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MTH 210 - Homework 3 - Due Wednesday, September 21, at the start of class
Exercise 1 (2 marks for part (a), 1 mark for part (b)). Let r1 be interest rate with compounding frequency m1 , and let r2 be the interest rate with compounding frequency m2 .
(a) Use a no-arbitrage argument to prove that
m1 T m2 T
r2
r1
= 1+
for all times T > 0.
1+
m1
m2
"
#
m2 /m1
r2
(b) Show r1 = m1
1+
−1 .
m2
Exercise 2 (Each part 1 mark). Assume the interest rate with compounding twice per year
is 4.3% = 0.043.
(a) Find the continuously compounded rate r.
(b) Find the annually compounded rate rA .
Exercise 3 (2 marks). Use the replication theorem to show that if the interest rate with
compounding frequency m from time t to time T has constant value r, then
Z(t, T ) = (1 + r/m)−m(T −t) .
Exercise 4 (1 mark). Consider an asset that pays N at maturity 3 years from now. Suppose
the annually compounded interest rate is 3% and the present value is 300. Find N .
Exercise 5 (2 marks). Consider an annuity starting at time 0 that pays 1 each year for M
years. Assume the annually compounded zero rate is rA for all maturities T = 1, . . . , M .
In an example in class, we found that the value of this annuity at its starting time 0 is
V0 =
M
X
i=1
Z(0, i) =
1 − (1 + rA )−M
.
rA
Find the value of the annuity at time t, where 0 < t < 1.
Exercise 6 (Each part 1 marks). (a) Consider an annuity that pays 1 every quarter for M
years. In other words, the payment times are T = t + 41 , t + 42 , . . . , t + 4M
. Show that the
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value at present time t is
1 − (1 + r4 /4)−4M
Vt =
,
r4 /4
assuming the quarterly compounded interest rate has constant value r4 .
(b) A fixed rate bond with notional N , coupon c, start date T0 , maturity Tn , and term length
α is an asset that pays N at time Tn and coupon payments αN c at times Ti for i = 1, . . . , n,
where Ti+1 = Ti + α. It is equivalent to an annuity plus N ZCBs. Consider a fixed rate
bond with notional N and coupon c that starts now, matures M years from now, and has
quarterly coupon payments. Show that the value at present time t is
Vt =
cN 1 − (1 + r4 /4)−4M
·
+ N (1 + r4 /4)−4M ,
4
r4 /4
assuming the quarterly compounded interest rate has constant value r4 .
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Solutions.
1.
(a) Seeking a contradiction, assume
r1
1+
m1
m1 T
m2 T
r2
> 1+
for some T > 0.
m2
(∗)
Fix t < T . Consider the portfolio
A: At time t, deposit 1 at interest rate r1 with compounding frequency m1 and borrow
(deposit negative) 1 at interest rate r2 with compounding frequency m2 .
Then V A (t) = 1 − 1 = 0 and
m1 T m2 T
r2
r1
− 1+
> 0 with probability one.
V (T ) = 1 +
m1
m2
A
Therefore A is an arbitrage portfolio. This contradicts the no-arbitrage assumption. Therefore assumption (∗) must be false.
By a similar argument, we can prove
m1 T m2 T
r1
r2
1+
< 1+
for some T > 0
m1
m2
(∗∗)
must also be false.
Therefore
r1
1+
m1
m1 T
m2 T
r2
= 1+
for all T > 0.
m2
(b) We solve (∗ ∗ ∗) for r1 .
m1 T m2 T
r1
r2
1+
= 1+
m1
m2
m1 m2
r1
r2
1+
= 1+
m1
m2
m2 /m1
r1
r2
1+
= 1+
m1
m2
m2 /m1
r1
r2
= 1+
−1
m1
m2
"
#
m2 /m1
r2
r1 = m1
1+
−1
m2
(∗ ∗ ∗)
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2.
(a) use the following result from class.
Result. If the interest rate with compounding frequency m is rm and the interest rate with
continuous compounding is r, then
rm mT
= erT for all T > 0
m
rm (b) r = m ln 1 +
m
(c) rm = m er/m − 1
(a)
1+
rm 0.043
We get r = m ln 1 +
= 2 ln 1 +
= 0.04254427 . . ..
m
2
(b) We use the result of Exercise 1 with m2 = 2, r2 = 0.043, m1 = 1, rA = r1 .
#
#
"
"
m2 /m1
2/1
0.043
r2
− 1 = 0.04346225 . . .
− 1 = (1) 1 +
rA = r1 = m1
1+
m2
2
3.
Consider two portfolios. At time t, they are
A: 1 ZCB with maturity T
B: N = (1 + r/m)−m(T −t) cash
We have V A (T ) = 1 and V B (T ) = N (1 + r/m)m(T −t) = (1 + r/m)−m(T −t) (1 +
r/m)m(T −t) = 1.
Therefore V A (T ) = V B (T ) with probability one.
By replication, V A (t) = V B (t).
But V A (t) = Z(t, T ) and V B (t) = (1 + r/m)−m(T −t) .
Therefore Z(t, T ) = (1 + r/m)−m(T −t) .
4.
The asset is equivalent to N ZCBs with maturity T = t + 3, where t is the current time.
The current value of the asset is
300 = N · Z(t, T ) = N (1 + r)−(T −t) = N (1 + 0.03)−3 .
Solving for N
N = 300(1 + 0.03)3 = 327.8181
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5.
The annuity is equivalent to
• 1 ZCB with maturity T=1
• 1 ZCB with maturity T=2
.
• ..
• 1 ZCB with maturity T=M
The value of the annuity at time 0 < t < 1 is
Vt =
M
X
M
M
X
X
−(T −t)
t
(1 + rA )
= (1 + rA )
Z(t, T ) =
1
(1 + rA )T
T =1
T =1
T =1
The last sum is equal to
M
X
1
1 − (1 + rA )−M
=
.
T
(1
+
r
r
A)
A
T =1
(For the proof, see the geometric sum trick on page 6 of these solutions.) Therefore
Vt = (1 + rA )t
1 − (1 + rA )−M
.
rA
6.
(a) The annuity is equivalent to the following collection of ZCBs.
• 1 ZCB with maturity T = t +
1
4
• 1 ZCB with maturity T = t +
2
4
.
• ..
• 1 ZCB with maturity T = t +
i
4
.
• ..
• 1 ZCB with maturity T = t +
4M
4
5
By the result of Exercise 3, the value at time t of the ZCB maturing at T = t +
i
is
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Z(t, t + i/4) = (1 + r4 /4)−4(t+i/4−t) = (1 + r4 /4)−i .
Therefore the value at time t of the annuity is
Vt =
4M
X
Z(t, t + i/4) =
i=1
4M
X
−4(t+i/4−t)
(1 + r4 /4)
i=1
=
4M
X
(1 + r4 /4)−i .
i=1
The final sum can be evaluated by the usual geometric sum trick (see page 6 of these
solutions), and we get
Vt =
1 − (1 + r4 /4)−4M
.
r4 /4
(b) The bond is equivalent to αcN = 14 cN copies of the annuity from part (a) (or the
annuity in part (a) paying αcN = 41 cN times as much) plus N ZCBs. Therefore the value
of the bond at time t is
Vt =
cN 1 − (1 + r4 /4)−4M
·
+ N (1 + r4 /4)−4M
4
r4 /4
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The Geometric Sum Trick. We often need
N
X
i=1
1
1 − (1 + r)−N
=
(1 + r)i
r
(G)
In Exercise 5, we used this with r = ra and N = M . In Exercise 6(a), we used this with
r = r4 /4 and N = 4M .
Here is the proof of (G). Define
S=
N
X
i=1
1
.
(1 + r)i
Then
rS = (1 + r)S − S = (1 + r)
N
X
i=1
=
N
−1
X
i=0
1
−
(1 + r)i
N
X
i=1
N
X
1
1
−
i
(1 + r)
(1 + r)i
i=1
1
= 1 − (1 + r)−N .
(1 + r)i
In summary
rS = 1 − (1 + r)−N .
Divide both sides by r to obtain (G).
Comment about 6(a). It is not correct logic to substitute rA by r4 /4 and M by 4M in the
following result from class.
Result. Consider an annuity starting at time t that pays C each year for M years. Assume
the annually compounded zero rate is rA for all maturities T = t + 1, . . . , t + M . The value
at time t is
1 − (1 + rA )−M
.
Vt = C ·
rA
To convince yourself, try the following exercise.
Exercise. Consider an annuity that pays 1 every quarter for M years. In other words, the
payment times are T = t + 14 , t + 24 , . . . , t + 4M
. Show that the value at present time t is
4
Vt =
1 − (1 + r8 /8)−8M
,
(1 + r8 /8)2 − 1
assuming the interest rate with compounding 8 times per year has constant value r8 .
Notice that the formula in this exercise is not what you obtain by substituting rA by r8 /8
and M by 8M in the result above.
Solution.
The annuity is equivalent to the following collection of ZCBs.
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• 1 ZCB with maturity T = t +
1
4
• 1 ZCB with maturity T = t +
2
4
.
• ..
• 1 ZCB with maturity T = t +
i
4
.
• ..
• 1 ZCB with maturity T = t +
4M
4
By the result of Exercise 3, the value at time t of the ZCB maturing at T = t +
i
is
4
Z(t, t + i/4) = (1 + r8 /8)−8(t+i/4−t) = (1 + r8 /8)−2i = ((1 + r/8)2 )−i
Therefore the value at time t of the annuity is
Vt =
4M
X
Z(t, t + i/4) =
i=1
4M
X
((1 + r8 /8)2 )−i .
i=1
We use a variation of the geometric sum trick.
((1 + r8 /8)2 − 1)Vt = (1 + r8 /8)2 Vt − Vt
= (1 + r8 /8)
2
4M
X
2 −i
((1 + r8 /8) )
i=1
=
4M
X
−
((1 + r8 /8)2 )−i
i=1
4M
−1
X
((1 + r8 /8)2 )−i −
i=0
4M
X
((1 + r8 /8)2 )−i
i=1
2 −4M
= 1 − ((1 + r8 /8) )
= 1 − (1 + r8 /8)−8M .
In summary
((1 + r8 /8)2 − 1)Vt = 1 − (1 + r8 /8)−8M .
Divide by (1 + r8 /8)2 − 1 to obtain
Vt =
1 − (1 + r8 /8)−8M
.
(1 + r8 /8)2 − 1