PROTECTING AGAINST TERM
STRUCTURE SHIFTS
(INVESTMENT MANAGEMENT WITH
LIABILITY STREAM)
September 1999

BASIC IDEA
1 Manager is concerned with Liability Stream.
2. Concern is to maximize net worth.
Major Tools
(1) Cash flow matching or dedication
(2) Sensitivity matching or Immunization
2

CASH FLOW MATCHING
Observe liability stream L
1
, L
2
,L
3,
..., L
T
3

Minimize Cost
Subject to
EXACT MATCHING
Cash flow
1. Meet Cash Flows
2. Do not issue Bonds
Let
1. P i
is Price of Bond i
2. N i
is Number of Bonds of Type i
Purchased
3. L t
is Liabilities in Time t
4. r is Short term Interest Rate
5. cf(it) is Cash Flow of Bond i in Period t.
Interest or principal and interest
6. S t
is Amount Invested in Short Term
Bond in Period t
4

MINIMIZE
∑ i
N i
P i
SUBJECT TO
1.
i
N i cf(it) + S t 1
(1 + r) L t
S t
= 0 all t
2.
N i
φ
0 all i
3.
S t
φ
0 all t
4.
S
1
====
0
5

RISKS OF CASH FLOW MATCHING
1. Reinvestment Risk
2. Risk of disappearing security a. Call b. Sinking fund c.
Default
6

Immunization
1.
Assume a manager is responsible for designing a portfolio that will have sufficient cash flows to meet a liability L t
is liability in period t.
Thus liabilities are L
1
, L
2
, L
3,
L
4
2.
Sensitivity matching involves finding a Portfolio that a) at equilibrium prices the Portfolio costs the same as the Liability b) the value of the asset portfolio and Liability portfolio move in tandem
3.
Need measures of how value of portfolio changes given change in factor or factors effecting it.
Choices
(1) Duration
(2) Coefficients of factor model
Types of Duration measures
(1) Analytical
(2) Numerical
(3) Time Series Estimation
7

ANALYTICAL
SENSITIVITY
MEASURES
8

ANALYTICAL
MODELS OF SENSITIVITY TO
INTEREST RATE CHANGE
1.
MACAULAY CHANGE
YIELD
2. MACAULAY/FISHER AND WEIL ADDITIVE
CHANGE
3.
BIERWAG AND KAUFMAN MULTIPLICATIVE
4.
BIERWAG ADDITIVE
MULTIPLICATIVE
5.
KHANG SHORT MORE
THAN LONG
6.
COX, INGERSOLL AND ROSS SHORT GAUSS
MARKOV AND
EXPECTATIONS
THEORY
7.
BRENNAN AND SCHWARTZ SHORT AND
LONG
MARKOV AND
EXPECTATIONS
THEORY
9

I. MAUCAULAY [D i
]
P
0
=
C
(1 +
C r)
+
(1 + r
C
)
2
+
(1 + r
C + 100
)
3
+ ...
+
(1 + r )
N dP
0 d(1 + r)
=
(1
C
+ r )
2
+
(1
2C
+ r )
3
+
(1
3C
+ r )
4
+ ...
+
(C + 100)(-N)
(1 + r )
N + 1 dP
0 d(1 + r)
=
1
-
+
1 r
∑ t
(1 tcf(t)
+ r ) t dP
0 tcf(t) d(1 + r)
=
P
0
1
-
+
1 r
[
∑ t
(1 +
P
0 r ) t
]
% price change = dP
P
0
0
= D
1 d(1
(1 +
+ r) r)
= MINUS DURATION
1 x %change in (1 + r)
10

Modified Duration dP
P o o
= D
M
1 dr wh ere D
1
M
=
1
D
+
1 r
Price change per interest rate change dP = D
1
M
P dr
•
Modified Duration useful when worried about riskiness of different positions.
•
Price change when hedging.
•
If use different r’s on different bonds such as yield to maturity, then duration is not additive.
11

General Principles
(1) The lower the interest rate, the longer the duration.
(2) The lower the coupon rate, the longer the duration.
(3) The greater the maturity, the longer the duration in general.
12

Estimated
Price
=
Original
Price dr
- (Duration)(P
0
) ----
1+r
P
0
++++ dP
====
P
0
−−−−
D
1 dr
1
++++ r
P
0
13

Interest Rate
Coupon
Principal
Maturity
Semi-Annual
Payment
Price
Duration
EXAMPLE 1
10%
10%
100
15 years
$100
8.07 years
14

CHANGE IN
INTEREST
RATE
-0.007
-0.006
-0.005
-0.004
-0.003
-0.002
-0.001
0
.001
0.002
0.003
0.004
0.005
-0.015
-0.014
-0.013
-0.012
-0.011
-0.01
-0.009
-0.009
0.006
0.007
0.008
0.009
0.01
0.011
0.012
0.013
0.014
0.015
TRUE
PRICE
16
127.588
125.429
123.322
121.264
119.254
117.292
115.376
113.504
111.675
109.889
108.144
106.44
104.774
103.146
101.555
100
98.48
96.994
95.542
94.122
92.733
91.375
90.047
88.748
87.478
86.235
85.019
83.83
82.666
81.527
80.412
INTEREST
RATE
0.043
0.044
0.045
0.046
0.047
0.048
0.049
0.05
0.051
0.052
0.053
0.054
0.055
0.035
0.036
0.037
0.038
0.039
0.04
0.041
0.042
0.056
0.057
0.058
0.058
0.06
0.061
0.062
0.063
0.064
0.065
ESTIMATED
PRICE
123.059
121.521
119.984
118.447
116.91
115.372
113.835
112.298
110.761
109.223
107.686
106.149
104.612
103.074
101.537
100
98.463
96.926
95.388
93.851
92.314
90.777
89.239
87.702
86.165
84.628
83.09
81.553
80.016
78.479
76.941

EXAMPLE 2
Coupon vs. Pure Discount:
Coupon
Maturity
Initial Price
Annual Payments
Interest Rate
10%
10 years
$100
10%
17

II. MACAULAY/FISHER AND WEIL [D
2
]
P
0
=
(1 +
C r
01
)
+
(1 + r
01
C
)(1 + r
12
C
)
+
(1 + r
01
)(1 + r
12
)(1 + r
23
)
+ ...
dp
0 d(1 + r
01
)
=
(1 +
C r
01
)
2
+
(1 + r
01
C
)
2
(1 + r
12 d(1 + r
12
) d(1 +
)
+ C
(1 + r
01
)(1 r
01
+
) r
12
)
2
+ ...
ASSUME d(1 +
(1 + r r
N
N
1N
1N
)
)
= d(1
(1 +
+ r
01 r
01
)
) dP
0 d(1 + r
01
)
=
(1 +
C r
01
)
2
+
(1 + r
01
2C
)
2
(1 + r
12
)
+
(1 + r
01
3C
)
2
(1 + r
12
)(1 + r
23
)
+
=
(1 +
1 r
01
) t t(CF)(t) i t
= 1
(1 + r i 1, i
)
19

dP
0
= -
P
0





  t i t
= 1 tCF(t)
(1
P
0
+ r i 1, i
)





  d(1
(1 +
+ r
01 r
01
)
) dP
0
= -
P
0
D
2 d(1
(1 +
+ r
01 r
01
)
)
% CHANGE
IN PRICE = [ MINUS DURATION ] X[ % CHANGE
IN ( 1 + r
01
)]
20

Coupon
Principal
Maturity
Duration
Initial Price
Yield Curve
(see attached)
EXAMPLE 3
10% (semi-annual)
$100
15 years
6.527
$740.27
21

INTEREST RATE
CHANGE
-0.004
-0.003
-0.002
-0.001
0
0.991
0.002
0.003
0.004
x
-0.01
-0.009
-0.008
-0.007
-0.006
-0.005
0.005
0.006
0.007
0.008
0.009
0.01
TRUE
77.909
76.908
75.928
74.968
74.027
73.106
72.204
71.32
70.453
p(x)
84.381
83.244
82.131
81.041
79.975
78.931
69.604
68.772
67.957
67.157
66.373
65.605
24
DURATION
69.334
68.404
67.474
66.544
65.615
64.685
77.702
76.772
75.842
74.912
73.983
73.053
72.123
71.193
70.264
newprice (x)
83.28
82.351
81.421
80.491
79.561
78.632

EXAMPLE 4
Coupon
Principal
Maturity
Annual Interest
Duration
5%
$100
30 years
13.054
25

Adding additional terms (convexity)
The formula for the first three terms in a
MacLaurin series expansion of a function f(X + h) in the region of h as h approaches zero is f(X + h) = f(X) + df(X)(h)
1
+ df
2
(X)(h
2 : 1
)
2
+ ...
Where the prime denotes derivatives. Define P(r) as the price of a bond at an interest rate r. Then, writing the price of the bond at a new interest rate (r + h) using the series expansion results in
P(r + h) = P(r) + P
′′′′
(r)h + 1/2P" (r) h
2
The price of the bond is
P
0
=
∑ T t = 1 cf(t)
(1 + r ) t
27

Then the first derivative with respect to (1 + r) is dp d(1 +
0 r)
= ∑ T t = 1
-
(1 tcf(t)
+ r ) t
1
1 + r and the second derivative is d d
( 1
2
++++
P
0 r )
2
==== T t
====
1 t ( t
++++
( 1
1
++++
) cf r ) t
( t )
( 1
++++
1 r )
2
This second derivative is called convexity.
return =
P ( r
++++ h )
−−−−
P ( r )
P ( r )
==== −−−−
D i
∆∆∆∆ i
++++
1 / 2 C i
(
∆∆∆∆ i
)
2
D
==== ∑




T t
====
1 tC ( t )
((((
1
++++ r
)))) t

 

P
0
28

C i
====

∑

 t
T
====
1 t ( t
(
++++
1
++++
1 ) C ( t r ) t
)



÷÷÷÷
P
0
∆∆∆∆ i
==== d
(
( 1
1
++++
++++ r r )
)
→
Note convexity is positive gives higher return under all changes.
→
Consider two bonds with same duration and different convexity assume C
A
> C
B
. Does A give higher return?
→
Answer: If it did, we would have dominance.
Thus return without yield curve shift must be higher for B.
29

EXAMPLE 5
Coupon
Maturity
Principal
Price
Yield curve flat
Duration
Convexity
10%
15 years
$100
$100
10%
8.07
96.59
30

R (x)
0.059
0.06
0.061
0.062
0.063
0.064
0.065
0.051
0.052
0.053
0.054
0.055
0.056
0.057
0.058
0.043
0.044
0.045
0.046
0.047
0.048
0.049
0.05
0.035
0.036
0.037
0.038
0.039
0.04
0.041
0.042
p (x)
98.48
96.994
95.542
94.122
92.733
91.375
90.047
88.748
87.478
86.235
85.019
83.83
82.666
81.527
80.412
127.588
125.429
123.322
121.264
119.254
117.292
115.376
113.504
111.674
109.889
108.144
106.44
104.774
103.146
101.555
100
32 new price (x)
98.463
96.926
95.388
93.851
92.314
90.777
89.239
97.702
86.165
84.628
83.09
81.553
80.016
78.479
76.941
122.059
121.521
119.984
118.447
116.91
115.372
113.835
112.298
110.761
109.223
107.686
106.149
104.612
103.074
101.537
100 newprice2 (x)
98.48
96.996
95.546
94.131
92.752
91.407
90.098
88.824
87.584
86.38
85.211
84.076
92.977
91.913
80.884
127.001
124.956
122.946
120.97
119.03
117.125
115.255
113.419
111.619
109.854
108.124
106.429
104.769
103.145
101.555
100

MEASURING
SENSITIVITY
NUMERICALLY
33

1
2
3
4
5 t
Numerical Estimation of Duration r
00 t
10
11
12
13
14 r
00 t
11
12
13
14
15
P
D dr
P
P
====
5
( 1 .
05 )
++++
5
( 1 .
055 )
2
++++
5
( 1 .
06 )
3
++++
5
( 1 .
065 )
4
++++
105
( 1 .
07 )
5
P
′′′′ ====
5
( 1 .
055 )
++++
5
( 1 .
06 )
2
++++
5
( 1 .
065 )
3
++++
5
( 1 .
07 )
4
++++
105
( 1 .
075 )
5
34

By definition:
P
P
fraction change in price
P
P = 92.202
P
====
90.2817
P
P
==== −−−−
.021
P dr = .01
Duration = 2.1
Calculation Duration
35

Two Factor Example
Assume two key rates a.
six-month rate b.
ten-year rate
36

2
3
4
5 t
1
Estimate Functional Relationship
∆ r
001
====
1
∆ r
001
++++
0
∆ r
020
∆ r
002
====
.8
∆ r
001
++++
.05
∆ r
020
∆ r
003
====
.4
∆ r
001
++++
.1
∆ r
020
∆ r
004
====
.1
∆ r
001
++++
.15
∆ r
020
∆ r
005
====
.05
∆ r
001
++++
.20
∆ r
020 r
00t
10
11
12
13
14 r
00t
′
11
11.8
12.4
13.1
14.05
r
00 t
″
10
11.05
12.10
13.15
14.20
37

∆
P
==== −−−−
D
1 dr
1
−−−−
D
2 dr
2
P
P
====
5
( 1 .
05 )
++++
5
( 1 .
055 )
2
++++
5
( 1 .
06 )
3
++++
5
( 1 .
065 )
4
++++
105
( 1 .
07 )
5
P
′′′′ ====
5
( 1 .
055 )
++++
5
( 1 .
059 )
2
++++
5
( 1 .
062 )
3
++++
5
( 1 .
0655 )
4
++++
105
( 1 .
07025 )
5
P
′′′′′′′′
=
5
(1.05)
+
5
(1.05525
)
2
+
5
(1.0605
5
)
3
+
(1.06575
)
4
105
+
(1.071
)
5
38

→ P = 92.202
P
′′′′ ====
92.028
P
′′′′′′′′ ====
91.835
P
′′′′
P
= 0.0019
P
P
′′′′′′′′
P
= .0040
P
D
1
= .19
D
2
= .40
39

COMMENTS
1) The advantage of Numerical calculation of duration lies in determining the duration for bonds with option features. One can numerically value the options at both sets of spot rates and then determine the price changes including the change in the value of the option.
2) The other consideration is that charges in spot rates are linked. The above assumption of a constant 1% change for all spots is unrealistic.
3) If one believed two factors one would obtain a
D
1
and D
2
for each factor
42

TIME SERIES ESTIMATION
Total
Return =
Expected
Return +
Due to
Passage of
Time
Sensitivity to Unexpected
Term
Structure
Shifts
Unexpected
Term +
Structure
Shifts
Random
Error dr r it
= r i
d i
(
1 + r
) +
εεεε i
Note:
(1) In stock area estimate directly
(2) In bonds D changes over time but can do for spots(unchanged duration), and build up to coupon bonds. Above is a "factor model."
43

Duration on portfolio is weighted average of duration bonds comprising it.
empirical duration =
ΣΣΣΣ
X i
D i where D i
for each pure discount bond
44

Immunization
(1) Protecting against shifts in spot rates a. D
A
= D
L b. D
A
= D
L
C
A
= C
L c. D
A1
= D
L1
D
A2
= D
L2
(2) If PV (assets) > PV (Liabilities) immunization involves scaling
D
A
=
L
A
D
L
Note:
Exact match portfolios are of course immunized.
46

Sensitivity Cash flow
Matching vs Matching
1. If all bonds fairly priced always cash flow match.
2. Must have sufficient mispricing to justify sensitivity.
3. Often optimum to do both.
47

I.
Terms a.
Duration b.
Cash flow matching c.
Sensitivity matching
II. Concepts a.
Risks of cash flow matching b.
Risks of sensitivity matching c.
Analytical duration measures d.
Numerical duration measures e.
What affects duration f.
Convexity
III. Calculations a.
Various duration and convexity measures
48

Problem
1. Assume a bond has been semi-annual payments of $5 and a life of four years.
What is its duration with a flat yield curve of 8%?
Answer:
Price of bond is $106.73, which is the present value of cash flows.
Duration =
1/2
××××
5
+
(1.04)
1
××××
5
(1.04
)
3/2
××××
5
2
+
(1.04
)
3
+
106.73
.
.
.
+
4
××××
105
(1.04
)
8
= 3.42
49