Analytical Finance II
Term structure of interest rates
An Qi
910402-9013
Introduction
The plot of the yield to maturity against time is called the yield curve. For the moment,
assume that this has been calculated from zero-coupon bonds and that these bonds have
been issued by a perfectly creditworthy source. The shape of the yield curve can be
concave or convex. A number of theories are used to explain the shape of the term
structure. The most common are:
1.Expectation value theory
2.Market segment theory
3.Liquidity theory
The most common is the last, which use arguments that the forward rate always should be
above the expected zero coupon rates and that investors are dependent on the liquidity
and borrowers mostly want fixed rates.
Term structure models are based on the assumption that the whole term structure of
interest rates can be derived from the stochastic behaviour of one or many variables. The
reason for modelling the entire term structure is to make all model prices internally
consistent.
In categorising these models, two properties are significant:
1.Number of state variables
Most models lack analytical solutions and have to be solved using numerical
methods. The computing time increases dramatically for each new state variable.
2. External consistency.
By external consistency, we mean coherence between the model term structure
and the observed term structure. When the model is used to price derivative
instruments, it is essential that the underlying instrument is priced in accordance
with observed market prices.
The rates with a lifetime less than a year are given by so called Treasure-Bills. Rates with
lifetime between one and ten years are given by Treasury-Notes and rates with lifetimes
more than ten years by Treasury-Bonds. In parallel to the curve above, we have the
municipal and corporate bonds. Closest to the curve above, i.e., with lowest spread are
those curves with the best (highest) ranking.
Examples
Spot rate
We assume that all of the investors believe one year spot rate in the next 5 years following the
table below:
Years
Spot rate
1
6%
2
8%
3
9%
4
9.5%
5
9.5%
Now we calculate the present value of zero-coupon bonds . Again we have some assumptions ,
the face value of zero-coupon bonds are 100 SEK . So we have
Time to maturity
Present value
1 year
100/(1+6%)=94.340
2 years
100/[(1+8%)(1+6%)]=87.352
3 years
100/[(1+9%)(1+8%)(1+6%)]=80.139
4 years
100/[(1+9.5%)(1+9%)(1+8%)(1+6%)]=73.186
5 years
100/[(1+9.5%)(1+9.5%)(1+9%)(1+8%)(1+6%)]=66.837
Using the formula
P
n
N

1  ytm
T
i 1
C
(1  ytm) ti
And the present value above , we could calculate the yield-to-maturity
Since our bond is zero-coupon bond , so C=0 ,and we have
Time to maturity
Yield to maturity
1 year
6%
2 years
6.7%
3 years
7.66%
4 years
8.12%
5 years
8.39%
10%
8%
yield to 6%
maturity 4%
2%
0%
1
2
3
4
time to maturity
5
Forward rate
With the spot rate curve known, we can (as we have seen above) calculate the forward
rates as:


 1 r
t2

rt2forward

t1
 1  r spot
t1



spot t 2
t1




1
t 2 t1
1
So we could have
Time to maturity
Forward rate
1 year
6%
2 years
10.0377%
3 years
11.0279%
4 years
11.0138%
5 years
9.5%
12%
10%
8%
6%
4%
2%
0%
11.03% 11.01%
10.04%
9.50% 9.50%
9%
8%
7.66% 8.12% 8.39%
6.70%
6%
1
2
3
4
5
time to maturity
spot rate
yield to maturity
forward rate
Application
I find some data of Swedish government bonds and bills on the NasdaqOMX
We will use bootstrapping to find the zero coupon curve. Swedish bonds are quoted in
YTM with day-count conversion: 30/360. The coupon frequency is 1, i.e., there is one
coupon per year for bonds.
By using the data on the lecture notes, we have following rates
r(1m) = r(30d) = 1.25 %
r(2m) = r(60d) = 1.22 %
r(3m) = r(90d) = 1.30 %
r(4m) = r(120d) = 1.31 %
r(6m) = r(180d) = 1.49 %
We now start the bootstrap with the bond RGKB 1041. This bond have a coupon rate
c = 6.75 %, ytm = 2.415 % and maturity at T = 2014-5-5 . (Today is 2012-12-19.) I.e.,
Time to maturity=1y4m16d = 360 + 120 +16 = 496 days. Therefore, we also have a
coupon payment of 6.75 at 136 days from now. We start by calculate the present value of
this coupon. We do this by extrapolation using 120 and 90 days:
1.31  1.30
r (136d )  1.31 
(136  120)  1.3153%
120  90
The value of the coupon is therefore:
PV 
6.75
 6.71676
(1  0.013153)136 / 360
The price of the bond is given by the quoted yield:
P
6.75
106.75

 6.6894  103.2974  109.9868
136 / 360
(1  0.02415)
(1  0.02415)1136 / 360
This means that a zero coupon bond with maturity T=496 days from now, have the price
P=109.9868-6.71676=103.2700. This gives the zero-coupon rate at time t=496 days:
103.2700 
106.75
, we get r(496d)=2.4347%
1  r 1136 / 360
The bond RGKB 1047. This bond have a coupon rate c = 5 %, ytm = 3.219 % and
maturity at T = 2020-12-1 . (Today is 2012-12-19.) I.e., Time to maturity=7y11m11d =
360*7 + 330 +11 = 2861 days. Therefore, we also have a coupon payment of 5 at
2501,2141,1781,1421,1061,701,341 days from now.
We start by calculate the present value of the interpolated coupons (341, 701 days
from today):
1.49 - 1.3153
r (341d )  1.49 
（341 - 180） 2.1292%
180 - 136
2.4347  1.49
701  496  3.0476%
r (701d )  2.4347 
496  180
We continue by calculate the present value of the extrapolated coupons
(1061,1421,1781,2141,2501 days from today):
r(1061d)=2.4347+0.002989556*(1061-496)=4.1238%
r(1421d)=2.4347+0.002989556*(1421-496)=5.2000%
r(1781d)=2.4347+0.002989556*(1781-496)=6.2763%
r(2141d)=2.4347+0.002989556*(2141-496)=7.3525%
r(2501d)=2.4347+0.002989556*(2501-496)=8.4288%
The value of the coupons is therefore:
PV 
5
1  0.021292
341/ 360

5
1  0.030476
701/ 360

5
1  0.041238
1061/ 360
5
5
5


1781/360
2141/ 360
(1  0.062763)
(1  0.073525)
(1  0.084288) 2501/ 360
 4.9012  4.7161  4.4386  4.0933  3.6998  3.2789  2.8498
 27.9777


5
(1  0.052000)1421/ 360
The present value if the bond is:
8


1
1  0.03219  1
P
100  5
  112.5755
0.03219
(1  0.03219) 2861/ 360 

Therefore, a zero coupon bond with maturity at T = 2861 days from today have the
present value of P =112.5755-27.9777=84.5978. This gives the zero-coupon rate at t
= 2861 days as:
105
84.5978 
（1  r）2861/ 360
r(2861d)=2.7559%
Now we can have a graph like this:
9
8
7
6
5
4
3
2
1
0
time to maturity
Conclusion
There are many methods to calculate the term structure of interest rates and pricing
bonds. And there are some very interesting connections between interest rates and time
to maturity. There are mainly three assumptions about the shape of the term structure. We
also have found a way to determine the value of zero coupon rates and use them to get a
curve of zero coupon bond.
Reference
1. http://www.nasdaqomxnordic.com/bonds/sweden/ at 10:10 2012-12-19
2. lecture notes by Jan Röman (MDH 2012)
3. http://voices.yahoo.com/the-cox-ingersoll-ross-cir-interest-rate-model-practice-14055
07.html?cat=4 13:00 2012-12-15
4. http://en.wikipedia.org/wiki/Cox%E2%80%93Ingersoll%E2%80%93Ross_model
14:00 2012-12-15
5. http://www.yelinsky.com/blog/archives/261.html 11:00 2012-12-19