2 COURSE OUTLINE Theory of Interest Life Insurance Interest: the basic theoryInterest: basic applicationsAnnuitiesAmortization and sinking fundsBondsLife InsurancePreparation for life contingenciesLife tables and population problemsLife annuitiesLife insurance
3 Chapter 1 INTEREST: THE BASIC THEORY Accumulation FunctionSimple InterestCompound InterestPresent Value and DiscountNominal Rate of InterestForce of Interest
4 1.1 ACCUMULATION FUNCTION DefinitionsThe amount of money initially invested is called the principal.The amount of money principal has grown to after the time period is called the accumulated value and is denoted by A(t) – amount function.t ≥0 is measured in years (for the moment)Define Accumulation function a(t)=A(t)/A(0)A(0)=principala(0)=1A(t)=A(0)∙a(t)
5 Natural assumptions on a(t) increasing(piece-wise) continuousa(t)a(t)a(t)(0,1)(0,1)(0,1)tttNote: a(0)=1
6 Definition of Interest and Rate of Interest Interest = Accumulated Value – Principal: Interest = A(t) – A(0)Effective rate of interest i (per year):Effective rate of interest in nth year in:
7 a(t)=t2+t+1 Example (p. 5) Verify that a(0)=1 Show that a(t) is increasing for all t ≥ 0Is a(t) continuous?Find the effective rate of interest i for a(t)Find in
8 ( ≡ Two Types of Accumulation Functions) Two Types of Interest( ≡ Two Types of Accumulation Functions)Simple interest:only principal earns interestbeneficial for short term (1 year)easy to describeCompound interest:interest earns interestbeneficial for long termthe most important type of accumulation function
9 a(t)=1+it, t ≥0 1.2 SIMPLE INTEREST Amount function: A(t)=A(0) ∙a(t)=A(0)(1+it)Effective rate is iEffective rate in nth year:(0,1)ta(t) =1+it11+i
10 a(t)=1+it i=0.15 A(t)=A(0)(1+it) =1000(1+0.15t) t=? A(0)=1000 Example (p. 5)SolutionJack borrows 1000 from the bank on January 1, 1996 at a rate of 15% simple interest per year. How much does he owe on January 17, 1996?A(0)=1000i=0.15A(t)=A(0)(1+it) =1000(1+0.15t)t=?
11 How to calculate t in practice? Exact simple interest number of daysOrdinary simple interest (Banker’s Rule) number of dayst =t=Number of days: count the last day but not the first
12 A(t)=1000(1+0.15t) Exact simple interest Number of days (from Jan 1 to Jan 17) = 16Exact simple interestt=16/365A(t)=1000( ∙ 16/365) =Ordinary simple interest (Banker’s Rule)t=16/360A(t)=1000( ∙ 16/360) =
13 1.3 COMPOUND INTEREST Interest earns interest After one year: a(1) = 1+iAfter two years: a(2) = 1+i+i(1+i) = (1+i)(1+i)=(1+i)2Similarly after n years: a(n) = (1+i)n
14 a(t)=(1+i)t COMPOUND INTEREST Accumulation Function Amount function: A(t)=A(0) ∙a(t)=A(0) (1+i)tEffective rate is iMoreover effective rate in nth year is i (effective rate is constant):1+it(0,1)ta(t)=(1+i)t11+i
15 How to evaluate a(t)?If t is not an integer, first find the value for the integral values immediately before and afterUse linear interpolationThus, compound interest is used for integral values of t and simple interest is used between integral values1ta(t)=(1+i)t1+i2(1+i)2
16 Example (p. 8) a(t)=(1+i)t A(t)=A(0)(1+i)t A(t)=1000(1+0.15)t Jack borrows 1000 at 15% compound interest.How much does he owe after 2 years?How much does he owe after 57 days, assuming compound interest between integral durations?How much does he owe after 1 year and 57 days, under the same assumptions as in (b)?How much does he owe after 1 year and 57 days, assuming linear interpolation between integral durationsIn how many years will his principal have accumulated to 2000?a(t)=(1+i)tA(t)=A(0)(1+i)tA(0)=1000, i=0.15A(t)=1000(1+0.15)t
17 1.4 PRESENT VALUE AND DISCOUNT The amount of money that will accumulate to the principal over t years is called the present value t years in the past-ttPRESENT VALUEPRINCIPALACCUMULATED VALUE
18 Calculation of present value t=1, principal = 1Let v denote the present valuev (1+i)=1v=1/(1+i)
19 v=1/(1+i) In general: t is arbitrary a(t)=(1+i)t [the present value of 1 (t years in the past)]∙ (1+i)t = 1the present value of 1 (t years in the past) = 1/ (1+i)t = vt
20 (0,1)ta(t)=(1+i)ta(t)=(1+i)tgives the value of one unit (at time 0) at any time t, past or future
21 If principal is not equal to 1… present value = A(0) (1+i)tt < 0t = 0t > 0PRESENT VALUE A(0) (1+i)tPRINCIPAL A (0)ACCUMULATED VALUE A(0) (1+i)t
22 Example (p. 11) Solution a(t)=(1+i)t The Kelly family buys a new house for 93,500 on May 1, How much was this house worth on May 1, 1992 if real estate prices have risen at a compound rate for 8 % per year during that period?Find the present value of A(0) = 93,500= 4 years in the pastt = - 4, i = 0.08Present value = A(0) (1+i)t = 93,500 (1+0.8) -4 = 68,725.29
23 If simple interest is assumed… a (t) = 1 + itLet x denote the present value of one unit t years in the pastx ∙a (t) = x (1 + it) =1x = 1 / (1 + it)NOTE:In the last formula, t > 0
24 Thus, unlikely to the case of compound interest, we cannot use the same formula for present value and accumulated value in the case of simple interest1a(t) =1+it1 / (1 + it)a(t) =1+it1t1 / (1 - it)
25 Discount Look at 112 as a basic amount We invest 100 Alternatively:We invest 100After one year it accumulates to 112The interest 12 was added at the end of the termLook at 112 as a basic amountImagine that 12 were deducted from 112 at the beginning of the yearThen 12 is amount of discount
26 Rate of Discount d = i = Definition Effective rate of discount d accumulated value after 1 year – principal accumulated value after 1 yearA(1) – A(0) A(1)d ==A(0) ∙a(1)– A(0) A(0) ∙a(1)a(1) – 1 a(1)==Recall:accumulated value after 1 year – principal principali =a(1) – 1 a(0)=
29 Present and accumulated values in terms of d: Present value = principal * (1-d)tAccumulated value = principal * [1/(1-d)t]If we consider positive and negative values of t then:a(t) = (1 - d)-t
30 Examples (p. 13)1000 is to be accumulated by January 1, 1995 at a compound rate of discount of 9% per year.Find the present value on January 1, 1992Find the value of i corresponding to dJane deposits 1000 in a bank account on August 1, If the rate of compound interest is 7% per year, find the value of this deposit on August 1, 1994.
31 1.5 NOMINAL RATE OF INTEREST Example (p. 13)A man borrows 1000 at an effective rate of interest of 2% per month. How much does he owe after 3 years?Note: t is the number of effective interest periods in any particular problem
32 More examples… (p. 14)You want to take out a mortgage on a house and discover that a rate of interest is 12% per year. However, you find out that this rate is “convertible semi- annually”. Is 12% the effective rate of interest per year?Credit card charges 18% per year convertible monthly. Is 18% the effective rate of interest per year?In both examples the given rates of interest (12% and 18%) were nominal rates of interest
33 …yet another example You have two credit card offers: Which is better? 17% convertible semi-annually16% convertible monthlyWhich is better?
34 Definition Suppose we have interest convertible m times per year The nominal rate of interest i(m) is defined so that i(m) / m is an effective rate of interest in 1/m part of a year
35 Note: If i is the effective rate of interest per year, it follows that In other words, i is the effective rate of interest convertible annually which is equivalent to the effective rate of interest i(m) /m convertible mthlyEquivalently:
36 Examples (p. 15)Find the accumulated value of 1000 after three years at a rate of interest of 24 % per year convertible monthlyIf i(6)=15% find the equivalent nominal rate of interest convertible semi-annually
38 Nominal rate of discount The nominal rate of discount d(m) is defined so that d(m) / m is an effective rate of discount in 1/m part of a yearFormula:
39 Formula relating nominal rates of interest and discount
40 ExampleFind the nominal rate of discount convertible semiannualy which is equivalent to a nominal rate of interest of 12% convertible monthly
41 1.6 FORCE OF INTERESTWhat happens if the number m of periods is very large?One can consider mathematical model of interest which is convertible continuouslyThen the force of interest is the nominal rate of interest, convertible continuously
42 Definition Nominal rate of interest equivalent to i: Let m approach infinity:We define the force of interest δ equal to this limit:
43 FormulaForce of interest δ = ln (1+i)Therefore eδ = 1+iand a (t) = (1+i)t =eδtPractical use of δ: the previous formula gives good approximation to a(t) when m is very large
44 ExampleA loan of 3000 is taken out on June 23, If the force of interest is 14%, find each of the following:The value of the loan on June 23, 2002The value of iThe value of i(12)
45 RemarkThe last formula shows that it is reasonable to define force of interest for arbitrary accumulation function a(t)
46 Definition The force of interest corresponding to a(t): Note: in general case, force of interest depends on tit does not depend on t ↔ a(t)= (1+i)t !
47 Example (p. 19)Find in δt the case of simple interestSolution
48 How to find a(t) if we are given by δt ? We have:Consider differential equation in which a = a(t) is unknown function:Since a(0) = 1 its solution is given by
49 ApplicationsProve that if δt = δ is a constant then a(t) = (1+i)t for some iProve that for any amount function A(t) we have:Note: δt dt represents the effective rate of interest over the infinitesimal “period of time” dt . Hence A(t)δt dt is the amount of interest earned in this period and the integral is the total amount
50 Remarks Do we need to define the force of discount? It turns out that the force of discount coincides with the force of interest! (Exercise: PROVE IT)Moreover, we have the following inequalities:and formulas:
51 Chapter 2 INTEREST: BASIC APPLICATIONS Equation of ValueUnknown Rate of InterestTime-Weighted Rate of Return
52 2.1 Equation of Value Four numbers: Time diagram principal A(0)accumulated value A(t) = A(0) ∙ a(t)period of investment t (determine effective period to find t)rate of interest iTime diagramBring all entries of the diagram to the same point in time and write the equation of value
53 ExamplesFind the accumulated value of 500 after 173 months at a rate of compound interest of 14% convertible quarterly (p. 30)Alice borrows 5000 from FF Company at a rate of interest 18% per year convertible semi-annually. Two years later she pays the company Three years after that she pays the company How much does she owe seven years after the loan is taken out? (p. 31)Eric deposits 8000 on Jan 1, and 6000 on Jan 1, 1997 and withdraws on Jan 1, Find the amount in Eric’s account on Jan 1, 2004 if the rate of compound interest is 15% per year (p. 31)
54 More Examples…Unknown timeJohn borrows 3000 from FFC. Two years later he borrows another Two years after that he borrows an additional At what point in time would a single loan of be equivalent if i = 0.18 ? (p. 32)Unknown rate of interestFind the rate of interest such that an amount of money will triple itself over 15 years (p. 32)
55 2.2 UNKNOWN RATE OF INTEREST We need to find the rate of interest iSet up equation of value and solve it for iVery often the resulting equation is a polynomial equation in i of degree higher than 2In general, there is no formula for solutions of equation of degree ≥ 5 (and the formulas for degrees 3 or 4 are very complicated)Use approximations (numerical methods)
56 ExamplesJoan deposits 2000 in her bank account on January 1, 1995, and then deposits 3000 on January 1, If there are no other deposits or withdrawals and the amount of money in the account on January 1, 2000 is 7100, find the effective rate of interest.Obtain a more exact answer to the previous question
57 2.3 TIME-WEIGHTED RATE OF RETURN Let B0, B1, … , Bn-1, Bn denote balances in a fund such that precisely one deposit or withdrawal (denoted by Wt) is made immediately after Bt starting from t=1Let W1, … , Wn-1 denote the amounts of deposits (Wt > 0) or withdrawals (Wt > 0) and let W0 = 0Determine rate of interest earned in the time period between balances:The time-weighted rate of return is defined by i = (1+i1 ) (1+i2 ) … (1+in ) - 1
58 Example (p. 35)On January 1, 1999, Graham’s stock portfolio is worth 500,000. On April 30, 1999, the value has increased to 525,000. At that point, Graham adds 50,000 worth of stock to this portfolio. Six month later, the value has dropped to 560,000, and Graham sells 100,000 worth of stock. On December 31, 1999, the portfolio is again worth 500,000. Find the time-weighted rate of return for Graham’s portfolio during 1999.
59 Unknown Time and Unknown Rate of Interest Continuous Annuities Chapter 3 ANNUITIESBasic ResultsPerpetuitiesUnknown Time and Unknown Rate of InterestContinuous AnnuitiesVarying Annuities
60 3.2 Basic ResultsDefinition: Annuity is a series of payments made at regular intervalsPractical applications: loans, mortgages, periodic investmentsBasic model: consider an annuity under which payments of 1 are made at the end of each period for n equal periods
61 Series of n payments of 1 unit FormulasSeries of n payments of 1 unitpresent value of annuityaccumulated value of annuitysn|an|1111…..n123The present (accumulated) value of the series of payments is equal to the sum of present (accumulated) values of each payment
62 Examples (p. 46)(Loan) John borrows 1500 and wishes to pay it back with equal annual payments at the end of each of the next ten years. If i = 17% determine the size of annual payment(Mortgage) Jacinta takes out 50,000 mortgage. If the mortgage rate is 13% convertible semiannually, what should her monthly payment be to pay off the mortgage in 20 years?(Investments) Eileen deposits 2000 in a bank account every year for 11 years. If i = 6 % how much has she accumulated at the time of the last deposit?
63 One more example… (p. 47)Elroy takes out a loan of 5000 to buy a car. No payments are due for the first 8 months, but beginning with the end of 9th month, he must make 60 equal monthly payments. If i = 18%, find:the amount of each paymentthe amount of each payment if there is no payment-free period
64 Annuity-immediate and Annuity-due Annuity-immediate (payments made at the end)sn|1111…..n123Annuity-due (payments made at the beginning)än|.. sn|1111…..nn+1123
65 Example (p. 50)Henry takes out a loan of 1000 and repays it with 10 equal yearly payments, the first one due at the time of the loan. Find the amount of each payment if i = 16%
66 3.3 PerpetuitiesDefinition: Perpetuity is an annuity whose payments continue foreverPractical applications: perpetual bondsBasic model: consider perpetuity under which payments of 1 are made forever
67 Formulaspresent value of perpetuitya∞|111…..123
68 3.4 Unknown Time and Unknown Rate of Interest Examples – Unknown TimeA fund of 5000 is used to award scholarships of amount 500, one per year, at the end of each year for as long as possible. If i=9% find the number of scholarships which can be awarded, and the amount left in the fund one year after the last scholarship has been awardedA trust fund is to be built by means of deposits of amount 5000 at the end of each year, with a terminal deposit, as small as possible, at the end of the final year. The purpose of this fund is to establish monthly payments of amount 300 into perpetuity, the first payment coming one month after the final deposit. If the rate of interest is 12% per year convertible quarterly, find the number of deposits required and the size of the final deposit
69 Example – Unknown Rate of Interest At what effective yearly rate of interest is the present value of 300 paid at the end of every month, for the next 5 years, equal to 15,000?1st method: linear interpolation2nd method: successive approximations
70 Series of n∙m payments of 1/m 3.5 Continuous AnnuitiesLet effective period be 1/m part of the year and i(m)/mbe the effective rate of interest: (1+ i(m)/m)m = 1 + iSeries of n∙m payments of 1/m1/m1/m1/m1/m…..1/m2/m3/mn = nm(1/m)Formula:1/m3/m2/mn = nm(1/m)…..
71 Let m approach infinity… Present value of continuous annuityAccumulated value of continuous annuityFormulas:
73 Arithmetic annuityDefinition: Arithmetic annuity is a series of payments made at regular intervals such thatthe first payment equals Ppayments increase by Q every yearThus payments form arithmetic progression:P, P+Q, P+2Q, …, P+(n-1)Q
74 Series of n payments k-th payment = P+ (k-1)Q FormulasSeries of n payments k-th payment = P+ (k-1)Qk=1,2,…, naccumulated valueSpresent valueAPP+QP+2QP+(n-1)Q…..n123
75 1) Increasing annuity with P = 1, Q = 1 Series of n payments k-th payment = kk=1,2,…, npresent value(Ia)n|accumulated value (Is)n|123nTwo Special Cases…..n123We have:P = 1Q = 1
76 2) Decreasing annuity with P = n, Q = -1 Series of n payments k-th payment = n – k + 1k=1,2,…, npresent value(Da)n|accumulated value (Ds)n|nn-1n-21…..n123We have:P = nQ = -1
77 Increasing perpetuity k-th payment = k continues foreverpresent value(Ia)∞|123…..123
78 Examples (p )Find the value, one year before the first payment, of a series of payments 200, 500, 800,… if i = 8% and the payments continue for 19 yearsFind the present value of an increasing perpetuity which pays 1 at the end of the 4th year, 2 at the end of the 8th year, 3 at the end of the 12th year, and so on, if i = 0.06Find the value one year before the first payment of an annuity where payments start at 1, increase by annual amounts of 1 to a payment of n, and then decrease by annual amounts of 1 to a final payment of 1Show both algebraically and verbally that (Da)n| = (n+1)an| - (Ia)n|
79 More examples: Geometric annuities Geometric annuity An annuity provides for 15 annual payments. The first is 200, each subsequent is 5% less than the one preceding it. Find the accumulated value of annuity at the time of the final payment if i = 9%Inflation and real rate of interest In settlement of a lawsuit, the provincial court ordered Frank to make 8 annual payments to Fred. The first payment of 10,000 is made immediately, and future payments are to increase according to an assumed rate of inflation of 0.04 per year. Find the present value of these payments assuming i = .07
81 4.1 AmortizationAmortization method: repay a loan by means of instalment payments at periodic intervalsThis is an example of annuityWe already know how to calculate the amount of each paymentOur goal: find the outstanding principalTwo methods to compute it:prospectiveretrospective
82 Two MethodsProspective method: outstanding principal at any point in time is equal to the present value at that date of all remaining paymentsRetrospective method: outstanding principal is equal to the original principal accumulated to that point in time minus the accumulated value of all payments previously madeNote: of course, this two methods are equivalent. However, sometimes one is more convenient than the other
83 Examples (p )(prospective) A loan is being paid off with payments of 500 at the end of each year for the next 10 years. If i = .14, find the outstanding principal, P, immediately after the payment at the end of year 6.(retrospective) A 7000 loan is being paid of with payments of 1000 at the end of each year for as long as necessary, plus a smaller payment one year after the last regular payment. If i = 0.11 and the first payment is due one year after the loan is taken out, find the outstanding principal, P, immediately after the 9th payment.
84 One more example… (p. 77)(Renegotiation) John takes out 50,000 mortgage at 12.5 % convertible semi-annually. He pays off the mortgage with monthly payments for 20 years, the first one is due one month after the mortgage is taken out. Immediately after his 60th payment, John renegotiates the loan. He agrees to repay the remainder of the mortgage by making an immediate cash payment of 10,000 and repaying the balance by means of monthly payments for ten years at 11% convertible semi-annually. Find the amount of his new payment.
85 4.2 Amortization Schedule Goal: divide each payment (of annuity) into two parts – interest and principalAmortization schedule – table, containing the following columns:paymentsinterest part of a paymentprincipal part of a paymentoutstanding principal
86 Outstanding Principal Example5,000 at 12 % per year repaid by 5 annual paymentsAmortization schedule:DurationPaymentInterestPrincipal RepaidOutstanding Principal1600.00787.052505.55881.503399.77987.284281.305148.61
87 Outstanding principal P Interest earned during interval (t-1,t) is iPTherefore interest portion of payment X is iP and principal portion is X - iPOutstanding principal PPayment Xt - 1tRecall: in practical problems, the outstanding principal P can be found by prospective or retrospective methodsExampleA 1000 loan is repaid by annual payments of 150, plus a smaller final payment. If i = .11, and the first payment is made one year after the time of the loan, find the amount of principal and interest contained in the third payment
88 an-t| an| ….. ….. If each payment is X then General methodoutstanding principal at tpresent value = outst. principal at 0an-t|an|11111…..…..n12tt+1interest portion of (t+1)-st payment = i a n-t| = 1 – vn-tprincipal portion of (t+1)-st payment = 1 – (1 – vn-t ) = vn-tIf each payment is X theninterest part of kth payment = X (1 – vn-k+1 )principal part of kth payment = X∙vn-k+1
89 Example (p. 79)A loan of 5000 at 12% per year is to be repaid by 5 annual payments, the first due one year hence. Construct an amortization schedule
90 General rules to obtain an amortization schedule DurationPaymentInterestPrincipal RepaidOutstanding Principal1600.00787.052505.55881.503399.77987.284281.305148.61i = 12 %Take the entry from “Outs. Principal” of the previous row, multiply it by i, and enter the result in “Interest”“Payment” – “Interest” = “Principal Repaid”“Outs. Principal” of prev. row - “Principal Repaid” = “Outs. Principal”Continue
91 Example (p. 80)A 1000 loan is repaid by annual payments of 150, plus a smaller final payment. The first payment is made one year after the time of the loan and i = .11. Construct an amortization schedule
92 4.3 Sinking FundsAlternative way to repay a loan – sinking fund method:Pay interest as it comes due keeping the amount of the loan (i.e. outstanding principal) constantRepay the principal by a single lump-sum payment at some point in the future
93 ….. iL iL lump-sum payment L interest iL 12nLoan LLump-sum payment L is accumulated by periodic deposits into a separate fund, called the sinking fundSinking fund has rate of interest j usually different from (and usually smaller than) iIf (and only if) j is greater than i then sinking fund method is better (for borrower) than amortization method
94 ExamplesThe View Royal Fire District needs a new fire truck which will cost $300,000. The district is able to arrange the necessary financing provided a sinking fund is established to provide for repayment of the debt. The loan must be repaid in four years. Monies for repayment will come from a tax increase on the land owners who are part of the fire district. The interest cost on the loan must be paid every six months as per the loan agreement with the province. If the district earns 8% compounded annually on the sinking fund, if the interest rate on the loan is 12% compounded semi-annually and if the payment must be made annually to the sinking fund, answer the following questions.the annual interest paymentthe annual sinking fund paymentthe total annual outlaythe annual amortization payment which would pay off this loan in 4 years
95 ExamplesA loan of 10,000 is being repaid with annual payments for 10 years using the sinking fund method. The loan charges 10% interest and the sinking fund earns 8%. If the loan was repaid using the amortization method, find the interest rate under the amortization method that makes to methods equally attractive for the borrower.
96 ExamplesKathy can take out a loan of 50,000 with Bank A or Bank B. With Bank A, she must repay the loan with 60 monthly payments using the amortization method with interest at 7% compounded monthly. With Bank B, she can repay the loan with 60 monthly payments using the sinking fund method. The sinking fund will earn 6.5% compounded monthly. What monthly interest rate can Bank B charge on the loan so that Kathy's payment will be the same under either option?
97 Yield RatesInvestor:makes a number of payments at various points in timereceives other payments in returnThere is (at least) one rate of interest for which the value of his expenditures will equal the value of the payments he received (at the same point in time)This rate is called the yield rate he earns on his investmentIn other words, yield rate is the rate of interest which makes two sequences of payments equivalentNote: to determine yield rate of a certain investor, we should consider only payments made directly to, or directly by, this investor
98 ExamplesAt what yield rate are payments of 500 now and 600 at the end of 2 years equivalent to a payment of 1098 at the end of 1 year?Herman borrows 5000 from George and agrees to repay it in 10 equal annual instalments at 11%, with first payment due in one year. After 4 years, George sells his right to future payments to Ruth, at a price which will yield Ruth 12% effectiveFind the price Ruth pays.Find George’s overall yield rate.
99 Amortization Schedule Other Topics Chapter 5 BONDSPrice of a BondBook ValueAmortization ScheduleOther Topics
100 5.1 Price of a BondBonds:certificates issued by a corporation or governmentare sold to investorsin return, the borrower (i.e. corporation or government) agrees:to pay interest at a specified rate (the coupon rate) until a specified date (the maturity date)and, at that time, to pay a fixed sum (the redemption value)Usually:the coupon rate is a nominal rate convertible semiannually and is applied to the face (or par) value stated on the front of the bondthe face and redemption values are equal (not always)Thus we have:regular interest paymentslump-sum payment at the end
101 Example of a bond Face amount = 500 Redemption value = 500 Redeemable in 10 years with semiannual coupons at rate 11%, compounded semiannuallyThen in return investor receives:20 half-yearly payments of (.055)(500) = interesta lump-sum payment of 500 at the end of the 10 years
102 Notations F = the face value (or the par value) r = the coupon rate per interest period (we assume that the quoted rate will be a nominal rate 2r convertible semiannually)Note: the amount of each interest payment (coupon) is FrC = the redemption value (often C = F, i.e. bond “redeemable at par”)i = the yield rate per interest periodn = the number of interest periods until the redemption date (maturity date)P = the purchase price of a bond to obtain yield rate i
103 Equation to find yield rate i redemption value CNote: time ismeasured inhalf-yearscoupon (interest) FrFrFr…..12npurchase price PEquation to find yield rate iNote: often C = F
104 ExamplesA bond of 500, redeemable at par after 5 years, pays interest at 13% per year convertible semiannually. Find the price to yield an investor8% effective per half-year16% effective per yearRemarksP < C since the yield rate is higher than the coupon rate, i > rtherefore the investor is buying the bond at a discountotherwise (if i < r) we would have P > C and then the investor would have to buy the bond at a premium of P - C
105 A 20 year bond matures for its par value of 10,000 A 20 year bond matures for its par value of 10,000. The coupon rate is 8% convertible semi-annually. Calculate the price of the bond at a 6% annual yield rateA 100 par-value 15-year bond with coupon rate 9% convertible semi-annually is selling for 94. Find the yield rate.
106 5.2 Book ValueThe book value of a bond at a time t is an analog of an outstanding balance of a loanThe book value Bt is the present value of all future paymentsAt time t where the tth coupon has just been paid we have:RemarksUsually C = FIn the last formula, an-t and v are computed using the yield rate iP = B0 < Bt < Bt+1< Bn = C or P = B0 > Bt > Bt+1 > Bn = C
107 ExamplesFind the book value immediately after the payment of 14th coupon of a 10-year 1,000 par-value bond with semiannual coupons, if r=.05 and the yield rate is 12% convertible semiannually.Let Bt and Bt+1 be the book values just after the tth and (t+1)th coupons are paid. Show that Bt+1 = Bt (1+i) – FrFind the book value in 1) exactly 2 months after the 14th coupon is paid.
108 How do we find the book value between coupon payment dates? ExampleFind the book value exactly 2 months after the 14th coupon is paid of a 10-year 1,000 par-value bond with semiannual coupons, if r=.05 and the yield rate is 12% convertible semiannually.Assume simple interest at rate i per period between adjacent coupon payments
109 Alternative approach (1+i) Bt Bt Bt+1 Since Bt+1 = Bt (1+i) – Fr we can view Bt (1+i) as the book value just before next (i.e. (t+1)th) coupon is paidBook value calculated using simple interest between coupon dates is called the flat price of a bondUsing linear interpolation between Bt+1 and Bt we obtain the market price (or the amortized value) of the bondClearly market price ≤ flat price at any given moment(1+i) Btflat priceFrBtBt+1market pricett+1
110 Example (p. 98)Find the market price exactly 2 months after the 14th coupon is paid of a 10-year 1,000 par-value bond with semiannual coupons, if r =.05 and the yield rate is 12% convertible semiannually.
111 5.3 Bond Amortization Schedule Goal: trace changes of the book valueBond amortization schedule – table, containing the following columns:timecouponinterestprincipal adjustmentbook valueTimeCouponInterestPrincipal adjustmentBook Value14031.128.88230.859.15330.579.43430.299.71Example1000 par value two-year bond which pays interest at 8% convertible semiannually; yield rate is 6% convertible semiannually
112 Algorithm Book value at time t is Bt Amount of coupon at time t+1 is FrThe amount of interest contained in this coupon is iBtFr – iBt represents the change in the book value between these datesTimeCouponInterestPrincipal adjustmentBook Value14031.128.88230.859.15330.579.43430.299.71
113 Example (p. 99)Consider 1000 par-value 10-year bond with semiannual coupons, r = .05 and the yield rate i = 0.06 effective semiannually. Find the amount of interest and change in book value contained in the 15th coupon of the bond.
114 Example (p. 99)Construct a bond amortization schedule for a 1000 par-value two-year bond which pays interest at 8% convertible semiannually, and has a yield rate of 6% convertible semiannually
115 5.4 Other Topics Different frequency of coupon payments Increasing or decreasing coupon paymentsDifferent yield ratesCallable bonds
116 Examples(Different frequency) Find the price of a 1000 par-value 10-year bond which has quarterly 2% coupons and is bought to yield 9% per year convertible semiannually(Increasing coupon payments) Find the price of a 1000 par-value 10-year bond which has semiannual coupons of 10 the first half-year, 20 the second half-year,…, 200 the last half-year, bought to yield 9% effective per year(Different yield rates) Find the price of a 1000 par-value 10-year bond with coupons at 11% convertible semiannually, and for which the yield rate is 5% per half-year for the first 5 years and 6% per half-year for the last 5 years
117 Callable bondsA borrower (i.e. corporation, government etc.) has the right to redeem the bond at any of several time pointsThe earliest possible date is the call date and the latest is the usual maturity dateOnce the bond is redeemed, no more coupons will be paidpossible redemptionMaturity DatePurchase DateCall Date
118 ExamplesConsider a 1000 par-value 10-year bond with semiannual 5% coupons. Assume this bond can be redeemed at par at any of the last 4 coupon dates. Find the price which will guarantee an investor a yield rate of6% per half-year4% per half-yearConsider a 1000 par-value 10-year bond with semiannual 5% coupons. This bond can be redeemed for 1100 at the time of the 18th coupon, for 1050 at the time of the 19th coupon, or for 1000 at the time of 20th coupon. What price should an investor pay to be guaranteed a yield rate of
119 Chapter 6 PREPARATION FOR LIFE CONTINGENCIES IntroductionContingent Payments
120 6.1 Introduction Ideal situation: all payments are made Real-life situations:failure to make a paymentdefault on a loanbad credit ratingslife contingencies and life insuranceContingent payments (we need to combine the theory of interest and elementary probability)
121 6.3 Contingent PaymentsAssume that for each payment of the loan (annuity, bond etc.) there is a probability that this payment is madeFinding the present value of such sequence of payments we need to take into account these probabilitiesWe need to find present value of expected values of all payments
122 Contingent Payments - Example Example Henry borrows 1000 from Amicable Trust and agrees to repay the loan in one year. If payment were certain, the company would charge 13% interest. From prior experience, however, it is determined that there is a 5% chance that Henry will not repay any money at all. What should Amicable Trust ask Henry to repay?
123 Examples(Loan) The All-Mighty Bank lends 50,000,000 to a small Central American country, with the loan to be repaid in one year. It is felt that there is a 20% chance that a revolution will occur and that no money will be repaid, a 30% chance that due to inflation only half the loan will be repaid, and a 50% chance that the entire loan will be repaid. If payments were certain, the bank would charge 9%. What rate of interest should the bank charge?(Payments contingent upon survival) Mrs. Rogers receives 1000 at the end of each year as long as she is alive. The probability is 80% she will survive one year, 50% she will survive 2 years, 30% she will survive 3 years, and negligible that she will survive longer than 3 years. If the yield rate is 15%, what should Mrs. Rogers place on these payments now?(Life insurance) An insurance company issues a policy which pays 50,000 at the end of the year of death, if death should occur during the next two years. The probability that a 25-year-old will live for one year is , and the probability he will live for two years is What should the company charge such a policy holder to earn 11% on its investments?
124 (Annuity) Alphonse wishes to borrow some money form Friendly Trust (Annuity) Alphonse wishes to borrow some money form Friendly Trust. He promises to repay 500 at the end of each year for the next 10 years, but there is a 5% chance of default in any year. Assume that once default occurs, no further payments will be received. How much can Friendly Trust lend Alphonse if it wishes to earn 9% on its investments?Redo the last example, without the restriction that once default occurs, no further payments will be received(Bond) A 20-year 1000 face value bond has coupons at 14% convertible semiannually and is redeemable at par. Assume a 2% chance that, in any given half-year, the coupon is not issued, and that once default occurs, no further payments are made. Assume as well that a bond can be redeemed only if all coupons have been paid. Find the purchase price to yield on investor 16% convertible semiannulllay.
125 Chapter 7 LIFE TABLES AND POPULATION PROBLEMS IntroductionLife TablesThe Stationary PopulationExpectation of Life
126 7.1 Introduction How do we find probabilities? Data obtained from practiceData required to find probabilities of surviving to certain ages (or, equivalently, of dying before certain ages) are contained in life tables
127 lx dx 7.2 Life Tables lx – number of lives survived to age x 1000 qx1,000,00015801.581998,420680682997,740485.493997,255435.44lx – number of lives survived to age xThus l0 = 1,000,000 is a starting populationSurvival function S(x) = lx / l0 is the probability of surviving to age xdx = lx – lx+1 – number of lives who died between (x, x+1)qx = dx /lx – probability that x – year-old will not survive to age x + 1
128 Examples (p. 129 – p. 130) lx dx Find Agelxdx1000 qx1,000,00015801.581998,420680682997,740485.493997,255435.44Findthe probability that a newborn will live to age 3the probability that a newborn will die between age 1 and age 3
129 Find an expression for each of the following: the probability that an 18-year-old lives to age 65the probability that a 25-year-old dies between ages 40 and 45the probability that a 25-year-old does not die between ages 40 and 45the probability that a 30-year-old dies before age 60There are four persons, now aged 40, 50, 60 and 70. Find an expression for the probability that both the 40-year-old and the 50-year-old will die within the five-year period starting ten years from now, but neither the 60-year-old nor the 70-year-old will die during that five-year period
130 Note:If we are already given by probabilities qx and starting population l0 we can construct the whole life table step-by-step since dx = qx lx and lx+1 = lx - dxExampleGiven the following probabilities of deaths q0 = .40, q1 = .20, q2 = .30, q3 = .70, q4 = 1 and starting with l0 = 100 construct a life table
131 More notations…qx = dx / lx – probability that x – year-old will not survive to age x + 1px = 1- qx – probability that x – year-old will survive to age x + 1Note: qx = (lx – lx+1) / lx and px = lx+1 / lxnqx – probability that x – year-old will not survive to age x + nnpx = 1- nqx – probability that x – year-old will survive to age x + nNote: nqx = (lx – lx+n) / lx and n px = lx+n / lxFormulas: lx – lx+n = dx + dx+1 + …+ dx+n n+mpx = mpx ∙ npx+m
132 What is mPx when m is not integer? tPx where 0 < t <1Assuming that deaths are distributed uniformly during any given year we can use linear interpolation to find tpx :
133 Examples (p. 132 – p. 133)30% of those who die between ages 25 and 75 die before age 50. The probability of a person aged 25 dying before age 50 is 20%. Find 25P50Using the following life table and assuming a uniform distribution of deaths over each year, find:4/3P1The probability that a newborn will survive the first year but die in the first two months thereafterAgelxdx1000 qx1,000,00015801.581998,420680682997,740485.493997,255435.44
134 7.3 Analytic Formulas for lx Properties of lxde Moivre’s LawForce of mortalityGompertz and Makeham formulas
135 Properties of lx and S(x) = lx / l0 DecreasingVanishes after terminal age of the population ωDecreases more rapidly near 0 and ω-∆ (population-dependant, e.g )l0ωlxω-∆
136 de Moivre’s Law First approximation – linear Derivative is constant Hence, the function does not satisfy the last propertyHowever, is reasonable in the middle range of ages
137 ExamplesFind probabilities of survival tPx and of dying tqx for de Moivre’s lawLetDoes it satisfy all required properties?Find tPx and t qxHenry and Henrietta are 19 years old. Find the probability that Henry survives at least 17 years, Henrietta lives at most 45 more years, and at least one of them survives for 32 years
138 Force of mortalityGiven that a person survived to age x, what is the probability that a person does not survive to age x+∆x?Dividing by ∆x and letting ∆x → 0 we get the force of mortality, that can be interpreted as conditional density of dying at age x given that person survived to age x
139 Force of mortalityUnconditional density of dying at age x is μx Sx = - S’x (recall: Sx is the probability of surviving to age x)Indeed, if a random variable X denotes age, then Sx = P(X ≥ x ) = 1 – P(X < x) = 1 – F(x) and therefore -S’x = F’(x), the density of dyingExpected density of deaths is μx lx
140 Force of mortalityIf we know force of mortality, we can find Sx and lx as follows:
142 Makeham formula (introduces age-independent term)
143 ExerciseCheck whether Gompertz and Makeham formulas satisfy properties of survival function. Which requirements should we impose on the constants in theses formulas (A, B, c, g, etc.)?
144 7.4 The Stationary Population Assume that in every given year (or, more precisely, in any given 12-months period) the number of births and deaths is the same and is equal to l0Then after a period of time the total population will remain stationary and the age distribution will remain constantpx and qx are defined as beforelx denote the number of people who reach heir xth birthday during any given yeardx = qx lx represent the number of people who die before reaching age x+1Also, dx represent the number of people who die during any given year between ages x and x + 1Similarly lx – lx + n represent the number of people who die during any given year between ages x and x + n
145 Number of people aged xLet Lx denote the number of people aged x (last birthday) at any given momentNote: Lx ≠ lxAssuming uniform distribution of deaths we obtain:More precisely:
146 Number of people aged x and over Let Tx denote the number of people aged x and over at any given momentThenAssuming uniform distribution we obtain:More precisely:
147 Example (p. 139)An organization has a constant total membership. Each year 500 new members join at exact age % leave after 10 years, 10% of those remaining leave after 20 years, and the rest retire at age 65. Express each of the following in terms of life table functions:The number who leave at age 40 each yearThe size of the membershipThe number of retired people alive at any given timeThe number of members who die each year
148 7.5 Expectation of LifeWhat is the average future life time ex of a person aged x now?The answer is given by expected value (or mathematical expectation) and is called the curtate expectation:
149 Complete expectation: (4) (1)(2)(2)&(3)&(4) imply:(3)Approximation:Sincewe get
150 RemarksThus Tx can be viewed as the total number of years of future life of those who form group lxNote: this interpretation makes sense in any population (stationary or not)
151 Average age at deathAverage age at death of a person currently aged x is given byLetting x = 0 we obtain the average age at death for all death among l0 individuals
152 Interpret verbally the expression Tx-Tx+n – nlx+n Examples (p. 142 – 143)If tp35 = (.98)t for all t, find e35 and eo35 without approximation. Compare the value for eo35 with its approximate valueInterpret verbally the expression Tx-Tx+n – nlx+nFin the average age at death of those who die between age x and age x+n
153 Chapter 8 LIFE ANNUITIES Basic ConceptsCommutation FunctionsAnnuities Payable mthlyVarying Life AnnuitiesAnnual Premiums and Premium Reserves
154 8.1 Basic ConceptsWe know how to compute present value of contingent paymentsLife tables are sources of probabilities of survivingWe can use data from life tables to compute present values of payments which are contingent on either survival or death
155 Example (pure endowment), p. 155 Yuanlin is 38 years old. If he reaches age 65, he will receive a single payment of 50,000. If i = .12, find an expression for the value of this payment to Yuanlin today. Use the following entries in the life table: l38 = 8327, l65 = 5411
156 t Ex = (t px ) (1 + t) – t = v t t px Pure EndowmentPure endowment: 1 is paid t years from now to an individual currently aged x if the individual survivesProbability of surviving is t pxTherefore the present value of this payment is the net single premium for the pure endowment, which is:t Ex = (t px ) (1 + t) – t = v t t px
157 Example (life annuity), p. 156 Aretha is 27 years old. Beginning one year from today, she will receive 10,000 annually for as long as she is alive. Find an expression for the present value of this series of payments assuming i = .09Find numerical value of this expression if px = .95 for each x
158 Series of payments of 1 unit as long as individual is alive Life annuitySeries of payments of 1 unit as long as individual is alivepresent value (net single premium) of annuity ax111…..…..agexx + 1x + 2x + npx2pxnpxprobability
159 Temporary life annuity Series of n payments of 1 unit (contingent on survival)present value ax:n|last payment111…..agexx + 1x + 2x + npx2pxnpxprobability
160 n - years deferred life annuity Series of payments of 1 unit as long as individual is alive in which the first payment is at x + n + 1present value n|axfirst payment11……agexx + 1x + 2x + nx + n +1x + n + 2n+1pxn+2pxprobabilityNote:
163 8.2 Commutation Functions Recall: present value of a pure endowment of 1 to be paid n years hence to a life currently aged xDenote Dx = vxlxThen nEx = Dx+n / Dx
164 Life annuity and commutation functions Since nEx = Dx+n / Dx we haveDefine commutation function Nx as follows:Then:
165 Identities for other types of life annuities temporary life annuityn-years delayed l. a.temporary l. a.-due
166 Accumulated values of life annuities temporary life annuitysinceandwe havesimilarly for temporary life annuity-due:and
167 Examples (p. 162 – p. 164)(life annuities and commutation functions) Marvin, aged 38, purchases a life annuity of 1000 per year. From tables, we learn that N38 = 5600 and N39 = Find the net single premium Marvin should pay for this annuityif the first 1000 payment occurs in one yearif the first 1000 payment occurs nowStay verbally the meaning of (N35 – N55) / D20(unknown rate of interest) Given Nx = 5000, Nx+1=4900, Nx+2 = 4810 and qx = .005, find i
168 Select groupSelect group of population is a group with the probability of survival different from the probability given in the standard life tablesSuch groups can have higher than average probability of survival (e.g. due to excellent health) or, conversely, higher mortality rate (e.g. due to dangerous working conditions)
169 NotationsSuppose that a person aged x is in the first year of being in the select groupThen p[x] denotes the probability of survival for 1 year and q[x] = 1 – p[x] denotes the probability of dying during 1 year for such a personIf the person stays within this group for subsequent years, the corresponding probabilities of survival for 1 more year are denoted by p[x]+1, p[x]+2, and so onSimilar notations are used for life annuities: a[x] denotes the net single premium for a life annuity of 1 (with the first payment in one year) to a person aged x in his first year as a member of the select groupA life table which involves a select group is called a select-and-ultimate table
170 Examples (p. 165 – p. 166)(select group) Margaret, aged 65, purchases a life annuity which will provide annual payments of 1000 commencing at age 66. For the next year only, Margaret’s probability of survival is higher than that predicted by the life tables and, in fact, is equal to p , where p65 is taken from the standard life table. Based on that standard life table, we have the values D65 = 300, D66 = 260 and N67 = If i = .09, find the net single premium for this annuity(select-and-ultimate table) A select-and-ultimate table has a select period of two years. Select probabilities are related to ultimate probabilities by the relationships p[x] = (11/10) px and p[x]+1 = (21/20) px+1. An ultimate table shows D60 = 1900, D61 = 1500, and ä 60:20| = 11, when i = .08. Find the select temporary life annuity ä:20|
171 The following values are based on a unisex life table: N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865, N42 = It is assumed that this table needs to be set forward one year for males and set back two years for females. If Michael and Brenda are both age 40, find the net single premium that each should pay for a life annuity of 1000 per year, if the first payment occurs immediately.
172 8.3 Annuities Payable mthly Payments every mth part of the yearProblem: commutation functions reflect annual probabilities of survivalFirst, we obtain an approximate formula for present valueAssume for a moment that the values Dy are also given for non-integer values of y
173 Annuity payable every 1/m part of the year Usual life annuityax111…..…..agexx + 1x + 2x + nAnnuity payable every 1/m part of the yearax1111…..…..agexx + 1/mx + 2/mx + (m-1)/mx + n