Valuation of insurance benefits .1 Whole life insurance: the continuous case, A¯ x

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 variableTx. 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

4.4 Valuation of insurance benefits 107 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=v^Tx = 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 theExpected Present Valueor EPV. It is also commonly referred to as theActuarial Value or theActuarial Present Value.

The EPV of the whole life insurance benefit payment with sum insured $1 is E[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^−δttpxμx+tdt. (4.1)

It is worth looking at the intuition behind this formula. InFigure 4.1we use the time-line format that was introduced inSection 2.4.

Consider timet, wherex≤ x+t < ω. The probability that(x)is alive at timetistpx, and the probability that(x)dies between agesx+tandx+t+dt, having survived to agex+t, is, loosely,μx+tdt, provided thatdtis very small.

In this case, the present value of the death benefit of $1 ise^−δt. Note that we regard the period fromttot+dtas so short that any payment in the interval can effectively be treated as occurring at timetfor discounting.

Now we can integrate (that is, sum the infinitesimal components of) the product of present value and probability over all the possible death intervals ttot+dtto obtain the EPV of the death benefit that will be paid in exactly

Time 0

x

. . . .

t

x+t

t+dt

x+t+dt (x) survivestyears

tpx

(x) dies μx+tdt Age

Event Probability

Present Value e^−δt

Figure 4.1 Time-line diagram for continuous whole life insurance.

one of these intervals. The result is the integral in (4.1). 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^−2δttpxμx+tdt

=²A¯x, (4.2)

where the superscript² indicates that calculation is at force of interest 2δ, or, equivalently, at rate of interestj, where 1+j=e^2δ=(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]=S²₂A¯x− ¯A²_x .

In fact we can calculate any probabilities associated with the random variable Zfrom 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 Valuation of insurance benefits 109 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 inChapter 2.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=v^Kx+1.

The EPV of the benefit, E[Z], is denoted byAxin actuarial notation.

InChapter 2we derived the probability function forKx, Pr[Kx=k]= k|qx, so the EPV of the benefit is

Ax=E v^Kx+1

=^∞

k=0

v^k+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].

In fact, we can always express the EPV of a life-contingent benefit by considering each time point at which the benefit could 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 inSection 4.6.We illustrate the process for the whole life insurance example inFigure 4.2.

The second moment of the present value is ∞

k=0

v^2(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

v^2(k+1)k|qx, (4.5)

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.

and so the variance of the present value of a benefit ofSpayable at the end of the year of death is

V v^Kx+1

=S²₂ Ax−

Ax

2

. (4.6)

4.4.3 Whole life insurance: the1/mthly case, A^(m)x

InChapter 2we introduced the random variableKx, representing the curtate future lifetime of(x), and we saw inSection 4.4.2that 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 the1/mthly curtate future lifetime random variable,K⁽_x^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,K_x⁽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 K_x^(m)= 1

mmTx. (4.7)

For example, suppose(x)lives exactly 23.675 years. Then

Kx=23, Kx⁽²⁾=23.5, K⁽_x⁴⁾=23.5, and K⁽_x¹²⁾=23₁₂⁸ =23.6667.

Note thatK_x^(m)is a discrete random variable.K_x^(m) =kindicates that the life (x)dies in the interval

k,k+_m¹

, fork=0,_m¹,_m²,. . ..

The probability function for K_x^(m) can be derived from the associated probabilities forTx. Fork=0,_m¹,_m²,. . .,

Pr[K⁽_x^m)=k]=Pr

k≤Tx<k+_m¹

=k|¹

mqx= kpx− _k+¹

mpx. InFigure 4.3we show the time-line for the 1/mthly 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.

4.4 Valuation of insurance benefits 111

Time 0 1/m 2/m 3/m

. . . .

Amount $1 $1 $1

Discount v^1/m v^2/m v^3/m

Probability 1

m

1 m

1 m

2 m

1

qx | qx |mqx

Figure 4.3 Time-line diagram for 1/mthly whole life insurance.

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=v^Kx(12)+1/12. We letA⁽¹²⁾_x denote the EPV of this benefit, so that

E[Z]=A⁽¹²⁾_x =v¹²¹ 1

12qx+v¹²² 1 12|¹

12qx+v¹²³ 2 12|¹

12qx+v¹²⁴ 3 12|¹

12qx+ · · ·. Similarly, for anym,

A^(m)_x =v^{m1 1}

mqx+v^{m2 1}

m|¹

mqx+v^{m3 2}

m|¹

mqx+v^{m4 3}

m|¹

mqx+ · · ·

=^∞

k=0

v^k+m1 k m|¹

mqx.

As for the continuous and annual cases, we can derive the variance of the present value of the 1/mthly whole life benefit by adjusting the interest rate for the first term in the variance. We have

E[Z²]=E[v^{2(K(x12)+1/12)}]=E[(v²)^K(x12)+1/12]=²A⁽¹²⁾_x , so the variance is

V

v^K(12)+1/12

=²A⁽¹²⁾_x − A⁽¹²⁾_x 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 Chapter 3,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 Axusingbackward recursion. To do this, we start from the highest age in the table,ω. We assume all lives expire by ageω, so thatq_ω−1 = 1. If the life table does not have a limiting age, we choose a suitably high value forωso thatq_ω−1 is as close to 1 as we like. This means that any life attaining age ω−1 may be treated as certain to die before ageω, in which case we know thatK_ω−1=0 and so

A_ω−1=E[v^Kω−1+1]=v.

Now, working from the summation formula forAx, we have Ax=^ω−

x−1

k=0

v^k+1kpxqx+k

=v qx+v²pxqx+1+v³2pxqx+2+ · · ·

=v qx+v px

v qx+1+v²px+1qx+2+v³2px+1qx+3+ · · · , giving the important recursion formula

Ax=v qx+v pxAx+1. (4.8) This formula can be used in spreadsheet format to calculateAxbackwards from A_ω−1back toAxo, wherexois the minimum age in the table.

The intuition forequation (4.8)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 for 1/mthly benefits; now the recursion will giveA⁽_x^m) 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⁽_x^m)=v^m1 1

mqx+v^m2 1 mpx ¹

mq_x+1 m +v^m3 2

mpx ¹

mq_x+2 m + · · ·

=v^m1 1

mqx+v^m1 1 mpx

v^m1 1 mq_x+1

m +v^m2 1 mp_x+1

m 1 mq_x+2

m + · · · , giving the recursion formula

A⁽_x^m)=v^m1 1

mqx+v^m1 1 mpxA^(m)

x+m¹

.

Example 4.1 Using the Standard Ultimate Survival Model fromSection 4.3, and an interest rate of 5% per year effective, construct a spreadsheet of values ofAxforx=20, 21,. . ., 100. Assume thatA129=v.

4.4 Valuation of insurance benefits 113 Table 4.1 Sample values of Axusing the Standard Ultimate

Survival Model,Example 4.1.

x A_x x A_x x A_x

30 0.07698 50 0.18931 98 0.85177

31 0.08054 51 0.19780 99 0.86153

32 0.08427 52 0.20664 100 0.87068

... ... ... ... ... ...

Solution 4.1 The survival model for the Standard Ultimate Survival Model is the ultimate part of the model used inExample 3.13and so values oftpx can be calculated as explained in the solution to that example. The value ofq129is 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,. . ., 80, are given inAppendix D Table D.3. Some excerpts are

shown inTable 4.1.

Example 4.2 Using the Standard Ultimate Survival Model fromSection 4.3, 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 4.2 For this example, we follow exactly the same process as for the previous example, except that we let the ages increase by 1/12 years in each row. We construct a column of values of ¹

12pxusing

1

12px=exp

−A/12−Bc^x(c^1/12−1)/logc . We again use 130 as the limiting age of the table. Then setA⁽¹²⁾

129¹¹₁₂ =v^1/12, and for all the other values ofA⁽_x¹²⁾use the recursion

A⁽_x¹²⁾=v^1/12 1

12qx+v^1/12 1 12pxA⁽¹²⁾

x+₁₂¹.

The first and last few lines of the spreadsheet are reproduced inTable 4.2.

It is worth making a remark about the calculations inExamples 4.1and4.2.

In Example 4.1 we saw that q129 = 0.99996, which is sufficiently close to 1 to justify us starting our recursive calculation by setting A129 = v.

InExample 4.2, our recursive calculation started fromA₁₂₉11

12 =v^1/12. If we calculate 1

12q₁₂₉11

12we find its value is 0.58960, which is certainly not close to 1.

Table 4.2 Sample values of A⁽¹²⁾_x using the Standard Ultimate Survival Model,Example 4.2.

x 1

12

p_x 1

12

q_x A⁽¹²⁾_x

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 0.99594

What is happening in these calculations is that, for Example 4.1, we are replacing the exact calculation

A129=v(q129+p129A130)

byA129 = v, which is justifiable becauseA130 is close to 1, meaning that v(q129+p129A130)is very close tov. Similarly, forExample 4.2,we replace the exact calculation

A⁽¹²⁾

129¹¹₁₂ =v^1/12

1 12q₁₂₉¹¹

12 + ¹

12p₁₂₉¹¹

12A⁽¹²⁾₁₃₀ byA⁽¹²⁾

129¹¹₁₂ =v^1/12. As the value ofA⁽¹²⁾₁₃₀is very close to 1, it follows that v^1/12

1 12q₁₂₉11

12+ ¹

12p₁₂₉11 12A⁽₁₃₀¹²⁾

can by approximated byv^1/12.

Example 4.3 Using the Standard Ultimate Survival Model, 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 4.3 For part (a), we must calculate 100 000A¯xand 100 000

2A¯x−(A¯x)²

4.4 Valuation of insurance benefits 115 Table 4.3 Mean and standard deviation of the present value of a whole life

insurance benefit of$100 000, forExample 4.3.

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

forx=20, 40, 60 and 80, where²A¯xis calculated at effective rate of interest j=10.25%. For parts (b) and (c) we replace each A¯x by A⁽_x¹²⁾ and Ax, respectively. The values are shown in Table 4.3. 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 Examples 4.1and4.2.

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. Third, as xincreases, 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.

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=

v^Tx =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 (4.9)

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,A_x: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 random variable for the benefit is now

Z=

v^Kx+1 if Kx≤n−1, 0 if Kx≥n.

The EPV of the benefit is denotedA_x:n¹ so that

A_x:n¹ =

n−1

k=0

v^k+1k|qx. (4.10)

The 1/mthly case,A^(m)_x:n¹

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 withinn years. The present value random variable for the benefit is

Z=

v^K(xm)+m1 if K_x^(m) ≤n−_m¹, 0 if K_x^(m) ≥n.

4.4 Valuation of insurance benefits 117 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

The EPV of the benefit is denotedA^(m)_x:n¹ so that

A^(m)_x:n¹ =

mn−1 k=0

v^{(k+1)/m k}

m|¹

mqx. (4.11)

Example 4.4 Using the Standard Ultimate Survival Model as specified in Section 4.3,with interest at 5% per year effective, calculateA¯_x:¹

10,A⁽⁴⁾¹_x:

10 and A¹

x:10 forx=20, 40, 60 and 80 and comment on the values.

Solution 4.4 We use formula (4.9) with n = 10 to calculate A¯_x:¹

10 (using numerical integration), and formulae (4.11) and (4.10), withm=4 andn=10 to calculateA⁽⁴⁾¹

x:10 andA¹

x:10.

The values are shown inTable 4.4,and we observe that values in each case increase as xincreases, 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 Example 4.3,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.

4.4.6 Pure endowment

Pure endowment benefits are conditional on the survival of the policyholder at a policy maturity date. 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 age x+10. Pure endowment benefits are not sold as stand-alone policies, but may be sold in conjunction with term insurance benefits to create an endowment insurance, which is covered in the next section. However, pure endowment valuation functions turn out to be very useful when we value other benefits.

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 A_x: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 ofZabove, we have Z=

0 with probability 1−npx,

vⁿ with probability npx. (4.12) Then we can see that the EPV is

A_x:n¹ =nEx=vⁿnpx. (4.13) Note that because the only possible payment date for the pure endowment is at timen, there is no need to specify continuous and discrete time versions;

there is only a discrete time version.

It is sometimes useful to have an expression for E[Z²], which we denote by

n2Ex. As

Z²=

0 with probability 1−npx,

v²ⁿ with probability npx, (4.14) we have

2

nEx=v²ⁿnpx=vⁿnEx. (4.15) So the variance of the present value is

V[Z]=n²Ex−nE²_x =nExvⁿ(1−npx)=nExvⁿnqx. (4.16) We will generally usevⁿnpxornExfor the pure endowment function, rather than theA_x: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)if(x)dies within a fixed term, saynyears, but if(x)survives fornyears, the sum insured is payable at the end of thenth year.

4.4 Valuation of insurance benefits 119 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 emerging insurance markets, for example in the context of microinsurance.

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=

v^Tx =e^−δTx if Tx<n,

vⁿ if Tx≥n

=v^min(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

= ¯A_x:n¹ +A_x:n¹ and in actuarial notation we write

A¯_x:n = ¯A_x:n¹ +A_x:n¹ = ¯A_x:n¹ +nEx. (4.17) 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.

In the situation when the death benefit is payable at the end of the year of death, the present value of the benefit is

Z=

v^Kx+1 if Kx≤n−1, vⁿ if Kx≥n

=v^min(Kx+1,n).