1 10.2 affine-yield models 劉彥君. 2 bond prices according to the risk-neutral pricing formula,...
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1
10.2 Affine-Yield Models
劉彥君
2
Bond Prices
• According to the risk-neutral pricing formula, the price at time t of a zero-coupon bond paying 1 at a latter time T is
•
R(t) 由 Y1(t) 、 Y2(t) 組成
TttFeETtBT
tduuR
0,)(~
),()(
)()()( 22110 tYtYtR (10.2.6)(10.2.6)
3
• And the solution of the system of stochastic differential equations
Is Markov
There must be some function f(t,y1,y2) such that
)(~
)()( 1111 tWddttYtdY
)(~
)()()( 2221212 tWddttYdttYtdY
))(()](|))(([ sXgsFtXhE
)(),(,),( 21 tYtYtfTtB
(pp.74)
(10.2.4)(10.2.4)
(10.2.5)(10.2.5)
4
• The discount factor
satisfies (see (5.2.18))
迭代上面的條件式：D(t)B(t, T) is a martingale under
d(D(t)B(t,T)) 的 dt 項為 0
tduuR
etD 0)(
)(
dttDtRtdD )()()(
P~
5
222221211111
2211
2121
21
2
1
2
1
)(),(,)()(),(,)()(
)(),(,)(),()(
YdYfdYdYfYdYf
dYfdYfdtfRfdtD
tYtYtdftDdttYtYtftDtR
tYtYtftDdTtBtDd
yyyyyy
yyt
)(~
)()( 1111 tWddttYtdY )(
~)()()( 2221212 tWddttYdttYtdY
(10.2.4)(10.2.4)
(10.2.5)(10.2.5)
)()()( 22110 tYtYtR (10.2.6)(10.2.6)
22112211222
212111122110
~~
2
1
2
1WdfWdfDdtfffY
fYfYffYYD
yyyyyyy
yyt
6
令 dt 項為 0 ，則可得
0,,2
1,,
2
1
,,,,,,
,,,,
21222111
2122221212121111
212122110
yytfyytf
yytfyyytfyyytfy
yytfyytfyy
yyyy
yyy
t
RyRyandTt 21 , ,0
Terminal condition ： RyRyyyTf 2121 , 1),,(
To solve this equation, we seek a solution of the affine-yield form
)()()(21
2211),,( tTAtTCytTCyeyytf
for some functions C1(τ), C2(τ), and A(τ).
Define τ=T-t to be the relative maturity (i.e., the time until maturity)
(10.2.18)(10.2.18)
(10.2.20)(10.2.20)
7
由於模型參數並不相依於 t ，而 zero-coupon bond prices 將透過 τ 相依於 t 。
由 Terminal condition 可得： 0)0()0()0( 21 ACC
Then compute derivatives, where ` denotes differentiation with respect to τ .
fCffCCffCf
fCffCffACyCyf
AAdt
d
iCdt
dCC
dt
d
yyyyyy
yyt
iii
22222121
2111
22112211
, ,
, , ,
)()(
2,1 ),()()(
微 (10.2.20)(10.2.20)
8
02
1
2
1
0,,2
1,,
2
1
,,,,,,
,,,,
022
21
2222211221111
21222111
2122221212121111
212122110
fCCA
yCCyCCC
yytfyytf
yytfyyytfyyytfy
yytfyytfyy
yyyy
yyy
t
則 (10.2.18)
(10.2.18)(10.2.18)
Since it must hold for all y1 and y2, thus
02
1
2
1
0
0
022
21
2222
1221111
CCA
CC
CCC
022
21
2222
1221111
)(2
1)(
2
1)(
)()(
)()()(
CCA
CC
CCC
(10.2.23)(10.2.23)
(10.2.24)(10.2.24)
(10.2.25)(10.2.25)
9
2222 )()( CC 代入 Terminal Condition C2(0) = 0 可得
21)(
2
22
eC (10.2.26)(10.2.26)
2
222
22
22
1)(
0)0(
0)0(
)()(
)(
)()()()(
2
22
2
211
2
22
2
12
22
2
212
22
22222222
eC
ddC
C
edCedCe
dedCe
eCCeCC
代入
計算
10
In particular, (10.2.23)(10.2.23) implies
1
2
221
1221
1111
21
1
11
1
)(
ee
Ce
CCeCed
d
1221111 )()()( CCC
21)(
2
22
eC (10.2.26)(10.2.26)
(10.2.23)(10.2.23)
將 (10.2.26)(10.2.26) 代入 (10.2.23)(10.2.23) 並以 initial condition C1(0)=0 解方程式
11
1211
211
12111
211
2
221
2
22111
2
221
2
2211
12
221
2
2211
12
2211
)(
)(
1)(
ededeC
dede
deeeeCe
eeCed
d
dxxPdxxP
dxxP
eCdxexQyxQ
CeyxQ
xQyxPdx
dy
)()(
)(
)(,0)(
,0)(
)()(
為齊次方程，時當
為齊次方程，時當
Note 一階線性微分方程：
12
If λ1≠ λ2, integration from 0 to τ
121
212
221
2
2211
11 1
1)(
eeeC
If λ1= λ2, we obtain instead
11
1
221
2
2211
11 1
1)(
eeC
(10.2.27)(10.2.27)
(10.2.28)(10.2.28)
Finally, (10.2.25) and the initial condition A(0)=0 imply
0 022
21 )(
2
1)(
2
1)( duuCuCA
and this can be obtained in closed form by a lengthy but straightforward computation.
(10.2.29)(10.2.29)
13
Short Rate and Long Rate
• Fix a positive relative maturity (say, 3 years)• Long rate L(t):
– The yield at time t on the zero-coupon bond with relative maturity (i.e., 到期日 )
• 一旦我們有一個模型，可以在風險中立測度下計算 short rate R(t)For all t>=0 ， the price of zero-coupon bond is
determined by the risk-neutral pricing formulaThe short-rate model alone determines the long rate.
所以我們不可以隨意為 Long rate 寫下 stochastic differential equation
但在任一 affine-yield model ， long rate 仍然符合某些 stochastic differential equation.
t
maturityt )(
14
Consider the canonical two-factor Vasicek model. As in the previous discussion, Zero-coupon bond prices in this model are of the form
)()()()()( 2211),( tTAtTCtYtTCtYeTtB where C1(τ), C2(τ) and A(τ) are given by (10.2.26)-(10.2.29)(10.2.26)-(10.2.29).
Thus, the long rate at time t is
)()()()()(1
),(log1
)( 2211
AtYCtYCttBtL
which is an affine function of the canonical factors Y1(t) and Y2(t) at time t.
(10.2.30)(10.2.30)
15
因為 canonical factors 沒有經濟意涵，我們希望使用 R(t) 與 L(t) 作為 model factor
)(~
)()()( 1121211111 tBddttXbtXbatdX
)(~
)()()( 2222212122 tBddttXbtXbatdX
two-factor Vasicek model
)()()( 22110 tXtXtR
(10.2.1)(10.2.1)
(10.2.2)(10.2.2)
(10.2.3)(10.2.3)
目標： X1(t) R(t) , X2(t) L(t)
16
)()()( 22110 tYtYtR
)()()()()(1
),(log1
)( 2211
AtYCtYCttBtL
(10.2.6)(10.2.6)
(10.2.30)(10.2.30)
)()(
)(
)()()(
)(1
0
2
1
21
11
11
AtY
tY
CCtL
tR
首先將 (10.2.6) 和 (10.2.30) 寫成 vector notation:
(10.2.31)(10.2.31)
We wish to solve this system for (Y1(t), Y2(t))
17
Lemma 10.2.2
• The matrix
is nonsingular if and only if
)()( 21
11
21
CCD
0
0
221121
2
and
(10.2.32)(10.2.32)
18
Consider xx xeexf 1)(
for which f(0)=0 and f ’(x)=xe-x > 0, for all x > 0
0,0)( xxf
Define )()()1(1
)( 2 xfxxhex
xh x
h’(x) is strictly negative , for all x > 0h(x) is strictly decreasing on (0, ∞)
To examine the nonsingular of D, consider the first case λ1≠ λ2
We can use (10.2.26)(10.2.26) and (10.2.27)(10.2.27)
121
212
221
2
2211
11 1
1)(
eeeC (10.2.27)(10.2.27)
21)(
2
22
eC (10.2.26)(10.2.26)
19
)()( 21
11
21
CCD
)()(
11
11
11)(
11
11
1)()(
11
)()(1
)det(
1221
22112
1221
22112
212121
2221
1221
11212121
2221
1
21
2
21
1221
12
1212
12112
hh
ee
eeee
eeeee
CCD
Because λ1 ≠λ2, h is strictly decreasing, h(λ1)≠ h(λ2)D is nonzero
if and only if δ2≠0 and 0221121 (10.2.32)(10.2.32)
20
Next consider the case λ1=λ2
21)(
2
22
eC (10.2.26)(10.2.26)
11
1
221
2
2211
11 1
1)(
eeC (10.2.28)(10.2.28)
imply
12
2221
1
2221
1
2211
1
2
1
21
1221
1
111 11
)()(1
)det(
f
eee
CCD
000)( ,0 22111 andf
0221121 (10.2.32)(10.2.32)is equivalent to
D is nonzero if and only if (10.2.32) holds.
21
Under the assumptions of Lemma 10.2.2, we can invert
)()(
)(
)()()(
)(1
0
2
1
21
11
11
AtY
tY
CCtL
tR(10.2.31)(10.2.31)
into
)()(
)(
)()()(
)(1
0
1
21
11
11
2
1
AtL
tR
CCtY
tY(10.2.33)(10.2.33)
Then we can compute
)(~
)(~
)()()(
)(
)()(
0
)()(
)()()(
0
)()(
)(~
)(~
)(
)(0
)()(
)(
)(
)()()(
)(
2
1
21
11
21
1
21
11
21
221
1
21
11
21
1
0
1
21
11
21
221
1
21
11
21
2
1
2
1
221
1
21
11
21
2
1
21
11
21
tWd
tWd
CCdt
L
tR
CCCC
dtACCCC
tWd
tWddt
tY
tY
CC
tdY
tdY
CCtdL
tdR
)(~
)()( 1111 tWddttYtdY
)(~
)()()( 2221212 tWddttYdttYtdY (10.2.4)(10.2.4)
(10.2.5)(10.2.5)
22
)(~
)()()( 1121211111 tBddttXbtXbatdX
)(~
)()()( 2222212122 tBddttXbtXbatdX
two-factor Vasicek model
)()()( 22110 tXtXtR
(10.2.1)(10.2.1)
(10.2.2)(10.2.2)
(10.2.3)(10.2.3)
目標： X1(t) R(t) , X2(t) L(t)
)(~
)(~
)()(
)(
)(
)()(
0
)()(
)()()(
0
)()()(
)(
2
1
21
11
21
1
21
11
21
221
1
21
11
21
1
0
1
21
11
21
221
1
21
11
21
tWd
tWd
CC
dtL
tR
CCCC
dtACCCCtdL
tdR
23
)()()(
0
)()( 1
0
1
21
11
21
221
1
21
11
21
2
1
ACCCCa
a
1
21
11
21
221
1
21
11
21
2221
1211
)()(
0
)()(
CCCCbb
bb
The matrix B is
The a1, a2 in (10.2.1) and (10.2.2) is
and the eigenvalues of B are λ1 > 0, λ2 > 0. With
)()(1
, 22
212
22
211
CC
The processes
)(~
)()(~
)(1
)(~
,)(~
)(~1
)(~
22112
222111
1 tWCtWCtBtWtWtB
are the Brownian motions appearing in (10.2.1) and (10.2.2)
1,0..
)(0)(10)(
120 ei
tLtRtR
24
Gaussian Factor Processes
)(~
)()( tWdtYtdY
Reform (10.2.4)~(10.2.5)
where
)(~
)(~
)(~
,0
,)(
)()(
2
1
221
1
2
1
tW
tWtW
tY
tYtY
the canonical two-factor Vasicek model in vector notation is
)(~
)()( 1111 tWddttYtdY
)(~
)()()( 2221212 tWddttYdttYtdY
(10.2.4)(10.2.4)
(10.2.5)(10.2.5)
Recall that λ1 > 0, λ2 > 0There is a closed-form solution to this matrix differential equation.
(10.2.34)(10.2.34)
25
To derive this solution, we first form the matrix exponential
0
)(!
1
n
nt tn
eeΛt defined by
where (Λt)=I, the 2 x 2 identity matrix.
If λ1 ≠λ2 ,then
ttt
t
t
eee
ee 221
1
21
21
0
If λ1 =λ2 ,then
tt
tt
ete
ee
11
1
21
0
In either case, ttt eeedt
d
(10.2.35)(10.2.35) (10.2.36)(10.2.36)
(10.2.37)(10.2.37)
Where the derivate is defined componentwise, and
Lemma 10.2.3
1 tt ee (10.2.38)(10.2.38)
where e-Λt is obtained by replacing λ1, λ2, and λ21 in the formula eΛt by -λ1, -λ2, and -λ21, respectively.
26
We consider first the case λ1 ≠λ2 .We claim that in this case
,....1,0,
0
221
1
21
21
n
tt
tt nn
nn
nn
(10.2.39)(10.2.39)
10
010t
使用數學歸納法n=0:
Assume (10.2.39) is true for some value of n.
12
121
11
1221
121
11
221
1
221
11
21
12
11
21
21
21
21
00
00
nn
n
nn
n
nn
nnn
tt
t
tt
t
tt
t
tt
tttt
nnnn
nn
which is (10.2.39) with n replaced by n+1.
27
ttt
t
n
n
nn n
n
n
n
n
n
n
n
n
nt
eee
e
ttt
t
tn
e
111
21
21
1
21
21
0
0
)(!
1
0 2!1
0 0 2!1
1!1
0 1!1
0
Having thus established (10.2.39) for all values of n, we have
This is (10.2.35)
28
We next consider the case λ1 =λ2 .We claim in this case that
,....2,1,0
11
121
1
n
ttn
tt
nnn
nn
(10.2.40)(10.2.40)
10
010t
使用數學歸納法n=0:
Assume (10.2.40)(10.2.40) is true for some value of n.
11
1121
11
11
1121121
11
11
121
1
121
11
1
00
00
nnn
n
nnnn
n
nnn
nnn
ttn
t
ttn
t
ttn
t
tt
tttt
which is (10.2.40)(10.2.40) with n replaced by n+1.
29
0 1!1
0
11!21
0 1!1
0
0)(
!
1
n
n
nn
nnnn
n
n
n
n
nt
tt
tt
ne
Having thus established (10.2.40)(10.2.40) for all values of n, we have
(10.2.41)(10.2.41)
But
tt
n
n
n
nn teed
dt
nd
dt
n
n11
211
210
11
210
1121 !
1
!
Substituting this into (10.2.41)(10.2.41), we obtain (10.2.36)(10.2.36).
tt
tt
ete
ee
11
1
21
0
(10.2.36)(10.2.36)
30
Now prove
and
ttt eeedt
d(10.2.37)(10.2.37)
1 tt ee (10.2.38)(10.2.38)
When λ1 ≠λ2, we have
ttt
t
t
eee
ee
dt
d221
1
22121
21
1 0
and
ttt
t
t
eee
ee 221
1
21
21
0
When λ1 =λ2,
tt
tt
eet
ee
dt
d11
1
1121
1
1
0
tt
tt
ete
ee
11
1
21
0
and
The verification of (10.2.37) and (10.2.38) can be done by straightforward matrix multiplications.
31
)(~
)(~
)()(
)()(
)()()(
tWdetWdtYtYe
tdYtYe
tdYetYetYed
tt
t
ttt
代入 (10.2.34)(10.2.34) )(~
)()( tWdtYtdY
Integration form 0 to t yields
t ut uWdeYtYe0
)(~
)0()(
We solve for
t utt
t utt
uWdeYe
uWdeeYetY
0
)(
0
)(~
)0(
)(~
)0()(
(10.2.42)(10.2.42)
課本有誤
Now we have define
0
)(!
1
n
nt tn
e then
32
If λ1 ≠λ2, equation (10.2.42) may be written componentwise as
t utt uWdeYetY0 1
)(11 )(
~)0()( 11
t utt utut
ttt
uWdeuWdee
YeYeetY
0 2)(
0 1)()(
21
21
2121
212
)(~
)(~
)0()0()(
221
221
If λ1 ＝ λ2, then the componentwise form of (10.2.42) is
t utt uWdeYetY0 1
)(11 )(
~)0()( 11
t utt ut
tt
uWdeuWdeut
YeYtetY
0 2)(
0 1)(
21
21212
)(~
)(~
)(
)0()0()(
11
11
(10.2.44)(10.2.44)
(10.2.43)(10.2.43)
(10.2.45)(10.2.45)
(10.2.46)(10.2.46)
33
nonrandom quantities + Ito integrals of nonrandom integrandsthe process Y1(t) and Y2(t) are Gaussian.R(t) = δ0+δ1Y1(t)+δ2Y2(t) is normally distributed.
The statistics of Y1(t) and Y2(t) are provided in Exercise 10.1Exercise 10.1
34
10.2.2 Two-Factor CIR Model
• In the two-factor Vasicek model, the canonical factors Y1(t) and Y2(t) are jointly normally distributed.
– Because Y1(t) and Y2(t) are driven by independent Brownian motions, they are not perfectly correlated.
for all t>0is a normal random variable with variance– >0 ： δ1 ≠ 0, δ2 ≠ 0
– =0 ： δ1 = δ2 = 0
• In particular, for each t>0,
)()()( 22110 tYtYtR
00)( tRPthere is a positive probability that R(t) is strictly negative.
(10.2.47)(10.2.47)
35
In the two-factor Cox-Ingersoll-Ross model (CIR), both factors are guaranteed to be nonnegative at all times almost surely.
We again define the interest rate by (10.2.47) but now assume that
0,0,0 210 (10.2.48)(10.2.48)
)()()( 22110 tYtYtR (10.2.47)(10.2.47)
R(0) ≥ 0, and R(t) ≥ 0 for all t ≥ 0 almost surely.
2,1,10)( itYP i
36
The evolution of the factor processes in the canonical two-factor CIR model is given by
)(
~)()()()(
)(~
)()()()(
2222212122
1121211111
tWdtYdttYtYtdY
tWdtYdttYtYtdY
(10.2.49)(10.2.49)
(10.2.50)(10.2.50)
In addition to (10.2.48), we assume
0,0,0,0,0,0 2112221121 (10.2.51)(10.2.51)
≥0 ≤0 >0
≥0 >0 ≤ 0
although the drift term of (10.2.49) can be negative but these conditions guarantee that
0)(and 0)(0)()(
0)(and 0)(0)()(
122221212
212121111
tYtYtYtY
tYtYtYtY
Starting with Y1(0) ≥ 0 and Y2(0) ≥ 0, we have Y1(t) ≥ 0 and Y2(t) ≥ 0 for all t ≥ 0 almost surely.
37
The Brownian motions and in (10.2.49) and (10.2.50) are assumed to be independent.
We do not need this assumption to guarantee nonnegativity of Y1(t) and Y2(t) but rather to obtain the affine-yield result below; see Remark 10.2.4.
)(~
1 tW )(~
2 tW
如果兩個 Brownian motions 的關係係數 ρ≠0, 在令 dt 項係數 =0 的偏微分方程中會多出一項
2121 yyfyy
將會使得 )(),(),( 21 ACC (10.2.56)-(10.2.58) 無法計算。
Remark 10.2.4
所以在一開始即假設兩個 Brownian motions 互相獨立。
38
Bond Prices
)(),(,),( 21 tYtYtfTtB
The price at time t of a zero-coupon bond maturing at a later time T must be of the form
for some function f(t, y1, y2). The discounted bond price has differential
222111
22211122221212
1212111122110
2222212111112211
2121
21
~~2
1
2
1
2
1
2
1
)(),(,)()(),(,)()(
)(),(,)(),()(
WdfYWdfY
dtfYfYfYY
fYYffYYD
dYdYfdYdYfdYdYfdYfdYfdtfRfdtD
tYtYtdftDdttYtYtftDtR
tYtYtftDdTtBtDd
yy
yyyyy
yt
yyyyyyyyt
39
Setting the dt term =0, 我們得到偏微分方程式
0,,2
1,,
2
1
,,,,
,,,,
2122221111
21222212122112121111
212122110
yytfYyytfY
yytfYYyytfYY
yytfyytfYY
yyyy
yy
t
(10.2.52)(10.2.52)
0,0,0 21 yyandTt
To solve this equation, we seek a solution of the affine-yield form
)()()(21
2211),,( tTAtTCytTCyeyytf (10.2.53)(10.2.53)
for some C1(τ), C2(τ) and A(τ), where τ=T-t.
40
The terminal condition
0)0()0()0(
1),()(),(,
21
21
ACC
TTBTYTYTf
With ‘ denoting differentiation with respect to τ, we have
)()(,2,1),()( AAdt
diCC
dt
dii
Then compute ft, fy1, fy2, fy1y1, fy2y2
(10.2.54)(10.2.54)
41
(10.2.52) becomes
0
2
1
2
1
02211
2222222112211
212211111
CCA
yCCCCyCCCC
(10.2.55)(10.2.55)
Because (10.2.52) must hold for all y1 ≥ 0 and y2 ≥ 0,
12
121
2211111
12
121
2211111
)()()()(
0
CCCC
CCCC
2222
12221122
2222
12221122
)()()()(
0
CCCC
CCCC
02221
02221
)(
0
CCA
CCA
(10.2.56)(10.2.56)
(10.2.57)(10.2.57)
(10.2.58)(10.2.58)
42
The solution to these equations satisfying the initial condition (10.2.54) can be found numerically.
Solving this system of ordinary differential equations numerically is simpler than solving the partial differential equation (10.2.52).
43
10.2.3 Mixed Model
• in the two-factor CIR model, – Both factors are always nonnegative.
• In the two-factor Vasicek model, – Both factors can become negative.
• In the two-factor mixed model, – one of the factors is always nonnegative – the other can become negative.
44
The Brownian motions and are independent.
The canonical two-factor mixed model is
)(~
)(
)(~
)()()(
)(~
)()()(
21
1121222
11111
tWdtY
tWdtYdttYtdY
tWdtYdttYtdY
(10.2.59)(10.2.59)
(10.2.60)(10.2.60)
We assume
R 2121 ,0,0,0,0,0
)(~
1 tW )(~
1 tW
Y1(0) ≥ 0, and we have Y1(t) ≥ 0 for all t ≥ 0 almost surely.
≥0 >0
>0
≥0 ≥0
On the other hand, even if Y1(t) > 0, Y2(t) can take negative values for t > 0
45
The interest rate is defined by
)()()( 22110 tYtYtR (10.2.61)(10.2.61)
In this model, zero-coupon bond prices have the affine-yield form
)()()()()( 2211),( tTAtTCtYtTCtYeTtB
Just as in the two-factor Vasicek model and the two-factor CIR model, the functions C1(τ), C2(τ) and A(τ) must satisfy the terminal condition
0)0()0()0( 21 ACC
(10.2.62)(10.2.62)
(10.2.63)(10.2.63)
Exercise 10.2Exercise 10.2 derives the system of ordinary differential equations that determine the functions C1(τ), C2(τ) and A(τ) .