Interest Rates
Outline
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Types of rates
Measuring interest rates
Zero rates
Bond pricing
Determining Treasury zero rates
Forward rates
Forward rate agreements
Duration
Theories of the term structure of interest rates
Summary
Types of Rates(1/3)
(1)
Treasury Rates(公債利率)

A government to borrow in its own currency.

An investor earns on Treasury bills and Treasury bonds.

It is usually assumed “risk-free rates”.
American Treasury Rates
Types of Rates(2/3)
(2) LIBOR(倫敦銀行間放款利率)

The London Interbank Offered Rate is a daily reference
rate based on the interest rates at which banks borrow funds
from other banks in the London wholesale money market .

LIBOR is short term .

Derivatives traders regard LIBOR rates as a better indication
of the “true” risk-free rate than Treasury rates .
Large banks also quote LIBID rates ( London Interbank Bid
Rate ) and is the rate at which they will accept deposits from
other banks.

LIBOR
Types of Rates(3/3)
(3) Repo Rates(附買回利率)

A Repurchase agreement (also known as a repo or Sale
and Repurchase Agreement) allows a borrower to use a
financial security as collateral for a cash loan at a fixed rate
of interest.

The difference between the price is the interest it earns. We
called the interest rate is repo rate.

The most common type of repo is an overnight repo.
Measuring interest rates(1/4)


Suppose that an amount A is invested for n years at an interest rate
of R per annum.
If the rate is compounded m times per annum, the terminal value
of the investment is A (1+ R/m)mn
(1)
Compound times
Future value (R=10%)
Every year (m=1)
110.00
Half year (m=2)
110.25
Season
110.38
(m=4)
Monthly (m=12)
110.47
Weekly
110.51
(m=51)
Example
1. A pension fund manager invests $10 million in a debt
obligation that promises to pay 7.3% per year for four
years. What is the future value of the $10 million?
To determine the future value of any sum of money invested
today, we can use the future value equation, which is:
R mn
A (1 
)
m
0.073 4
10,000,000(1 
)  $13,255,584.66
1
Measuring interest rates(2/4)
With continuous compounding, it can be shown that
an amount A invested for n years at rate R grows
to
Ae
Rn
(2)
Compounding a sum of money at a continuously compounded rate R for n years
involves multiplying it by eRn. Discounting it at a continuously compounded
rate R for n years involves multiplying e-Rn.
Which of the following amounts is closest to the end value of investing $3000 for ¾ years at s continuously compounded rate of 12%?
3000  e
1 2%
3
4
 3283
Measuring interest rates(3/4)
Suppose that Rc is a rate of interest with continuous
compounding and Rm is the equivalent rate with
compounding m times per annum. From the results in
equations (1) and (2), we have
Rc
e = ( 1+ Rm/m )
mn
Rc = m ln(1+ Rm/m)
Rm=m (e
Rc/m
-1)
(3)
(4)
These equations can be used to convert a rate with a compounding frequency of m times
per annum to a continuously compounded rate and vice versa.
Measuring interest rates(4/4)
Rm mn
A(1  )  Ae Rc n
m
Rm mn
Rc n
e  (1  )
m
Rm
Rc n  mn ln(1  )
m
Rm
 Rc  m ln(1  )(3)
m
or
Rm
Rc n  mn ln(1 
)
m
Rc n
R
 ln(1  m )
mn
m
e
e
Rc
m
Rc
m
e
ln(1
Rm
)
m
Rm
 1
m
Rm  m(e
Rc
m
 1)(4)
Example
(1) Consider an interest rate is quoted as 10% per annum with semiannual
compounding . From equation (3) with m=2 and Rm=0.1 ,the equivalent
rate with continuous compounding is
2ln (1+ 0.1/2) = 0.09758
Or
9.758% per annum
(2)Suppose that a lender quotes the interest rate on loan as 8% per
annum with continuous compounding , and that interest is actually paid
quarterly . From equation (1) with m=4 and Rc=0.08 , the equivalent
rate with quarterly compounding is
4(e0.08/4 -1)=0.0808
Or 8.08 per annum. This means that on a $1,000 loan ,interest payments of
$20.20 would be required each quarter
Zero rates
Definition:
1. The n-year zero-coupon interest rate is the rate of
interest earned on an investment that starts today
and lasts for n years. All the interest and principal
is realized at the end of n years.
2. Sometimes also referred to as the n-year spot
rate.
Bond Pricing(1/4)

Suppose that a 2-year Treasury bond with a principal
of $100 provides coupons at the rate of 6% per annum
semiannually.
Maturity
(years)
0.5
Zero Rate
(% cont comp)
5.0
1.0
5.8
1.5
6.4
2.0
6.8
Bond Pricing(2/4)
(1)Theoretical price of the bond (理論價格)
3e
-0.05*0.5
＋ 3e
-0.058*1.0
＋ 3e
-0.064*1.5
＋ 103e
Maturity
(years)
0.5
Zero Rate
(% cont comp)
5.0
1.0
5.8
1.5
6.4
2.0
6.8
-0.068*2.0
＝98.39
Continuous
Compounding
Ae-Rn
Bond Pricing(3/4)
(2) Bond Yield (債券報酬率)
A bond’s yield is the single discount rate that, when
applied to all cash flows, gives a bond price equal to
its market price. Assumed: y is the yield on the bond.
3e
-y*0.5
＋ 3e
-y*1.0
＋3e
-y*1.5
＋ 103e
-y*2.0
y＝6.76% (trial and error )
＝98.39
Bond Pricing(4/4)
(3) Par Yield (面額報酬率)
The par yield for a certain bond maturity is the coupon
rate that causes the bond price to equal its par value.
Suppose that the coupon on a 2-year bond in our
example is c per annum.
c/2e-0.05*0.5＋c/2e-0.058*1.0＋c/2e-0.064*1.5＋(100+c/2)e-0.068*2.0＝100
c＝6.87
The 2-year par yield is 6.87% per annum with
semiannual compounding.
Determining Treasury zero rates(1/5)

The most popular approach is known as the
bootstrap method(拔靴法or 導引法).
Bond
Principal
(dollars)
Time to
Maturity
(years)
Annual
Coupon
(dollars)
Bond Cash
Price
(dollars)
100
0.25
0
97.5
100
0.50
0
94.9
100
1.00
0
90.0
100
1.50
8
96.0
100
2.00
12
101.6
＊Half the stated coupon is assumed to be paid every 6 months.
Determining Treasury zero rates(2/5)
The bootstrap method

The 3-month bond provides a return of 2.5 in 3 months on an initial
investment of 97.5.

Six 3-month
months zero rate is (4×2.5)/97.5＝
With quarterly compounding, •the
•(2x5.1)/94.9=10.748%
10.256％(per annum).

2 ln （1＋0.10748/2）＝0.10469
One year
The rate is expressed with continuous compounding, it becomes 4 ln
ln （1＋10/90）＝0.10536
（1＋0.10256/4）＝0.10127

Similarly the 6 month and 1 year rates are 10.469％ and 10.536％ with
continuous compounding.
Determining Treasury zero rates(3/5)
The bootstrap method


To calculate the 1.5 year zero rate, the payments are as follows:
6 months : ＄4 ; 1 year:＄4 ; 1.5 years :＄104
4e-0.10469*0.5 ＋ 4e-0.10536*1.0 ＋ 104e-y*1.5 ＝96
Suppose
the 1.5-year zero rate is denoted by R.
e-y*1.5 =0.85196
R=-ln(0.85196)/1.5
= 0.10681
6e-0.10469*0.5 ＋
6e-0.10536*1.0 ＋ 6e-0.10681*1.5 ＋ 106e-R*2.0 ＝
R＝10.681％
101.6
e-R*2.0 =0.80561
 Similarly the two-year rate is 10.808％
R=-ln(0.80561)/2.0 = 0.10808
Determining Treasury zero rates(4/5)
Zero Curve Calculated from the Data
(Figure 1)
12
Zero
Rate (%)
11
10.469
10
10.127
10.53
6
10.68
1
10.808
Maturity (yrs)
9
0
0.5
1
1.5
2
2.5
Determining Treasury zero rates(5/5)

A common assumption is that the zero curve is linear
between the points determined using the bootstrap
method.

The zero curve is horizontal prior to the first point and
horizontal beyond the last point.

By using longer maturity bonds, the zero curve would be
more accurately determined beyond 2 years.
Forward rates(1/5)
Definition: the rates of interest implied by current
zero rates for periods of time in the future.
Forward rates(2/5)
Calculation of Forward Rates (Table 5)
Zero Rate for
Forward Rate
an n-year Investment for n th Year
Year (n)
(% per annum)
(% per annum)
1
3.0
2
4.0
5.0
3
4.6
5.8
4
5.0
6.2
5
5.3
6.5
Forward rates(3/5)
Calculation of Forward Rates (Table 5)
The 4％ per annum rate for 2 years mean that, in return for an
investment of ＄100 today, the receive
100e0.04*2＝＄108.33 .
Suppose that ＄100 is invested. A rate of 3％ for the first year
and 5％ for the second year gives at the end of the second year.
100e0.03*1e0.05*1＝＄108.33
When interest rates are continuously compounded and in
successive time periods are combined, the overall equivalent
rate is simply the average rate during the whole period.
Forward rates(4/5)
Formula of Forward Rates


The forward rate for year 3 is the rate of interest that is implied
by a 4％ per annum 2-year zero rate and a 4.6％ per annum 3year zero rate.
If R1and R2 are the zero rates for maturities T1 and
T2 ,respectively, and RF is the forward interest rate for the
period of time between T1 and T2 ,then
RF
R2T2  R1T1

T2  T1
RF  R2  ( R 2  R1 )
T1
T2  T1
(5)
(6)
Example
Find the forward rate 3f4
Rf= R4 + (R4-R3) x T3 / (T4-T3)
= 5% + 0.4% x 3
= 6.2%
(1+ 0f4 )4 = (1+ 0f3)3 x (1+ 3f4 )
(1+5%)4 = (1+4.6%)3 x (1+ 3f4 )
= 0.0620920 ≒ 0.062
Forward rate agreements(1/4)

Definition: FRA is an over-the-counter agreement
that a certain interest rate will apply to either
borrowing or lending a certain principal during a
specified future period of time.
Forward rate agreements(2/4)
Notation Define
Rk: The rate of interest agreed to in the FRA
RF: The forward LIBOR interest rate for the period
between times T1 and T2 calculated today.
RM: The actual LIBOR interest rate observed in the
market at times T1 for the period between times
T1 and T2
L: The principal underlying the contract.
Forward rate agreements(3/4)
Cash flow formula
If X company could earn RM from the LIBOR loan ,The FRA
means that it will earn RK. The extra interest rate (which may be
negative) that is earns as a result of entering into the FRA is RKRM. The interest rate set at time T1 and paid at time T2. The
extra interest rate therefore leads to a cash flow to company at
time T2 of
C  L( RK  RM )(T2  T1 )
Similarly there is a cash flow to the Y company at time T2
of
C  L( RM  RK )(T2  T1 )
Example
Suppose that a company enters into an FRA that specifies it will
receive a fixed rate of 4% on a principal of $1 -million for a 3-month
period starting in 3 years . If 3-month LIBOR proves to be 4.5% for
the 3-month period the cash flow to the lender will be
1,000,000  (0.04  0.045)  0.25  $1,250
at the 3.25-year point. This is equivalent to a cash flow of

1,250
 $1,236.09
1  0.045  0.25
at the 3-year point . The cash flow to the party on the opposite
side of the transaction will be +$1,250 at the 3.25-year point or
+1,236.09 at the 3-year point (All the example interest rates in
this example are expressed with quarterly compounding . )
Forward rate agreements(4/4)
Valuation Formulas
Value of FRA where a fixed rate RK will be
received on a principal L between times T1 and
T2 is
V  L( RK  RF )(T2  T1 )e  R2T2
Value of FRA where a fixed rate is paid is
V  L( RF  RK )(T2  T1 )e R2T2
RF is the forward rate for the period and R2 is the
zero rate for maturity T2
Example
Suppose the LIBOR zero and forward rates are as in Table 4.5 .
Consider an FRA where we will receive a rate of 6% ,measured with
annual compounding , on a principal of $1-millon between the end
of year 1 and the end of year 2. In this case , the forward rate is 5%
with continuous compounding or 5.127% with annual compounding .
The value of the FRA is
1,000,000  (0.06  0.05127)e 0.042  $8,058
We see that an FRA can be valued if we
1.Caculate the payoff on the assumption that forward rates are realized ,that
is on the assumption that RM=RF . (RM= The actual LIBOR interest rate RF
=The forward LIBOR interest )
2.Discount this payoff at the risk-free rate .
Duration(1/3)

Definition: A measure of how long on average the
holder of the bond has to wait before receiving
cash payment.

A zero-coupon bond that lasts n years has a
duration of n years.

A coupon-bearing bond lasting n years has a
duration of less than n years.
Duration(2/3)
• Duration of a bond that provides cash flow c i at time t i is
 ci e
D   ti 
i 1
 B
n
 yti



n
(12)
B   Ci e  yti
i 1
where B is its price and y is its yield (continuously
compounded) This leads to
B
  D y
B
Example

For the bond in table 4.6 , the bond price B is 94.213 and the duration D is
2.653 ,so that equation give
B  94.213  2.653y
 249.95y
When yield on the bond increase by 10 basis point (=0.1%) , it follows that
y  0.001 .The duration relationship predicts that B  249.95  0.001
= -0.250 ,so that the bond price goes down to 94.213-0.250=93.963 . How
accurate is this ? When the bond yield increase by 10 basis point to 12.1% , the
bond is
5e 0.1210.5  5e 0.1211.0  5e 0.1211.5  5e 0.1212.0
 5e 0.1212.5  105e 0.1213.0  93.963
Which is (to three decimal place) the same as that predicted by the duration
relationship .
Duration(3/3)
Modified Duration
The preceding analysis is based on the assumption
that y is expressed with continuous compounding. If
y is expressed with a compounding frequency of m
times per year, then
BD y
B  
1 y / m
A variable D*, defined by
D
D*  
1 y / m
Example
The bond in table 4.6 has a price of 94.213 and a duration of 2.653. The
yield , express with semiannual compounding is 12.3673%. The
modified duration , D* is given by
2.653
D* 
 2.4985
1  0.123673 / 2
B  94.213  2.4985y
 235.39y
When the yield (semiannually compound) on the bond increase by 10 basis
point (=0.1%) , we have y  0.001 The duration relationship predicts that
We expect B to be -235.39 x 0.001= -0.235 , so that the bond price goes
down to 94.213-0.235= 93.978. How accurate is this ? When the bond yield
increase by 10 basis point to 12.4673% , an exact calculation similar to that
previous example shows that the bond price becomes 93.978. This shows that
the modified duration calculation gives good accuracy .
Theories of the term structure of interest
rates

Expectations Theory: forward rates equal
expected future zero rates.

Market Segmentation: short, medium and long
rates determined independently of each other.

Liquidity Preference Theory: forward rates
higher than expected future zero rates.
Summary(1/2)

Treasury rates are the rates paid by a government on
borrowings in its own currency.

LIBOR rates are short-term lending rates offered by
banks in the interbank market.

The n-year zero or spot rate is the applicable to an
investment lasting for n years when all of the return
is realized at the end.

Forward rates are the rates applicable to future
periods of time implied by today’s zero rates.
Summary(2/2)

FRA is an over-the-counter agreement that a certain
interest rate will apply to either borrowing or
lending a certain principal at LIBOR during a
specified future period of time.

Duration measures the sensitivity of the value of a
bond portfolio to a small parallel shift in the zerocoupon yield curve.

Liquidity preference theory can be used to explain
the interest rate term structures that are observed in
practice.