st334 actuarial methods

214 Pages·2016·1.31 MB·English

ST334 ACTUARIAL METHODS
version2016/03
These notes are for ST334 Actuarial Methods. The course covers Actuarial CT1 and some related financial
topics.
Actuarial CT1 is called ‘Financial Mathematics’ by the Institute of Actuaries. However, we reserve the term
‘financial mathematics’ for the study of stochastic models of derivatives and financial markets which is con-
siderablymoreadvancedmathematicallythanthematerialcoveredinthiscourse.
There are 2 versions of these notes—one is in monochrome and is more suitable for printing and the other
containsbookmarksandhyperlinksandisintendedforviewinginaPDFreadersuchasSumatraPDF.
Contents
1 SimpleandCompoundInterest................................................................ 1
1.1 Simple interest 1. 1.2 Compound interest 3. 1.3 Exercises 8. 1.4 Discount rate and discount
factor10. 1.5 Exercises15. 1.6 Timedependentinterestrates16. 1.7 Exercises18.
2 CashFlows,EquationsofValueandProjectAppraisal......................................... 21
2.1Cashflows21. 2.2Netpresentvalueanddiscountedcashflow22. 2.3Exercises27. 2.4Equa-
tionsofvalueandtheinternalrateofreturn30. 2.5 Comparingtwoprojects33. 2.6 Exercises35.
2.7 Measuringinvestmentperformance38. 2.8 Exercises39.
3 Perpetuities,AnnuitiesandLoans............................................................. 43
3.1 Perpetuities43. 3.2 Annuities44. 3.3 Exercises51. 3.4 Loanschedules57. 3.5 Loansrepaid
byspecifiedinstalments61. 3.6 Exercises62.
4 BasicFinancialInstruments .................................................................. 69
4.1 Markets, interest rates and financial instruments 69. 4.2 Fixed interest government borrow-
ings 70. 4.3 Other fixed interest borrowings 71. 4.4 Investments with uncertain returns 73.
4.5 Derivatives74. 4.6 Exercises79.
5 Bonds,EquitiesandInflation ................................................................. 81
5.1 Bondcalculations81. 5.2 Exercises87. 5.3 Equitycalculations90. 5.4 Realandmoneyrates
ofinterest91. 5.5 Exercises93.
6 InterestRateProblems...................................................................... 101
6.1 Spot rates, forward rates and the yield curve 101. 6.2 Exercises 107. 6.3 Vulnerability to
interestratemovements111. 6.4Exercises116. 6.5Stochasticinterestrateproblems123. 6.6Ex-
ercises125.
7 Arbitrage................................................................................... 131
7.1 Arbitrage131. 7.2 Forwardcontracts133. 7.3 Exercises138.
Appendix1: SomeMathematicalPrerequisites.................................................. 143
Appendix2: Answers.......................................................................... 145
Exercises 1.3145. Exercises 1.5147. Exercises 1.7148. Exercises 2.3149. Exercises 2.6152.
Exercises 2.8155. Exercises 3.3158. Exercises 3.6166. Exercises 4.6173. Exercises 5.2174.
Exercises 5.5179. Exercises 6.2187. Exercises 6.4193. Exercises 6.6202. Exercises 7.3206.
ST334 Actuarial Methods (cid:13)c R.J. Reed Sep 27, 2016(9:51) Page i
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Prologue
OrderofTopics.
Thereareargumentsforconsideringtheformulæforperpetuitiesandannuitiescoveredinchapter3before the
discussion of project appraisal in chapter 2. This has not been done in order to avoid starting with a long list
of formulæ on nominal and effective interest rates and annuities. However, this does mean that the solutions
of some of the problems on project appraisal in exercises 6 of chapter 2 (page 35) can be simplified by using
resultsfromsection2ofchapter3(page44).
Exercises
Anenormousnumberofexercisesisincluded. Youarenotexpectedtodothemall—but, itisveryimportant
that you practice solving exercises quickly and accurately. Only you can decide how many you need to do in
ordertoachievetherequiredproficiencytopassthetestsandtheexamination.
Books
References.
Brealey,R&S.Myers(2005,eighthedition)PrinciplesofCorporateFinance McGrawHill
Brett,M.(2003,fifthedition)HowtoReadtheFinancialPages RandomHouse
Donald,D.W.A.(1975)CompoundInterestandAnnuities-Certain Heinemann
McCutcheon,J.J.&W.F.Scott(1986)AnIntroductiontotheMathematicsofFinance Butterworth
Ross, S.M. (2002, second edition) An Elementary Introduction to Mathematical Finance CUP (Chapter 4)
Steiner,R.(2007,secondedition)MasteringFinancialCalculations PrenticeHall
Vaitilingam,R.(2001,fourthedition)TheFinancialTimesGuidetoUsingtheFinancialPages PrenticeHall
Most of the exercises have been taken from past examination papers for the Faculty and Institute of Actuar-
ies. The web reference is http://www.actuaries.org.uk/students/pages/past-exam-papers and they are
copyrightTheInstituteandFacultyofActuaries.
[e-mail].
Index of Notation
symbol page meaning
a ora 43 §1.1 presentvalueofperpetuity
∞ ∞,i
a¨ ora¨ 43 §1.1 valueattime1ofaperpetuity
∞ ∞,i
a(m) anda¨(m) 52 §3(22) value at time 0 and at time t = 1/ of a perpetuity payable m times
∞ ∞ m
peryear
a ora 44 §2.1 presentvalueofanannuitywithnpayments
n n,i
a¨ ora¨ 44 §2.1 valueattime1ofanannuitywithnpayments
n n,i
a(m) ora(m) 45 §2.3 valueattime0ofanannuitywithmpaymentsperyearfornyears
n n,i
a¨(m) ora¨(m) 45 §2.3 valueattime 1/ ofanannuitywithmpaymentsperyearfornyears
n n,i m
a 49 §2.7 valueattime0ofacontinuouslypayableperpetuitywithρ(t) = 1
∞
a 49 §2.7 value at time 0 of a continuously payable annuity with ρ(t) = 1 for
n
t ∈ (0,n)
a and a¨ 47 §2.4 annuitypayableannuallyfornyearsanddelayedbyk years
k| n k| n
a(m) and a¨(m) 47 §2.4 annuitypayablemtimesperyearfornyearsanddelayedbyk years
k| n k| n
a 49 §2.7 value at time 0 of a continuously payable annuity with ρ(t) = 1 for
k| n
t ∈ (k,k+n)
ACT/365 1 §1.4 actualnumberofdaysdividedby360
A(p) 4 §2.6 accumulatedvalueattimepofaninvestmentof1attime0
C 81 §1.1 redemptionvalueofabond
CD 69 §1.3 certificateofdeposit
c(i) 113 §3.5 convexity
CP 69 §1.3 commercialpaper
d = 1−ν 10 §4.1 discountrate
d(m) 12 §4.5 nominaldiscountratepayablemtimesperyear
d 12 §4.5 nominaldiscountratepayablepertimeperiod,p
p
(Da) 49 §2.6 presentvalueofadecreasingannuityfornyears
n
DCF 22 §2.1 discountedcashflow
d(i) 113 §3.3 effectivedurationorvolatilityormodifiedduration
d (i) 111 §3.2 durationorMacaulaydurationordiscountedmeanterm
M
DPP 33 §5.3 discountedpaybackperiod
δ = ln(1+i) 6 §2.8 nominalrateofinterestcompoundedcontinuouslyor
forceofinterestcorrespondingtotheeffectiveannualinterest,i
δ : (0,∞) → [0,∞) 17 §6.1 forceofinterestfunction
f 81 §1.1 faceorparvalueofabond
FRA 76 §5.5 forwardrateagreement
f 103 §1.5 forwardinterestratep.a.formoneyattimeT inthefutureforlength
T,k
oftimek
F = ln(1+f ) 103 §1.6 forwardforceofinterest
T,k T,k
f 104 §1.7 forward interest rate p.a. at time t for investment starting at time T
t,T,k
forlengthoftimek
F = ln(1+f ) 105 §1.7 forwardforceofinterest
t,T,k t,T,k
F = lim F 105 §1.7 instantaneousforwardrateattimetforinvestmentstartingattimeT
t,T k→0 t,T,k
g = fr/ 83 §1.5 annualcoupondividedbyredemptionvalueofabond
C
ST334 Actuarial Methods (cid:13)c R.J. Reed Sep 27, 2016(9:51) Page iii
i 1 §1.2 simpleinterest
i 3 §2.1 compoundinterest
i 4 §2.5 theeffectiveinterestrateovertheperiodp
ep
i(m) = i 4 §2.5 nominalrateofinterestcompoundedmtimesperyear
1/m
i andi 91 §4.1 moneyandrealrateofinterest
M R
i 4 §2.5 nominalrateofinterestperperiodp
p
(Ia) and(Ia¨) 48 §2.5 valueattime0andattime1ofanincreasingannuityfornyears
n n
(Ia¯) 50 §2.8 increasingcontinuouslypayableannuitywithastepfunctionρ
n
(Ia) 50 §2.8 increasingcontinuouslypayableannuitywithρ(t) = tfort ∈ (0,n)
n
(Is) and(Is¨) 48 §2.5 valueattimenandattimen+1ofanincreasingannuityfornyears
n n
(Is¯) 50 §2.8 increasingcontinuouslypayableannuitywithastepfunctionρ
n
(Is) 50 §2.8 increasingcontinuouslypayableannuitywithρ(t) = tfort ∈ (0,n)
n
IRR 30 §4.1 internalrateofreturn
LIBID 70 §1.4 LondonInter-BankBidRate
LIBOR 70 §1.4 LondonInter-BankOfferedRate
LIMEAN 70 §1.4 averageofLIBOR andLIBID
LIRR 38 §7.3 linkedinternalrateofreturn
MWRR 38 §7.1 moneyweightedrateofreturn
NPV 22 §2.1 netpresentvalue
ν = 1/(1+i) 10 §4.1 discountfactor
ν : [0,∞) → [0,∞) 17 §6.2 presentvaluefunction
P(n,i) 81 §1.1 currentpriceofabondwithnyearstomaturityandyieldi
r 81 §1.1 couponrateperyearofabond
ρ : (0,∞) → R 25 §2.5 continuouspaymentstream
s ors 45 §2.2 valueattimenofanannuitywithnpayments
n n,i
s¨ ors¨ 45 §2.2 valueattimen+1ofanannuitywithnpayments
n n,i
s(m) ors(m) 45 §2.3 valueattimenofanannuitywithmpaymentsperyearfornyears
n n,i
s¨(m) ors¨(m) 45 §2.3 value at time n + 1/ of an annuity with m payments per year for
n n,i m
nyears
s 49 §2.7 value at time n of a continuously payable annuity with ρ(t) = t for
n
t ∈ (0,n)
TWRR 38 §7.1 timeweightedrateofreturn
y 101 §1.2 t-yearspotrateofinterest
t
Y = ln(1+y ) 101 §1.2 t-yearspotforceofinterest
t t
CHAPTER 1
Simple and Compound Interest
1 Simple interest
1.1 Background. Theexistenceofamarketwheremoneycanbeborrowedandlentallowspeopletotransfer
consumptionbetweentodayandtomorrow.
Theamountofinterestpaiddependsonseveralfactors,includingtheriskofdefaultandtheamountofdepre-
ciationorappreciationinthevalueofthecurrency.
Interestonshort-termfinancialinstrumentsisusuallysimpleratherthancompound.
1.2 Thesimpleinterestproblemoveronetimeunit. Thismeansthereisonedeposit(sometimescalledthe
principal)andonerepayment. Therepaymentisthesumoftheprincipalandinterestandissometimescalled
thefuturevalue.
Suppose an investor deposits c at time t and receives a repayment c at the later time t +1. Then the gain
0 0 1 0
(orinterest)ontheinvestmentc isc −c . Theinterestrate,iisgivenby
0 1 0
c −c
1 0
i =
c
0
Notethat
1
c = c (1+i) and c = c
1 0 0 1
1+i
The quantity c is often called the present value (at time t ) of the amount c at time t +1. The last formula
0 0 1 0
encapsulatesthemaxim“apoundtodayisworthmorethanapoundtomorrow”.
Example 1.2a. Suppose the amount of £100 is invested for one year at 8% per annum. Then the repayment in
poundsis
c =(1+0.08)×100=108
1
Thepresentvalueinpoundsoftheamount£108inoneyear’stimeis
1
c = ×108=100
0 1+0.08
Of course, the transaction can also be viewed from the perspective of the person who receives the money. A
borrowerborrowsc attimet andrepaysc attimet +1. Theinterestthattheborrowermustpayontheloan
0 0 1 0
isc −c .
1 0
1.3 The simple interest problem over several time units. Consider the same problem over n time units,
wheren > 0isnotnecessarilyintegralandmaybelessthan1. Thereturntotheinvestorwillbe
c = c +nic = c (1+ni)
n 0 0 0
Itfollowsthatpresentvalue(attimet )oftheamountc attimet +nis
0 n 0
1
c = c
0 n
1+ni
1.4 Day and year conventions. Interest calculations are usually based on the exact number of days as a
proportionofayear.
Domestic UK financial instruments usually assume there are 365 days in a year—even in a leap year. This
conventionisdenotedACT/365whichisshortforactualnumberofdaysdividedby365.
ST334 Actuarial Methods (cid:13)c R.J. Reed Sep 27, 2016(9:51) Section 1 Page 1
Page 2 Section 1 Sep 27, 2016(9:51) ST334 Actuarial Methods (cid:13)c R.J. Reed
Example1.4a. AssumingACT/365,howmuchinterestwillthefollowinginvestmentspay?
(a) Adepositof£1000for60daysat10%p.a.
(b) Adepositof£1000foroneyearwhichisnotaleapyearat10%p.a.
(c) Adepositof£1000foroneyearwhichisaleapyearat10%p.a.
Solution. (a)1000×0.1×60/365=16.44. (b) 1000×0.1=100. (c) 1000×0.1×366/365=100.27.
UnderACT/365,adepositat10%foroneyearwhichisaleapyearwillactuallypayslightlymorethan10%—it
willpay:
366
10%× = 10.027%
365
Mostoverseasmoneymarketsassumeayearhas360daysandthisconventionisdenotedACT/360.
1.5 The time value of money. In general, an amount of money has different values on different dates.
Supposeanamountc isinvestedatasimpleinterestrateofi. Thetimevalueisrepresentedinthefollowing
0
diagram:
nyears nyears
(cid:122) (cid:125)(cid:124) (cid:123) (cid:122) (cid:125)(cid:124) (cid:123)
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
pastvalue presentvalue futurevalue
.............................................................. .............................................................. ..............................................................
c
0 c0 c0(1+in)
(1+in)
Acalculationconcerningsimpleinterestcanonlybedonewithrespecttoonefixedtimepoint. Thisfixedtime
pointissometimescalledthefocalpointorvaluationdate.
Hereisanexampleofanincorrectvalueobtainedbyusingmorethanonefocalpoint:
Considertheamountc attoday’stimet andsimpleinterestip.a. Thisamountc hadvaluex = c /(1+2i)
0 0 0 0
twoyearsagoattimet −2. Butthevaluexattimet −2hasvaluex(1+i) = c (1+i)/(1+2i)oneyear
0 0 0
later at time t −1. This argument is incorrect because two different focal points (t −2 and t −1) have
0 0 0
beenused. Infact,theamountc attimet hadvaluec /(1+i)attimet −1.
0 0 0 0
Simpleinterestmeansthatinterestisnotgivenoninterest—henceusingmorethanonefocalpointwillgivean
error.
Example1.5a. Apersontakesoutaloanof£5000at9%p.a.simpleinterest.Herepays£1000after3months(time
t +1/4),£750afterafurthertwomonths(timet +5/12)andafurther£500afterafurthermonth(timet +1/2).
0 0 0
Whatistheamountoutstandingoneyearaftertheloanwastakenout(timet +1)?
0
Solution. Thequestionasksfortheamountoutstandingattimet +1. Hencewemusttaket +1asthefocalpoint.
0 0
Thisgives
(cid:20) (cid:18) (cid:19)(cid:21) (cid:20) (cid:18) (cid:19)(cid:21) (cid:20) (cid:18) (cid:19)(cid:21)
9 7 1
5000[1+0.09]−1000 1+0.09 −750 1+0.09 −500 1+0.09 =3070.625
12 12 2
Here is another solution: £1,000 is borrowed for 3 months, £750 for 5 months, £500 for 6 months and £2,750 for
12months. Hencethetotalinterestpaymentis
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
0.09 0.09 0.09
1000× ×3 + 750× ×5 + 500× ×6 +[2750×0.09]=320.625
12 12 12
The total capital outstanding after 12 months is £2,750 and hence the total amount due is £2,750 + £320.625 =
£3,070.625. Theissueroftheloanwouldroundthisupto£3,070.63.
1.6
Summary.
• Terminology: simpleinterest;timevalueofmoney;presentvalue;ACT/365
• Simpleinterestformulæ:
1
c = (1+ni)c and c = c
n 0 0 n
1+ni
1 Simple and Compound Interest Sep 27, 2016(9:51) Section 2 Page 3
2 Compound interest
2.1 Explanation of compound interest. Compound interest means that interest is given on any interest
whichhasalreadybeencredited.
Example 2.1a. Suppose aninvestor invests £720 fortwo yearsat acompound interestrate of 10%p.a. Findthe
amountreturned.
Solution. Theamountinpoundsisc =720(1+i)2 =720×1.12 =871.20.
2
In general, if c denotes the initial investment, i denotes the interest rate p.a. and c denotes the accumulated
0 n
valueafternyears,then
c = c (1+i)n
n 0
The accumulated interest is clearly c [(1 + i)n − 1]. This formula assumes the interest rate is constant and
0
hencedoesnotdependonthetimepointortheamountinvested.
2.2 Simpleversuscompoundinterest. Withsimpleinterest,thegrowthoftheaccumulatedvalueislinear;
withcompoundinterest,thegrowthisexponential. Exercise25showsthatfortheinvestor,simpleispreferable
tocompoundforlessthan1yearwhilstcompoundispreferabletosimpleformorethat1year.
Accounts which allow investors to withdraw and reinvest will pay compound interest—otherwise investors
wouldcontinuallywithdrawandreinvest.
2.3 Presentvalue. Supposet1 ≤ t2 thenaninvestmentofc/(1+i)t2−t1 attimet1 willproduceareturnofc
attimet . Itfollowsthat
2
c
(1+i)t2−t1
isthediscountedvalueattimet oftheamountcattimet .
1 2
In particular, the expression “the present value (time 0) of the amount c at time t” means exactly the same as
“thediscountedvalueattime0oftheamountcattimet.”
Thus,ifthecompoundinterestrateisiperunittime,thenthepresentvalueoftheamountcattimetis
c 1
= cνt where ν = iscalledthediscountfactor.
(1+i)t 1+i
Example2.3a. Inflationadjustedrateofreturn. Supposetheinflationrateisi perannum. Supposefurtherthatan
0
investmentproducesareturnofi perannum. Whatistheinflationadjusted rateofreturn,i ,oftheinvestment?
1 a
Inparticular,ifi =0.04(or4%perannum)andtheinflationrateisi =0.02(or2%perannum),findi .
1 0 a
Solution. Supposethecapitalattime0isc . After1year, theaccumulatedvaluewillbec (1+i ). Adjustingfor
0 0 1
inflation,thetime0valueoftheamountc (1+i )attime1is
0 1
c (1+i )
0 1
(1+i )
0
Hence
1+i i −i
i = 1 −1= 1 0
a
1+i 1+i
0 0
Clearly,ifi issmalltheni ≈i −i .
0 a 1 0
Forourparticularcasei =0.02/1.02=0.0196or1.96%. Theusualapproximationisi −i whichgives2%.
a 1 0
2.4 Nominal and effective rates of interest—a numerical example. Loan companies often quote a rate
suchas15%perannumcompoundedmonthly. Thismeansthataloanof£100foroneyearwillaccumulateto
(cid:18) 0.15(cid:19)12
£100 1+ = £100×1.1608 = £116.08.
12
Theeffectiveinterestrateperannumis16.08%.
Similarly, supposeaninvestmentpaysarateofinterestof12%p.a. compounded12timesperyear. Thenthe
correspondingeffectiveannualrateofinterestisdefinedtobetheratewhichwouldproducethesameamount
ofinterestperyear;i.e.
(cid:18) 0.12(cid:19)12
1+ −1 = (1+0.01)12−1 = 0.1268 or12.68%.
12
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Example2.4a. Suppose£1000isinvestedfortwoyearsat10%perannumcompoundedhalf-yearly. Whatisthe
amountreturned?
Solution. Theamountis£1000×(1.05)4 =£1215.51.
2.5 Nominalandeffectiveratesofinterest—thegeneralcase. Ingeneral,i(m)andi denotethenominal
1/m
rateofinterestperunittimewhichiscompoundedmtimesperunittimeandwhichcorrespondstotheeffective
interestrateofiperunittime. Therelationbetweenthetwoquantitiesisgivenby:
(cid:18) i(m)(cid:19)m
i = 1+ −1 (2.5a)
m
(cid:2) (cid:3)
Thequantityi(m) = m (1+i)1/m−1 isalsodescribedasanominalrateofinterestperunittimepaidmthly,
orconvertiblemthly,orwithmthlyrests. Itisleftasanexercisetoprovethat
i = i(1) > i(2) > i(3) > ···
If p = 1/m, the quantity i(m) is also denoted i and is called the nominal rate of interest per period p corre-
p
spondingtotheeffectiveinterestrateofiperunittime. Writingequation(2.5a)intermsofpgives
1
(1+i)p = 1+pi wherep = andi = i(m) (2.5b)
p p
m
Thequantityi = pi = i(m)/miscalledtheeffectiveinterestrateovertheperiodp. Inparticular
ep p
i(1) = i = i
e1
issimplycalledtheeffectiveinterestrateperunittime.
Clearly, equivalent effective annual interest rates should be used for comparing nominal interest rates with
differentcompoundingintervals.
Example2.5a. Supposethenominalrateofinterestis10%p.a.paidquarterly. Find
(a) theeffectiveannualrateofinterest; (b) theeffectiverateofinterestover3months.
Solution. In this case p = 1/4 and i = 0.1. The answer to part (b) is i = 0.1/4 = 0.025. Hence the effective
p ep
annualrateisi=(1+i )4−1=(1+0.025)4−1=1.1038−1whichis10.38%.
ep
2.6 Terminology. Herearetwospecialcases:
• iftherateofinterestofferedona6monthinvestmentis10%, thenthismeansthattheinvestorwillreceive
5%ofhiscapitalininterestattheendofthe6months.
Ingeneral,forasituationoflessthanoneyear,theequivalentsimpleinterestrateperannumisusuallyquoted.
• a3yearinvestmentat10%meansthattheinvestorwillreceiveanamountofinterestofc ×1.13wherec is
0 0
hisinitialinvestment. Thisassumestheinterestcanbereinvested. Inthiscase,thequotedrateistheoneyear
ratewhichiscompounded.
The basic time unit is always assumed to be one year unless it is explicitly specified otherwise. Thus the
phrase“a3yearinvestmentat10%”usedaboveisjustshorthandfor“a3yearinvestmentat10%p.a.”
In general, to specify an interest rate, we need to specify the basic time unit and the conversion period. An
interestrateiscalledeffective ifthebasictimeunitandtheconversionperiodareidentical—andtheninterest
iscreditedattheendofthebasictimeperiod. Whentheconversionperioddoesnotequalthebasictimeperiod,
theinterestrateiscallednominal. Twomoreexamples:
• An effective interest rate of 12% over 2 years is the same as a nominal interest rate of 6% p.a. convertible
every 2 years. There is one interest payment after 2 years and this payment has size equal to 12% of the
capital1.
• An effective interest rate of 3% per quarter is the same as a nominal interest rate of 12% p.a. payable
quarterly.
In general, for both p ∈ (0,1) and p > 1, an effective interest rate i over the period p is also described as a
ep
nominalinterestrateofi = i /pperannumconvertible1/ptimesayear.
p ep
1 Inpractice,thisisnotquitethesameas6%p.a.simpleinterestbecausethesimpleinterestreceivedafteryearonecould
beinvestedelsewhereforyeartwo.
1 Simple and Compound Interest Sep 27, 2016(9:51) Section 2 Page 5
Calculations should usually be performed on effective interest rates—the nominal rate is just a matter of pre-
sentation. Rememberthat:
nominalrate = frequency×effectiverate or i(m) = m×i .
ep
Formanyproblems,youshouldfollowthefollowingsteps(orpossiblyasubsetofthefollowingsteps):
• Fromi ,thenominalrateforperiodp,calculatei ,theeffectiverateforperiodp,byusingnominal
p ep
rate = frequency × effectiverate(wherefrequencyis1/p).
• Converttotheeffectiverateforperiodq byusing(1+i )1/q = (1+i )1/p.
eq ep
• Converttothenominalrateforperiodq byusingnominalrate = frequency × effectiveratewhere
frequencyis1/q.
For p ≥ 0, the accumulation factor2 A(p) = (1+i)p = 1+pi gives the accumulated value at time p of an
p
investmentof1attime0.
2.7 Workedexamplesonnominalandeffectiveinterestrates.
Example2.7a. Whatisthesimpleinterestrateperannumwhichisequivalenttoanominalrateof9%compounded
six-monthly,ifthemoneyisinvestedfor3years?
Solution. Now£1investedatanominalrateof9%compoundedsix-monthlywillaccumulateto(1.045)6 =1.302in
3years. Iftheannualrateofsimpleinterestisrthenweneed
1+3r =(1.045)6
andsor =0.100753givingasimpleinterestrateof10.08%.
Example2.7b. Supposetheeffectiveannualinterestrateis41/%. Whatisthenominalannualinterestratepayable
2
(a)quarterly;(b)monthly;(c)every2years?
Solution. (a)Now(1+i )4 =1+i=1.045impliesi =0.011065foronequarter. Henceanswer(a)isi(4) =4i =
ep ep ep
0.04426ornominal4.426%p.a.payablequarterly. (b)(1+i )12 =1.045impliesi(12) =12i =0.04410. Hence
ep ep
answer(b)isnominal4.410%p.a.payablemonthly. (c)Finally,(1+i )1/2 = 1.045impliesi = 0.0920. Hence
ep ep
answer(c)isi =i /2=0.0460,ornominal4.6%p.a.payableevery2years.
p ep
Example2.7c. Aninvestorinvestsandwithdrawsthefollowingamountsinasavingsaccountatthestatedtimes
date investment withdrawal
1April2000 £1,500
1April2001 £1,000
1April2002 £1,800
1April2003 £1,500
Between 1 April 2000 and 1 April 2001, the account paid nominal 6% per annum convertible quarterly. Between
1April2001and1October2002,theaccountpaid5%perannumeffective. From1October2002,theaccountpaid
4% per annum effective. Interest is added to the account after close of business on 31 March each year. Find the
accumulatedamountat1October2003.
Solution. Therearetwopossibleapproaches.
• Calculatetheamountintheaccountateverytimepointwherethereisatransaction.
At31March2001,amountintheaccountis1,500×1.0154. Henceattheendoftheday1April2001,theamount
is1,500×1.0154−1,000. Proceedinginthismannergivestheanswer
(((1500×1.0154−1000)×1.05+1800)×1.051/2×1.041/2+1500)×1.041/2 =4110.41 (2.7a)
• Converteveryamountintothevalueat1October2003andthenaddtheseamountstogether.
So the deposit on 1 April 2000 is worth 1,500×1.0154 ×1.053/2 ×1.04 on 1 October 2003. The withdrawal on
1April2001isworth −1,000×1.053/2×1.04on1October2003. Andsoon. Thisleadstothesameanswer. In
fact,thesecondapproachisjustthesameasexpandingoutequation(2.7a)inthefirstapproach.
2 Thetermfactor isusedbecausetheaccumulatedvalueisc A(p)iftheinitialcapitalisc . SomebooksuseA(t ,t )to
0 0 1 2
denotetheaccumulatedvalueattimet ofaninvestmentmadeattimet . WethenhaveA(0,t )=A(0,t )A(t ,t )andso
2 1 2 1 1 2
A(0,t ) A(t )
A(t ,t )= 2 = 2
1 2 A(0,t ) A(t )
1 1
HencetherearenorealadvantagesinusingthisalternativenotationA(t ,t )andwestickwithA(t).
1 2
Page6 Section2 Sep27,2016(9:51) ST334ActuarialMethods(cid:13)cR.J.Reed
Example2.7d. Supposeafinancialinstrumentpaysanominalinterestrateof21/%p.a.convertibleevery2years.
2
(a)Whatistheeffectiveinterestrateover2years. (b)Whatistheequivalenteffectiveannualinterestrate?
Solution. (a)Theeffectiveinterestrateover2yearsis5%. (b)Letidenotetheequivalenteffectiveannualinterest
rate. Then(1+i)2 =1.05andsoi=0.0247. Hencetheeffectiveannualinterestrateis2.47%.
Example2.7e. Thenominalinterestratesforlocalauthoritydepositsonaparticulardayareasfollows:
Term Nominalinterestrate(%)
overnight 103/
16
onemonth 93/
4
3months 97/
8
6months 10
Findtheaccumulationof£1000for(a)onemonthand(b)3months.
Solution. For(a),wehave
1000(1+0.0975/12)=1008.12
For(b),wehave
1000(1+0.09875/4)=1024.69
Notethattheonemonthnominalratecorrespondstoaneffectiveannualrateof
(cid:18) 0.0975(cid:19)12
1+ −1=0.1019772 or10.20%.
12
The3monthnominalratecorrespondstoaneffectiveannualrateof
(cid:18) 0.09875(cid:19)4
1+ −1=0.1024674 or10.25%.
4
Onereasonforthisapparentinconsistencyisthatthemarketallowsforthefactthatinterestrateschangeovertime.
Ifbothratescorrespondedtothesameeffectiveannualrateofinterest,thenthe3monthratecouldbeobtainedfrom
the1monthratebydoingthecalculation:
(cid:34)(cid:18) 0.0975(cid:19)3 (cid:35)
4× 1+ −1 =0.09829433
12
whichisslightlylessthantherateof97/%inthetableabove.
8
Example2.7f. Showthataconstantrateofsimpleinterestimpliesadecliningeffectiverateofinterest.
Solution. Letidenotetherateofsimpleinterestperunittime. Thentheeffectiverateofinterestfortimeperiodnis:
(1+in)−(1+i(n−1)) i
=
1+i(n−1) 1+i(n−1)
whichdecreaseswithn. Itisifortimeperiod1andi/(1+i)fortimeperiod2,etc.
Example2.7g. Showthataconstantrateofcompoundinterestimpliesaconstantrateofeffectiveinterest.
Solution. Letidenotetherateofcompoundinterestperunittime. Thentheeffectiverateofinterestfortimeperiod
nis:
(1+i)n−(1+i)n−1 (1+i)−1
= =i
(1+i)n−1 1
2.8 Continuouscompounding: theforceofinterest. Fromequation(2.5a)wehave
(cid:104) (cid:105)
i(m) = m (1+i)1/m−1 (2.8a)
Now lim n(x1/n −1) = lnx if x > 1 (see exercise 22). Using this result on equation (2.8a) shows that
n→∞
lim i(m) existsandequalsln(1+i). Denotethislimitbyδ. Hence
m→∞
δ = ln(1+i) or i = eδ −1
where δ = lim i(m) = lim i is called the nominal rate of interest compounded continuously or the
m→∞ p→0 p
forceofinterestcorrespondingtotheeffectiveannualrateofinterest,i.
Itfollowsthatifanamountc isinvestedatarateδ perannumcompoundedcontinuously,thenafter1
0
yearitwillhavegrowntoc = c eδ andafterpyearsitwillhavegrowntoc epδ,wherepisnotnecessarily
1 0 0
aninteger.