For a whole life insurance policy, the time at which the benefit will be paid is unknown until the policyholder actually dies and the policy becomes a claim.
Since the present value of a future payment depends on the payment date, the present value of the benefit payment is a function of the time of death, and is therefore modelled as a random variable. Given a survival model and an interest rate we can derive the distribution of the present value random variable for a
life contingent benefit, and can therefore compute quantities such as the mean and variance of the present value.
We start by considering the value of a benefit of amount $1 payable following the death of a life currently agedx. Using a benefit of $1 allows us to develop valuation functions per unit of sum insured, then we can multiply these by the actual sum insured for different benefit amounts.
We first assume that the benefit is payable immediately on the death of(x).
This is known as the continuous case since we work with the continuous future lifetime random variable Tx. Although in practice there would normally be a short delay between the date of a person’s death and the time at which an insurance company would actually pay a death benefit (due to notification of death to the insurance company and legal formalities) the effect is slight and we will ignore that delay here.
For our life(x), the present value of a benefit of $1 payable immediately on death is a random variable,Z, say, where
Z =vTx = e−δTx.
We are generally most interested in the expected value of the present value random variable for some future payment. We refer to this as the Expected Present Value or EPV. It is also commonly referred to as the Actuarial Value.
The EPV of the whole life insurance benefit payment with sum insured $1 isE[e−δTx]. In actuarial notation, we denote this expected value byA¯x, where the bar aboveAdenotes that the benefit is payable immediately on death.
AsTxhas probability density functionfx(t)= tpxµx+t, we have A¯x=E[e−δTx] =
∞
0
e−δt tpxµx+tdt. (4.1)
It is worth looking at the intuition behind this formula. We use the time-line format that was introduced in Section2.4in Figure4.1.
Consider times, wherex≤ x+s< ω. The probability that(x)is alive at timesisspx, and the probability that(x)dies between agesx+sandx+s+ds, having survived to agex+s, is, loosely,µx+sds, provided thatdsis very small.
In this case, the present value of the death benefit of $1 ise−δs.
Now we can integrate (that is, sum the infinitesimal components of) the product of present value and probability over all the possible death intervalss
4.4 Valuation of insurance benefits 77
Time 0
x
s
x+s
s+ds
x+s+ds ω
(x)survivessyears
spx
(x) dies
µx+sds Age
Probability
Present value e−δs
Figure 4.1 Time-line diagram for continuous whole life insurance.
tos+dsto obtain the EPV of the death benefit that will be paid in exactly one of these intervals.
Similarly, the second moment (about zero) of the present value of the death benefit is
E[Z2] =E[(e−δTx)2] =E[e−2δTx]
= ∞
0
e−2δt tpxµx+tdt
=2A¯x (4.2)
where the superscript2indicates that calculation is at force of interest 2δ, or, equivalently, at rate of interestj, where 1+j=e2δ=(1+i)2.
The variance of the present value of a unit benefit payable immediately on death is
V[Z] =V[e−δTx] =E[Z2] −E[Z]2=2A¯x− A¯x
2
. (4.3)
Now, if we introduce a more general sum insured,S, say, then the EPV of the death benefit is
E[SZ] =E[Se−δTx] =SA¯x
and the variance is
V[SZ] =V[Se−δTx] =S2
2A¯x− ¯A2x
.
In fact we can calculate any probabilities associated with the random variable Z from the probabilities associated withTx. Suppose we are interested in the
probability Pr[Z ≤0.5], for example. We can rearrange this into a probability forTx:
Pr[Z ≤0.5] =Pr[e−δTx ≤0.5]
=Pr[−δTx≤log(0.5)]
=Pr[δTx >−log(0.5)]
=Pr[δTx >log(2)]
=Pr[Tx>log(2)/δ]
=upx
whereu = log(2)/δ. We note that low values ofZ are associated with high values ofTx. This makes sense because the benefit is more expensive to the insurer if it is paid early, as there has been little opportunity to earn interest. It is less expensive if it is paid later.
4.4.2 Whole life insurance: the annual case,Ax
Suppose now that the benefit of $1 is payable at the end of the year of death of (x), rather than immediately on death. To value this we use the curtate future lifetime random variable,Kx, introduced in Chapter2. Recall thatKxmeasures the number of complete years of future life of(x). The time to the end of the year of death of(x)is thenKx+1. For example, if(x)lived for 25.6 years from the issue of the insurance policy, the observed value ofKxwould be 25, and the death benefit payable at the end of the year of death would be payable 26 years from the policy’s issue.
We again useZto denote the present value of the whole life insurance benefit of $1, so thatZis the random variable
Z =vKx+1.
The EPV of the benefit, E[Z], is denoted byAxin actuarial notation.
In Chapter2we derived the probability function forKx, Pr[Kx=k] = k|qx, so the EPV of the benefit is
Ax =E[vKx+1] =∞
k=0
vk+1k|qx=vqx+v21|qx+v32|qx+ · · ·. (4.4)
Each term on the right-hand side of this equation represents the EPV of a death benefit of $1, payable at timekconditional on the death of(x)in(k−1,k].
4.4 Valuation of insurance benefits 79
Time 0 1 2 3
……
Amount $1 $1 $1
Discount v v2 v3
Probability qx 1|qx 2|qx
Figure 4.2 Time-line diagram for discrete whole life insurance.
In fact, we can always express the EPV of a life-contingent benefit by con- sidering each time point at which the benefit can be paid, and summing over all possible payment times the product of
(1) the amount of the benefit,
(2) the appropriate discount factor, and
(3) the probability that the benefit will be paid at that time.
We will justify this more rigorously in Section4.6. We illustrate the process for the whole life insurance example in Figure4.2.
The second moment of the present value is ∞
k=0
v2(k+1)k|qx = ∞ k=0
(v2)(k+1)k|qx=(v2)qx+(v2)21|qx+(v2)32|qx+ · · · .
Just as in the continuous case, we can calculate the second moment about zero of the present value by an adjustment in the rate of interest fromito(1+i)2−1.
We define
2Ax=∞
k=0
v2(k+1)k|qx, (4.5)
and so the variance of the present value of a benefit ofS payable at the end of the year of death is
S2 2
Ax−(Ax)2
. (4.6)
4.4.3 Whole life insurance: the1/mthly case, A(xm)
In Chapter2we introduced the random variableKx, representing the curtate future lifetime of(x), and we saw in Section4.4.2that the present value of an
insurance benefit payable at the end of the year of death can be expressed in terms ofKx.
We now define the random variableKx(m), wherem > 1 is an integer, to be the future lifetime of(x)in years rounded to the lower m1th of a year. The most common values ofmare 2, 4 and 12, corresponding to half years, quarter years and months. Thus, for example,Kx(4)represents the future lifetime of(x), rounded down to the lower 1/4.
Symbolically, if we let denote the integer part (or floor) function, then Kx(m)= 1
mmTx . (4.7)
For example, suppose(x)lives exactly 23.675 years. Then
Kx=23,Kx(2)=23.5,Kx(4)=23.5, andKx(12)=23128 =23.6667.
Note thatKx(m)is a discrete random variable.Kx(m)=kindicates that the life (x)dies in the interval[k,k+1/m), for(k=0,m1,m2,. . .).
The probability function for Kx(m) can be derived from the associated probabilities forTx. Fork=0,m1,m2,. . . ,
Pr[Kx(m)=k] =Pr
k≤Tx <k+m1
=k|1
mqx = kpx− k+1
mpx. In Figure4.3we show the time-line for themthly benefit. At the end of each 1/myear period, we show the amount of benefit due, conditional on the death of the insured life in the previous 1/myear interval, the probability that the insured life dies in the relevant interval, and the appropriate discount factor.
Suppose, for example, thatm=12. A whole life insurance benefit payable at the end of the month of death has present value random variableZwhere
Z =vKx(12)+1/12.
Time 0 1/m 2/m 3/m
……
Amount $1 $1 $1
Discount v1/m v2/m v3/m
Probability 1
mqx 1
m|1
mqx 2
m|1 mqx
Figure 4.3 Time-line diagram formthly whole life insurance.
4.4 Valuation of insurance benefits 81 We letA(x12)denote the EPV of this benefit, so that
E[Z] =A(x12)=v1/12 1
12qx+v2/12 1
12|1
12qx+v3/12 2
12|1
12qx+v4/12 3
12|1
12qx
+ · · · . Similarly, for anym,
A(xm) =v1/m 1
mqx+v2/m 1
m|1
mqx+v3/m 2
m|1
mqx+v4/m 3
m|1
mqx+ · · · (4.8)
= ∞ k=0
vk+m1 k
m|1
mqx. (4.9)
As for the continuous and annual cases, we can derive the variance of the present value of themthly whole life benefit by adjusting the interest rate for the first term in the variance. We have
E[Z2] =E[v2(Kx(12)+1/12)] =E[(v2)Kx(12)+1/12] =2A(x12), so the variance is
2A(x12)−(A(x12))2.
4.4.4 Recursions
In practice, it would be very unusual for an insurance policy to provide the death benefit at the end of the year of death. Nevertheless, the annual insurance functionAxis still useful. We are often required to work with annual life tables, such as those in Chapter3, in which case we would start by calculating the annual function Ax, then adjust for a more appropriate frequency using the relationships and assumptions we develop later in this chapter.
Using the annual life table in a spreadsheet, we can calculate the values of Axusingbackwards recursion. To do this, we start from the highest age in the table,ω. We assume all lives expire by ageω, so thatqω−1=1. If the life table does not have a limiting age, we choose a suitably high value forωso thatqω−1
is as close to 1 as we like. This means that any life attaining ageω−1 may be treated as certain to die before ageω, in which case we know thatKω−1 =0 and so
Aω−1=E[vKω−1+1] =v.
Now, working from the summation formula forAxwe have
Ax =ω−
x−1
k=0
vk+1kpxqx+k
=vqx+v2pxqx+1+v32pxqx+2+ · · ·
=vqx+vpx
vqx+1+v2px+1qx+2+v32px+1qx+3+ · · · , giving the important recursion formula
Ax=vqx+vpxAx+1. (4.10) This formula can be used in spreadsheet format to calculateAxbackwards from Aω−1back toAx0, wherex0is the minimum age in the table.
The intuition for equation (4.10) is that we separate the EPV of the whole life insurance into the value of the benefit due in the first year, followed by the value at agex+1 of all subsequent benefits, multiplied bypxto allow for the probability of surviving to agex+1, and byvto discount the value back from agex+1 to agex.
We can use the same approach formthly benefits; now the recursion will give A(xm)in terms ofA(m)
x+m1. Again, we split the benefit into the part payable in the first period – now of length 1/myears, followed by the EPV of the insurance beginning after 1/myears. We have
A(xm)=v1/m 1
mqx+v2/m 1
mpx 1
mqx+1
m +v3/m 2
mpx 1
mqx+2 m + · · ·
=v1/m 1
mqx+v1/m 1
mpx
v1/m 1
mqx+1
m +v2/m 1
mpx+1
m 1 mqx+2
m + · · · , giving the recursion formula
A(xm)=v1/m 1
mqx+v1/m 1
mpxA(m)
x+m1 .
Example 4.1Using the Standard Ultimate Survival Model from Section4.3, and an interest rate of 5% per year effective, construct a spreadsheet of values ofAxforx=20, 21,. . ., 100. Assume thatA129=v.
Solution 4.1The survival model for the Standard Ultimate Survival Model is the ultimate part of the model used in Example3.13and so values oftpx can be calculated as explained in the solution to that example. The value ofq129is
4.4 Valuation of insurance benefits 83 Table 4.1. Spreadsheet results for Example4.1, for the calculation
of Axusing the Standard Ultimate Survival Model.
x Ax x Ax x Ax
20 0.04922 47 0.16577 74 0.49215
21 0.05144 48 0.17330 75 0.50868
22 0.05378 49 0.18114 76 0.52536
23 0.05622 50 0.18931 77 0.54217
24 0.05879 51 0.19780 78 0.55906
25 0.06147 52 0.20664 79 0.57599
26 0.06429 53 0.21582 80 0.59293
27 0.06725 54 0.22535 81 0.60984
28 0.07034 55 0.23524 82 0.62666
29 0.07359 56 0.24550 83 0.64336
30 0.07698 57 0.25613 84 0.65990
31 0.08054 58 0.26714 85 0.67622
32 0.08427 59 0.27852 86 0.69229
33 0.08817 60 0.29028 87 0.70806
34 0.09226 61 0.30243 88 0.72349
35 0.09653 62 0.31495 89 0.73853
36 0.10101 63 0.32785 90 0.75317
37 0.10569 64 0.34113 91 0.76735
38 0.11059 65 0.35477 92 0.78104
39 0.11571 66 0.36878 93 0.79423
40 0.12106 67 0.38313 94 0.80688
41 0.12665 68 0.39783 95 0.81897
42 0.13249 69 0.41285 96 0.83049
43 0.13859 70 0.42818 97 0.84143
44 0.14496 71 0.44379 98 0.85177
45 0.15161 72 0.45968 99 0.86153
46 0.15854 73 0.47580 100 0.87068
0.99996, which is indeed close to 1. We can use the formula Ax=vqx+vpxAx+1
to calculate recursivelyA128,A127,. . .,A20, starting fromA129=v. Values for x=20, 21,. . ., 100, are given in Table4.1. 2 Example 4.2Using the Standard Ultimate Survival Model from Section4.3, and an interest rate of 5% per year effective, develop a spreadsheet of values ofA(x12)forxstarting at age 20, in steps of 1/12.
Solution 4.2For this example, we follow exactly the same process as for the previous example, except that we let the ages increase by 1/12 year in each
Table 4.2. Spreadsheet results for Example4.2, for the calculation of A(x12)using the Standard
Ultimate Survival Model.
x 1
12
px 1
12
qx A(x12)
20 0.999979 0.000021 0.05033
20121 0.999979 0.000021 0.05051 20122 0.999979 0.000021 0.05070 20123 0.999979 0.000021 0.05089
... ... ... ...
50 0.999904 0.000096 0.19357
50121 0.999903 0.000097 0.19429
... ... ... ...
1291012 0.413955 0.586045 0.99427
1291112 1
row. We construct a column of values of 1
12pxusing
1
12px=exp
−A/12−Bcx(c1/12−1)/log(c) . We again use 130 as the limiting age of the table. Then setA(12)
1291112 =v1/12, and for all the other values ofA(x12)use the recursion
A(x12)=v1/12 1
12qx+v1/12 1
12pxA(12)
x+121
.
The first and last few lines of the spreadsheet are reproduced in Table4.2. 2 It is worth making a remark about the calculations in Examples4.1and4.2. In Example4.1we saw thatq129 = 0.99996, which is sufficiently close to 1 to justify us starting our recursive calculation by settingA129=v. In Example4.2, our recursive calculation started fromA12911
12 =v1/12. If we calculateq12911
12
we find its value is 0.58960, which is certainly not close to 1.
What is happening in these calculations is that, for Example 4.1, we are replacing the exact calculation
A129=v(q129+p129A130)
4.4 Valuation of insurance benefits 85 byA129 =v, which is justifiable becausep129is small andA130is close to 1, meaning thatv(q129+p129A130)is very close tov. Similarly, for Example4.2, we replace the exact calculation
A(12)
1291112 =v1/12
1 12q12911
12 + 1
12p12911 12A(13012)
byA(12)
1291211 =v1/12. As the value ofA(13012)is very close to 1, it follows that v1/12
1 12q12911
12 + 1
12p129A(13012)
can by approximated byv1/12.
Example 4.3Using the Standard Ultimate Survival Model specified in Section 4.3, and an interest rate of 5% per year effective, calculate the mean and standard deviation of the present value of a benefit of $100 000 payable (a) immediately on death, (b) at the end of the month of death, and (c) at the end of the year of death for lives aged 20, 40, 60, 80 and 100, and comment on the results.
Solution 4.3For part (a), we must calculate 100 000A¯xand 100 000
2A¯x−(A¯x)2
forx=20, 40, 60 and 80, where2A¯xis calculated at effective rate of interest j = 10.25%, and for parts (b) and (c) we replace eachA¯x byA(x12) andAx, respectively. The values are shown in Table4.3. The continuous benefit values in the first column are calculated by numerical integration, and the annual and monthly benefit values are calculated using the spreadsheets from Examples4.1 and4.2.
We can make the following observations about these values. First, values for the continuous benefit are larger than the monthly benefit, which are larger than the annual benefit. This is because the death benefit is payable soonest under (a) and latest under (c). Second, asxincreases the mean increases for all three cases. This occurs because the smaller the value ofx, the longer the expected time until payment of the death benefit is. Third, asxincreases, the standard deviation decreases relative to the mean, in all three cases. And further, as we get to very old ages, the standard deviation decreases in absolute terms, as the possible range of payout dates is reduced significantly.
It is also interesting to note that the continuous and monthly versions of the whole life benefit are very close. That is to be expected, as the difference arises from the change in the value of money in the period between the moment of death and the end of the month of death, a relatively short period. 2
Table 4.3. Mean and standard deviation of the present value of a whole life insurance benefit of $100 000, for Example4.3.
Continuous (a) Monthly (b) Annual (c) Age,x Mean St. Dev. Mean St. Dev. Mean St. Dev.
20 5 043 5 954 5 033 5 942 4 922 5 810
40 12 404 9 619 12 379 9 600 12 106 9 389
60 29 743 15 897 29 683 15 865 29 028 15 517 80 60 764 17 685 60 641 17 649 59 293 17 255
100 89 341 8 127 89 158 8 110 87 068 7 860
4.4.5 Term insurance The continuous case,A¯x:n1
Under a term insurance policy, the death benefit is payable only if the policyholder dies within a fixed term of, say,nyears.
In the continuous case, the benefit is payable immediately on death. The present value of a benefit of $1, which we again denote byZ, is
Z =
vTx =e−δTx if Tx≤n,
0 if Tx>n.
The EPV of this benefit is denotedA¯x:n1 in actuarial notation. The bar aboveA again denotes that the benefit is payable immediately on death, and the 1 above xindicates that the life(x)must die before the term ofnyears expires in order for the benefit to be payable.
Then
A¯x:n1 = n
0
e−δttpxµx+tdt (4.11) and, similarly, the expected value of the square of the present value is
2A¯x:n1 = n
0
e−2δttpxµx+tdt
which, as with the whole life case, is calculated by a change in the rate of interest used.
The annual case,Ax:n1
Next, we consider the situation when a death benefit of 1 is payable at the end of the year of death, provided this occurs withinn years. The present value
4.4 Valuation of insurance benefits 87 random variable for the benefit is now
Z =
vKx+1 if Kx≤n−1, 0 if Kx≥n.
The EPV of the benefit is denotedAx:n1 so that
Ax:n1 =
n−1
k=0
vk+1k|qx. (4.12)
The 1/mthly case,A(m)1x:n
We now consider the situation when a death benefit of 1 is payable at the end of the 1/mth year of death, provided this occurs withinnyears. The present value random variable for the benefit is
Z=
vKx(m)+m1 if Kx(m)≤n−m1, 0 if Kx(m)≥n.
The EPV of the benefit is denotedA(m)1x:n so that
A(m)1x:n =
mn−1 k=0
v(k+1)/m k
m|1
mqx. (4.13)
Example 4.4Using the Standard Ultimate Survival Model as specified in Section4.3, with interest at 5% per year effective, calculate A¯1
x:10,A(4)1
x:10
andA1
x:10 forx=20, 40, 60 and 80 and comment on the values.
Solution 4.4We use formula (4.11) with n = 10 to calculate A¯1
x:10 (using numerical integration), and formulae (4.13) and (4.12), withm=4 andn=10 to calculateA(4)1
x:10 andA1
x:10.
The values are shown in Table4.4, and we observe that values in each case increase asxincreases, reflecting the fact that the probability of death in a 10- year period increases with age for the survival model we are using. The ordering of values at each age is the same as in Example4.3, for the same reason – the ordering reflects the fact that any payment under the continuous benefit will be paid earlier than a payment under the quarterly benefit. The end year benefit is paid later than the quarterly benefit, except when the death occurs in the final quarter of the year, in which case the benefit is paid at the same time. 2
Table 4.4. EPVs of term insurance benefits.
x A¯1
x:10 A(4)1
x:10 A1
x:10
20 0.00214 0.00213 0.00209
40 0.00587 0.00584 0.00573
60 0.04356 0.04329 0.04252
80 0.34550 0.34341 0.33722
4.4.6 Pure endowment
Pure endowment benefits are conditional on the survival of the policyholder. For example, a 10-year pure endowment with sum insured $10 000, issued to(x), will pay $10 000 in 10 years if(x)is still alive at that time, and will pay nothing if (x)dies before agex+10. Pure endowment benefits are not sold as stand-alone policies, but may be sold in conjunction with term insurance benefits to create the endowment insurance benefit described in the following section. However, pure endowment valuation functions turn out to be very useful.
The pure endowment benefit of $1, issued to a life agedx, with a term ofn years has present valueZ, say, where:
Z =
0 if Tx <n, vn if Tx≥n.
There are two ways to denote the EPV of the pure endowment benefit using actuarial notation. It may be denoted Ax:n1. The ‘1’ over the term subscript indicates that the term must expire before the life does for the benefit to be paid. This notation is consistent with the term insurance notation, but it can be cumbersome, considering that this is a function which is used very often in actuarial calculations. A more convenient standard actuarial notation for the EPV of the pure endowment isnEx.
If we rewrite the definition ofZ above, we have Z =
0 with probability 1−npx,
vn with probability npx. (4.14) Then we can see that the EPV is
Ax:n1 =vnnpx. (4.15)
Note that because the pure endowment will be paid only at timen, assuming the life survives, there is no need to specify continuous and discrete time versions;
there is only a discrete time version.
4.4 Valuation of insurance benefits 89 We will generally use the more direct notation vnnpx ornEx for the pure endowment function, rather than theAx:n1 notation.
4.4.7 Endowment insurance
An endowment insurance provides a combination of a term insurance and a pure endowment. The sum insured is payable on the death of(x)should(x)die within a fixed term, saynyears, but if(x)survives fornyears, the sum insured is payable at the end of thenth year.
Traditional endowment insurance policies were popular in Australia, North America and the UK up to the 1990s, but are rarely sold these days in these markets. However, as with the pure endowment, the valuation function turns out to be quite useful in other contexts. Also, companies operating in these territories will be managing the ongoing liabilities under the policies already written for some time to come. Furthermore, traditional endowment insurance is still relevant and popular in some other insurance markets.
We first consider the case when the death benefit (of amount 1) is payable immediately on death. The present value of the benefit isZ, say, where
Z =
vTx =e−δTx if Tx <n, vn if Tx ≥n
=vmin(Tx,n)=e−δmin(Tx,n). Thus, the EPV of the benefit is
E[Z] = n
0
e−δttpxµx+tdt+ ∞
n
e−δntpxµx+tdt
= n
0
e−δttpxµx+tdt+e−δnnpx
= ¯Ax:n1 +Ax:n1 and in actuarial notation we write
A¯x:n = ¯A1x:n +Ax:n1 . (4.16) Similarly, the expected value of the squared present value of the benefit is
n
0
e−2δttpxµx+tdt+e−2δnnpx
which we denote2A¯x:n.