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

Column (7) shows the difference between actual market price and no-arbitrage price (i. e. (3)-(6)).

If this difference is positive, i. e. market price is greater than no-arbitrage price, we can sell short the bond and buy replicating portfolio. Since they have the same returns, our gain will be equal to price difference. If the difference is negative, we do the other way round.

Problem 2.

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

	Maturity (yrs)	Price	D_mac	D_mod
	(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 P_port(Twist scenario)=90, 1822 and P_port(Steeping scenario)=89, 5671.

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