival. Apparently,qyz≡0 if(y, z) /∈P J.
Fig. 4. Life insurance combined with deferred life annuity: Prospective reserve in state “active” (upper plot) and in states “retired with age 60+m”,m=0,. . .,5 (lower plot).
As a concrete example, we consider the following constellation of parameters:x =30, the life table DAV 1994 R with year of birth 1967, withdrawal probabilities taken from DAV (1995) (entries for insurance period 45 years and longer),r0 = · · · = r4 = 0.4, M = 5, D = 1, α = 0.7,R5 = 0.5 and v = 1.04−1. Fig. 4 exhibits the corresponding prospective reserves. Note that by the Theorem of Cantelli (see e.g. Milbrodt and Stracke (1997), Corollary 6.4), the reserves, and hence also the equivalence premium, do not depend on the hardly available probabilitiesrm, m∈ {0, . . . ,4}.
Var(Lm(k)−Lm(k−1))=
the conditional probability that the insured leaves the stateawith exact age 60+m, provided he was active imme-diately before. Then
Together with the Chapman-Kolmogorov equations, (19) and (21) this yields forτ ∈(k−1, k]:
pam(0, τ−0)=p59+m
Inserting this into (23) and using (17) and (18), we get
Var(Lm(k)−Lm(k−1))=
For the variances of the annual losses in stateaCorollary 2.7 1 of Part I gives
By (20), (24) and (25),
paa(0, τ−0)(1−qaw({τ}))qaw(dτ )|B((k−1,k])= k Y
j=1
(px+j−1−sj−1)
sk−1 px+k−1
εk(dτ )|B((k−1,k]),
k∈ {1, . . . ,59−x}, (27)
whereεkdenotes the Dirac-measure sitting atk, and by (16), (17) and (25),
paa(0, τ−0)qad(dτ )|B((k−1,k])= k Y
j=1
(1−uj) px+j−1qx+k−1dτ|B((k−1,k]), k∈ {1, . . . ,65−x}.
Inserting these relations into (26), respectively, into (27) and regarding (24) yields
Fig. 6. Life insurance combined with deferred life annuity: variances of total annual losses.
Now we come back to our numerical example. Figs. 5 and 6 show the variances of annual losses in all non-absorbing states and of the total annual loss. At the beginning, the variance of the annual loss in stateadecreases, since the absolute value of the amount at risk for the transition(a, d)decreases at first because of the increase ofVa and D=1. Then the variance increases, until the age 60, the first possible age of retirement, is reached; this is due to the increase of the amount at risk for transitions(a, w)and(a, d). Finally, in the last five years of potential activity the variance decreases rapidly, a phenomenon which basically comes from the fact that the probability of still being active at the beginning of the respective year rapidly decreases. The jump upwards of the variance between the years 12 and 13 of the insurance contract reflects a drastic jump upwards in the underlying withdrawal probabilities resulting from the German taxation laws. The variances of the annual losses in statesm∈ {0, . . . ,5}show two effects:
• They decrease as time proceeds, since for everymthe amount at risk for the transition(m, d)decreases (see the lower plot in Fig. 5 exhibiting the corresponding prospective reserves).
Fig. 7. Deferred life annuity: variances of annual losses if withdrawal is possible (triangles) and excluding the option to withdraw (bullets).
According to Corollary 2.7, the “curve” in Fig. 6 showing the variances of the total annual losses is obtained by aggregating all “curves” of Fig. 5. In contrast to what is true for premiums and reserves, the variances of the annual losses beyond the age of 60 do depend on the probabilities of retirementrm, m∈ {0, . . . ,4}, which are unknown in general. Roughly speaking, earlier retirement leads to smaller variances. For the present set of parameters the variance of the total loss over the whole time insured is 0.226, modifying the probabilitiesrmsuch thatr0= · · · =r4=0.75 leads to a variance of 0.195, and forr0= · · · =r4=0.25 we obtain the value 0.254.
Figs. 7 and 8 illustrate that even if the full prospective reserve is repaid upon withdrawal (which by the Theorem of Cantelli entails that the prospective reserves do not depend on the withdrawal probabilities), the variances of the total annual losses do depend on the withdrawal probabilities. For these figures the parameters are altered as follows. We putα=1, D=0 andr0= · · · =r4=0, i.e. we assume that the whole prospective reserve is repaid, that no death benefits are paid and that we have a fixed age of retirement, namely 65. In Fig. 7 the bullets show the variances of the total annual losses with withdrawal probabilites as above, the triangles mark the variances if the withdrawal probabilities are set to 0. The variance in the presence of withdrawal is smaller throughout: the
variance of the total loss over the whole time insured then is 0.418, whereas the corresponding value is 0.462 if the withdrawal probabilities are assumed to be 0. At first sight it appears to be strange that the additional option to withdraw from the insurance contract leads to a reduction of the variance of the loss. However, there is a simple and intuitively plausible explanation for that: by the assumption thatα=1 the loss realized upon withdrawal is 0, and hence equals the expectation of the total loss over the whole time insured. The same phenomenon may also be viewed at from a slightly different angle. Fig. 8 exhibits the variances of the total annual losses with all withdrawal probabilities set to 0 (bullets, this is the same “curve” as in Fig. 7), compared to all withdrawal probabilities except the one at 65 set to zero, the latter is 0.3 for the “curve” marked with triangles and 0.5 for the “curve” marked with crosses. This corresponds to the situation of a deferred life annuity with the option of refund of the prospective reserve at the end of the deferment period (at age 65), the probability that the insured opts for refunding instead of the annuity being 0.3 and 0.5, respectively. Again, the variance decreases as this probability increases (the overall variance of the total loss is 0.462, respectively 0.372 and 0.312), the explanation being as above.
Acknowledgements
A major part of the programming work connected with these examples was done by Patric Holubeck and Andrea Tirpitz. The author acknowledges with gratitude the efforts of both referees which led to significant improvements of both parts of this paper.
References
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