MATH4210 Financial Mathematics (2015-2016)
Tutorial 4
Different types of interest rates
1. n-year interest rate
2. Bond yield
3. Forward rate
n-year Interest Rate
The interest rate on an investment that is made for a period of time starting today and lasting for n years.
Question: How can we find the n-year interest rate?
Simple case: No coupon
ln V
n
where V is the current price of a zero-coupon bond to be mature exactly n years from now.
rn =
Coupon-bearing case
Boostrap method
Face Value
100
100
100
100
100
1 year: r1
2 years: r2
3 years: r3
4 years: r4
5 years: r5
Time to Maturity
1
2
3
4
5
100 ∗ e−r1 = 97,
Annual Coupon
0
2
5
8
10
Bond Price
97
95
95
98
100
r1 = ln(100/97) = 3.05%
2 ∗ e−r1 + 102 ∗ e2∗r2 = 95,
r2 = 4.59%
5e−r1 + 5e−2r2 + 105e−3r3 = 95,
r3 = 6.81%
8e−r1 + 8e−2r2 + 8e−3r3 + 108e−4r4 = 98,
r4 = 8.65%
10e−r1 + 10e−2r2 + 10e−3r3 + 10e−4r4 + 110e−5r5 = 100,
1
r5 = 10.23%
Bond Yield
It is also called the yield-to-maturity. It is the percentage rate of return paid on the bond if it is bought and held
to its maturity date. More precisely, for a bond, it is the rate R that, when used to discount all future payouts
from the bond to their present value, will yield the present price of the bond. Once we know the present market
value V (t) for the bond at the present time t, R can be solved from
X
V (t) =
Π(Ti )e−R(Ti −t) ,
i
where Π(Ti ) is the payoff size to occur at some future time Ti > t.
Face Value
100
Time to Maturity
5
Annual Coupon
10
Bond Price
100
For the 5-year bond in the table, we calculate the bond yield R by
10e−R + 10e−2R + 10e−3R + 10e−4R + 110e−5R = 100,
R = 9.53%
Note that this is a nonlinear equation, we need to use some numerical methods (such as Bisection method or
Newton’s method) to find the approximate solution.
Forward Interest Rate
It is the interest rates implied by zero rates for periods of time which begins in the future and lasts for n years.
Notice that the forward rate is determined at current time.
Face Value
100
100
100
100
100
Time to Maturity
1
2
3
4
5
Annual Coupon
0
2
5
8
10
Bond Price
97
95
95
98
100
r(1, 2):
er1 ∗ er(1,2) = e2∗r2 ,
r(1, 2) = 2 ∗ r2 − r1 = 6.13%
r(2, 3):
e2∗r2 ∗ er(2,3) = e3∗r3 ,
r(2, 3) = 3 ∗ r3 − 2 ∗ r2 = 11.27%
e3∗r3 ∗ er(3,4) = e4∗r4 ,
r(3, 4) = 4 ∗ r4 − 3 ∗ r3 = 14.15%
e4∗r4 ∗ er(4,5) = e5∗r5 ,
r(4, 5) = 5 ∗ r5 − 4 ∗ r4 = 16.57%
r(3, 4):
r(4, 5):
If we change r5 from 10.23% to 11.23%, then the forward interest rate r(4, 5) will change from 16.57% to 21.57%.
Small change in n-year interest rate will cause big changes in forward rate. Forward interest rate is sensitive to
n-year interest rate.
Example 1
The following table gives the prices of four bonds:
Face Value($)
100
100
100
100
Time to Maturity(years)
0.5
1
1.5
2
Annual Coupon*($)
0
0
6.2
8
Bond Price($)
98
95
101
104
∗ Half the stated coupon is assumed to be paid every 6 months.
1. Calculate the zero rates for maturities of 6-month, 12-month, 18-month and 24-month.
2
2. What is the yield-to-maturity (in percentage) of the fourth bond?
Answer:
1. Let r1 , r2 , r3 and r4 be the zero rates for maturities of 6-month, 12-month, 18-month and 24-month respectively.
100e−r1 ×0.5 = 98,
100e−r2 ×1 = 95,
r1 = 4.04%
r2 = 5.13%
3.1er1 ×0.5 + 3.1e−r2 ×1 + 103.1e−r3 ×1.5 = 101,
4e
−r1 ×0.5
+ 4e
−r2 ×1
+ 4e
−r3 ×1.5
+ 104e
−r4 ×2
r3 = 5.44%
= 104,
r4 = 5.81%
2. Let R be the yield to maturity, then
4e−R×0.5 + 4e−R×1 + 4e−R×1.5 + 104e−R×2 = 104
Let f (R) = 4e−R×0.5 + 4e−R×1 + 4e−R×1.5 + 104e−R×2 − 104, and use the bisection method to solve the
nonlinear equation f (R) = 0. We set the initial R0 ∈ [5%, 6%].
f (5%) = 1.52 > 0,
f (6%) = −0.46 < 0,
R ∈ (5%, 6%),
f (5.5%) = 0.53 > 0,
R ∈ (5.5%, 6%)
Keep repeating this process, we are able to get an approximate yield-to-maturity 5.77%.
Put-call Parity
c(S(t), E, τ, r) − p(S(t), E, τ, r) = S(t) − Ee−rτ ,
where τ = T − t
Note that the call option and the put option are written on the same underlying S(t), have the same strike price E
and the same expiry date T .
Example 2
Given the bounds for the European call options:
max(S(t) − Ee−rτ , 0) ≤ c(S(t), E, τ, r) ≤ S(t),
prove the bounds for the European put options:
max(Ee−rτ − S(t), 0) ≤ p(S(t), E, τ, r) ≤ Ee−rτ
Answer: From the given inequality and the Put-call parity, we have
Ee−rτ − S(t) = p(S(t), E, τ, r) − c(S(t), E, τ, r)
Therefore, we have
Ee−rτ − S(t) ≥ p(S(t), E, τ, r) − S(t) ⇒ Ee−rτ ≥ p(S(t), E, τ, r)
Similarly, we have
Ee−rτ − S(t) ≤ p(S(t), E, τ, r) − max(S(t) − Ee−rτ , 0) ⇒ max(0, Ee−rτ − S(t)) ≤ p(S(t), E, τ, r)
3
Example 3
Given
p(S(t), E, τ1 , r) ≤ p(S(t), E, τ2 , r)
where the time to maturity τ1 < τ2 . Prove that for European call options c(S(t), E, τ, r) on non-dividend-paying
stocks, they also become more valuable as the time to expiration increases,
c(S(t), E, τ1 , r) ≤ c(S(t), E, τ2 , r).
Answer:
We have
c(S(t), E, τ1 , r) − p(S(t), E, τ1 , r) = S(t) − Ee−rτ2
and
c(S(t), E, τ2 , r) − p(S(t), E, τ2 , r) = S(t) − Ee−rτ2
Then we have
c(S(t), E, τ2 , r) − c(S(t), E, τ1 , r)
=S(t) − Ee−rτ2 + p(S(t), E, τ2 , r) − (S(t) − Ee−rτ2 + p(S(t), E, τ1 , r))
=p(S(t), E, τ2 , r) − p(S(t), E, τ1 , r) + Ee−rτ2 − Ee−rτ1 ≤ 0
Example 4
A European gap call option pays off S(T ) − E1 when S(T ) > E2 , or nothing when S(T ) ≤ E2 at maturity T . A
European gap put option pays off E1 − S(T ) when S(T ) ≤ E2 , or nothing when S(T ) > E2 at maturity T . Here
E1 is known as the strike price and E2 the trigger price.
Let cgap (t) and pgap (t) denote the price of a European gap call option and a European gap put option respectively. Suppose two gap options have the same maturity T and the underlying stock pays no dividend, prove the
following put-call parity for European gap options:
−r(T −t)
S(t) + pgap (t) − cgap (t) = E1
,
t≤T
Answer: Recall the Proposition 3 in Chapter 3: Let Π1 (t) and Π2 (t) be two portfolios such that the combinations
of products within the portfolios cannot be changed between now and the expiry time T . If Π1 (T ) = Π2 (T ), then
we have Π1 (t) = Π2 (t) for all t ≤ T . Therefore, we only need to prove the parity for t = T . That is, if we have
Π1 (t) = S(t) + pgap (t) − cgap (t)
and
Π2 (t) = E1 e−r(T −t)
be the two portfolios, we only need to show that Π1 (T ) = Π2 (T ). Since
E1 − S(T ), if S(T ) ≤ E2 ,
0,
if S(T ) ≤ E2
Π1 (T ) = S(T ) +
−
= E1
0,
if S(T ) > E2 .
S(T ) − E1 , if S(T ) > E2
and Π2 (T ) = E1 e−r(T −t) er(T −t) = E1 , we have Π1 (T ) = Π2 (T ) and therefore we can conclude that Π1 (t) = Π2 (t)
for all t ≤ T .
4