¡Descarga Ejercicios de Renta Fija: Soluciones Detalladas y más Ejercicios en PDF de Finanzas solo en Docsity!
Fixed Income Markets & Securities
Solutions to Problem Set 2 Prof. Aurelio Vasquez
- Synthetic Bonds, Arbitrage, and Pricing Replicate bond C with bonds A and B, matching their cash ows:
Cash Flow Year 1: 4 NA + 6NB = 12 (1) Cash Flow Year 2: 4 NA + 6NB = 12 (2) Cash Flow Year 3: 104 NA + 106NB = 112 (3)
Equation (1) is redundant. Solving (2) and (3):
NA = 3
NB = 4
So selling 3 bond A and buying 4 bond B gives the same cash ows of bond C. The cost of this investment is 3(94) + 4(100) = 118 > 112 the cost of bond C. To earn an arbitrage pro t of $6 you buy a unit of C, buy 3 units of A and sell 4 units of B.
- Synthetic Bonds, Arbitrage, Yield to Maturity, and Pricing (i) A: 10 d 1 + 10 d 2 + 10 d 3 + 110 d 4 = 127. B: 20 d 2 = 18. C: 30 d 2 + 30 d 3 + 10 d 4 = 55. D: 10 d 1 + 10 d 3 + 10 d 4 = 27.
From B: d 2 = 0: 94 From A-D: 9 :4 + 100d 4 = 99: 4! d 4 = 0: 90 From C: 30(0:94) + 30d 3 + 10(0:90) = 55: 80! d 3 = 0: 62 From D: 10 d 1 + 6:2 + 9:0 = 27: 80! d 1 = 1: 26 Negative interest!
Price of bond E = 50(1.26)+20(0.94)+20(0.62)+150(0.90)= 229.
(1 + y)
(1 + y)^2
(1 + y)^3
(1 + y)^4
y=1.
(ii)
Price of bond F = (^) (1:02828688)^1001 + (^) (1:02828688)^1002 + (^) (1:02828688)^4003 + (^) (1:02828688)^21004 = 2438
Synthetic Price = 100(1.26)+100(0.94)+400(0.62)+2100(0.90)=2358 < 2438
(di erence of $80)
Set the matrix equation:
2 (^66) 6 4
3 (^77) 7 5
2 (^66) 6 4
NA
NB
NC
ND
3 (^77) 7 5
2 (^66) 6 4
3 (^77) 7 5
Inverting the matrix and solving:
2 (^66) (^64)
NA
NB
NC
ND
3 (^77) 75 =
2 (^66) (^64)
3 (^77) (^75)
Cash Flow Today Buy 19 units of A at 127.20 -2416. Sell 19.50 units of B at 18.80 366. Buy 10 units of C at 55.80 -558. Sell 9 units of D at 27.80 250. Total -2,358. Sell 1 unit of F at 2438 2,438. Net Pro t 80.
To make $10 pro t, you do the same dividing by 8.
2 (^66) (^66) (^64)
NA
NB
NC
ND
NF
3 (^77) (^77) 75 =
2 (^66) (^66) (^64)
3 (^77) (^77) (^75)
Similarly, if we:
Short 1 bond A and long 2 bond Es, we can replicate bond C
Short 2 bond As and long 4 bond Es, we can replicate bond D
Hence, only bond positions A and E are needed to attempt to price positions F and G.
Attempt to price F:
Since we are only using two bonds for pricing, all we can do is to match two cash ows and hope the other three also match. We will try to match the rst two cash ows:
2 6 4
K A^1 K E^1
K A^2 K E^2
3 7 5
" NA NE
=
" K F^1 K F^2
)
2 6 4
3 7 5
" NA NE
=
" 169 157
Solving this system, we get NA = 3 and NE = 1.
This only matches the rst two cash ows for bond F. Do the other cash ows match?
K 3 K 4 K 5 3 bond A 144 198 330 -1 bond E -76 -90 - 68 108 252
It matches other three cash ows of F as well. Hence,
PF = NA PA + NE PE
Attempt to price G:
2 (^64)
K A^1 K E^1
K A^2 K E^2
3 (^75)
" NA NE
=
" K^1 G K^2 G
)
2 (^64)
3 (^75)
" NA NE
=
" 18 124
Solving this system, we get NA = 2 and NE = 3.
This only matches the rst two cash ows for bond G. Do the other cash ows match?
K 3 K 4 K 5 -2 bond A -96 -132 - 3 bond E 228 270 234 132 138 14
But K 3 of position G is 129. This does not match cash ow 3 for position G, but it does give an upper bound for the price of G:
PG < NA PA + NE PE = 2 309 + 3 325 = 357
Now is this bound the minimum bound? One way to check is to vary which two years you use to compare the weights in A and E to replicate the \closest" strategy to G. This method is time consuming. Another method is to simply solve the following linear constrained optimization prob- lem:
Min NA PA + NB PB + NC PC + ND PD + NE PE s.t. K^1 A NA + K^1 B NB + K^1 C NC + K D^1 ND + K E^1 NE K G^1 K^2 A NA + K^2 B NB + K^2 C NC + K D^2 ND + K E^2 NE K G^2 K^3 A NA + K^3 B NB + K^3 C NC + K D^3 ND + K E^3 NE K G^3 K^4 A NA + K^4 B NB + K^4 C NC + K D^4 ND + K E^4 NE K G^4 K^5 A NA + K^5 B NB + K^5 C NC + K D^5 ND + K E^5 NE K G^5
The constraints guarantee the strategy created always pays at least in each period as position G. Solving this using Excel's solver function gives the minimum price as $357, as we computed earlier.
- Arbitrage, Pricing, and Bond Yields
Let Pi (i=A,B, or C) be the current price of bond i with coupon Ki and dt be the price of a zero with maturity t. Then we can write:
PA = KA
X^26
t=
dt + 100 d 26 (1)
PB = KB
X^26
t=
dt + 100 d 26 (2)
where
P 26 t=1 dt^ is the price of an annuity with annual coupon $1 and maturity of 26 years.
(1) (2) ) (PA PB ) = (KA KB )
X^26
t=
dt
X^26
t=
dt =
PA
KA KB
PB
KA KB
Now divide by P and factor out (^) (1+cy)
c (1 + y)
1 +
(1 + y)
(1 + y)T^ ^1
(1 + y)T
Use the fact that PN i=0 a i (^) = 1 aN^ + 1 a for^ jaj^ <^1
c (1 + y)
" 1 ( (^) 1+^1 y )T 1 (^) 1+^1 y
(1 + y)T
= c (1 + y)
1 + y y
1
(1 + y)T
(1 + y)T 1
c y
(1 + y)T^
c y(1 + y)T y c y
y c y(1 + y)T^ ) y = c
(b) As in part (a), here we prove that the yield is the annual coupon divided by price.
P =
CF
(1 + y)
CF
(1 + y)^2
CF
(1 + y)t^
CF
(1 + y)
1 +
(1 + y)
(1 + y)^2
CF
(1 + y)
" 1 1 (^) 1+^1 y
CF
(1 + y)
1 + y y
CF
y ) y = CFP
- Synthetic Bonds, Arbitrage, and Pricing
There are four bonds involved in this problem. We can assume any three of these four bonds are correctly priced. Therefore, there are four scenarios to this problem: Scenario 1: Assume bond B, C, D are correctly priced. From bond D, we have 100 0 d 1 = 95: 23 ) 0 d 1 = 0: 9523 From bond C, we have 7 0 d 1 + 107 0 d 2 = 100: 12 ) 0 d 2 = 0: 8734 From bond B, we have 5 0 d 1 + 10 0 d 2 + 115 0 d 3 = 99: 89 ) 0 d 3 = 0: 7513
Therefore, the value of bond A is:
15 0 d 1 + 20 0 d 2 + 125 0 d 3 = 15(0:09523) + 20(0:8734) + 125(0:7513) = 125.
However, the actual price of A is only $124.67. A is underpriced. Therefore, we buy underpriced bond A and nance the purchase by selling the combination of bond B, C, and D to eliminate future net cash ows and lock in arbitrage pro t today.
2 6 4
3 7 5
2 6 4
NB
NC
ND
3 7 5 =
2 6 4
3 7 5
2 (^64)
NB
NC
ND
3 (^75) =
2 (^64)
3 (^75)
1 2 (^64)
3 (^75) =
2 (^64)
3 (^75)
Note: negative sign denotes shorting.
Scenario 2: Assume bond A, C, D are correctly priced.
From bond D, we have 100 0 d 1 = 95: 23 ) 0 d 1 = 0: 9523 From bond C, we have 7 0 d 1 + 107 0 d 2 = 100: 12 ) 0 d 2 = 0: 8734 From bond A, we have 15 0 d 1 + 20 0 d 2 + 125 0 d 3 = 124: 67 ) 0 d 3 = 0: 7433
Therefore, the value of bond B is:
5 0 d 1 + 10 0 d 2 + 115 0 d 3 = 5(0:09523) + 10(0:8734) + 115(0:7433) = 98.
However, the actual price of B is only $99.89. B is overpriced. Therefore, we sell overpriced bond B and nance the purchase by selling the combination of bond A, C, and D.
2 (^64)
3 (^75)
2 (^64)
NA
NC
ND
3 (^75) =
2 (^64)
3 (^75)
2 6 4
NA
NC
ND
3 7 5 =
2 6 4
3 7 5
1 2 6 4
3 7 5 =
2 6 4
3 7 5
Scenario 3: Assume bond A, B, C are correctly priced.
Note: We will use a slightly di erent method.
First, we want to replicate the cash ows of D.
2 (^64)
3 (^75)
2 (^64)
NA
NB
NC
3 (^75) =
2 (^64)
3 (^75)
K =
2 (^66) (^66) (^66) (^66) (^66) 4
3 (^77) (^77) (^77) (^77) (^77) 5
P =
" 84 : 0932 99 : 414 113 : 3659 178 : 4267 175 : 9827 107 : 324
K ^1 =
2 (^66) (^66) (^66) (^66) (^66) 4
3 (^77) (^77) (^77) (^77) (^77) 5
W =
2 (^66) (^66) (^66) (^64)
3 (^77) (^77) (^77) (^75)
K N = W ) N = K ^1 W
N =
2 (^66) (^66) (^66) (^64)
3 (^77) (^77) (^77) (^75)
We sell 2 : 31376 shares of bond 1 buy 8 : 20088 shares of bond 2 sell 2 : 96124 shares of bond 3 sell 0 : 7616 shares of bond 4 buy 0 : 002584 shares of bond 5 buy 0 : 004393 shares of bond 6
The price has to be the same as the value of the above portfolio. Otherwise, arbitrage opportunity will exist.
V = P N = 150.
(c) Now the fourth column of matrix K becomes a linear combination of column one and two. Hence, K becomes singular and noninvertible. Rank(K)=5 and we can not span the six-dimensional space. Market is not complete. We can not achieve our desired cash ows.
- Bond Yields and Pricing
(a) For bond A:
1 + yA
(1 + yA)^2 ) 110(1 + yA)^2 = 10(1 + yA) + 110 ) 110 + 220yA + 110y A^2 = 10 + 10yA + 110 ) 110 y^2 A + 210yA 10 = 0 ) 11 y A^2 + 21yA 1 = 0
) yA =
p 212 4(11)( 1) 2(11)
For bond B:
1 + yB
(1 + yB )^2
(1 + yB )^3
(1 + yB )^4
(1 + yB )^5 Solving numerically ) yB = 6%
Bond B has a higher yield. (b) In order to nd the true value of these two bonds, we have to nd the proper discount rate. We have the par yield: maturity (in years) par yield 1 3% 2 4% 3 5% 4 6% 5 7%
Therefore, we know r 1 = 0: 03
