HullOFOD11eSolutionsCh32.pdf - CHAPTER 32 No-Arbitrage...

CHAPTER 32
No-Arbitrage Models of the Short Rate
Practice Questions
32.1
Equilibrium models usually start with assumptions about economic variables and derive the
behavior of interest rates. The initial term structure is an output from the model. In a no-
arbitrage model, the initial term structure is an input. The behavior of interest rates in a no-
arbitrage model is designed to be consistent with the initial term structure.
32.2
No. The approach in Section 32.2 relies on the argument that, at any given time, all bond prices
are moving in the same direction. This is not true when there is more than one factor.
32.3
Using the notation in the text,
3 s 
, 1 T  ,
100 L 
,
87 K 
, and
2 0 1 1
2 0 1
0 015 1
(1 ) 0 025886
0 1 2 0 1
P
e
e 
   
  
 
   
  
From equation (31.6),
(0 1) 0 94988 P   
,
(0 3) 0 85092 P   
, and 1 14277 h   so that equation
(31.20) gives the call price as call price is
100 0 85092 (1 14277) 87 0 94988 (1 11688) 2 59 N N           
or $2.59.
32.4
As mentioned in the text, equation (32.10) for a call option is essentially the same as Bl ack’s
model. By analogy with Black’s formulas corresponding expression for a put option is
(0 ) ( ) (0 ) ( )
P
KP T N h LP s N h       
In this case, the put price is
87 0 94988 ( 1 11688) 100 0 85092 ( 1 14277) 0 14 N N             
Since the underlying bond pays no coupon, put – call parity states that the put price plus the
bond price should equal the call price plus the present value of the strike price. The bond price
is 85.09 and the present value of the strike price is
87 0 94988 82 64    
. Put – call parity is
therefore satisfied:
82 64 2 59 85 09 0 14       
32.5
As explained in Section 32.2, the first stage is to calculate the value of r at time 2.1 years
which is such that the value of the bond at that time is 99. Denoting this value of r by r

, we
must solve
(2 1 2 5) (2 1 3 0) 2 5 (2 1 2 5) 102 5 (2 1 3 0) 99 B r B r A e A e
 
                 
where the
A
and
B
functions are given by equations (31.7) and (31.8). In this case,
A (2.1, 2.5) = 0.999685, A (2.1,3.0) = 0.998432, B( 2.1,2.5) = 0.396027, and B (2.1, 3.0) =
0.88005, and Solver shows that 065989 . 0 *  r . Since
434745 . 2 ) 5 . 2 , 1 . 2 ( 5 . 2
* ) 5 . 2 , 1 . 2 (

  r B
e A
and