Term Structure of Interest Ratesx

```Determinants of Interest Rates
• Higher Rates, More money!!!
1
Learning Objectives
• Know what specific factors determine interest
rates.
• Know what “term structure of interest rates”
means.
• Understand how forward rates of interest can be
derived from the term structure of interest rates.
Copyright 2014 by Diane S. Docking
2
What affects interest rates
on bonds?
3
The Determinants of Interest Rates
• Factors that may influence interest rates:
 Inflation
 Current economic activity
 Expectations of future growth
 Term to maturity
4
Interest Rates on Different Maturity Bonds Move Together
Copyright 2014 by Diane S. Docking
http://www.federalreserve.gov/releases/h15/data.htm
5-5
Term to Maturity
required by investors on long-term bonds to
compensate investors for the greater price
fluctuations as market interest rates change.
6
Scott Docking
The Determinants of Interest Rates (cont.)
• The Yield Curve and Discount Rates
 The relationship between the investment term and
the interest rate is called the term structure of
interest rates
 We can plot this relationship on a graph called the
yield curve
7
Yield Curve
Yield
%
Time to Maturity
An upward-sloping yield curve indicates that Treasury
Securities with longer maturities offer higher annual yields
8
3 Theories of Term Structure
1. Pure Expectations Theory (aka Unbiased
Expectations Theory)
3. Market Segmentation Theory
9
Pure Expectations Theory
(aka Unbiased Expectations Theory)
Key Assumption:
Bonds of different
maturities are perfect
substitutes
In words:
Rates are a function of
investors’ expectations.
10
Derivation – Pure Expectations Theory
• Investment strategies for a two-period horizon
 1. Buy \$1 of one-year bond and when matures buy another one-year
bond
 2. Buy \$1 of two-year bond and hold it
• The current spot rate on 1 year Treasury securities is 1R1%, while the
current spot rate on 2 year Treasury securities is 1R2%.
Year 1
time(beginning of year t) Rmaturity (yrs)
Option I:
1R1%,
Option II:
1R2%.
Year 2
2
f1 %
1R2%.
R = observable spot rate
f = expected forward rate
11
Derivation – of Geometric Average for Predicting
Current L-T Rates under Pure Expectations Theory
•
Returns should be equal under both options
E  R I   E  R II 
E  R I   1 1 R1 1 2 f 1    1




E  R II   1 1 R 2  1 1 R 2   1  1 1 R 2   1
1 R  1
1
root
1
1 R 1
E 1 R 2 
1
1
f
2
   1  1 1 R 2 
1
  1  1
1/2
2
f

1
 1 R 2  1 1 R 1 1 2 f
1
R2
1

 1
  1
2
2
1
2
2
Cancel common terms, square
1
Solve for current 2-year rate
12
Derivation – of Geometric Average for Predicting
Current L-T Rates under Pure Expectations Theory
(cont.)
In words: Interest rates on long bond (L-T rates) = geometric average of current and
expected short-term rates over life of long bond
In formula:
Predicting Current L-T rates


E  t R n    1 t R1 1 t 1 f 1 1 t  2 f 1  1 t  n 1 f 1 

where :
R  t he observed market rat e,
f  t he expected forward rat e,
t  t ime period for which t he rat e is applicable ,
n  maturit y of t he bond.
1
n
 1


Arithmetic average for predicting current L-T rates:
E  t Rn  
t
R1  t 1 f 1  t  2 f 1  ... t  n 1 f 1
n
13
Example:
Predicting current long-term rates under Pure
Expectations Theory
The one-year interest rates over the next five years are
expected to be 5%, 6%, 7%, 8% and 9%. Based upon the
pure expectations theory, what should be today’s interest
rate on a 2-year bond?, 3-year bond? 4-year bond? 5-year
bond?
So Given:
1R1= 5%, 2f1= 6%, 3f1= 7%, 4f1= 8%, 5f1= 9%,
Find:
1R2, 1R3,
1R4,
1R5
14
Example (cont.):
Predicting current long-term rates under Pure
Expectations Theory
Year 1
Option I:
1 R1 %
Option II:
1 R2 %
Option III:
1 R3 %
Option IV:
1 R4 %
Option V:
1 R5 %
Year 2
2
f1 %
Year 3
Year 4
Year 5
3
f1 %
4
f1 %
5
f1 %
3
f1 %
4
f1 %
5
f1 %
4
f1 %
5
f1 %
5
f1 %
15
Solution:
Predicting current long-term rates under Pure
Expectations Theory
Year 1
Option I:
1R1%=5%
Option II:
1 R2 %
Year 2
2
Year 3
Year 4
Year 5
f1  6 %
(1+1R2)2= (1+1R1) (1+2f1)
1/2 -1
R
=
[(1+
R
)
(1+
f
)]
1 2
1 1
2 1
1

E  1 R 2    1 . 05 1 . 06  2   1  5 . 499 %


16
Solution:
Predicting current long-term rates under Pure
Expectations Theory
Year 1
Option I:
1R1%=5%
Option III:
1 R3 %
Year 2
2
f1  6 %
Year 3
3
Year 4
Year 5
f1  7 %
(1+1R3)3= (1+1R1) (1+2f1) (1+3f1)
1/3 -1
R
=
[(1+
R
)
(1+
f
)
(1+
f
)]
1 3
1 1
2 1
3 1
1

E  1 R 3    1 . 05 1 . 06 1 . 07  3   1  5 . 997 %


17
Solution: Predicting current long-term rates under
Pure Expectations Theory – Geometric Average
1

E  1 R 2    1 . 05 1 . 06  2   1  __________ _


1

E  1 R 3    1 . 05 1 . 06 1 . 07  3   1  _________


1

E  1 R 4    1 . 05 1 . 06 1 . 07 1 . 08  4   1  ________


1

E  1 R 5    1 . 05 1 . 06 1 . 07 1 . 08 1 . 09  5   1  _______


18
Solution: Predicting current long-term rates under
Pure Expectations Theory – Arithmetic Average
E 1 R2  
E 1 R3  
56
 ________
2
567
 ________
3
E 1 R4  
5678
E 1 R5  
56789
 _________
4
 _________
5
19
Derivation – Pure Expectations Theory:
of Geometric Average for Predicting Future S-T Rates
• Investment strategies for a two-period horizon
 1. Buy \$1 of one-year bond and when matures buy another one-year
bond
 2. Buy \$1 of two-year bond and hold it
• The current spot rate on 1 year Treasury securities is 1R1%, while the
current spot rate on 2 year Treasury securities is 1R2%.
Year 1
time(beginning of year t) Rmaturity (yrs)
Option I:
1R1%,
Option II:
1R2%.
Year 2
2
f1 %
1R2%.
R = observable spot rate
f = expected forward rate
Copyright 2014 by Diane S. Docking
20
Derivation – of Geometric Average for Predicting
Future S-T Rates under Pure Expectations Theory
•
Returns should be equal under both options
E  R I   E  R II 
E  R I   1 1 R1 1 2 f 1    1




E  R II   1 1 R 2  1 1 R 2   1  1 1 R 2   1
1 R  1
1 R 1
1
1
2
1
1
f1 
2
f
   1  1 1 R 2 
1
2
f
1
  1  1
1 1 R 2 2
1
1 1 R1 
1
1
R2
 1
  1
2
2
2
Cancel common terms
Solve for “Implied” 1-year forward 1-year rate
(i.e. 1-yr “Implied” forward rate at beginning of
year 2)
21
Derivation – of Geometric Average for Predicting
Future S-T Rates under Pure Expectations Theory
(cont.)
In words: Future S-T interest rates are a function of current S-T and L-T rates
In formula:
Predicting Future S-T rates
E  t  n 1
 1 t R n  n 
f1   
1
n 1 
 1 t R n 1 

where :
R  t he observed market rat e,
f  t he expected forward rat e,
t  t ime period for which t he rat e is applicable ,
n  maturit y of t he bond.
22
Example:
Forecasting Future Short-term Interest Rates under
the Pure Expectations Theory
Current rates on bonds are as follows:
1-year bond = 5%
4-year bond = 6.5%
2-year bond = 5.5%
5-year bond = 7%
3-year bond = 6%
What is the expected future rate on a 1-year bond at the beginning
of year 2 (i.e., 1 year from today)? At the beginning of year 3
(i.e., 2 years from today)? At the beginning of year 4? At the
beginning of year 5?
So Given: 1R1= 5%, 1R2 = 5.5%, 1R3 = 6%, 1R4 = 6.5%, 1R5 = 7%
Find: 2f1, 3f1, 4f1, 5f1
23
Solution:
Forecasting Future Short-term Interest Rates under the
Pure Expectations Theory
Year 1
Option I:
Option II:
1R1=5%
1R2
Year 2
2
Year 3
Year 4
Year 5
f1 %
5.5%
(1+1R2)2= (1+1R1) (1+2f1)
2 /(1+ R )] -1
f
=
[(1+
R
)
2 1
1 2
1 1
E2
 1 . 055 2 
f1   
 1  6 . 002 %
1 
 1 . 05  
24
Solution (cont.):
Forecasting Future Short-term Interest Rates under the
Pure Expectations Theory
Year 1
Option II:
1R2=5.5%
Option III:
1R3=6%
Year 2
Year 3
3
Year 4
Year 5
f1 %
(1+1R3)3= (1+1R2)2 (1+3f1)
3
2
3f1 = [(1+1R3) /(1+1R2) ] -1
E 3
 1 . 06 3 
f1   
 1  7 . 0071 %
2 
 1 . 055  
25
Solution: Forecasting future Short-term Interest
Rates under the Pure Expectations Theory
E2
E3
 1 . 055 2
f1   
1
 1 . 05 
 1 . 06 
f1   
2
 1 . 055 
3

  1  __________


  1  __________ ___

26
Solution: Forecasting future Short-term Interest
Rates under the Pure Expectations Theory
E4
 1 . 065 4 
f1   
 1  ________
3 
 1 . 06  
E5
 1 . 07 5
f1   
4
 1 . 065 

  1  _______

27
3 Theories of Term Structure
1. Pure Expectations Theory (aka Unbiased
Expectations Theory)
28
• Key Assumption:
• Implication:
Bonds of different maturities
are substitutes, but are not
perfect substitutes
Modifies Pure Expectations
Theory with features of Market
Segmentation Theory
• Investors prefer short-term rather than long-term bonds. This implies that
investors must be paid positive liquidity premium, ℓnt, to hold long term
bonds.
29
• Investment strategies for a two-period horizon
 1. Buy \$1 of one-year bond and when matures buy another one-year
bond
 2. Buy \$1 of two-year bond and hold it
Long-term interest rates are geometric averages of current and expected
future short-term interest rates plus liquidity risk premiums that increase
with maturity
Year 1
Option I:
Option II:
1R1%,
1R2%.
Year 2
2
f1 %   2 %
1R2%.
30
Derivation – of Geometric Average for Predicting
Current L-T Rates under Liquidity Premium Theory
•
Returns should be equal under both options
E  R I   E  R II 
E  R I   1 1 R1 1 2 f 1   2    1




E  R II   1 1 R 2  1 1 R 2   1  1 1 R 2   1
1 R  1
1
1
1 R 1
1
E 1 R 2 
1


  1  1 R   1

2
f 1   2   1  1 1 R 2   1
2
2
f 1  2

2
1/2
2
1
 1 R 2  1 1 R 1 1 2 f 1   2 
2

1
2
Cancel common terms, square root
1
Solve for current 2-year rate
31
Derivation – of Geometric Average for Predicting Current L-T
Rates under Liquidity Premium Theory (cont.)
• Long-term interest rates are geometric averages of current and
expected future short-term interest rates plus liquidity risk
Geometric average for predicting current L-T rates:
E  t R n   1 t R1 1 t 1 f 1   t 1 ... 1 t  n 1 f 1   t  n 1 
1
n
1
ℓt = liquidity premium for period t
ℓ2 < ℓ3 < …< ℓn
32
Example:
Predicting current long-term rates under
The one-year interest rates over the next four years are expected to be 5%,
6%, 7%, 8% and 9%.
Investors' preferences for holding short-term bonds, so liquidity premiums
for one- to five-year bonds are 0%, 0.25%, 0.5%, 0.75%, and 1%.
Based upon the liquidity premium theory, what should be today’s interest
rate on a 2-year bond?, 3-year bond? 4-year bond?, 5-year bond?
So Given:
1R1= 5%, 2f1= 6%, 3f1= 7%, 4f1= 8%, 5f1= 9%,
ℓ1= 0%, ℓ2= 0.25%, ℓ3= 0.5%, ℓ4= 0.75%, ℓ5= 1%,
Find:
1R2, 1R3,
1R4,
1R5
33
Example (cont.):
Predicting current long-term rates under Liquidity
Year 1
Option I:
1 R1 %
Option II:
1 R2 %
Option III:
1 R3 %
Option IV:
1 R4 %
Option V:
1 R5 %
Year 2
2
f1 %   2 %
Year 3
Year 4
Year 5
3
f1 %   3 %
4
f1 %   4 %
5
f1 %   5 %
3
f1 %   3 %
4
f1 %   4 %
5
f1 %   5 %
4
f1 %   4 %
5
f1 %   5 %
5
f1 %   5 %
34
Solution:
Predicting current long-term rates under Liquidity
Year 1
Option I:
1 R1 %
Option II:
1 R2 %
Year 2
2
Year 3
Year 4
Year 5
f1 %   2 %
(1+1R2)2= (1+1R1) (1+2f1+ℓ2)
1/2 -1
R
=
[(1+
R
)
(1+
f
+ℓ
)]
1 2
1 1
2 1
2
1
R 2  1 1 R1 1 2 f 1   2 
1
 1 . 05 1  . 06  . 0025
 1 . 05 1 . 0625

1
2
2
1

1
2
1
 1  5 . 6232 %
35
Solution: Predicting current long-term rates
R 2  1 1 R1 1 2 f 1   2 
1
1
2
 1 . 05 1  . 06  . 0025
 1 . 05 1 . 0625
1

1
2
1

1
2
1
 1  _________
R 3  1 1 R1 1 2 f 1   2 1 3 f 1   3 
 1 . 05 1 . 0625
1  . 07  . 005 
 1 . 05 1 . 0625
1 . 075 
1
3
1
1
3
3
1
1
 1  _________
36
Solution: Predicting current long-term rates
1
1
R 4  1 1 R1 1 2 f 1   2 1 3 f 1   3 1 4 f 1   4 
 1 . 05 1 . 0625
1 . 075 1  . 08  . 0075 
 1 . 05 1 . 0625
1 . 075 1 . 0875 
1
4
1
4
1
4
1
1
 1  ________
R 5  1 1 R1 1 2 f 1   2 1 3 f 1   3 1 4 f 1   4 1 5 f 1   5 
 1 . 05 1 . 0625
1 . 075 1  . 0875 1  . 09  . 01 
 1 . 05 1 . 0625
1 . 075 1 . 0875 1 . 10 
1
5
1
5
1
5
1
1
 1  ________
37
Relationship Between the Liquidity
Theory
38
Relationship Between the Liquidity Premium and Pure
Expectations Theory
Our Example
39
Predicting Future S-T Rates
• Investment strategies for a two-period horizon
 1. Buy \$1 of one-year bond and when matures buy another one-year
bond
 2. Buy \$1 of two-year bond and hold it
Year 1
Option I:
Option II:
1R1%
- ℓ1%
1R2%
- ℓ2%
Year 2
2
1 R2 %
f1 %
- ℓ2%
40
Derivation – of Geometric Average for Predicting
Future S-T Rates under Liquidity Premium Theory
•
•
Returns should be equal under both options
E  R I   E  R II 
E  R I   1 1 R1   1 1 2 f 1    1




E  R II   1 1 R 2   2  1 1 R 2   2   1  1 1 R 2   2   1
1 R
1
1 R
1
1
1
  1  1 2 f
  1 1 2 f
1
   1  1 1 R 2   2 
1
  1  1 R
1
1 1 R 2   2 
1 1 R1   1 1
2
2
f1 
1

 1
 2  1
2
2
2
2
Cancel common terms, square root
Solve for “Implied” 1-year forward 1-year rate
(i.e. 1-yr “Implied” forward rate at beginning of
year 2)
41
Formula for Predicting Future S-T Rates under
Liquidity Premium Theory: tRn- ℓnt = same as pure expectations theory; replace
tRn by tRn- ℓnt in above equation to get adjusted forward-rate forecast
E  t  n 1
 1 t R n   nt n

f1   
1
n 1 
 1 t R n 1   nt 1 

where :
R  t he observed market rat e,
f  t he expected forward rat e,
t  t ime period for which t he rat e is applicable ,
n  maturit y of t he bond.
42
Example:
Forecasting future Short-term Interest Rates under
Current rates on bonds are as follows:
1-year bond = 5%
4-year bond = 6.5%
2-year bond = 5.5%
5-year bond = 7%
3-year bond = 6%
Investors' preferences for holding short-term bonds, so liquidity premiums for
one- to five-year bonds are 0%, 0.25%, 0.5%, 0.75%, and 1%
What is the expected future rate on a 1-year bond at the beginning of year 2 (i.e.,
1 year from today)? At the beginning of year 3 (i.e., 2 years from today)? At
the beginning of year 4? At the beginning of year 5?
So Given: 1R1= 5%, 1R2 = 5.5%, 1R3 = 6%, 1R4= 6.5%, 1R5 = 7%
ℓ1= 0%, ℓ2= 0.25%, ℓ3= 0.5%, ℓ4= 0.75%, ℓ5= 1%,
Find:
2f1,
3f1,
4f1,
5f1
43
Solution:
Forecasting future Short-term Interest Rates under the
Year 1
Year 2
Option I:
1R1- ℓ1=5%-0%
Option II:
1R2- ℓ2=5.5%-0.25%
2
Year 3
Year 4
Year 5
f1 %
(1+1R2-ℓ2)2= (1+1R1-ℓ1) (1+2f1)
2
2f1 = [(1+1R2-ℓ2) /(1+1R1-ℓ1)] -1
E2
 1  0 . 055  0 . 0025
f1   
1


1

0
.
05

0

2 
 1

 1 . 0525  2 
 
  1  5 . 5006 %
1
 1 . 05  
44
Solution:
Forecasting future Short-term Interest Rates under the
Year 1
Year 2
Option II:
1R2- ℓ2=5.5%-0.25%
Option III:
1R3- ℓ3=6%-0.5%
(1+1R3-ℓ3)3= (1+1R2-ℓ2)2 (1+3f1)
3
2
3f1 = [(1+1R3-ℓ3) /(1+1R2-ℓ2) ] -1
Year 3
3
E3
Year 4
Year 5
f1 %
 1  0 . 06  0 . 005 3 
f1   
1
2 
 1  0 . 055  0 . 0025  
 1 . 055 3 
 
 1  6 . 0018 %
2 
 1 . 0525  
45
Solution: Forecasting future Short-term Interest
Rates under the Liquidity Premium Theory
E2
E3
 1  0 . 055  0 . 0025
f1   
1
1 . 05 


2

  1  ________

 1  0 . 06  0 . 005 3
f1   
2
 1  0 . 055  0 . 0025 

  1  ________

46
Solution: Forecasting future Short-term Interest
Rates under the Liquidity Premium Theory
E4
 1  0 . 065  0 . 0075 4
f1   
3
 1  0 . 06  0 . 005 
E 5

  1  _______

 1  0 . 07  0 . 01 5
f1   
4
 1  0 . 065  0 . 0075 

  1  _______

47
3 Theories of Term Structure
1. Pure Expectations Theory (aka Unbiased
Expectations Theory)
3. Market Segmentation Theory
48
Market Segmentation Theory
• Key Assumption:
Bonds of different maturities are
not substitutes at all
• Implication:
Markets are completely segmented;
interest rate at each maturity are
determined separately
•
•
Short- and long-term markets are independent of one another and yields are determined
by supply and demand in each maturity market sector.
Assumes borrowers and lenders prefer certain maturities and cannot be induced to
substitute between maturities by small yield changes