Lecture 5
The term structure of interest rates
Lecture 5
1 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
Lecture 5
2 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
This is a strong assumption for two reasons:
Lecture 5
2 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
This is a strong assumption for two reasons:
We assume that the same constant interest rate holds for borrowing
as well as for lenders. This is called a perfect bank in the book.
Lecture 5
2 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
This is a strong assumption for two reasons:
We assume that the same constant interest rate holds for borrowing
as well as for lenders. This is called a perfect bank in the book.
It is a fact that if we want to borrow money from a bank, then the
interest rate depends on the time at which we should pay back our
money.
Lecture 5
2 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
This is a strong assumption for two reasons:
We assume that the same constant interest rate holds for borrowing
as well as for lenders. This is called a perfect bank in the book.
It is a fact that if we want to borrow money from a bank, then the
interest rate depends on the time at which we should pay back our
money.
In this lecture and the next we will look at implications if we do not
assume a constant rate for all times.
Lecture 5
2 / 14
Introduction
So far we have assumed that regardless of the time at which a cash flow
occurs we can use the same interest rate.
This is a strong assumption for two reasons:
We assume that the same constant interest rate holds for borrowing
as well as for lenders. This is called a perfect bank in the book.
It is a fact that if we want to borrow money from a bank, then the
interest rate depends on the time at which we should pay back our
money.
In this lecture and the next we will look at implications if we do not
assume a constant rate for all times.
We will still assume that the lending rate is equal to the borrowing rate.
Lecture 5
2 / 14
Spot rates (1)
Definition
The yearly spot rate st is the interest rate we get if we invest our
money for t years:
(1 + st )t .
Here we assume that t = 1, 2, . . ..
Lecture 5
3 / 14
Spot rates (1)
Definition
The yearly spot rate st is the interest rate we get if we invest our
money for t years:
(1 + st )t .
Here we assume that t = 1, 2, . . ..
The m periods per year spot rate is:
st mt
.
1+
m
Here mt has to be an integer.
Lecture 5
3 / 14
Spot rates (1)
Definition
The yearly spot rate st is the interest rate we get if we invest our
money for t years:
(1 + st )t .
Here we assume that t = 1, 2, . . ..
The m periods per year spot rate is:
st mt
.
1+
m
Here mt has to be an integer.
The continuous spot rate is:
e st ·t .
Here t ∈ R+
Lecture 5
3 / 14
Spot rates (2)
Definition
The yearly discount factor is
dt =
1
.
(1 + st )t
Here t = 1, 2, . . . .
Lecture 5
4 / 14
Spot rates (2)
Definition
The yearly discount factor is
dt =
1
.
(1 + st )t
Here t = 1, 2, . . . .
The m periods per year discount factor is
dt =
1
1+
.
st mt
m
Here mt has to be an integer.
Lecture 5
4 / 14
Spot rates (2)
Definition
The yearly discount factor is
dt =
1
.
(1 + st )t
Here t = 1, 2, . . . .
The m periods per year discount factor is
dt =
1
1+
.
st mt
m
Here mt has to be an integer.
The continuous discount factor is
dt = e −st ·t .
Here t ∈ R+
Lecture 5
4 / 14
The term structure of interest rates (1)
Definition
The function
t 7→ st
is called the term structure of interest rate (TSIR).
Lecture 5
5 / 14
The term structure of interest rates (1)
Definition
The function
t 7→ st
is called the term structure of interest rate (TSIR).
In real life this curve is typically increasing but it can also be decreasing or
hump-shaped.
Lecture 5
5 / 14
The term structure of interest rates (2)
How do we calculate the TSIR?
Lecture 5
6 / 14
The term structure of interest rates (2)
How do we calculate the TSIR?
The key to solve this problem is to note that the price Pt of a ZCB with
face value F and maturity in t years from now is equal to
Pt = F · dt =
F
.
(1 + st )t
Lecture 5
6 / 14
The term structure of interest rates (2)
How do we calculate the TSIR?
The key to solve this problem is to note that the price Pt of a ZCB with
face value F and maturity in t years from now is equal to
Pt = F · dt =
F
.
(1 + st )t
We can solve this to get
st = (F /Pt )1/t − 1.
Lecture 5
6 / 14
The term structure of interest rates (3)
Example
Assume that the following ZCBs exists in a market:
Time to maturity
1
2
5
Face value
100 000
50 000
125 000
Lecture 5
Price
99 065
48 652
107 267
7 / 14
The term structure of interest rates (3)
Example
Assume that the following ZCBs exists in a market:
Time to maturity
1
2
5
Face value
100 000
50 000
125 000
Price
99 065
48 652
107 267
Using these ZCBs we can calculate the spot rates
s1 = (100 000/99 065)1/1 − 1 = 0.994%
Lecture 5
7 / 14
The term structure of interest rates (3)
Example
Assume that the following ZCBs exists in a market:
Time to maturity
1
2
5
Face value
100 000
50 000
125 000
Price
99 065
48 652
107 267
Using these ZCBs we can calculate the spot rates
s1 = (100 000/99 065)1/1 − 1 = 0.994%
s2 = (50 000/48 652)1/2 − 1 = 1.38%
Lecture 5
7 / 14
The term structure of interest rates (3)
Example
Assume that the following ZCBs exists in a market:
Time to maturity
1
2
5
Face value
100 000
50 000
125 000
Price
99 065
48 652
107 267
Using these ZCBs we can calculate the spot rates
s1 = (100 000/99 065)1/1 − 1 = 0.994%
s2 = (50 000/48 652)1/2 − 1 = 1.38%
s5 = (125 000/107 267)1/5 − 1 = 3.11%
Lecture 5
7 / 14
The term structure of interest rates (4)
Sometimes we want to use coupon bonds to calculate the TSIR.
Lecture 5
8 / 14
The term structure of interest rates (4)
Sometimes we want to use coupon bonds to calculate the TSIR.
We can then use the relationship (here we assume yearly paid coupons)
n
P=
X
F
C
+
.
(1 + sn )n
(1 + sk )k
k=1
Lecture 5
8 / 14
The term structure of interest rates (4)
Sometimes we want to use coupon bonds to calculate the TSIR.
We can then use the relationship (here we assume yearly paid coupons)
n
P=
X
F
C
+
.
(1 + sn )n
(1 + sk )k
k=1
Note that while a bond’s yield is unique to that bond,
n
P=
X
F
C
+
,
n
(1 + λ)
(1 + λ)k
k=1
the discount factors calculated from the spot rate curve are universal in
the sense that they can be applied to any cash flows.
Lecture 5
8 / 14
The term structure of interest rates (5)
Example
Assume that we have the following two bonds:
Time to maturity
1
2
Face value
100 000
50 000
Lecture 5
Coupon rate
0%
3%
Price
99 065
48 652
9 / 14
The term structure of interest rates (5)
Example
Assume that we have the following two bonds:
Time to maturity
1
2
Face value
100 000
50 000
Coupon rate
0%
3%
Price
99 065
48 652
As in the previous example we get s1 = 0.994%.
Lecture 5
9 / 14
The term structure of interest rates (5)
Example
Assume that we have the following two bonds:
Time to maturity
1
2
Face value
100 000
50 000
Coupon rate
0%
3%
Price
99 065
48 652
As in the previous example we get s1 = 0.994%.To find s2 in this case we
use the fact that
F + RF
RF
+
P2 =
1 + s1 (1 + s2 )2
Lecture 5
9 / 14
The term structure of interest rates (5)
Example
Assume that we have the following two bonds:
Time to maturity
1
2
Face value
100 000
50 000
Coupon rate
0%
3%
Price
99 065
48 652
As in the previous example we get s1 = 0.994%.To find s2 in this case we
use the fact that
F + RF
RF
+
P2 =
1 + s1 (1 + s2 )2
or
0.03 · 50 000 1.03 · 50 000
.
48 652 =
+
1.00994
(1 + s2 )2
Lecture 5
9 / 14
The term structure of interest rates (5)
Example
Assume that we have the following two bonds:
Time to maturity
1
2
Face value
100 000
50 000
Coupon rate
0%
3%
Price
99 065
48 652
As in the previous example we get s1 = 0.994%.To find s2 in this case we
use the fact that
F + RF
RF
+
P2 =
1 + s1 (1 + s2 )2
or
0.03 · 50 000 1.03 · 50 000
.
48 652 =
+
1.00994
(1 + s2 )2
we get
s2 = 4.49%
Lecture 5
9 / 14
The term structure of interest rates (6)
Let us return to the first example.
Lecture 5
10 / 14
The term structure of interest rates (6)
Let us return to the first example.
How do we determine st for t = 3, 4, 6, 7, . . . in this case?
Lecture 5
10 / 14
The term structure of interest rates (6)
Let us return to the first example.
How do we determine st for t = 3, 4, 6, 7, . . . in this case?
If we do not have access to any more bonds, then we need to interpolate
and extrapolate using the given bonds.
Lecture 5
10 / 14
The term structure of interest rates (6)
Let us return to the first example.
How do we determine st for t = 3, 4, 6, 7, . . . in this case?
If we do not have access to any more bonds, then we need to interpolate
and extrapolate using the given bonds.
In general we generate a curve for every t ∈ R+ instead of only finding
suitable rates for every year.
Lecture 5
10 / 14
The term structure of interest rate (7)
Suggested solutions to the problem include
Use linear interpolation.
This usually works fine, but we get problems when we try to
extrapolate. There could also be a problem with the kinks that
appear in linear interpolation.
Lecture 5
11 / 14
The term structure of interest rate (7)
Suggested solutions to the problem include
Use linear interpolation.
This usually works fine, but we get problems when we try to
extrapolate. There could also be a problem with the kinks that
appear in linear interpolation.
Use some more elaborate interpolation method such as splines.
Lecture 5
11 / 14
The term structure of interest rate (7)
Suggested solutions to the problem include
Use linear interpolation.
This usually works fine, but we get problems when we try to
extrapolate. There could also be a problem with the kinks that
appear in linear interpolation.
Use some more elaborate interpolation method such as splines.
In this way we can get a smooth term structure curve. One problem
with using splines is that we do not have a guarantee that the curve
will be increaing even if the interest rates we start with is an
increasing sequence.
Lecture 5
11 / 14
The term structure of interest rate (8)
Use a functional form and find the curve in the family of curves
that best fit observed data.
Lecture 5
12 / 14
The term structure of interest rate (8)
Use a functional form and find the curve in the family of curves
that best fit observed data.
In general, if the observed spot rates are denoted s̄t and the
functional form s(t, θ) for some parameter vector θ, then we want to
solve the problem
n
X
min
(s(t; θ) − s̄t )2 .
θ
t=1
Lecture 5
12 / 14
The term structure of interest rate (8)
Use a functional form and find the curve in the family of curves
that best fit observed data.
In general, if the observed spot rates are denoted s̄t and the
functional form s(t, θ) for some parameter vector θ, then we want to
solve the problem
n
X
min
(s(t; θ) − s̄t )2 .
θ
t=1
Problems with this approach are that the minimization problem need
not be well defined and that the curve need not pass through the
observed spot rates.
Lecture 5
12 / 14
The term structure of interest rate (9)
When we create a TSIR we must use bonds with the same credit rating.
We also need to use bonds from the same country.
Lecture 5
13 / 14
The term structure of interest rate (9)
When we create a TSIR we must use bonds with the same credit rating.
We also need to use bonds from the same country.
This means that every country has a number of TSIRs, one for each credit
rating class.
Lecture 5
13 / 14
The term structure of interest rate (9)
When we create a TSIR we must use bonds with the same credit rating.
We also need to use bonds from the same country.
This means that every country has a number of TSIRs, one for each credit
rating class.
If the issuer of a bond can no longer pay the coupons and/or the face
value of a bond, then the bond is said to default.
Lecture 5
13 / 14
The term structure of interest rates (10)
In order to help investers, there are rating organisations that put a credit
rating on bonds.
Lecture 5
14 / 14
The term structure of interest rates (10)
In order to help investers, there are rating organisations that put a credit
rating on bonds.
A government bond issued by a large and stable country (e.g. the US or
the UK) is considered to be non-defaultable.
Lecture 5
14 / 14
The term structure of interest rates (10)
In order to help investers, there are rating organisations that put a credit
rating on bonds.
A government bond issued by a large and stable country (e.g. the US or
the UK) is considered to be non-defaultable.
Large well run companies have a low probability of deafulting on their
bonds, and get a high rating, while companies with a more uncertain
future get a lower grading.
Lecture 5
14 / 14
The term structure of interest rates (10)
In order to help investers, there are rating organisations that put a credit
rating on bonds.
A government bond issued by a large and stable country (e.g. the US or
the UK) is considered to be non-defaultable.
Large well run companies have a low probability of deafulting on their
bonds, and get a high rating, while companies with a more uncertain
future get a lower grading.
Bonds that have a low probaiblity of default are called investment grade,
while bonds that have a higher risk of default are called junk bonds.
Lecture 5
14 / 14