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Fixed Income and Derivatives. Home Assignment 1. Problem 1.. Treasury Bond Prices on February 15, 2009. Price (% FV and ticks). Treasury Bond Prices on February 15, 2009. Problem 2.. Maturity (yrs). ZCBond 2. ZCBond 2. ZCBond 2. Portfolio's value



 

Natalia Remizova

Fixed Income and Derivatives

Home Assignment 1

Problem 1.

a) To convert bond prices into decimal figures we simply divide ticks by 32 and add to the numbers before hyphens, i. e. (4)+(5)/32, e. g. for the first price: 101+12, 75/32=101, 3984 (see column (6)).

b) Using bootstrap method we get:

· for r1: 101, 3984=(100+7, 874/2)/(1+ r1/2) => r1= 0, 0501

· for r2: 108, 9844=14, 25/2/(1+ r1/2) +(100+14, 25/2)/(1+ r2/2)2 => r2= 0, 0493

· for r3: 102, 1563=6, 375/2/(1+ r1/2) +6, 375/2/(1+ r2/2)2 +(100+6, 375/2)/(1+ r3/2)3 => r2= 0, 0486

· for r4: 102, 5664=6, 25/2/(1+ r1/2) +6, 25/2/(1+ r2/2)2 +6, 25/2/(1+ r3/2)3 +(100+6, 25/2)/(1+ r4/2)4 => r4= 0, 0489

· for r5: 100, 8438=5, 25/2/(1+ r1/2) +5, 25/2/(1+ r2/2)2 +5, 25/2/(1+ r3/2)3 +5, 25/2/(1+ r4/2)4 +(100+5, 25/2)/(1+ r5/2)5 => r5= 0, 0489

Treasury Bond Prices on February 15, 2009

  Coupon Maturity Maturity (periods)

Price (% FV and ticks)

Price (decimal figures) Coupon per period
  (1)

(2)

(3)

(4)

(5)

(6)

(7)

7, 875

15. 08. 2009

12, 75

101, 3984

3, 9375

14, 250

15. 02. 2010

31, 5

108, 9844

7, 125

6, 375

15. 08. 2010

102, 1563

3, 1875

6, 250

15. 02. 2011

18, 125

102, 5664

3, 125

5, 250

15. 08. 2011

100, 8438

2, 625

 

c) To find out whether there are any arbitrage opportunities we calculate no-arbitrage prices (column (6)).

· P1=(100+6, 6875)/(1+ r1/2)= 104, 0813

· P2=5, 375/(1+ r1/2)+ (100+5, 375/2)/(1+ r2/2)2 = 104, 0813

· P3=2, 875/2/(1+ r1/2) +2, 875/2/(1+ r2/2)2 +(100+2, 875/2)/(1+ r3/2)3= 102, 0069

· P4=6, 25/2/(1+ r1/2) +6, 25/2/(1+ r2/2)2 +6, 25/2/(1+ r3/2)3 +(100+6, 25/2)/(1+ r4/2)4= 111, 7471

 

Treasury Bond Prices on February 15, 2009

  Coupon Maturity Price Maturity (periods) Coupon per period No-arbitrage price Gain/Loss
  (1) (2) (3) (4) (5) (6) (7)

13, 375

15. 08. 2009

104, 0800

6, 6875

104, 0813

-0, 0013

10, 750

15. 02. 2011

110, 9380

5, 3750

111, 0409

-0, 1029

5, 750

15. 08. 2011

102, 0200

2, 8750

102, 0069

0, 0131

11, 125

15. 02. 2011

114, 3750

5, 5625

111, 7471

2, 6279

 


Problem 2.

a) Macaulay duration of zero-coupon bonds is equal to their maturity. Modified duration = Dmod=DMac/(1+r) for annual compounding. Pricei=100/(1+ri) (see columns (2) and (3)).

Maturity (yrs)

Price

Dmac

Dmod

(1)

(2)

(3)

(4)

ZCBond 1

96, 1538

0, 9804

ZCBond 2

88, 8996

2, 9412

ZCBond 3

67, 5564

9, 8039

 

To immunize the ZCBond 2 we need to use ZCBond 1 as a hedging instrument.

q=8, 8996*2, 9412/(96, 1538*0, 9804)=2, 7737

Thus we need to sell 2, 7737 units of ZCBond 1 to hedge against small parallel changes in the spot rates.

 

b) To construct a barbell portfolio we take ZCBond 1 (as the shortest maturity) and ZCBond 3 (as the longest maturity) with such weights that the duration of the portfolio would be equal to duration of ZCBond 2. Barbell portfolio would generate additional profits if long spot rates would change more significant than short rates.

To get the same duration of the portfolio as ZCBond 2, ZCBond 1 and ZCBond 3 should have the following weights:

q(short)=q(ZCBond 1)=(10-3)/(10-1)=7/9= 0, 7778

q(long)=q(ZCBond 3)=1-7/9=2/9= 0, 2222

c) First compute new returns for both scenarios

Bonds' returns

Twist scenario

Steeping scenario

ZCBond 1

0, 9569

0, 9662

ZCBond 2

0, 8890

0, 8890

ZCBond 3

0, 7089

0, 6439

 

Then compute bonds’ prices:

Bonds' prices

Twist scenario

Steeping scenario

ZCBond 1

95, 6938

96, 6184

ZCBond 2

88, 8996

88, 8996

ZCBond 3

70, 8919

64, 3928

Portfolio's value

90, 1822

89, 4571

 

Portfolio’s value is the weighted average of ZCBond 1 and ZCBond 3, the weighs are 7/9 and 2/9 respectively.

Thus Pport(Twist scenario)=90, 1822 and Pport(Steeping scenario)=89, 5671.

 



  

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