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Affine	Yield	Curves:	Flexibility	versus
Incompleteness
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Affine Yield Curves: Flexibility versus Incompleteness
Dhia Eddine Barbouche and Bert Koehler
Affine interest rate models are amongst the most popular interest rate mod-
els due to their flexibility, analytic expressions for bond prices and analytic
approximations for some interest derivatives, see for example the seminal ar-
ticle of Duffie and Kan [Duf-Kan]. The subclass of affine models for which
interest rates are bounded from below is of the form rt = n
j=1 δjXjt − c
where
Xjt = Xj0 +
t
0
bj +
n
k=1
ajkXks ds + σj
t
0
XjtdWjs
Here rt is the short rate process, δj ≥ 0, c ≥ 0 is some negative displacement
term, ajk, bj ≥ 0 except for ajj < 0 and (W1t, ..., Wnt) are independent Brow-
nian motions. Here and in the following we always work in the risk neutral
measure Q unless otherwise indicated. The logzeroyields defined as
Y (T, X0) = −
1
T
log E e− T
0 rsds
|X0
can be written as an affine linear combination of the state X = (X1, ..., Xn)
Y (T, X0) = −
1
T
n
j=1
(bjAj(T) + Xj0Bj(T))
where the deterministic functions T −→ Aj(T), Bj(T) in maturity T ≥ 0
obey the system of ODEs
dBj
dT
=
1
2
σ2
j B2
j +
n
k=1
akjBk − δj
Bj(0) = 0 and Aj(T) =
T
0
Bj(s)ds
The shape of the curves T −→ − 1
T
Bj(T) determines the flexibility of the
model in reproducing the great variety of yield curves seen in the history.
1
From this point of view affine models with a cascade structure are promising.
Here we deal with affine cascade models with 3 factors:
X1t = X10 +
t
0
(b1 + X2s − a1X1s)ds + σ1
t
0
X1sdW1s
X2t = X20 +
t
0
(b2 + X3s − a2X2s)ds + σ2
t
0
X2sdW2s
X3t = X30 +
t
0
(b3 − a3X3s)ds + σ3
t
0
X3sdW3s
The corresponding system of ODEs simplifies to (normalizing δ1 = 1)
dB1
dT
=
1
2
σ2
1B2
1 − a1B1 − 1
dB2
dT
=
1
2
σ2
2B2
2 − a2B2 + B1 − δ2
dB3
dT
=
1
2
σ2
3B2
3 − a3B3 + B2 − δ3
One can easily see that T −→ Bj(T) ≤ 0 and falling in T: for that purpose
write for example
dB2
dT
= (
1
2
σ2
2B2 − a2) · B2 + B1 − δ2
which can be solved as
B2(t) = e
t
0 ( 1
2
σ2
2B2(s)−a2)ds
· B2(0) +
t
0
e− s
0 ( 1
2
σ2
2B2(q)−a2)dq
· (B1(s) − δ2)ds
As B2(0) = 0 and B1(t) − δ2 ≤ 0 we see that B2(t) ≤ 0. Similarly taking
derivatives in t we get
d2
B2
dT2
= (σ2
2B2 − a2) ·
dB2
dT
+
dB1
dT
and by the way
dB2
dt
(t) = e
t
0 (σ2
2B2(s)−a2)ds
·
dB2
dt
(0) +
t
0
e− s
0 (σ2
2B2(q)−a2)dq
·
dB1
dt
(s)ds
As dB2
dt
(0) = −δ2 ≤ 0 and dB1
dt
(t) ≤ 0 we get dB2
dt
(t) ≤ 0. On the other hand we
have ∃ limT→∞ Bj(T) and so limT→∞
1
T
Bj(T) = 0. What is interesting about
2
affine cascade structure models is that for suitably chosen parameters aj, σj
we can achieve that T −→ − 1
T
B2(T) attains its maximum in T2,max ≥ 10
and T −→ − 1
T
B3(T) attains its maximum in T3,max ≥ 25. This permits a
great flexibility of yield curves
Y (T, X0) = −
1
T
3
j=1
(bjAj(T) + Xj0Bj(T))
For example the choice of (risk-neutral) parameters


b1 a1 σ1 δ1
b2 a2 σ2 δ2
b3 a3 σ3 δ3

 =


0.00005 0.3 0.05 1
0 0.05 0.05 0
0.00005 0.01 0.01 0


and c = 0.015 can represent all historically seen yield triples with tenors 3M,
10Y and 30Y by admissible state vectors X = (X1, X2, X3), Xj ≥ 0 and can
even cope with EUR-atm-Swaption volatility surface from 31.03.2016. Affine
cascade structure models with
d
dT
−
B2(T)
T
(T = 10) ≥ 0
d
dT
−
B3(T)
T
(T = 25) ≥ 0
are called flexible affine cascade structure models. The purpose of this
note is to elaborate in how far a 4 factor affine model can generate an
incomplete bond market together with the flexibility of a 3 factor flexible
affine cascade structure model. This means we have a 4 factor affine model
Xt = (X1t, ..., X4t) with negatively bounded interest rates, but logzeroyields
reduce to a 3 factor state dynamic
Y (T, X0) = −
1
T
4
j=1
(bjAj(T) + Xj0Bj(T)) =
−
1
T
4
j=1
bjAj(T) −
1
T
3
j=1
Bj(T) ·
4
k=1
˜cjkXk0
where Bj(T) obey a system of ODEs coming from a 3 factor flexible affine
cascade structure model. So from the bond-prices alone it is not possible to
3
recover the full 4-factor state but we require furthermore that the dynamics
of Xt does not reduce generally to a 3-dimensional process. So if we add some
interest-optionprices to bondprices the full 4-state Xt can be determinated.
The motivation for such an ”unspanned volatility” phenomenon derives from
the conjecture that option prices may have a real-world-dynamic not fully ex-
plained by the real-world-dynamic of yields. In general unspanned volatility
in an n-dimensional affine model postulates a nontrivial linear relation
0 =
n
j=1
cjBj(T) for all T ≥ 0
Defining a (polynomial) vector field on the ring of polynomials P ∈ C[B1, ..., Bn]
by
Z =
n
j=1
1
2
σ2
j B2
j +
n
k=1
akjBk − δj
∂
∂Bj
we get a sequence of polynomials Pm by setting
P1(B1, ..., Bn) =
n
j=1
cjBj
Pm+1(B1, ..., Bn) = Z(Pm)
Let I = I(Pm, m ∈ N) be the ideal generated by the full sequence. By
Hilberts Basis Theorem I is finitely generated, so there are P1, ..., PN which
generate the whole ideal. In fact it is sufficient to find a N ∈ N with PN+1 ∈
I(P1, ..., PN ) because then all further PN+2, PN+3, ... are also in I(P1, ..., PN ).
Let V (I) = V (P1, ..., PN ) ⊂ Cn
be the algebraic variety of the ideal, that
means
V (P1, ..., PN ) = {(B1, ..., Bn) with Pm(B1, ..., Bn) = 0 for m = 1, ..., N}
Obviously the (real analytic) curve T −→ (B1(T), ..., Bn(T)) lies inside
V (P1, ..., PN ). On the other hand if (0, ..., 0) ∈ V (P1, ..., PN ), then we have
n
j=1
cj
dm
Bj
dtm
(t = 0) = 0
4
first for 0 ≤ m ≤ N − 1 but then due to the generation property also for all
m. Because T −→ (B1(T), ..., Bn(T)) is real analytic we have
n
j=1
cjBj(t) = 0
identically in this case, so the two conditions are equivalent. This simple
observation reduces the search for unspanned affine models to a polynomial
basis problem which can in principle be solved by algorithm. Up to now the
only known example stems from Colin-Dufresne and Goldstein for n = 3 and
results in a rational curve as V (P1, P2) ⊂ C3
. So there are no ”transcenden-
tal” unspanned affine models known. Maybe this is a general property due to
the fact that the intersection variety carries a holomorphic vector field with
meromorphic pole of first order in the projective extension.
Coming back to our 4-factor setting it is instructive to consider an exam-
ple: Assume that Bj(T) = Bj(T) for j = 1, 2, 3, so we have
dB1
dT
=
1
2
σ2
1B2
1 − a1B1 − 1
dB2
dT
=
1
2
σ2
2B2
2 − a2B2 + B1 − δ2
dB3
dT
=
1
2
σ2
3B2
3 − a3B3 + B2 − δ3
and furthermore
dB4
dt
=
1
2
σ2
4B2
4 − a44B4 + a41B1 + a42B2 + a43B3 − δ4
The corresponding process looks like
X1t = X10 +
t
0
(b1 + a41X4s + X2s − a1X1s)ds + σ1
t
0
X1sdW1s
X2t = X20 +
t
0
(b2 + a42X4s + X3s − a2X2s)ds + σ2
t
0
X2sdW2s
X3t = X30 +
t
0
(b3 + a43X4s − a3X3s)ds + σ3
t
0
X3sdW3s
X4t = X40 +
t
0
(b4 − a44X4s)ds + σ4
t
0
X4sdW4s
5
Now assume we have a linear relation
B4(T) =
3
j=1
cjBj(T) for all T ≥ 0
Taking derivatives we obtain
1
2
σ2
4
3
j=1
cjBj
2
− a44
3
j=1
cjBj + a41B1 + a42B2 + a43B3 − δ4 =
3
j=1
cj
1
2
σ2
j B2
j − ajBj + Bj−1 − δj
If this quadratic relation vanishes identically in C[B1, B2, B3] we must have
B4(T) = cjBj(T) for some j ∈ {1, 2, 3} and
σ2
4c2
j = cjσ2
j
cjaj − a44cj + a4j = 0
a4,j−1 − cj = 0
a4,j−2 = 0
cjδj − δ4 = 0
If for example j = 3 then
c3 =
σ2
3
σ2
4
a43 = (a44 − a3)c3
a42 = c3
a41 = 0
δ4 = c3δ3
6
Now we define X5t = X3t + c3X4t. Then we get
X1t = X10 +
t
0
(b1 + X2s − a1X1s)ds + σ1
t
0
X1sdW1s
X2t = X20 +
t
0
(b2 + X5s − a2X2s)ds + σ2
t
0
X2sdW2s
X5t = X50 +
t
0
(b3 + c3b4 + a43X4s − a3X3s − c3a44X4s)ds +
σ3
t
0
X3sdW3s + c3σ4
t
0
X4sdW4s
Inserting a43 = (a44 − a3)c3 and using c3 =
σ2
3
σ2
4
we get
σ3
t
0
X3sdW3s + c3σ4
t
0
X4sdW4s ∼ σ3
t
0
X5sdW5s
and
X1t = X10 +
t
0
(b1 + X2s − a1X1s)ds + σ1
t
0
X1sdW1s
X2t = X20 +
t
0
(b2 + X5s − a2X2s)ds + σ2
t
0
X2sdW2s
X5t = X50 +
t
0
(b3 + c3b4 − a3X5s)ds + σ3
t
0
X5sdW5s
so the process collapses to a 3-factor dynamic.
The first situation we want to examine more detailed is when there is a
polynomial relation between B1(T) and B2(T): so we have
dB1
dT
=
1
2
σ2
1B2
1 − a1B1 − 1
dB2
dT
=
1
2
σ2
2B2
2 − a2B2 + B1 − δ2
with the corresponding vector field
Z =
1
2
σ2
1B2
1 − a1B1 − 1
∂
∂B1
+
1
2
σ2
2B2
2 − a2B2 + B1 − δ2
∂
∂B2
7
The initial conditions may vary so we set B1(T = 0, z1) = z1 and B2(T =
0, z1, z2) = z2. Let
P(B1, B2) =
L
k,m=0
ckmBk
1 Bm
2
be a polynomial and let P(B1(T, 0), B2(T, 0, 0)) = 0 for all T ≥ 0. We may
assume that this polynomial is prime. The zeroset of P in C2
is a (possibly
singular) algebraic curve. As the real curve T −→ (B1(T, 0), B2(T, 0, 0)) lies
inside this curve and all further derivatives {(B1, B2) with (Zq
)(P)(B1, B2) =
0} and because P is prime we must have Z(P) ∈ (P), so there is some linear
polynomial with
Z(P) = (˜c1B1 + ˜c2B2 + ˜c3) · P
From this we obtain a recurrence relation
(˜c1 −
1
2
σ2
1(k − 1))ck−1,m − (m + 1)ck−1,m+1 = (
1
2
σ2
2(m − 1) − ˜c2)ck,m−1 −
(a1k + a2m + ˜c3)ckm − (m + 1)δ2ck,m+1 − (k + 1)ck+1,m
Let
k0 = max{k ∈ N with ∃m ∈ N with ckm = 0}
m0 = max{m ∈ N with ck0,m = 0}
We set k = k0 + 1 in the recurrence relation and get
(˜c1 −
1
2
σ2
1k0)ck0,m − (m + 1)ck0,m+1 = 0
Putting further m = m0 we get
(˜c1 −
1
2
σ2
1k0)ck0,m0 = 0
or because of ck0,m0 = 0
˜c1 =
1
2
σ2
1k0
But then we conclude 0 = (m + 1)ck0,m+1 for all m = −1, 0, 1, 2, ..., m0 − 1,
so we must have m0 = 0.
8
A similar reasoning shows ckm = 0 for m > k0 − k. So all polynomials P
which carry an invariant vector field Z coming from a 2-dimensional cascade
model must have the form
P(B1, B2) = Bk0
1 +
k0−1
k=0
Bk
1 ·
k0−k
m=0
ckmBm
2
For a further specification of polynomial relations among B1(t) and B2(t)
we make use of some explicit representations. The ODE for B1(t, z1) can be
solved explicitly, its solution is
B1(t, z1) = −
2 − z1(h1 − a1) − (2 + z1(h1 + a1))e−h1t
h1 + a1 − σ2
1z1 + (h1 − a1 + σ2
1z1)e−h1t
where h1 = a2
1 + 2σ2
1 > a1. Set for abbreviation
w1 =
σ2
1z1 + h1 − a1
σ2
1z1 − h1 − a1
Then we may rewrite B1(t, z1) as
B1(t, z1) = −
h1 − a1 + (h1 + a1)w1e−h1t
σ2
1(1 − w1e−h1t)
The second ODE
dB2
dt
=
1
2
σ2
2B2
2 − a2B2 + B1 − δ2
can be transformed to a hypergeometric ODE: Set
B2(t, z) = −
2
σ2
2
·
∂
∂t
C2(t, z) /C2(t, z)
then we obtain a linear ODE of second order
∂2
C2
∂t2
+ a2
∂C2
∂t
+
1
2
σ2
2(B1 − δ2)C2 = 0
Define
h2 = a2
2 + 2σ2
2 δ2 +
2
h1 + a1
9
and two sequences (˜cn)n, (ˆcn)n
˜cn+1
˜cn
=
n2
+ h2
h1
n + 1
h1
σ2
σ1
2
(n + 1)2 + h2
h1
(n + 1)
ˆcn+1
ˆcn
=
n2
− h2
h1
n + 1
h1
σ2
σ1
2
(n + 1)2 − h2
h1
(n + 1)
with initial ˜c0, ˆc0 = 1. If h2
h1
= q ∈ N, then the second sequence has to be
defined for n ≥ q with initial ˆcq = 1. They satisfy
|ˆcn| <
C
n2
|˜cn| <
C
n2
In the following we make the assumption: h2
h1
/∈ N.
We define two power series
R2(w) =
∞
n=0
˜cnwn
R2(w) =
∞
n=0
ˆcnwn
They fulfill hypergeometric ODEs:
w(1 − w)
d2
R2
dw2
+ 1 −
h2
h1
(1 − w)
dR2
dw
−
1
h1
σ2
σ1
2
R2 = 0
w(1 − w)
d2
R2
dw2
+ 1 +
h2
h1
(1 − w)
dR2
dw
−
1
h1
σ2
σ1
2
R2 = 0
Two linearly independent solutions of
∂2
C2
∂t2
+ a2
∂C2
∂t
+
1
2
σ2
2(B1 − δ2)C2 = 0
can now be constructed by setting
C2(t, z1) = e−1
2
(a2+h2)t
R2 e−h1t
w1
C2(t, z1) = e
1
2
(h2−a2)t
R2 e−h1t
w1
10
The Wronskian can be calculated to
∂C2
∂t
(t, z1)C2(t, z1) −
∂C2
∂t
(t, z1)C2(t, z1) = h2
So the general solution C2(t, z1, z2) with initial conditions C2(0, z1, z2) = 1
and ∂C2
∂t
(0, z1, z2) = −1
2
σ2
2z2 is given by
C2(t, z1, z2) = −
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) C2(t, z1) +
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) C2(t, z1)
So far we constructed the general solution space not regarding any special
polynomial restriction. Note that R2(w) can reduce to a polynomial, if there
is an n0 ∈ N with
n2
0 −
h2
h1
n0 +
1
h1
σ2
σ1
2
= 0
Then ˆcn = 0 for n ≥ n0 + 1 and so − 2
σ2
2
· ∂
∂t
C2(t, z) /C2(t, z) will be a
rational function in w1 · e−h1t
and so a rational function in B1(t, z1). Now
we prove that there are no different polynomial relations among B1(t) and
B2(t). So we go back to our polynomial
P(B1, B2) = Bk0
1 +
k0−1
k=0
Bk
1 ·
k0−k
m=0
ckmBm
2
which satisfies Z(P) ∈ (P). Let (z1, z2) be an arbitrary point in the al-
gebraic curve V (P(z1, z2) = 0) ⊂ C2
. If we take this point as an initial
point for the dynamical system t −→ (B1(t, z1), B2(t, z1, z2)), then due to
the invariance property of the ideal (P) under Z the real analytic curve
t −→ (B1(t, z1), B2(t, z1, z2)) lies completely inside the algebraic curve. Now
the explicit formulas for B1(t, z1), B2(t, z) also allow for complex values of t
in a suitable half-plane and by identity they fulfill P(B1(t, z1), B2(t, z)) = 0
for these complex t as well. So we fix a point (z1, z2) with Re(z1) ≤ a1
σ2
1
and
P(z1, z2) = 0 and define for ε > 0
tε =
1
h1
· log(w1) + ε
11
Using
B1(t, z1) = −
h1 − a1 + (h1 + a1)w1e−h1t
σ2
1(1 − w1e−h1t)
we get an estimate
1
Cε
< |B1(tε, z1)| <
C
ε
Because of
|ˆcn| <
C
n2
|˜cn| <
C
n2
the series R2(1), R2(1) are absolutely convergent and so we have
∃ lim
ε→0
C2(tε, z1, z2) = −
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · w
h2−a2
2h1
1 R2(1) +
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · w
−
h2+a2
2h1
1 R2(1)
We remind that
B2(t, z1, z2) = −
2
σ2
2
·
∂C2
∂t
(t, z)
C2(t, z)
and
∂
∂t
C2(t, z1, z2) = −
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1)
∂
∂t
C2(t, z1) +
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1)
∂
∂t
C2(t, z1)
Now using the asymptotic of ˜cn, ˆcn for n −→ ∞ one can easily show
∂C2
∂t
(tε, z1) < C| log(ε)|
∂C2
∂t
(tε, z1) < C| log(ε)|
12
and so
∂C2
∂t
(tε, z1, z2) < C| log(ε)|
where the constant C certainly depends on (z1, z2). If limε→0 C2(tε, z1, z2) =
0, then we would have |B2(tε, z1, z2)| < C| log(ε)|. But now we remind
P(B1(tε, z1), B2(tε, z1, z2)) = Bk0
1 (tε, z1) +
k0−1
k=0
Bk
1 ·
k0−k
m=0
ckmBm
2 = 0
and
1
Cε
< |B1(tε, z1)| <
C
ε
If |B2(tε, z1, z2)| would only grow logarithmically as ε −→ 0 the term |Bk0
1 (tε, z1)| ∼
1
εk0
would dominate all other terms in P and so P(B1(tε, z1), B2(tε, z1, z2)) =
0 for all sufficiently small ε > 0 in contradiction to our assumption. So we
conclude:
0 = lim
ε→0
C2(tε, z1, z2) = −
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · w
h2−a2
2h1
1 R2(1) +
1
h2
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · w
−
h2+a2
2h1
1 R2(1)
respectively
0 =
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · w
h2
h1
1 R2(1) −
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) · R2(1)
for all (z1, z2) ∈ C2
in the algebraic curve P(z1, z2) = 0. (0, 0) must belong
to the curve by assumption and we may start the dynamic t ∈ (−ε0, ε0) −→
(B1(t, 0), B2(t, 0, 0)), so for all z1 real and small positive or negative there is a
(real) z2 with P(z1, z2) = 0. For these z1 we have w1 = w1(z1) =
σ2
1z1+h1−a1
σ2
1z1−h1−a1
<
0. For this choice all the quantities R2(1), R2(1), ∂C2
∂t
(0, z1) + 1
2
σ2
2z2C2(0, z1)
13
and ∂C2
∂t
(0, z1) + 1
2
σ2
2z2C2(0, z1) are real but w
h2
h1
1 has a nontrivial imag-
inary part because of h2
h1
/∈ N. By identity we may conclude
R2(1) ·
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) = 0
R2(1) ·
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) = 0
for all (z1, z2) inside the algebraic curve P = 0. We remind that because of
˜cn > 0 we have R2(1) > 1 positive so we end up with
R2(1) ·
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) = 0
∂C2
∂t
(0, z1) +
1
2
σ2
2z2C2(0, z1) = 0
If R2(1) = 0 then would get
∂C2
∂t
(0, z1)C2(0, z1) −
∂C2
∂t
(0, z1)C2(0, z1) = 0
in contradiction to the Wronskian = h2 > 0. So we have R2(1) = 0. Now
we remind that w −→ R2(w) is a hypergeometric function whose auxiliary
parameters can be read of the ODE
w(1 − w)
d2
R2
dw2
+ 1 −
h2
h1
(1 − w)
dR2
dw
−
1
h1
σ2
σ1
2
R2 = 0
so
γ = 1 + α + β = 1 −
h2
h1
αβ =
1
h1
σ2
σ1
2
ˆcn+1
ˆcn
=
(n + α) · (n + β)
(n + 1) · (n + γ)
14
This implies γ − α − β = 1 > 0 and so Gaussian-summation-formula is
applicable:
R2(1) =
Γ(γ) · Γ(γ − α − β)
Γ(γ − α) · Γ(γ − β)
with
Γ(z) =
1
z
· e−const·z
·
1
∞
m=1(1 + z
m
) · e− z
m
As R2(1) = 0 and Γ(z) = 0 we must have Γ(γ − α) = ∞ or Γ(γ − β) = ∞.
In either case we get α = −q − 1 or β = −q − 1 with q ∈ N and then the
series (ˆcn) terminates at q +1. So we have shown that R2(w) is a polynomial
and further for all (z1, z2) in the algebraic curve
z2 = −
2
σ2
2
·
∂C2
∂t
(0, z1)
C2(0, z1)
= −
2
σ2
2
·

1
2
(h2 − a2) − h1w1
dR2
dw
(w1)
R2(w1)


Now (B1(t, 0), B2(t, 0, 0)) are by assumption in the algebraic curve, so setting
z1 = B1(t, 0) and z2 = B2(t, 0, 0) we get
w1 = w1(z1) =
σ2
1B1(t, 0) + h1 − a1
σ2
1B1(t, 0) − h1 − a1
= −
h1 − a1
h1 + a1
· e−h1t
and
B2(t) = −
2
σ2
2
·

1
2
(h2 − a2) + h1
h1 − a1
h1 + a1
e−h1t
dR2
dw
(−h1−a1
h1+a1
e−h1t
)
R2(−h1−a1
h1+a1
e−h1t)


So we end up with the Proposition:
Proposition 1 All polynomial relations among the first two cascade struc-
ture model coefficient functions arise from termination of (ˆcn) and in this
case B2(t) is a rational function of B1(t).
The next step is to explore wether there are parameter combinations with
B2(t) rational in B1(t) and the cascade model is flexible in the sense of
d
dt
−B2(t)
t
(t = 10) ≥ 0. For that purpose we first derive some estimates.
15
We know that t −→ B2(t) < 0 is falling and converges against limt→∞ B2(t) =
−h2−a2
σ2
2
. So we introduce B2(t) = B2(t) + h2−a2
σ2
2
> 0 and get from the ODE
of B2(t)
dB2
dt
=
1
2
σ2
2B2
2 − h2B2 + B1 +
2
h1 + a1
with initial B2(0) = h2−a2
σ2
2
. Now we have
0 < B1(t) +
2
h1 + a1
<
4h1e−h1t
(h1 + a1)2
The ODE can be written as
dB2
dt
=
1
2
σ2
2B2 − h2 B2 + B1 +
2
h1 + a1
Using that t −→ B2(t) is falling from h2−a2
σ2
2
> 0 we see
1
2
σ2
2B2 − h2 ≤
1
2
(h2 − a2) − h2 = −
1
2
(a2 + h2) < 0
So we can get an upper bound for B2(t) by solving
dB2
dt
= −
1
2
(a2 + h2)B2 +
4h1e−h1t
(h1 + a1)2
which yields
B2(t) =
h2 − a2
σ2
2
e−1
2
(a2+h2)t
+
8h1
(h1 + a1)2(a2 + h2 − 2h1)
(e−2h1t
− e−1
2
(a2+h2)t
)
and
0 < B2(t) +
h2 − a2
σ2
2
< B2(t)
On the other hand we have
dB2
dt
=
dB2
dt
=
1
2
σ2
2B2
2 − h2B2 + B1 +
2
h1 + a1
> −h2B2 > −h2B2
16
In the end we come up with
∆(t) = t2
·
d
dt
−
B2(t)
t
= B2(t) − t
dB2
dt
≤
−
h2 − a2
σ2
2
+ (1 + h2t) · B2(t)
Now the flexibility condition ∆(t = 10) ≥ 0 can be tested numerically against
the termination condition
∃q ∈ N with
h2
h1
= q +
1
qh1
σ2
σ1
2
and the initial condition
0 =
1
2
(h2 − a2) + h1
h1 − a1
h1 + a1
dR2
dw
(−h1−a1
h1+a1
)
R2(−h1−a1
h1+a1
)
We could not find any parameter combinations fulfilling all three conditions.
In the next step we return to the 3-factor cascade setting with a quadratic
relation coming from the reduction of a fourth factor. We remind the linear
relation
B4(T) =
3
j=1
cjBj(T) for all T ≥ 0
by assumption and from that we obtained by applying the vector field the
following relation
1
2
σ2
4
3
j=1
cjBj
2
− a44
3
j=1
cjBj + a41B1 + a42B2 + a43B3 − δ4 =
3
j=1
cj
1
2
σ2
j B2
j − ajBj + Bj−1 − δj
or
P2(B1, B2, B3) =
1
2
c1(σ2
4c1 − σ2
1)B2
1 +
1
2
c2(σ2
4c2 − σ2
2)B2
2 +
1
2
c3(σ2
4c3 − σ2
3)B2
3 +
σ2
4(c1c2B1B2 + c1c3B1B3 + c2c3B2B3) + B1(a41 + c1a1 − c2 − c1a44) +
B2(a42 + c2a2 − c3 − c2a44) + B3(a43 + c3a3 − c3a44) − δ4 + c1δ1 + c2δ2 + c3δ3 = 0
17
Due to the cascade structure of B1(t), B2(t), B3(t) there is a natural restric-
tion of the 4-vector field to a 3-dimensional or further 2-dimensional setting.
So we can apply the vector field
Z3 =
3
j=1
1
2
σ2
j B2
j − ajBj + Bj−1 − δj
∂
∂Bj
to P2 and gain a polynomial P3(B1, B2, B3) = Z3(P2) of degree 3. In the
generic case P2 and P3 intersect transversely, so elimination of B3 will in the
generic case result in a nontrivial polynomial relation between B1, B2. But
we have shown that this is in conflict with the desired flexibility condition.
There are two further special cases possible: first the ideal generated by P2, P3
is induced by one single linear polynomial in B1, B2, B3. But by applying Z3
once more this will result in a polynomial relation among B1, B2. Or second
P3 is in the ideal generated by P2 which means
P3 = (˜c1B1 + ˜c2B2 + ˜c3B3 + ˜δ) · P2
Comparing coefficients of highest order 3 in this equation yields under the
assumption cj = 0 for j = 1, 2, 3
(σ2
j − ˜cj) · (σ2
4cj − σ2
j ) = 0 for j = 1, 2, 3
σ2
4cj(2˜ck − σ2
k) + ˜cj(σ2
4ck − σ2
k) = 0 for j = k
˜c1c2c3 + ˜c2c1c3 + ˜c3c2c1 = 0
The first case is ˜cj = σ2
j for all j = 1, 2, 3. Then we get
σ2
j ck + σ2
kcj =
σ2
j σ2
k
σ2
4
and
0 = σ2
1c2c3 + σ2
2c1c3 + σ2
3c2c1
From the first three equations we infer
cj =
σ2
j
2σ2
4
18
Inserting these expressions in the fourth equation leads to the contradiction
0 =
3σ2
1σ2
2σ2
3
2σ2
4
The second case is
˜c1 = σ2
1
˜c2 = σ2
2
σ2
4c3 = σ2
3
The further equations then imply ˜c3 = 1
2
σ2
3 and
˜c3 =
1
2
σ2
3
c1 = −
σ2
1
σ2
4
c2 = −
σ2
2
σ2
4
which is in contradiction to
σ2
4c2σ2
1 + σ2
2(σ2
4c1 − σ2
1) = 0
The last case where
σ2
4cj = σ2
j for j = 1, 2, 3
implies
˜ck =
1
2
σ2
k
which is in contradiction to
˜c1c2c3 + ˜c2c1c3 + ˜c3c2c1 = 0
So we end up with the
Proposition 2 There are no flexible 4-factor cascade structure affine models
which allow for unspanned volatility.
19
References
Collin-Dufresne,P. and R.Goldstein(2002), ”Do bonds span the fixed income
markets? Theory and evidence for unspanned volatility”, Journal of Finance
57, 1685-1730
Cox,D.A., J.B.Little and D.O’Shea(1992), Ideals, Varieties and Algorithms,
Springer
Duffie,D. and R.Kan(1996), ”A yield factor model of interest rates”, Mathe-
matical Finance 6, 379-406
20
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Affine Yield Curves: Flexibility versus Incompleteness

  • 2. Affine Yield Curves: Flexibility versus Incompleteness Dhia Eddine Barbouche and Bert Koehler Affine interest rate models are amongst the most popular interest rate mod- els due to their flexibility, analytic expressions for bond prices and analytic approximations for some interest derivatives, see for example the seminal ar- ticle of Duffie and Kan [Duf-Kan]. The subclass of affine models for which interest rates are bounded from below is of the form rt = n j=1 δjXjt − c where Xjt = Xj0 + t 0 bj + n k=1 ajkXks ds + σj t 0 XjtdWjs Here rt is the short rate process, δj ≥ 0, c ≥ 0 is some negative displacement term, ajk, bj ≥ 0 except for ajj < 0 and (W1t, ..., Wnt) are independent Brow- nian motions. Here and in the following we always work in the risk neutral measure Q unless otherwise indicated. The logzeroyields defined as Y (T, X0) = − 1 T log E e− T 0 rsds |X0 can be written as an affine linear combination of the state X = (X1, ..., Xn) Y (T, X0) = − 1 T n j=1 (bjAj(T) + Xj0Bj(T)) where the deterministic functions T −→ Aj(T), Bj(T) in maturity T ≥ 0 obey the system of ODEs dBj dT = 1 2 σ2 j B2 j + n k=1 akjBk − δj Bj(0) = 0 and Aj(T) = T 0 Bj(s)ds The shape of the curves T −→ − 1 T Bj(T) determines the flexibility of the model in reproducing the great variety of yield curves seen in the history. 1
  • 3. From this point of view affine models with a cascade structure are promising. Here we deal with affine cascade models with 3 factors: X1t = X10 + t 0 (b1 + X2s − a1X1s)ds + σ1 t 0 X1sdW1s X2t = X20 + t 0 (b2 + X3s − a2X2s)ds + σ2 t 0 X2sdW2s X3t = X30 + t 0 (b3 − a3X3s)ds + σ3 t 0 X3sdW3s The corresponding system of ODEs simplifies to (normalizing δ1 = 1) dB1 dT = 1 2 σ2 1B2 1 − a1B1 − 1 dB2 dT = 1 2 σ2 2B2 2 − a2B2 + B1 − δ2 dB3 dT = 1 2 σ2 3B2 3 − a3B3 + B2 − δ3 One can easily see that T −→ Bj(T) ≤ 0 and falling in T: for that purpose write for example dB2 dT = ( 1 2 σ2 2B2 − a2) · B2 + B1 − δ2 which can be solved as B2(t) = e t 0 ( 1 2 σ2 2B2(s)−a2)ds · B2(0) + t 0 e− s 0 ( 1 2 σ2 2B2(q)−a2)dq · (B1(s) − δ2)ds As B2(0) = 0 and B1(t) − δ2 ≤ 0 we see that B2(t) ≤ 0. Similarly taking derivatives in t we get d2 B2 dT2 = (σ2 2B2 − a2) · dB2 dT + dB1 dT and by the way dB2 dt (t) = e t 0 (σ2 2B2(s)−a2)ds · dB2 dt (0) + t 0 e− s 0 (σ2 2B2(q)−a2)dq · dB1 dt (s)ds As dB2 dt (0) = −δ2 ≤ 0 and dB1 dt (t) ≤ 0 we get dB2 dt (t) ≤ 0. On the other hand we have ∃ limT→∞ Bj(T) and so limT→∞ 1 T Bj(T) = 0. What is interesting about 2
  • 4. affine cascade structure models is that for suitably chosen parameters aj, σj we can achieve that T −→ − 1 T B2(T) attains its maximum in T2,max ≥ 10 and T −→ − 1 T B3(T) attains its maximum in T3,max ≥ 25. This permits a great flexibility of yield curves Y (T, X0) = − 1 T 3 j=1 (bjAj(T) + Xj0Bj(T)) For example the choice of (risk-neutral) parameters   b1 a1 σ1 δ1 b2 a2 σ2 δ2 b3 a3 σ3 δ3   =   0.00005 0.3 0.05 1 0 0.05 0.05 0 0.00005 0.01 0.01 0   and c = 0.015 can represent all historically seen yield triples with tenors 3M, 10Y and 30Y by admissible state vectors X = (X1, X2, X3), Xj ≥ 0 and can even cope with EUR-atm-Swaption volatility surface from 31.03.2016. Affine cascade structure models with d dT − B2(T) T (T = 10) ≥ 0 d dT − B3(T) T (T = 25) ≥ 0 are called flexible affine cascade structure models. The purpose of this note is to elaborate in how far a 4 factor affine model can generate an incomplete bond market together with the flexibility of a 3 factor flexible affine cascade structure model. This means we have a 4 factor affine model Xt = (X1t, ..., X4t) with negatively bounded interest rates, but logzeroyields reduce to a 3 factor state dynamic Y (T, X0) = − 1 T 4 j=1 (bjAj(T) + Xj0Bj(T)) = − 1 T 4 j=1 bjAj(T) − 1 T 3 j=1 Bj(T) · 4 k=1 ˜cjkXk0 where Bj(T) obey a system of ODEs coming from a 3 factor flexible affine cascade structure model. So from the bond-prices alone it is not possible to 3
  • 5. recover the full 4-factor state but we require furthermore that the dynamics of Xt does not reduce generally to a 3-dimensional process. So if we add some interest-optionprices to bondprices the full 4-state Xt can be determinated. The motivation for such an ”unspanned volatility” phenomenon derives from the conjecture that option prices may have a real-world-dynamic not fully ex- plained by the real-world-dynamic of yields. In general unspanned volatility in an n-dimensional affine model postulates a nontrivial linear relation 0 = n j=1 cjBj(T) for all T ≥ 0 Defining a (polynomial) vector field on the ring of polynomials P ∈ C[B1, ..., Bn] by Z = n j=1 1 2 σ2 j B2 j + n k=1 akjBk − δj ∂ ∂Bj we get a sequence of polynomials Pm by setting P1(B1, ..., Bn) = n j=1 cjBj Pm+1(B1, ..., Bn) = Z(Pm) Let I = I(Pm, m ∈ N) be the ideal generated by the full sequence. By Hilberts Basis Theorem I is finitely generated, so there are P1, ..., PN which generate the whole ideal. In fact it is sufficient to find a N ∈ N with PN+1 ∈ I(P1, ..., PN ) because then all further PN+2, PN+3, ... are also in I(P1, ..., PN ). Let V (I) = V (P1, ..., PN ) ⊂ Cn be the algebraic variety of the ideal, that means V (P1, ..., PN ) = {(B1, ..., Bn) with Pm(B1, ..., Bn) = 0 for m = 1, ..., N} Obviously the (real analytic) curve T −→ (B1(T), ..., Bn(T)) lies inside V (P1, ..., PN ). On the other hand if (0, ..., 0) ∈ V (P1, ..., PN ), then we have n j=1 cj dm Bj dtm (t = 0) = 0 4
  • 6. first for 0 ≤ m ≤ N − 1 but then due to the generation property also for all m. Because T −→ (B1(T), ..., Bn(T)) is real analytic we have n j=1 cjBj(t) = 0 identically in this case, so the two conditions are equivalent. This simple observation reduces the search for unspanned affine models to a polynomial basis problem which can in principle be solved by algorithm. Up to now the only known example stems from Colin-Dufresne and Goldstein for n = 3 and results in a rational curve as V (P1, P2) ⊂ C3 . So there are no ”transcenden- tal” unspanned affine models known. Maybe this is a general property due to the fact that the intersection variety carries a holomorphic vector field with meromorphic pole of first order in the projective extension. Coming back to our 4-factor setting it is instructive to consider an exam- ple: Assume that Bj(T) = Bj(T) for j = 1, 2, 3, so we have dB1 dT = 1 2 σ2 1B2 1 − a1B1 − 1 dB2 dT = 1 2 σ2 2B2 2 − a2B2 + B1 − δ2 dB3 dT = 1 2 σ2 3B2 3 − a3B3 + B2 − δ3 and furthermore dB4 dt = 1 2 σ2 4B2 4 − a44B4 + a41B1 + a42B2 + a43B3 − δ4 The corresponding process looks like X1t = X10 + t 0 (b1 + a41X4s + X2s − a1X1s)ds + σ1 t 0 X1sdW1s X2t = X20 + t 0 (b2 + a42X4s + X3s − a2X2s)ds + σ2 t 0 X2sdW2s X3t = X30 + t 0 (b3 + a43X4s − a3X3s)ds + σ3 t 0 X3sdW3s X4t = X40 + t 0 (b4 − a44X4s)ds + σ4 t 0 X4sdW4s 5
  • 7. Now assume we have a linear relation B4(T) = 3 j=1 cjBj(T) for all T ≥ 0 Taking derivatives we obtain 1 2 σ2 4 3 j=1 cjBj 2 − a44 3 j=1 cjBj + a41B1 + a42B2 + a43B3 − δ4 = 3 j=1 cj 1 2 σ2 j B2 j − ajBj + Bj−1 − δj If this quadratic relation vanishes identically in C[B1, B2, B3] we must have B4(T) = cjBj(T) for some j ∈ {1, 2, 3} and σ2 4c2 j = cjσ2 j cjaj − a44cj + a4j = 0 a4,j−1 − cj = 0 a4,j−2 = 0 cjδj − δ4 = 0 If for example j = 3 then c3 = σ2 3 σ2 4 a43 = (a44 − a3)c3 a42 = c3 a41 = 0 δ4 = c3δ3 6
  • 8. Now we define X5t = X3t + c3X4t. Then we get X1t = X10 + t 0 (b1 + X2s − a1X1s)ds + σ1 t 0 X1sdW1s X2t = X20 + t 0 (b2 + X5s − a2X2s)ds + σ2 t 0 X2sdW2s X5t = X50 + t 0 (b3 + c3b4 + a43X4s − a3X3s − c3a44X4s)ds + σ3 t 0 X3sdW3s + c3σ4 t 0 X4sdW4s Inserting a43 = (a44 − a3)c3 and using c3 = σ2 3 σ2 4 we get σ3 t 0 X3sdW3s + c3σ4 t 0 X4sdW4s ∼ σ3 t 0 X5sdW5s and X1t = X10 + t 0 (b1 + X2s − a1X1s)ds + σ1 t 0 X1sdW1s X2t = X20 + t 0 (b2 + X5s − a2X2s)ds + σ2 t 0 X2sdW2s X5t = X50 + t 0 (b3 + c3b4 − a3X5s)ds + σ3 t 0 X5sdW5s so the process collapses to a 3-factor dynamic. The first situation we want to examine more detailed is when there is a polynomial relation between B1(T) and B2(T): so we have dB1 dT = 1 2 σ2 1B2 1 − a1B1 − 1 dB2 dT = 1 2 σ2 2B2 2 − a2B2 + B1 − δ2 with the corresponding vector field Z = 1 2 σ2 1B2 1 − a1B1 − 1 ∂ ∂B1 + 1 2 σ2 2B2 2 − a2B2 + B1 − δ2 ∂ ∂B2 7
  • 9. The initial conditions may vary so we set B1(T = 0, z1) = z1 and B2(T = 0, z1, z2) = z2. Let P(B1, B2) = L k,m=0 ckmBk 1 Bm 2 be a polynomial and let P(B1(T, 0), B2(T, 0, 0)) = 0 for all T ≥ 0. We may assume that this polynomial is prime. The zeroset of P in C2 is a (possibly singular) algebraic curve. As the real curve T −→ (B1(T, 0), B2(T, 0, 0)) lies inside this curve and all further derivatives {(B1, B2) with (Zq )(P)(B1, B2) = 0} and because P is prime we must have Z(P) ∈ (P), so there is some linear polynomial with Z(P) = (˜c1B1 + ˜c2B2 + ˜c3) · P From this we obtain a recurrence relation (˜c1 − 1 2 σ2 1(k − 1))ck−1,m − (m + 1)ck−1,m+1 = ( 1 2 σ2 2(m − 1) − ˜c2)ck,m−1 − (a1k + a2m + ˜c3)ckm − (m + 1)δ2ck,m+1 − (k + 1)ck+1,m Let k0 = max{k ∈ N with ∃m ∈ N with ckm = 0} m0 = max{m ∈ N with ck0,m = 0} We set k = k0 + 1 in the recurrence relation and get (˜c1 − 1 2 σ2 1k0)ck0,m − (m + 1)ck0,m+1 = 0 Putting further m = m0 we get (˜c1 − 1 2 σ2 1k0)ck0,m0 = 0 or because of ck0,m0 = 0 ˜c1 = 1 2 σ2 1k0 But then we conclude 0 = (m + 1)ck0,m+1 for all m = −1, 0, 1, 2, ..., m0 − 1, so we must have m0 = 0. 8
  • 10. A similar reasoning shows ckm = 0 for m > k0 − k. So all polynomials P which carry an invariant vector field Z coming from a 2-dimensional cascade model must have the form P(B1, B2) = Bk0 1 + k0−1 k=0 Bk 1 · k0−k m=0 ckmBm 2 For a further specification of polynomial relations among B1(t) and B2(t) we make use of some explicit representations. The ODE for B1(t, z1) can be solved explicitly, its solution is B1(t, z1) = − 2 − z1(h1 − a1) − (2 + z1(h1 + a1))e−h1t h1 + a1 − σ2 1z1 + (h1 − a1 + σ2 1z1)e−h1t where h1 = a2 1 + 2σ2 1 > a1. Set for abbreviation w1 = σ2 1z1 + h1 − a1 σ2 1z1 − h1 − a1 Then we may rewrite B1(t, z1) as B1(t, z1) = − h1 − a1 + (h1 + a1)w1e−h1t σ2 1(1 − w1e−h1t) The second ODE dB2 dt = 1 2 σ2 2B2 2 − a2B2 + B1 − δ2 can be transformed to a hypergeometric ODE: Set B2(t, z) = − 2 σ2 2 · ∂ ∂t C2(t, z) /C2(t, z) then we obtain a linear ODE of second order ∂2 C2 ∂t2 + a2 ∂C2 ∂t + 1 2 σ2 2(B1 − δ2)C2 = 0 Define h2 = a2 2 + 2σ2 2 δ2 + 2 h1 + a1 9
  • 11. and two sequences (˜cn)n, (ˆcn)n ˜cn+1 ˜cn = n2 + h2 h1 n + 1 h1 σ2 σ1 2 (n + 1)2 + h2 h1 (n + 1) ˆcn+1 ˆcn = n2 − h2 h1 n + 1 h1 σ2 σ1 2 (n + 1)2 − h2 h1 (n + 1) with initial ˜c0, ˆc0 = 1. If h2 h1 = q ∈ N, then the second sequence has to be defined for n ≥ q with initial ˆcq = 1. They satisfy |ˆcn| < C n2 |˜cn| < C n2 In the following we make the assumption: h2 h1 /∈ N. We define two power series R2(w) = ∞ n=0 ˜cnwn R2(w) = ∞ n=0 ˆcnwn They fulfill hypergeometric ODEs: w(1 − w) d2 R2 dw2 + 1 − h2 h1 (1 − w) dR2 dw − 1 h1 σ2 σ1 2 R2 = 0 w(1 − w) d2 R2 dw2 + 1 + h2 h1 (1 − w) dR2 dw − 1 h1 σ2 σ1 2 R2 = 0 Two linearly independent solutions of ∂2 C2 ∂t2 + a2 ∂C2 ∂t + 1 2 σ2 2(B1 − δ2)C2 = 0 can now be constructed by setting C2(t, z1) = e−1 2 (a2+h2)t R2 e−h1t w1 C2(t, z1) = e 1 2 (h2−a2)t R2 e−h1t w1 10
  • 12. The Wronskian can be calculated to ∂C2 ∂t (t, z1)C2(t, z1) − ∂C2 ∂t (t, z1)C2(t, z1) = h2 So the general solution C2(t, z1, z2) with initial conditions C2(0, z1, z2) = 1 and ∂C2 ∂t (0, z1, z2) = −1 2 σ2 2z2 is given by C2(t, z1, z2) = − 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) C2(t, z1) + 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) C2(t, z1) So far we constructed the general solution space not regarding any special polynomial restriction. Note that R2(w) can reduce to a polynomial, if there is an n0 ∈ N with n2 0 − h2 h1 n0 + 1 h1 σ2 σ1 2 = 0 Then ˆcn = 0 for n ≥ n0 + 1 and so − 2 σ2 2 · ∂ ∂t C2(t, z) /C2(t, z) will be a rational function in w1 · e−h1t and so a rational function in B1(t, z1). Now we prove that there are no different polynomial relations among B1(t) and B2(t). So we go back to our polynomial P(B1, B2) = Bk0 1 + k0−1 k=0 Bk 1 · k0−k m=0 ckmBm 2 which satisfies Z(P) ∈ (P). Let (z1, z2) be an arbitrary point in the al- gebraic curve V (P(z1, z2) = 0) ⊂ C2 . If we take this point as an initial point for the dynamical system t −→ (B1(t, z1), B2(t, z1, z2)), then due to the invariance property of the ideal (P) under Z the real analytic curve t −→ (B1(t, z1), B2(t, z1, z2)) lies completely inside the algebraic curve. Now the explicit formulas for B1(t, z1), B2(t, z) also allow for complex values of t in a suitable half-plane and by identity they fulfill P(B1(t, z1), B2(t, z)) = 0 for these complex t as well. So we fix a point (z1, z2) with Re(z1) ≤ a1 σ2 1 and P(z1, z2) = 0 and define for ε > 0 tε = 1 h1 · log(w1) + ε 11
  • 13. Using B1(t, z1) = − h1 − a1 + (h1 + a1)w1e−h1t σ2 1(1 − w1e−h1t) we get an estimate 1 Cε < |B1(tε, z1)| < C ε Because of |ˆcn| < C n2 |˜cn| < C n2 the series R2(1), R2(1) are absolutely convergent and so we have ∃ lim ε→0 C2(tε, z1, z2) = − 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · w h2−a2 2h1 1 R2(1) + 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · w − h2+a2 2h1 1 R2(1) We remind that B2(t, z1, z2) = − 2 σ2 2 · ∂C2 ∂t (t, z) C2(t, z) and ∂ ∂t C2(t, z1, z2) = − 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) ∂ ∂t C2(t, z1) + 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) ∂ ∂t C2(t, z1) Now using the asymptotic of ˜cn, ˆcn for n −→ ∞ one can easily show ∂C2 ∂t (tε, z1) < C| log(ε)| ∂C2 ∂t (tε, z1) < C| log(ε)| 12
  • 14. and so ∂C2 ∂t (tε, z1, z2) < C| log(ε)| where the constant C certainly depends on (z1, z2). If limε→0 C2(tε, z1, z2) = 0, then we would have |B2(tε, z1, z2)| < C| log(ε)|. But now we remind P(B1(tε, z1), B2(tε, z1, z2)) = Bk0 1 (tε, z1) + k0−1 k=0 Bk 1 · k0−k m=0 ckmBm 2 = 0 and 1 Cε < |B1(tε, z1)| < C ε If |B2(tε, z1, z2)| would only grow logarithmically as ε −→ 0 the term |Bk0 1 (tε, z1)| ∼ 1 εk0 would dominate all other terms in P and so P(B1(tε, z1), B2(tε, z1, z2)) = 0 for all sufficiently small ε > 0 in contradiction to our assumption. So we conclude: 0 = lim ε→0 C2(tε, z1, z2) = − 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · w h2−a2 2h1 1 R2(1) + 1 h2 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · w − h2+a2 2h1 1 R2(1) respectively 0 = ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · w h2 h1 1 R2(1) − ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) · R2(1) for all (z1, z2) ∈ C2 in the algebraic curve P(z1, z2) = 0. (0, 0) must belong to the curve by assumption and we may start the dynamic t ∈ (−ε0, ε0) −→ (B1(t, 0), B2(t, 0, 0)), so for all z1 real and small positive or negative there is a (real) z2 with P(z1, z2) = 0. For these z1 we have w1 = w1(z1) = σ2 1z1+h1−a1 σ2 1z1−h1−a1 < 0. For this choice all the quantities R2(1), R2(1), ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) 13
  • 15. and ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) are real but w h2 h1 1 has a nontrivial imag- inary part because of h2 h1 /∈ N. By identity we may conclude R2(1) · ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) = 0 R2(1) · ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) = 0 for all (z1, z2) inside the algebraic curve P = 0. We remind that because of ˜cn > 0 we have R2(1) > 1 positive so we end up with R2(1) · ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) = 0 ∂C2 ∂t (0, z1) + 1 2 σ2 2z2C2(0, z1) = 0 If R2(1) = 0 then would get ∂C2 ∂t (0, z1)C2(0, z1) − ∂C2 ∂t (0, z1)C2(0, z1) = 0 in contradiction to the Wronskian = h2 > 0. So we have R2(1) = 0. Now we remind that w −→ R2(w) is a hypergeometric function whose auxiliary parameters can be read of the ODE w(1 − w) d2 R2 dw2 + 1 − h2 h1 (1 − w) dR2 dw − 1 h1 σ2 σ1 2 R2 = 0 so γ = 1 + α + β = 1 − h2 h1 αβ = 1 h1 σ2 σ1 2 ˆcn+1 ˆcn = (n + α) · (n + β) (n + 1) · (n + γ) 14
  • 16. This implies γ − α − β = 1 > 0 and so Gaussian-summation-formula is applicable: R2(1) = Γ(γ) · Γ(γ − α − β) Γ(γ − α) · Γ(γ − β) with Γ(z) = 1 z · e−const·z · 1 ∞ m=1(1 + z m ) · e− z m As R2(1) = 0 and Γ(z) = 0 we must have Γ(γ − α) = ∞ or Γ(γ − β) = ∞. In either case we get α = −q − 1 or β = −q − 1 with q ∈ N and then the series (ˆcn) terminates at q +1. So we have shown that R2(w) is a polynomial and further for all (z1, z2) in the algebraic curve z2 = − 2 σ2 2 · ∂C2 ∂t (0, z1) C2(0, z1) = − 2 σ2 2 ·  1 2 (h2 − a2) − h1w1 dR2 dw (w1) R2(w1)   Now (B1(t, 0), B2(t, 0, 0)) are by assumption in the algebraic curve, so setting z1 = B1(t, 0) and z2 = B2(t, 0, 0) we get w1 = w1(z1) = σ2 1B1(t, 0) + h1 − a1 σ2 1B1(t, 0) − h1 − a1 = − h1 − a1 h1 + a1 · e−h1t and B2(t) = − 2 σ2 2 ·  1 2 (h2 − a2) + h1 h1 − a1 h1 + a1 e−h1t dR2 dw (−h1−a1 h1+a1 e−h1t ) R2(−h1−a1 h1+a1 e−h1t)   So we end up with the Proposition: Proposition 1 All polynomial relations among the first two cascade struc- ture model coefficient functions arise from termination of (ˆcn) and in this case B2(t) is a rational function of B1(t). The next step is to explore wether there are parameter combinations with B2(t) rational in B1(t) and the cascade model is flexible in the sense of d dt −B2(t) t (t = 10) ≥ 0. For that purpose we first derive some estimates. 15
  • 17. We know that t −→ B2(t) < 0 is falling and converges against limt→∞ B2(t) = −h2−a2 σ2 2 . So we introduce B2(t) = B2(t) + h2−a2 σ2 2 > 0 and get from the ODE of B2(t) dB2 dt = 1 2 σ2 2B2 2 − h2B2 + B1 + 2 h1 + a1 with initial B2(0) = h2−a2 σ2 2 . Now we have 0 < B1(t) + 2 h1 + a1 < 4h1e−h1t (h1 + a1)2 The ODE can be written as dB2 dt = 1 2 σ2 2B2 − h2 B2 + B1 + 2 h1 + a1 Using that t −→ B2(t) is falling from h2−a2 σ2 2 > 0 we see 1 2 σ2 2B2 − h2 ≤ 1 2 (h2 − a2) − h2 = − 1 2 (a2 + h2) < 0 So we can get an upper bound for B2(t) by solving dB2 dt = − 1 2 (a2 + h2)B2 + 4h1e−h1t (h1 + a1)2 which yields B2(t) = h2 − a2 σ2 2 e−1 2 (a2+h2)t + 8h1 (h1 + a1)2(a2 + h2 − 2h1) (e−2h1t − e−1 2 (a2+h2)t ) and 0 < B2(t) + h2 − a2 σ2 2 < B2(t) On the other hand we have dB2 dt = dB2 dt = 1 2 σ2 2B2 2 − h2B2 + B1 + 2 h1 + a1 > −h2B2 > −h2B2 16
  • 18. In the end we come up with ∆(t) = t2 · d dt − B2(t) t = B2(t) − t dB2 dt ≤ − h2 − a2 σ2 2 + (1 + h2t) · B2(t) Now the flexibility condition ∆(t = 10) ≥ 0 can be tested numerically against the termination condition ∃q ∈ N with h2 h1 = q + 1 qh1 σ2 σ1 2 and the initial condition 0 = 1 2 (h2 − a2) + h1 h1 − a1 h1 + a1 dR2 dw (−h1−a1 h1+a1 ) R2(−h1−a1 h1+a1 ) We could not find any parameter combinations fulfilling all three conditions. In the next step we return to the 3-factor cascade setting with a quadratic relation coming from the reduction of a fourth factor. We remind the linear relation B4(T) = 3 j=1 cjBj(T) for all T ≥ 0 by assumption and from that we obtained by applying the vector field the following relation 1 2 σ2 4 3 j=1 cjBj 2 − a44 3 j=1 cjBj + a41B1 + a42B2 + a43B3 − δ4 = 3 j=1 cj 1 2 σ2 j B2 j − ajBj + Bj−1 − δj or P2(B1, B2, B3) = 1 2 c1(σ2 4c1 − σ2 1)B2 1 + 1 2 c2(σ2 4c2 − σ2 2)B2 2 + 1 2 c3(σ2 4c3 − σ2 3)B2 3 + σ2 4(c1c2B1B2 + c1c3B1B3 + c2c3B2B3) + B1(a41 + c1a1 − c2 − c1a44) + B2(a42 + c2a2 − c3 − c2a44) + B3(a43 + c3a3 − c3a44) − δ4 + c1δ1 + c2δ2 + c3δ3 = 0 17
  • 19. Due to the cascade structure of B1(t), B2(t), B3(t) there is a natural restric- tion of the 4-vector field to a 3-dimensional or further 2-dimensional setting. So we can apply the vector field Z3 = 3 j=1 1 2 σ2 j B2 j − ajBj + Bj−1 − δj ∂ ∂Bj to P2 and gain a polynomial P3(B1, B2, B3) = Z3(P2) of degree 3. In the generic case P2 and P3 intersect transversely, so elimination of B3 will in the generic case result in a nontrivial polynomial relation between B1, B2. But we have shown that this is in conflict with the desired flexibility condition. There are two further special cases possible: first the ideal generated by P2, P3 is induced by one single linear polynomial in B1, B2, B3. But by applying Z3 once more this will result in a polynomial relation among B1, B2. Or second P3 is in the ideal generated by P2 which means P3 = (˜c1B1 + ˜c2B2 + ˜c3B3 + ˜δ) · P2 Comparing coefficients of highest order 3 in this equation yields under the assumption cj = 0 for j = 1, 2, 3 (σ2 j − ˜cj) · (σ2 4cj − σ2 j ) = 0 for j = 1, 2, 3 σ2 4cj(2˜ck − σ2 k) + ˜cj(σ2 4ck − σ2 k) = 0 for j = k ˜c1c2c3 + ˜c2c1c3 + ˜c3c2c1 = 0 The first case is ˜cj = σ2 j for all j = 1, 2, 3. Then we get σ2 j ck + σ2 kcj = σ2 j σ2 k σ2 4 and 0 = σ2 1c2c3 + σ2 2c1c3 + σ2 3c2c1 From the first three equations we infer cj = σ2 j 2σ2 4 18
  • 20. Inserting these expressions in the fourth equation leads to the contradiction 0 = 3σ2 1σ2 2σ2 3 2σ2 4 The second case is ˜c1 = σ2 1 ˜c2 = σ2 2 σ2 4c3 = σ2 3 The further equations then imply ˜c3 = 1 2 σ2 3 and ˜c3 = 1 2 σ2 3 c1 = − σ2 1 σ2 4 c2 = − σ2 2 σ2 4 which is in contradiction to σ2 4c2σ2 1 + σ2 2(σ2 4c1 − σ2 1) = 0 The last case where σ2 4cj = σ2 j for j = 1, 2, 3 implies ˜ck = 1 2 σ2 k which is in contradiction to ˜c1c2c3 + ˜c2c1c3 + ˜c3c2c1 = 0 So we end up with the Proposition 2 There are no flexible 4-factor cascade structure affine models which allow for unspanned volatility. 19
  • 21. References Collin-Dufresne,P. and R.Goldstein(2002), ”Do bonds span the fixed income markets? Theory and evidence for unspanned volatility”, Journal of Finance 57, 1685-1730 Cox,D.A., J.B.Little and D.O’Shea(1992), Ideals, Varieties and Algorithms, Springer Duffie,D. and R.Kan(1996), ”A yield factor model of interest rates”, Mathe- matical Finance 6, 379-406 20 View publication statsView publication stats