Interest Rate Models:
An Introduction
CH3. Discrete-Time Binomial Models
Andrew J. G. Cairns
報告者:張國培
指導教授:戴天時
1
3.1 A Simple No-Arbitrage Model
• P(t,T) : the price at time t of a zero-coupon bond which
matures at time T
• Risk-free rate of interest
r(t+s)= -logP(t,t+1) for 0≤s<1
• Cash account
B(0)=1
B(t+1)=
𝐵(𝑡)
𝑃(𝑡,𝑡+1)
= exp
𝑡+1
𝑟
0
𝑠 𝑑𝑠 = exp[
𝑡
𝑠=0 𝑟(𝑠)]
2
3.1 A Simple No - Arbitrage Model
Is it possibile for us to develop a stochastic model for the dynamics
of these bond prices which is arbitrage free?
It is trivial to demonstrate if interest rates are deterministic.
Define the instantaneous risk - free rate, r(t), to be equal to F( 0 ,T,T  1 ) for T  t  T  1
ie. r (t )  F( 0 ,T,T  1 ) for T  t  T  1
T 1
T 1
P(t,T)  exp (   r(s) )  exp (   F( 0 ,s,s  1 ) )
s t
s t
T 1

exp (   F( 0 ,s,s  1 ) )
s 0
t 1
exp (   F( 0 ,s,s  1 ) )
s 0
Forward rate F( 0 ,t,T) 
1
P( 0 ,t)
ln
T  t P( 0 ,T)
proof :
let Z(t, T) is the zero - coupon rate at time t which matures at time T
Z(t,T) 
- ln P(t,T)
T t
e Z( 0 ,t)*t e F( 0 ,t,T)*(T t)  e Z( 0 ,T)*T
Z( 0 ,T)*T  Z( 0 ,t)*t
T t
- ln P( 0 ,T)
- ln P( 0 ,t)
*T 
*t
T
t

T t
1
P( 0 ,t)

ln
T  t P( 0 ,T)
 F( 0 ,t,T) 
P( 0 ,T)
- - - - - - - - - - - (*)
P( 0 ,t)
With this structure, the prices of all bonds grow at the risk - free rate,

and the model is arbitrage free.
3
Case1: When P(t,T)<
𝑃(0,𝑡)
units of the
𝑃(0,𝑇)
𝑃(0,𝑡)
獲得 $
𝑃(0, 𝑇)
𝑃(0,𝑇)
sell
𝑃(0,𝑇)
𝑃(0,𝑡)
T − bond
0
Buy 1units of the t-bond
花費 $𝑃(0, 𝑡)
Case2: When P(t,T)>
1
t
units of the T-bond
𝑃(𝑡,𝑇)
1
花費 $
P(t,T)=1
𝑃(𝑡,𝑇)
T
Buy
套利: $
1
𝑃(𝑡,𝑇)
−
𝑃(0,𝑡)
>0
𝑃(0,𝑇)
𝑃(0,𝑇)
𝑃(0,𝑡)
1
units of the T-bond
𝑃(𝑡,𝑇)
1
獲得 $
P(t,T)=1
𝑃(𝑡,𝑇)
Sell
Sell 1units of the t-bond
獲得 $𝑃(0, 𝑡)
0
𝑃(0,𝑡)
Buy
units of the
𝑃(0,𝑇)
𝑃(0,𝑡)
花費 $
𝑃(0, 𝑇)
𝑃(0,𝑇)
T − bond
t
T
套利: $
𝑃(0,𝑡)
1
𝑃(0,𝑇) 𝑃(𝑡,𝑇)
>0
4
3.2 The Ho and Lee No - Arbitrage Model
Suppose that at time 1 either all prices go up or they all go down relative to the risk - free
return on cash : for all T  1,
P(0, T)
is the forward price at time 0 for delivery at time 1 of the zero - coupon bond which
P(0,1)
matures at time T.
P(0, T)

u(0,
T)
if ' up'

P(0,1)

P (1, T )  
d(0, T) P(0, T)
if ' down'

P(0,1)
Note that if u(0, s)  d(0, s)  1 for all s, then price at time 1 are deterministic.
Repeat this step at all future times.
P(t, T)

u(t,
T
t)

P(t, t  1)

P (t  1, T )  
d(t, T - t) P(t, T)

P(t, t  1)
if ' up'
if ' down'
u(t, s) and d(t, s) are known at time t
Assume
(i) u(t, s) ≧ d(t, s) for all t, s
(ii) u(t, s) = u(s) (there is no dependence upon prices or upon t.)
(iii) u(1) = d(1) = 1 (ensure that P(t, t)  1 t)
5
Consider all price changes between ti mes 0 and 1.
Theorem 3.1
(i) Suppose that the model is arbitrage free. Then
u(T)>1>d(T)>0 for all T≧ 2
Proof :
by definition u(T)>d(T) for all T ≧ 2 ,
 prices are positive
 d(T)>0
 so u(T)>d(T)>0
(i ) Consider
u (T )  d (T )  1
P (1, T ) d (T ) P (0, T ) 1
1


P (0, T )
P (0,1)
P (0, T ) P(0,1)
(ii ) Consider 1  u (T )  d (T )
B (1) 
 P (1, T ) 
P(0, T )
P(0,1)
1
P (1, T ) P(0,1) 1
P(1, T )
P(0, T )


 P(1, T ) 
P (0,1)
P (0, T ) P(0,1) P(0, T )
P(0,1)
 (i )(ii )皆存在套利機會
 u (T )  1  d (T )  0
is true
6
(ii) Suppose that the model is arbitrage free .Define
1 d (T )
q (T ) 
for all T  2.
u (T )  d (T )
Then there exist q.0<q<1, such that q(T)=q for all T≧ 2 .
q defines the equivalent martingale measure Q ; that is , PQ ('up ')  q , PQ ('down ') 1 q .
Proof : ( 課本給的是special case,這邊證明我們從廣義的角度來證 )
EQT [ P ( t 1,T )| Ft ] q (T t )
P ( t ,T ) u (T t )
P ( t ,T ) d (T t )
 (1 q (T t ))
P ( t ,t 1)
P ( t ,t 1)
P ( t ,T )
[ q (T t ) u (T t )  (1 q (T t )) d (T t )]
P ( t ,t 1)
P ( t ,T )
1 d (T t )
u (T t ) 1

[
u (T  t ) 
d (T t )]
P ( t ,t 1) u (T t )  d (T t )
u (T  t )  d (T  t )
P ( t ,T )

P ( t ,t 1)

P ( t 1,T )
P ( t ,T )
1
P ( t ,T ) P ( t ,t 1) P ( t ,T )
| Ft ]


B ( t 1)
P ( t ,t 1) B ( t 1) P ( t ,t 1) B ( t )
B (t )
P(t, T)
By Tower property,under Q T  measure ,
is a martingale.
B(t)
proof :
 0s  t
P(t, T)
P(t, T)
E[
| Fs ] E[E[
| Ft ] | Fs ]
B(t)
B(t)
P(t, T)
 E [ E [... E [ E[
|Ft -1 ] | Ft -2 ]| ...| Fs1 ]| Fs ]
B(t)
P ( t 1,T )
 E [ E [... E [
| Ft -2 ]| ...| Fs1 ]| Fs ]
B ( t 1)
......
P(s,T)

B(s)
 EQT [
7
ex :
Replicate P(1,2) by using T - bond and cash
V (0)  xB(0)  yP(0, T )

V (1)  xB(1)  yP(1, T )
( we hold x units of cash and y units of P(0, T))
V(1) should be equal to P(1,2) regardless of whether prices go up or down.

xB(1)  yu (T ) P (0, T ) B (1)  u (2) P (0,2) B (1)
down 
xB(1)  yd (T ) P (0, T ) B (1)  d (2) P (0,2) B (1)
up
(u (2)  d (2)) P (0,2)
(u (T )d (2)  d (T )u (2)) P (0,2)
P (0,1)
x
u (T )  d (T )
(u (T )  d (T )) P (0, T )
P (0,1)
(u (T )d (2)  d (T )u (2)) P (0,2)  (u (2)  d (2)) P (0,2)
V (0)  x  yP(0, T )  [
]
u (T )  d (T )
1  d (T )
u (T )  1
 [u (2)
 d ( 2)
]P (0,2)
u (T )  d (T )
u (T )  d (T )
 arbitrage - free  V(0)  P(0,2) V(1)  P(1,2)
y
1  d (T )
u (T )  1
 d ( 2)
1
u (T )  d (T )
u (T )  d (T )
 u( 2 )q(T)  d( 2 )( 1-q(T))  1
 u ( 2)
1-d( 2 )
u( 2 )-d( 2 )
q(T)  q( 2 )  q for all T
 q(T) 
that is,
8
(iii) Suppose there exists an equivalent martingale measure, Q;
that is, a q such tht 0  q  1 and EQ [P( 1,T)/B( 1 )]  P( 0 ,T)/B( 0 ) for all T.
Then there is no arbitrage between time 0 and 1 in the binomial model.
proof :
N
N
T T 1
Take any portfolio {x }
Then
with net value 0; that is,  x T P (0, T )  0.
T 1
N
N
T 1
T 1
EQ [ x T P (1, T )]   x T B (1) EQ [
N
 B (1) xT
T 1
P (1, T )
]
B (1)
P (0, T )
B ( 0)
0
N
Hence, if we consider the random variable  x T P (1, T ), either both outcome are 0 or
T 1
one outcome is positive and one negative. So no arbitrage is possible between times 0 and 1.
Remark 3.2 The requirement that q(T)  q for all T and for some 0  q  1 imposes the relationship
u(T) 
[ 1-( 1-q)d(T)]
q
for all T.
u( 2 )q(T)  d( 2 )( 1-q(T))  1
9
3.3 Recombining Binomial Model
As in Section 3.2 we assume that u(t,T,Ft )  u(T) for all t,Ft .(Markov process) Furthermore,
we would like the price to be path independent.
But only depend upon the number of up-steps.
Denintion : P(t, T, i)  P(t, T) given that there have been i down-steps and
(t - i) up-steps between 0 and t ( i  0,1,..., t)
𝑃(2, 𝑇, 0)
𝑃(1, 𝑇, 0)
𝑃(2, 𝑇, 1)
𝑃(0, 𝑇, 0)
𝑃(1, 𝑇, 1)
𝑃(2, 𝑇, 2)
10
ex : Consider the two - year period t  0 to t  2. We require that all prices after
the up-down sequence are equal to the price after the down-up sequence
𝑃(2, 𝑇, 0)
for t  1
𝑃(1, 𝑇, 0)
P (0, T ,0)
P (0,1,0)
P (0, T ,0)
P (1, T ,1)  d (T )
P (0,1,0)
P (1, T ,0)  u (T )
𝑃(2, 𝑇, 1)
𝑃(0, 𝑇, 0)
𝑃(1, 𝑇, 1)
for t  2
P (1, T ,0)
 d (T  1)
P (1,2,0)
u (T )( P (0, T ,0)
𝑃(2, 𝑇, 2)
)
P (0,1,0)
up-down
P (1,2,0)
d (T )( P (0, T ,0)
)
P(1, T,1)
P (0,1,0)
 u(T - 1)
 u (T  1)
down-up
P(1,2,1)
P (1,2,1)
d (T  1)u (T ) u (T  1)d (T )


P (1,2,0)
P (1,2,1)
d(T)
d (T  1)
P (1,2,1) d (2)

k
where k 

0  k  1 - - - - - - - - - (**)
u(T)
u (T  1)
P (1,2,0) u (2)
d(T)
since u(1)  d(1)  1, so
 k T 1
u(T)
( q*u(T)+(1-q)*d(T)=1 )
form Theorem 3.1 and Remark 3.2
P (2, T ,1)  d (T  1)
1
u (T ) 
(1  q )k T 1  q
k T 1
and d (T ) 
(1  q )k T 1  q
- - - - - - - -(3.1)
11
綜上所述, 只要知道任一期間上漲下跌的值以及P(0, t) t  1,2,3...., T
我們就可代入下列式子將可得到一顆 binomial lattice.
d( 2 )
d (T )

 k T 1
u( 2 )
u (T )
1  d( 2 )
q
u( 2 )  d( 2 )
1
u(T) 
( 1  q)k T 1  q
k
k T 1
d(T) 
( 1  q)k T 1  q
12
Example3.3
Suppose P(0, T)  0.94, 0.9, 0.87, 0.84 for T  1,2,3,4 respectively.
Further, it is known that P(1,2)  0.94 or 0.965. It follows that
P (1,2,0) P (0,1)
 1.007889
P (0,2)
P (1,2,1) P (0,1)
d ( 2) 
 0.981778
P (0,2)
1  d ( 2)
q
 0.697872
u ( 2)  d ( 2)
d ( 2)
k
 0.974093
u ( 2)
By equation 3.1
u ( 2) 
T
P(0, T)

u(T) P(0,1)

P (1, T )  
d(T) P(0, T)

P(0,1)
if ' up'
if ' down'
(by Theorem 3.1 (ii))
(by (**))
1
2
3
4
u(T)
1.0000
1.007889
1.015694
1.023414
d(T)
1.0000
0.981778
0.963749
0.945917
Then we can then compute values for the P(t, T, x) using
P(t - 1, T, x)

u(T
t

1)

P (t  1, t , x)

P(t, T, x)  
d(T - t  1) P(t - 1, T, x - 1)

P (t  1, t , x  1)
x  0,1,...., t - 1
x  1,2,..., t
13
Then we can then compute values for the P(t, T, x) using
P(t - 1, T, x)

u(T - t  1) P (t  1, t , x)

P(t, T, x)  
d(T - t  1) P(t - 1, T, x - 1)

P (t  1, t , x  1)
table for P(t,1)
t
x
0
1
0
0.94
1
1
1
x
0
1
2
3
4
0
0.84
x  0,1,...., t - 1
x  1,2,..., t
x
0
1
2
table for P(t,2)
t
0
1
0.9
0.965
0.94
table for P(t,4)
t
1
2
3
0.91454 0.962583 0.988124
0.845287 0.913355 0.962525
0.866644 0.937589
0.913299
2
1
1
1
x
0
1
2
3
table for P(t,3)
t
0
1
2
0.87
0.940057 0.981838
0.891981 0.956401
0.931624
3
1
1
1
1
4
1
1
1
1
1
So, we can observe the lattice structure for P(t,4)、R(t,4)、r(t)
14
Remark 3.4. Under this model for u(T), d(T) let us consider t he forward - rate curve.
F(t , T  1, T)  ln
P (t , T  1)
u (T  t )
 F (0, T  1, T )  ln
 D(t ) ln k ,
P (t , T )
u (T )
t
where D(t)   I(s) is the number of down - steps and
s 1
1 if there is a down - step at time s
I(s)  
0 otherwise
proof : We prove the result by induction. And it is true for t  0.
Suppose it is true for t.
Case1 : Consider t he case I(t  1)  0. Then D(t  1)  D(t) and
F (t  1, T  1, T , D(t  1))  ln
 ln
P(t  1,T  1,D(t  1 ))
P(t  1,T,D(t  1 ))
u (T  t  1) P (t , T  1, D(t ))
u (T  t ) P (t , T , D(t ))
P (t , t  1, D (t ))
( D(t  1)  D(t))
P (t , t  1, D(t ))
u (T  t  1)
P (t , T  1, D (t ))
 ln(
)  ln
u (T  t )
P (t , T , D(t ))
u (T  t  1)
 ln(
)  F(t , T  1, T , D (t ) )
u (T  t )
u (T  t  1)
u (T  t )
 ln(
)  F (0, T  1, T )  ln
 D (t ) ln k
u (T  t )
u (T )
u (T  t  1)
 F (0, T  1, T )  ln
 D(t  1) ln k
u (T )
So, the result is true for t  1 if I(t  1)  0
15
Case2 : Consider t he case I(t  1)  1. Then D(t  1)  D(t)  1 and
F (t  1, T  1, T , D(t  1))  ln
 ln
P(t  1,T  1,D(t  1 ))
P(t  1,T,D(t  1 ))
d (T  t  1) P (t , T  1, D(t ))
P(t , t  1, D(t ))
d (T  t ) P (t , T , D (t ))
( D(t  1)  D(t)  1)
P(t , t  1, D(t ))
d (T  t  1)
P (t , T  1, D(t ))
 ln
 ln
d (T  t )
P(t , T , D(t ))
u (T  t  1)k T t  2
1
k T-1

F(t
,
T

1
,
T
,
D
(
t
)
)
(

u(T)

and
d(T)

)
u (T  t )k T t 1
( 1  q)k T 1  q
( 1  q)k T 1  q
u (T  t  1)
u (T  t )
 ln
 ln k  F (0, T  1, T )  ln
 D(t ) ln k
u (T  t )
u (T )
u (T  t  1)
 F (0, T  1, T )  ln
 D(t  1) ln k
u (T )
So, the result is true for t  1 if I(t  1)  1
 ln
Hence the result is true for t  1, and the result follows by induction.
16
Corrllary 3.5 The risk - free rate of interest is then
u (1)
 D(t ) ln k
u (t  1)
 F (0, t , t  1)  ln u (t  1)  D(t ) ln k
r (t )  F (t , t , t  1)  F (0, t , t  1)  ln
 F (0, t , t  1)  ln d (t  1)  U (t ) ln k
where U(t)  t - D(t) is the number of up - steps up to time t.
We observe that this model for r(t) is a random walk with constant volatility but time - varying drift.
Remark 3.6 It is necessary to put constraints on q and k to ensure that the risk - free rate remains positive
(that is, P(t, t  1)  1) over a specified period of time. However, for any admissible (q, k),
there will exist some t 0 which implies that, for all t  t 0 , P (t , t  1, i ) will be greater than 1
for some i, we cannot prevent interest rates from going negative eventually.
(ps. dr(t)  θ(t)dt  σdw(t) )
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