Modelling Home Equity Conversion Loans With Life Insurance Models

HOME EQUITY CONVERSION LOANSIt is known that life insurance mathematics has its background in probabilitytheory (probability that a certain person will die at some age, or probability thatsome person will be alive at some point in future). Also, it is known that lifeinsurance mathematics has some background in the theory of interest. Themain feature of any life insurance contract consists of the moment of policyissue and the moment of benefit payment. The time lag is determined by twomoments. One moment can be determined as the moment of death, but it alsocan be determined as the payment moment. These two moments can, but alsomust coincide. We can talk about moment of death T and moment of paymentΨ. So, it is necessary to include a present value of the payment in perspective. Ifwe consider the inconsistent financial market, then the present value of thepayment of a unit of money is given with a discount factor v (Paramenter, 1999,p. 10). If we look at a situation where Ψ assumes a value t, then our presentvalue of the payment of a unit of money is given by vt for an inconsistent market(Capinski & Zastawniak, 2003, p. 22). In the condition of a consistent marketthe present value is defined in a similar way. In that case we have continuoustime and we assume that interest is compounded continuously. It is alsonecessary to use the discount factor for a certain period of time, but now thediscount factor is defined slightly differently. In that case interest calculation isbased on the discount factor that is given by (Gerber, 1997):e t(0.1)Note that a case where compounding occurs a finite number of times per year isreferred to as discrete compounding, while if n it is referred to ascontinuous compounding (Hoy, Livernois, McKenna, Ress, & Stengos, 2001).So, by relation (0.1) we have defined the present value of the payment of a unitof money after t years, where the financial market is consistent and is theunit-time-interval force of interest (Merton, 1990). This approach allowsanalyzing the case where the interest rate can change more often than onceevery year (Moller & Steffensen, 2007, p. 49).133

Economic Annals, Volume LVIII, No. 199 / October – December 2013We presuppose that the interest rate (or force of interest) is certain (nonrandom) and does not change in time, or so-called ‘classical actuarialdiscounting’ (Mario, Bühlamann, & Furrer, 2008, p. 13).Let us observe a life insurance policy issued by a certain insurance company.The payment follows the death of the insured whenever the death occurs. Hisinheritors will receive that payment of a unit of money. This type of product iscalled whole life insurance. As we have said before, Ψ stands for time of benefitpayment. Age-at-death is the random variable. The insurance company mustdetermine the present value of the payment of a unit of money which followsthe death of the insured at an unknown moment, where the financial market isconsistent with force of interest δ. The difference between the present value ofthe payment and the payment itself depends on the force of interest and thetime lag between the policy issue and the moment of payment. Force of interestis given as the market value. We already have assumed that force of interest isnot random. On the other hand, the death of the insured can occur at any timein the future with certain probabilities. The probability depends on the age ofthe insured and various other factors. If the moment of payment Ψ is random,then the present value of the future payment of a unit of money is also random,and by (0.1) is equal to the random variable Z. Random variable Z could bewritten as follows:Z e .(0.2)We can only define the expected value of Z because Ψ is a random variable. Theexpected value for random variable Z is given as follows: A E Z E e .(0.3)The quantity A represents actuarial present value. Actuarial present valuerepresents both segments of the actuarial perspective on life insurance as a twosided problem. On one side there is the problem of defining the probability thatsome person will die at age x.134

HOME EQUITY CONVERSION LOANSThe second side of the problem is defining the present value of the sum payableat some moment in the future. An important fact is that the actuarial presentvalue is the moment-generating function of Ψ (Rotar, 2007), which is given by A E e M .(0.4)By Z e , the l th moment isl E Z l E e M l .(0.5)If we know the moment-generating function M s , then we know allmoments of Z. So, for l 1 we have E Z E e M ,(0.6)and for l 2: E Z 2 E e 2 M 2 .(0.7)Now, it is possible to define variance for our random variable:Var Z E Z 2 E Z M 2 M 2 .22(0.8)Let μ(x) μ, where μ stands for force of mortality. Then X is exponential, and bythe lack-of-memory property T(x) is also exponential (Ross, 1997).If the random variable has a distinct moment-generating function, then therandom variable has a distinct distribution. Also, if the random variable has adistinct distribution, then the random variable has a distinct momentgenerating function (Epps, 2009). It is known that the moment-generatingfunction for an exponential random variable with the parameter a and for z a is(Rotar, 2007)135

Economic Annals, Volume LVIII, No. 199 / October – December 2013 M z ezxaxae dx0a1. a z 1 za(0.9)For z a a moment-generating function does not exist. Therefore, we couldconclude that the moment-generating function for random variable T x is given by M T x z . So, by (0.3) and (0.9) we have as follows: zAx (0.10) and2 Var. Z 2 (0.11)It is obvious that to calculate the expected value of actuarial present value and itsvariance it is not necessary to use age-at-death x . It is enough that we knowforce of mortality and force of interest.3.1 Whole life insurance – benefits payable at the moment of deathIn this case, T T x for a life-age x . Then the present value of thepayment is Z e T x , and the actuarial present value is T x Ax E e M T x .(0.0.12)In (0.0.12) M T x is the moment-generating function of T x . The density ofT x (Klugman, Panjer, & Willmot, 2004) is given as follows:fT x x t t px . t x t t px136(0.0.13)

HOME EQUITY CONVERSION LOANSAccording to (0.0.12) and (0.0.13), it follows that Ax e t x t t px .(0.0.14)0We could see that for the exponential random variable the force of interest isconstant and Ax . , which corresponds to the actuarial presentvalue for the doubled interest rate and is given by A E Z E e . As we saw, E Z 2 E e 2 T x 22 2 T xxMaking use of this notation, we can write:Var Z 2Ax Ax .2(0.0.15)3.2. Whole life insurance – benefits payable at the end of the year of deathThe models below could be applied for any choice of time unit. The fact that inthe title a year has been chosen as the time unit does not restrict using the K x 1.models presented below for any time unit. In this case we have x K If we know that equality P K kpx qx k is true, than we can derivethe following expression for actuarial present value: Ax E e e k 1 P K x k e k 1 k px qx k .(0.0.16)k 0 k 0 Let us estimate the net premium A60 for Yugoslav demographic tables 1952–1954 and 0, 05 (Šain, 2009., p. 282) So we have:P K 50 k l60 k d60 k d60 k. l60 l60 kl60137

Economic Annals, Volume LVIII, No. 199 / October – December 2013It makes sense to take k 30. It is large enough. We can see that qx dxlx(Paramenter, 1999, p. 128). If we take the expression above and put it in (0.0.16), then it follows thatA60 40d60 ke l k 0k 10.242656.60Another way to compute actuarial present value is based on the followingrelation:Ax e qx px e Ax 1 e qx px Ax 1 .(0.0.17)If we know the value of An and px we can move 'backwards’ and computeAx n x . The proof for relation (0.0.17) is intuitive by nature. Let us setsome initial moment as the standpoint for the proccess which has beenobserved. Starting from that moment, the insured person can either die orsurvive during the next year. The probability that he will die is q x . The insuredsum is one unit of money and its present value is e . The probability that hewill survive is px . In that case the insured person will be x 1 years old and hislife is ‘continued’. The insurance process will start over and the present value ofthe insured sum is Ax 1 . We can conclude that the actuarial present value is thesum of the two summands. The first is e with probability q x and the second isAx 1 with probability px . So it follows that E Z T x 1 q E Z E Z T x 1 P T x 1 E Z T x 1 P T x 1 x E Z T x 1 px , only if E Z T x 1 e и E Z T x 1 e Ax 1 is true.138

HOME EQUITY CONVERSION LOANS Let us prove that E Z T x 1 e Ax 1 is true. In that case we have: K x 1E Z T x 1 E e T x 1 e E e K x T x 1 e E e 1 K x 1 e E e T x 1 1 K x 1 e Ax 1 .We have used the fact that life expectancy for an insured person who hassurvived the first year is equal to the sum of one year and life expectancy for anx 1 years-old person expressed in years.Analogous to the case when benefit is payable at the moment of death, we havethe case when benefit is payable at the end of the year of death and force ofmortality is constant net premium defined by Ax qx(Batten, 2005). Thisqx irelation could be written in another way, as follows: Axqx1 px1 px. qx i 1 px e 1 px e Let us look at the relation between Ax и Ax with more attention. The main toolfor this relation’s analysis is linear interpolation procedure. Here theassumption is made that lifetime is uniformly distributed within each year ofage. In that case we have as follows (Bowers, Gerber, Hickman, Jones, & Nesbit,1997):Ax i Ax ,(0.0.18)139

Economic Annals, Volume LVIII, No. 199 / October – December 2013where, as we have stated before, i e 1 . So i is effective annual rate (yield).It is important to notice that for small , correction coefficienti 1 (Mario,Bühlamann, & Furrer, 2008). Let us look at the situation where we have0.04081 1.02 . This0.04i e 0.04 1 0.04081 , and 0.04 . In that case becomes understandable if we recall that e x 1 x x2 o x2 .2So, if we use that fact we have as follows:1 e 1 1 o . 2iIn relation (0.0.19)i (0.0.19)differs from 1 by approximatelyLet us prove that Ax i 2.Ax . According to that we will observe two quantities,lifetime T T x and corresponding curtate time K K x . Let us introducethe third quantity. This quantity is defined as the difference between the firsttwo (the bigger one T T x and the smaller one K K x ), and it could be,named as a fractional part of lifetime T T x . We will denote this quantityby Tr x . As we have said, Tr x T x K x . K K x и Tr x areindependent and Tr x is uniform at 0,1 . We could write as follows: e E e E e Tr K Ax E E e K E e T E e e Tr 140 K 1 Tre Ax E e Tr .(0.0.20)

HOME EQUITY CONVERSION LOANS It is known that E e Tr M Tr where MTr z is the moment- generating function for Tr . Also, we know that E e Tr 1 e andaccording to (0.0.20) we could write it as follows:1 e e 1 Ax e AxAx . (0.0.21) The link between (0.0.21) and (0.0.18) is obvious.3.3. The case of benefits payable at the end of the m Let us denote by Axm -nthly periodthe actuarial present value of benefits payable at the m m nthly period. So here we will present an approximation formula for Axassuming that the lifetime is uniformly distributed in each year. According tothat we can write (Bowers, Gerber, Hickman, Jones, & Nesbit, 1997)Ax m ii(0.0.22)Ax m It is important to mention that in the relation above iRelation(0.0.22)followsfrom m (0.0.18) m 1 i 1mandwe 1 . knowthat lim i ln 1 i . According to that we can conclude that the m m icoefficient of correction m is increasing if m is increasing.iHence, for any m :1 ii m i 1 2 o .141

Economic Annals, Volume LVIII, No. 199 / October – December 2013According to what has been said above, we now can write the relation as follows:0 ii m 1 i 2 e2 26.(0.0.23)It is obvious that the coefficient of correction is increasing from 1 toi .Let us prove the expression for the actuarial present value where benefit ispayable at the end of m -nthly period of the year in which death occurs. Denotemby K the number of complete periods of the length 1 m that the insured m m survived. Let us call these periods m -ths. Set R K mK . This is thevariable which represents the number of complete m -ths lived in the year ofdeath. Let us look at a simple example where m 12 ; in other words, we areobserving1th of the year. We will assume that one month is equal to that121th part of the year, although it is not true that every month is the same12(number of days differ from month to month). Also, let’s say that we areobserving the person insured at age T 24.48 . This age is expressed in years. Ifwe express that age by months we have 293.6. We could write this as m follows: K 293 , and mK 288 . It is obvious that the person insured haslived 5 months during his 25th year of life and approximately 60% of the 6thmonth. We can write R m 5.Let us come back to the general case. Under the assumption made the randomR m and K are independent, and R m takes on values1, 2,., m 1 with the same probability. Variable R m takes the mentionedvariablesvalues because it is limited and it can be as large as the basic unit (in our casethat variable can be related to the 11-month period). We mentioned before thatwe have taken an annual interest rate. According to that we must take a relative142

HOME EQUITY CONVERSION LOANSinterest rate (we need to divide our annual interest rate by ( m ) and then discount benefit for K m 1 .Then we can write:Ax m m K 1E exp m m mK R 1 E exp m R m 1 E exp K m m R 1 E exp K E exp m m R 1 E exp K E exp m m R 1 e E exp K 1 E exp m Ax e E exp R 1 . m R 1 can be written down m According to (0.0.12) expression E exp m , where M R m 1 z is the moment-generating function for mrandom variable R 1 .as MmR 1143

Economic Annals, Volume LVIII, No. 199 / October – December 2013If we know that the random variable takes values 1, 2,., m with identicalprobabilities, then we can write it as follows:M R m 1 z 1 z 1 emzzk 1 ee. m m 1 ezk 1m(0.0.24)3.4. Deferred whole life insuranceThis type of insurance product provides a benefit only if the insured survives cyears. As we have said before, we set if the condition for benefitpayment has not been satisfied. So, for T x c we have , and theopposite, if T x c we have T x . The case where T x c isirrelevant, because we are observing the continuous variable. Because Z e , 0Z T x eT x cT x c.(0.0.25)Let us observe the situation where death occurs during the period of c yearsand 0.1 . In that case the present value of the payment of a unit of money is Z e e 0.1* 0 . Let us assume that we have no yield on invested capital, or 0 . So according to that we have ln 1 i ln 1 0 .Then, the exponent of Napier s number is an undefined expression, or 0 .However, the exponential function is ‘more powerful’ than the logarithmicfunction. So, according to that in exponent of Napier s number we will have and then our benefit equals zero.In this case we will denote actuarial present value bycAx c p x e c Ax c .(0.0.26)As we can see, the actuarial present value for the deferred whole life insurancecan be presented as the product of the actuarial present value for the whole life144

HOME EQUITY CONVERSION LOANSinsurance with the starting age x c and the probability that the insuredperson will survive c years. This product must be discounted for c years.So we can write (Rotar, 2007): E Z 0 P T x c E Z T x c P T x c p e E e E e T x T x c c px T x c ccxcc k T x c px E e (0.0.27)px e c Ax c .Let us observe the case where benefit is payable at the end of the period in whichdeath occurs. So, T x c Z 0 K x 1 K x 1 T x c Z e .(0.0.28)We will denote the actuarial present value in this case as follows: Ac x e * P K x k . k 1c kOr we could write it down as follows (Rotar, 2007):cAx c p x e c Ax c .(0.0.29)Because the number of discount periods is the integer in this case, we couldwrite it as follows:cAx c px v c Ax c .(0.0.30)145

Economic Annals, Volume LVIII, No. 199 / October – December 20133.5. Term insurance – the continuous-time caseThis type of insurance product, in the general case, provides payment of the unitof money if death occurs during n years. So, we can write: T x n T x T x n Z e T x , Z 0.(0.0.31)In this case we will denote actuarial present value bynA1x:n e t fT x t dt .(0.0.32)0Here, the actuarial present value depends on the distribution of the randomvariable in the interval 0, n and it does not depend on the mortality rateafter n years (Rotar, 2007).It is possible to use the following formula: Ax A1x:n e n n px Ax n .(0.0.33)Formula (0.0.33) can be written in another way, as follows: A1x:n e n n px Ax n Ax .(0.0.34)Proof for relation (0.0.33) is simple and it follows from relation (0.0.31), and the(0.0.25) equation reference goes here. If we denote by Z1 the random variable in(0.0.25) where c n , then Z is the same as in (0.0.31). Let us write:Z146 e T x , T x n, Z1 0,Txn , T x n 0 T x , T x n. e(0.0.35)

HOME EQUITY CONVERSION LOANS T x eIt is obvious that Z Z1 and that the relation in fact presents thepresent value for a payment of a unit of money for the whole life insurancecontract.Let us look the same insurance product but for the case where benefit is payableat the end of the year (or some other basic time period). So, for term insurancefor n years we have: e K x 1 ,Z , 0T x n(0.0.36)T x n.In that case we will denote actuarial present value by Ax:n . Also, we can write:Ax: nn 1 e k 1 P K x k n 1 e k 0 k 0 k 1 npx qx k .(0.0.37)In that sense, if we observe (0.0.33) then we have: Ax A1x:n e n n px Ax n ,(0.0.38)and it is possible to write: A1x:n e n n px Ax n Ax .(0.0.39)3.6. Pure endowmentIn this case benefit is payable only if the insured survives until a certain age. Tobe precise, the insured

With a home equity conversion loan the borrower does not have to make monthly payments to repay the loan. Home equity conversion loans do not have a predetermined maturity date, as do conventional loans. But it is a loan, and every loan must be repaid. Home equity conversion loans do not need to be repaid until the last surviving borrower dies.