is 20 1.01t−1
fort=1, 2, 3,. . .so that the EPV of renewal expenses is 20
∞ t=1
1.01t−1vttp[50]= 20 1.01
∞ t=1
1.01t vt tp[50]
= 20 1.01
∞ t=1
vtj tp[50]
= 20
1.01(¨a[50]j−1)
where the subscriptjindicates that the calculation is at rate of interestjwhere 1.01v=1/(1+j), that is j = 0.0396. The EPV of the deferred annuity is 80 00015|¨a([1250]), so the single premium is
1 000+ 20
1.01(¨a[50]j−1)+80 00015|¨a([1250]).
Asa¨[50]j =19.4550 and15|¨a[(5012)] =6.04129, the single premium is $484 669.
2 We end this section with a comment on the premiums calculated in Examples 6.6and6.7. In Example6.6, the annual premium is $2295.04 and the expenses at time 0 are $2 000 plus 50% of the first premium, a total of $3146.75, which exceeds the first premium. Similarly, in Example6.7the total premium in the first year is $227.88 and the total expenses in the first year are $500 plus 10%
of premiums in the first year. In each case, the premium income in the first year is insufficient to cover expenses in the first year. This situation is common in practice, especially when initial commission to agents is high, and is referred to as new business strain. A consequence of new business strain is that an insurer needs to have funds available in order to sell policies. From time to time insurers get into financial difficulties through pursuing an aggressive growth strategy without sufficient capital to support the new business strain. Essentially, the insurer borrows from shareholder (or participating policyholder) funds in order to write new business. These early expenses are gradually paid off by the expense loadings in future premiums. The part of the premiums that funds the initial expenses is called thedeferred acquisition cost.
6.7 Profit 155 policyholder funds, it is necessary for the business to be sufficiently profitable for the payment of a reasonable rate of return – in other words, to make a profit. In traditional insurance, we often load for profit implicitly, by margins in the valuation assumptions. For example, if we expect to earn an interest rate of 6% per year on assets, we might assume only 5% per year in the premium basis. The extra income from the invested premiums will contribute to profit. In participating business, much of the profit will be distributed to the policyholders in the form of cash dividends or bonus. Some will be paid as dividends to shareholders, if the company is proprietary.
We may also use margins in the mortality assumptions. For a term insurance, we might use a slightly higher overall mortality rate than we expect. For an annuity, we might use a slightly lower rate.
More modern premium setting approaches, which use projected cash flows, are presented in Chapter 11, where more explicit allowance for profit is incorporated in the methodology.
Each individual policy sold will generate a profit or a loss. Although we calcu- late a premium assuming a given survival model, for each individual policy the experienced mortality rate in any year can take only the values 0 or 1. So, while the expected outcome under the equivalence principle is zero profit (assuming no margins), the actual outcome for each individual policy will either be a profit or a loss. For the actual profit from a group of policies to be reliably close to the expected profit, we need to sell a large number of individual contracts, whose future lifetimes can be regarded as statistically independent, so that the losses and profits from individual policies are combined.
As a simple illustration of this, consider a life who purchases a one-year term insurance with sum insured $1000 payable at the end of the year of death. Let us suppose that the life is subject to a mortality of rate of 0.01 over the year, that the insurer can earn interest at 5% per year, and that there are no expenses.
Then, using the equivalence principle, the premium is P=1 000×0.01/1.05=9.52.
The future loss random variable is Ln0=
1 000v−P=942.86 ifTx≤1, with probability 0.01,
−P= −9.52 ifTx>1, with probability 0.99.
The expected loss is 0.01×942.86+0.99×(−9.52)=0, as required by the equivalence principle, but the probability of profit is 0.99, and the probability of loss is 0.01. The balance arises because the profit, if the policyholder survives
the year, is small, and the loss, if the policyholder dies, is large. Using the equivalence principle, so that the expected future loss is zero, makes sense only if the insurer issues a large number of policies, so that the overall proportion of policies becoming claims will be close to the assumed proportion of 0.01.
Now suppose the insurer were to issue 100 such policies to independent lives.
The insurer would expect to make a (small) profit on 99 of them. If the outcome from this portfolio is that all lives survive for the year, then the insurer makes a profit. If one life dies, there is no profit or loss. If more than one life dies, there will be a loss on the portfolio. LetDdenote the number of deaths in the portfolio, so thatD∼B(100, 0.01). The probability that the profit on the whole portfolio is greater than or equal to zero is
Pr[D≤1] =0.73576
compared with 99% for the individual contract. In fact, as the number of policies issued increases, the probability of profit will tend, monotonically, to 0.5. On the other hand, while the probability of loss is increasing with the portfolio size, the probability of very large aggregate losses (relative, say, to total premiums) is much smaller for a large portfolio, since there is a balancing effect from diversificationof the risk amongst the large group of policies.
Let us now consider a whole life insurance policy with sum insuredSpayable at the end of the year of death, initial expenses of I, renewal expenses of e associated with each premium payment (including the first) issued to a select life agedxby annual premiums ofP. For this policy
Lg0=SvK[x]+1+I+ea¨K
[x]+1 −Pa¨K
[x]+1 , whereK[x]denotes the curtate future lifetime of[x].
If death occurs shortly after the policy is issued, so that only a few premiums are paid, the insurer will make a loss, and, conversely, if the policyholder lives to a ripe old age, we would expect that the insurer would make a profit as the policyholder will have paid a large number of premiums, and there will have been plenty of time for the premiums to accumulate interest. We can use the future loss random variable to find the minimum future lifetime for the policyholder in order that the insurer makes a profit on this policy. The probability that the insurer makes a profit on the policy, Pr[Lg0<0], is given by
Pr[Lg0<0] =Pr
SvK[x]+1+I+ea¨K
[x]+1 −Pa¨K
[x]+1 <0
.
6.7 Profit 157 Rearranging and replacinga¨K
[x]+1 with(1−vK[x]+1)/d, gives Pr[Lg0<0] =Pr
vK[x]+1<
P−e d −I S+P−de
=Pr
K[x]+1> 1 δlog
P−e+Sd P−e−Id
. (6.4)
Suppose we denote the right-hand side term of the inequality in equation (6.4) byτ, so that the contract generates a profit for the insurer if K[x]+1 > τ.
Generally,τ is not an integer. Thus, ifτ denotes the integer part ofτ, then the insurer makes a profit if the life survives at leastτ years, the probability of which isτ p[x].
Let us continue this illustration by assuming thatx = 30,S = $100 000, I = 1 000, ande = 50. Then we find thatP =$498.45, and from equation (6.4) we find that there is a profit ifK[30]+1>52.57. Thus, there is a profit if the life survives for 52 years, the probability of which is52p[30]=0.70704.
Figure6.1shows the profits that arise should death occur in a given year, in terms of values at the end of that year. We see that large losses occur in the early years of the policy, and even larger profits occur if the policyholder dies at an advanced age. The probability of realizing either a large loss or profit is small. For example, if the policyholder dies in the first policy year, the loss to the insurer is $100 579, and the probability of this loss isq[30] =0.00027.
Similarly, a profit of $308 070 arises if the death benefit is payable at time 80,
-200000 -100000 0 100000 200000 300000 400000 500000 600000
0 10 20 30 40 50 60 70 80 90
Year
Profit
Figure 6.1 Profit at year-end if death occurs in that year for the whole life insurance described in Section6.7.
and the probability of this is79|q[30] =0.00023. It is important to appreciate that the premium has been calculated in such a way that the EPV of the profit from the policy is zero.
Example 6.9A life insurer is about to issue a 25-year endowment insurance with a basic sum insured of $250 000 to a select life aged exactly 30. Premiums are payable annually throughout the term of the policy. Initial expenses are
$1200 plus 40% of the first premium and renewal expenses are 1% of the second and subsequent premiums. The insurer allows for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each policy anniversary (including the last). The death benefit is payable at the end of the year of death.
Assume the Standard Select Survival Model with interest at 5% per year.
(a) Derive an expression for the future loss random variable,Lg0, for this policy.
(b) Calculate the annual premium for this policy.
(c) Let L0(k)denote the present value of the loss on the policy given that K[30]=kfork≤24 and letL0(25)denote the present value of the loss on the policy given that the policyholder survives to age 55. CalculateL0(k) fork=0, 1,. . ., 25.
(d) Calculate the probability that the insurer makes a profit on this policy.
(e) Calculate V[Lg0].
Solution 6.9 (a) First, we note that if the policyholder’s curtate future lifetime, K[30], iskyears wherek=0, 1, 2,. . ., 24, then the number of bonus addi- tions isk, the death benefit is payablek+1 years from issue, and hence the present value of the death benefit is 250 000(1.025)K[30]vK[30]+1. However, if the policyholder survives for 25 years, then 25 bonuses vest. Thus, ifP denotes the annual premium,
Lg0=250 000(1.025min(K[30], 25))vmin(K[30]+1, 25) +1 200+0.39P−0.99Pa¨min(K
[30]+1, 25). (b) The EPV of the premiums, less premium expenses, is
0.99Pa¨[30]:25 =14.5838P.
As the death benefit is $250 000(1.025t)if the policyholder dies in thetth policy year, the EPV of the death benefit is
250 000 24
t=0
vt+1t|q[30](1.025t)=250 000 1
1.025A[1
30]:25j
=3099.37
where 1+j=(1+i)/(1.025), so thatj=0.02439.
6.7 Profit 159 Table 6.3. Values of the future loss
random variable for Example6.9.
Value ofK[30], k
PV of loss, L0(k)
0 233 437
1 218 561
... ...
23 1 737
24 −4 517
≥25 −1 179
The EPV of the survival benefit is
250 000v2525p[30]1.02525=134 295.43, and the EPV of the remaining expenses is
1 200+0.39P.
Hence, equating the EPV of premium income with the EPV of benefits plus expenses we find thatP=$9 764.44.
(c) Given thatK[30]=k, wherek=0, 1,. . ., 24, the present value of the loss is the present value of the death benefit payable at timek+1 less the present value ofk+1 premiums plus the present value of expenses. Hence
L0(k)=250 000(1.025k)vk+1+1 200+0.39P−0.99Pa¨k+1. If the policyholder survives to age 55, there is one extra bonus payment, and the present value of the future loss is
L0(25)=250 000(1.02525)v25+1 200+0.39P−0.99Pa¨25 . Some values of the present value of the future loss are shown in Table6.3.
(d) The full set of values for the present value of the future loss shows that there is a profit if and only if the policyholder survives 24 years and pays the premium at the start of the 25th policy year. Hence the probability of a profit is24p[30]=0.98297.
Note that this probability is based on the assumption that future expenses and future interest rates are known and will be as in the premium basis.
(e) From the full set of values forL0(k)we can calculate E[(Lg0)2] =
24 k=0
(L0(k))2k|q[30]+(L0(25))225p[30]=12 115.552
which is equal to the variance as E[Lg0] =0. 2 Generally speaking, for an insurance policy, the longer a life survives, the greater is the profit to the insurer, as illustrated in Figure6.1. However, the converse is true for annuities, as the following example illustrates.
Example 6.10An insurance company is about to issue a single premium deferred annuity to a select life aged 55. The first annuity payment will take place 10 years from issue, and payments will be annual. The first annuity pay- ment will be $50 000, and each subsequent payment will be 3% greater than the previous payment. Ignoring expenses, and using the Standard Select Survival Model with interest at 5% per year, calculate
(a) the single premium,
(b) the probability the insurance company makes a profit from this policy, and (c) the probability that the present value of the loss exceeds $100 000.
Solution 6.10 (a) LetPdenote the single premium. Then P=50 000
∞ t=10
vt(1.03t−10)tp[55]=$546 812.
(b) LetL0(k)denote the present value of the loss given thatK[55] =k,k = 0, 1,. . .. Then
L0(k)=
−P fork=0, 1,. . ., 9,
−P+50 000v10a¨k−9j fork=10, 11,. . ., (6.5) wherej=1.05/1.03−1=0.019417.
Since a¨k−9j is an increasing function of k, formula (6.5) shows that L0(k)is an increasing function ofkfork ≥ 10. The present value of the profit will be positive ifL0(k) <0. Using formula (6.5), this condition can be expressed as
−P+50 000v10a¨k−9j <0, or, equivalently,
¨
ak−9j<1.0510P/50 000.
6.7 Profit 161 Writing a¨k−9j = (1−vk−9)/dj wheredj = j/(1+j), this condition becomes
vkj−9>1−dj1.0510P/50 000 , and asvj =exp{−δj}whereδj=log(1+j)this gives
k−9<−log
1−dj1.0510P/50 000 /δj.
Hence we find thatL0(k) <0 ifk<30.55, and so there will be a profit if the policyholder dies before age 86. The probability of this is 1−31p[55]= 0.41051.
(c) The present value of the loss will exceed 100 000 if
−P+50 000v10a¨k−9j >100 000 ,
and following through exactly the same arguments as in part (b) we find thatL0(k) > 100 000 ifk > 35.68. Hence the present value of the loss will be greater than $100 000 if the policyholder survives to age 91, and the probability of this is 36p[55]=0.38462.
Figure6.2showsL0(k)fork=1, 2,. . ., 50. We can see that the loss is constant for the first 10 years at−P and then increases due to annuity payments. In contrast to Figure6.1, longevity results in large losses to the insurer. We can also clearly see from this figure that the loss is negative ifktakes a value less
than 31, confirming our answer to part (b). 2
–600000 –500000 –400000 –300000 –200000 –100000 0 100000 200000 300000 400000
0 5 10 15 20 25 30 35 40 45 50
Year of death, k
Present value of loss
Figure 6.2 Present value of loss from Example6.10.