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7-3 R 0,1 Time Spot interest rates 102 R 0,2
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7-1
Chapter 7
Non Flat Term Structure
7-2
Notation
R0,1 = the spot interest rate observed at time 0 (first subscript) and lasting one period of time.
R0,2 = the spot interest rate observed at time 0 (first subscript) and lasting two periods.
7-3
R0,1Time
Spot interest rates
10 2
R0,2
7-4
One-period Present Value
PV1 =1
1 + R0,1
0.9615 =
11.04 1
10
$1
S1 = 96.15 =
1001.04 100
S1 = One-period strip
7-5
Two-period Present Value
PV2 =1
(1 + R0,2)2
S2 = Two-period strip
0.8573 =
1(1.08)2 1
10
$1
S2 = 85.73 =
100(1.08)2 100
2
7-6
n-period Present Value
PVn =1
(1 + R0,n)n
0.4665 =
1(1.10)8 1
0
$1
S8 = 46.65 =
100(1.10)8 100
n
7-7
1/(1 + R0,1) or 1/(1 + R0,2)2? or
Which of the following can be true?a. LHS > RHSb. LHS = RHSc. LHS < RHSd. It depends.e. None of the above.
11 + R0,1
1(1 + R0,2)2
7-8
PV1 PV2 . . . PVn
1 at timereceived $1 of
luePresent va
2 at timereceived $1 of
luePresent va
n at timereceived $1 of
luePresent va …
7-9
One-period Strip
Price ofone-period
strip= S1 = [PV1][PAR]
S1 = = 100 = 96.15.1,0R1
PAR
04.11
7-10
Two-period Strip
Price oftwo-period
strip= S2 = [PV2][PAR]
S2 = = 100 = 85.73.22,0 )R1(
PAR
2)08.1(
1
7-11
Cash flows for treasury strip with price Sn
Points in time
0 1 … n
Cash flows –Sn 0 … 0 … +PAR
7-12
strip. a of parof dollar per Price
PARstrip period-n of Price
n timeat received$1 of Value
present ThePARSPV n
n
7-13
Forward Interest Rates
7-14
Timing of a Forward Contract
Points in time
0 Delivery dateSign Pay $ and
contract receive commodityTransaction and or
Set Deliver commodity and price receive $
7-15
Forward Contract Cash Flows for Lender (Bond Buyer)
Points in time
0 1 2Delivery date Maturity date
Cash flows 0 -F +PAR(Lend $F) (Receive PAR)
7-16
Time 1 Value of $1 Received at Time 2
F =2,0f1
1
0.8917 =
11.1215 1
10
$1
2f0,2
7-17
Time 2 Value of $1 Received at Time 3
3,0f11
10
$1
3f0,3
2
= Time 2 value of $1 received at time 33,0f1
1
7-18
310 2
R0,1 f0,2 f0,3
Time 0 value = )f1)(f1)(R1(1
3,02,01,0
Time 1 value = )f1)(f1(1
3,02,0 of $1 received at time 3
of $1 received at time 3
7-19
Link between Spot and Forward
Interest Rates
7-20
210
R0,1 f0,2
1 + R0,1
R0,2
$1
$1 (1 + R0,1)(1 + f0,2)
(1 + R0,2)2
$1 1.04
$1
(1.04)(1.1215) = 1.1664
(1.08)2 = 1.1664
7-21
(1 + R0,2)2 = (1 + R0,1)(1 + f0,2).
Geometric mean of R0,1 and f0,2 is R0,2.
In terms of present value,
.)f1)(R1(1
)R1(1
2,01,02
2,0
7-22
.2fRR
fR1R21fRfR1RR21
2,01,02,0
2,01,02,0
2,01,02,01,02
2,02,0
Arithmetic Mean
Approximation
7-23
Arithmetic Mean
Approximation210
R0,1 f0,2
.04 .12+
2R0,2
2(.08) = .16
= .16
7-24
Three-period Case(1 + R0,3)3 = (1 + R0,1)(1 + f0,2)(1 + f0,3)(1 + R0,3)3 = (1 + R0,2)2 (1 + f0,3).
n-period Case(1 + R0,n)n = (1 + R0,1)(1 + f0,2) . . . (1+f0,n)
= (1 + R0,n-1)n-1 (1 + f0,n).
7-25
.R1)R1(f1
1,0
22,0
2,0
.)R1()R1(f1 2
2,0
33,0
3,0
.)R1()R1(f1 1n
1n,0
nn,0
n,0
In Terms of Forward Rates
7-26
Since the prices of one- and two-period strips are
20,2
20,1
1 )R(1PARS R1
PARS
.1215.173.8515.96
04.1)08.1(f1
SS
)R1(1R11
R1)R1(f1
2
2,0
2
1
22,0
1,0
1,0
22,0
2,0
7-27
.1
)1()1(
1
1,0
3
22
2,0
33,0
3,0
n
nn SSf
SS
RR
f
Similarly,
7-28
Shape of the Term Structure
(1 + R0,2)2 = (1 + R0,1)(1 + f0,2).
7-29Maturity1 2
Flat R0,1 R0,2
Flat Term Structure
= f0,2
7-30Maturity1 2
Rising R0,1
R0,2
R0,1 < R0,2
R0,1
Rising Term Structure
7-31Maturity1 2
Rising R0,1
R0,2
R0,1 < R0,2
R0,1
Rising Term Structure
f0,2
< f0,2
7-32Maturity1 2
Declining R0,1
R0,2
Declining Term Structure
R0,1
R0,1 > R0,2
f0,2
> f0,2
7-33
Annuities
7-34
.)R1(
1R11PVA Annuityof
luePresent Va2
2,01,02
.8188.18573.09615.0)08.1(
104.11
AnnuityofluePresent Va
2
.)R1(
1R11PVA Annuityof
luePresent Van
n,01,0n
7-35
Prices of Coupon-Bearing Bonds
Cash flows for Coupon-bearing bond
Points in time
0 1 2 through n–1 n
Cash flows –Pn c … c … +PAR
7-36
.PVPARcR1PARcP 1
1,01
One-period Bond
.92.10104.11006P1
7-37
.PVPARPVAc)R1(
PARcR1cP
22
22,01,0
2
Two-period Bond
.65.96)08.1(
100604.16P 22
7-38
.strip of Priceslopec
PVPARPVAc)R1(
PARcR1cP
nn
nn,01,0
n
n-period Bond
7-39
Yield to Maturity and Spot Rates
7-40
In general, y is a polynomial average of R0,1 and R0,2.
22,01,0
22 )R1(PARc
R1c
y)(1PARc
y1cP
7-41
Examples of Yield to Maturity0 1 2
Calculator: N = 2, PV = -96.65, PMT = 6, FV = 100.
y6 = 7.8754
96.65
+= 04.16
2)08.1(106
96.65
+= y16 2)y1(
106
7-42
0 1 2
y3 = 7.9360
Which bond is better?
91.19
+= 04.13
2)08.1(103
91.19
+= y13 2)y1(
103
7-43
Two-period Par Bond
%.84.71/(1.08)/(1.04)1(
)/(1.08)1(1yPAR
)R/(11)R/(11))R/(11(1yPAR
2
2
2
20,20,1
20,2
2
7-44
n-period Par Bond
. Annuityperiod-n of luePresent Van] Timeat $1 of luePresent Va[1
PVAPV1yPAR
)R/(11)R/(11))R/(11(1yPAR
n
nn
nn0,0,1
nn0,
n
7-45
Finding the Term Structure of
Interest Rates
7-46
STRIPS: One-period Strip
Since S1 = 1,0R1
PAR
1 + R0,1 = = = 1.041S
PAR15.96
100
R0,1 = 0.04 = 4%.
7-47
STRIPS: Two-period Strip
Since S2 = 22,0 )R1(
PAR
(1 + R0,2)2 = = = 1.16652S
PAR73.85
100
R0,2 = 0.08 = 8%.
7-48
STRIPS: n-period Strip
Since Sn = nn,0 )R1(
PAR
(1 + R0,n)n =nS
PAR
R0,n = – 1.nS
PARn
7-49
Bootstrapping—Recursive Method1-Period Bond
2-Period Bond
%5.4R 045.1
10452.99
R1ParcP
0,1
1,0
%5R )R(1
104.50045.150.409.99
)R1(Parc
R1cP
0,220,2
22,01,0
7-50
Disadvantages of Bootstrapping
Maturities may be missing.Errors in earlier maturities compound
and create bigger errors for longer maturities
7-51
Two Bonds with Same Maturity But Different Coupons
]PV[Par]PVA[cP
)R1(Parc
R1cP
22
22,01,0
7-52
P
c
S2
4.50
5.00
c99.415
100.35
P
0 21
104.504.5099.415105.005.00100.35
7-53
96.0PV 96S
10091
100S
87.1
ParS
ParS
PVA
00.91)87.1)(50.4(415.99S
87.150.400.5415.9935.100PVA
11
1
212
2
2
7-54
DISADVANTAGE
There may only be a small number of maturities with bond pairs or triples.