Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions
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irk-123456789-45642009-12-08T12:00:29Z Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions Banna, O. Mishura, Y. 2008 Article Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions / O. Banna, Y. Mishura // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 3-4. — С. 1-16. — Бібліогр.: 7 назв.— англ. 0321-3900 http://dspace.nbuv.gov.ua/handle/123456789/4564 en Інститут математики НАН України |
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Banna, O. Mishura, Y. Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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Banna, O. Mishura, Y. |
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Banna, O. |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions |
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approximation of fractional brownian motion with associated hurst index separated from 1 by stochastic integrals of linear power functions |
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Інститут математики НАН України |
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2008 |
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http://dspace.nbuv.gov.ua/handle/123456789/4564 |
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Approximation of fractional Brownian motion with associated Hurst index separated from 1 by stochastic integrals of linear power functions / O. Banna, Y. Mishura // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 3-4. — С. 1-16. — Бібліогр.: 7 назв.— англ. |
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AT bannao approximationoffractionalbrownianmotionwithassociatedhurstindexseparatedfrom1bystochasticintegralsoflinearpowerfunctions AT mishuray approximationoffractionalbrownianmotionwithassociatedhurstindexseparatedfrom1bystochasticintegralsoflinearpowerfunctions |
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2025-07-02T07:46:37Z |
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Theory of Stochastic Processes
Vol.14 (30), no.3-4, 2008, pp.1-16
OKSANA BANNA AND YULIYA MISHURA
APPROXIMATION OF FRACTIONAL
BROWNIAN MOTION WITH ASSOCIATED
HURST INDEX SEPARATED FROM 1 BY
STOCHASTIC INTEGRALS OF LINEAR POWER
FUNCTIONS
In this article we present the best uniform approximation of the frac-
tional Brownian motion in space L∞([0, T ];L2(Ω)) by martingales
of the following type
t∫
0
a(s)dWs, where W is a Wiener process, a
is a function defined by a(s) = k1 + k2s
α, k1, k2 ∈ R, s ∈ [0, T ],
α = H − 1/2, H is the Hurst index, separated from 1, associated
with the fractional Brownian motion.
1. Introduction
A fractional Brownian motion (fBm) with associated Hurst index H ∈
(0, 1) is a Gaussian process {BH
t , t ≥ 0} with mean EBH
t = 0 and co-
variance EBH
t B
H
s = 1
2
(t2H + s2H − |t − s|2H), such that BH
0 = 0. We
consider only the case when the Hurst index H ∈ (1
2
, 1). From [6] we
know that fBm {BH
t , t ∈ [0, T ]} admits the following representation
BH
t =
t∫
0
z(t, s)dWs, where {Wt, t ∈ [0, T ]} is a Wiener process, z(t, s) =
(
H − 1
2
)
cH s
1/2−H
t∫
s
uH−1/2(u−s)H−3/2du, cH =
(
2H·Γ
�
3
2
−H
�
Γ
�
H+
1
2
�
Γ(2−2H)
)1/2
, Γ(x),
x > 0 is the Gamma function. We will use the following notation α = H− 1
2
.
It is known that the fBm with index H ∈ (1
2
, 1) is not a semimartin-
gale, actually it is neither a martingale nor a process of bounded variation.
However in [1], [5], and [7] it is shown that in some metric space fBm can
be approximated by semimartingales and processes of bounded variation.
Hence in this article we present the approximation of the fractional Brow-
nian motion in the space L∞([0, T ];L2(Ω)) by martingales, namely, by the
2000 Mathematics Subject Classifications. 60G15, 60H05.
Key words and phrases. Fractional Brownian motion, Wiener integral, approxima-
tion.
1
2 OKSANA BANNA AND YULIYA MISHURA
stochastic integrals of the particular Wiener process
∫ t
0
a(s)dWs, where W
is a standard Wiener process, a is a non-random function a ∈ L2[0, T ]. In
fact, we consider only a particular case when a(s) is a linear power function,
and under this restriction we derive some interesting and even surprising
results. In the previous articles we presented the approximation of the fBm
with constant functions, power functions of the type a(s) = k · sα, k ∈ R
and functions a(s) = k · sγ , k ∈ R, γ > 0 [2], and power functions with a
negative power a(s) = k ·s−α, k > 0 [4]. Here we present the approximation
of fBm with functions a(s) = k1 + k2 · sα, k1, k2 ∈ R, for H < H0, where
H0 ≈ 0, 94005165. In [3] this approximation is derived in empirical way (be-
low it is shown how to define H0), as well as the minimum and the function
where this minimum is reached are presented. Note that this minimum is
not equal to zero. The best approximation of the fractional Brownian mo-
tion for H ≥ H0 is presented in [3]. Namely, it is proved that the minimum
of the function max
t∈[0,T ]
E(BH
t −Mt)
2, where Mt =
t∫
0
a(s)dWs, t ∈ [0, T ] is a
stochastic integral of Wiener process {Wt, t ∈ [0, T ]} (Wiener integral),
is reached at certain k1 and k2. We briefly outline an algorithm of how to
find this minimum. For a function a(s) = k1 + k2 · sα, k1, k2 ∈ R
E(BH
t −Mt)
2 = t2H
(
1 − k2 · cHH +
k2
2
2H
)
− 2k1 · tα+1
α+1
(c3(H) − k2) + k2
1t
=: f(t, k1, k2),
where c3(H) = c1(H)(α+ 1), c1(H) = α cH · 1
α+1
· B(1 − α, α).
First consider the derivative over t of f(t, k1, k2):
∂f
∂t
= t2α (2H − 2k2 · cH + k2
2) + 2tαk1 (k2 − c3(H)) + k2
1. (1)
This quadratic polynomial of t has its discriminant
D
4
= k2
1
(
c23(H) − 2H − 2k2 (c3(H) − cH)
)
. (2)
We equate this discriminant with zero and find k2:
k2 =
2H−c23(H)
2(cH−c3(H))
=: k0
2.
Discriminant D in (2) is linearly dependent on k2, in fact, it is decreasing
with respect to k2, hence:
a) if k2 > k0
2 then D < 0;
b) if k2 = k0
2 then D = 0;
c) if k2 < k0
2 then D > 0, and, in this case, if k1 ≥ 0 then ∂f
∂t
= 0 at
points t1 = t1(k1, k2) and t2 = t2(k1, k2), where
t1(k1, k2) =
(
|k1|
�
c3(H)−k2−
√
c23(H)−2H−2k2(c3(H)−cH)
�
2H−2k2cH+k2
2
) 1
α
=: |k1|
1
α · c1(k2),
APPROXIMATION OF FBM 3
t2(k1, k2) =
(
|k1|
�
c3(H)−k2+
√
c23(H)−2H−2k2(c3(H)−cH)
�
2H−2k2cH+k2
2
) 1
α
.
These roots exist and are positive if and only if k1(c3(H)−k2) > 0. However
we consider only the case when k2 < k0
2, and k0
2 < 0, so the roots are positive
if and only if k1 > 0. Further, we investigate the behavior of f(t, k1, k2) with
respect to k1, k2. If k1 ≤ 0 then ∂f
∂t
> 0 for all t.
If k1 > 0 then ∂f
∂t
> 0 for 0 < t < x
1
α
1 , and the function f(t, k1, k2) is
increasing; ∂f
∂t
< 0 for x
1
α
1 < t < x
1
α
2 , f(t, k1, k2) is decreasing; ∂f
∂t
> 0 for
t > x
1
α
2 , f(t, k1, k2) is increasing.
Consider the following cases:
I. For k1 ≤ 0, k2 ∈ R
min
k1≤0,k2∈�
max
t∈[0,T ]
f(t, k1, k2) = min
k1≤0,k2∈�
f(T, k1, k2) = T 2H
(
1 − c2H
2H
)
.
It is proved that the minimum is reached at k1 = 0, k2 = cH .
Use the proposition in [3], showing that function f(T, k1, k2) has its
minimum at (k∗1, k
∗
2), where
k∗1 = (α+1)(c3(H)−cH)Tα
α2 ,
k∗2 = (α+1)(cH(α+1)−c1(H)(2α+1))
α2 .
(3)
Moreover in [3] there are graphs displaying the dependence of k0
2 and k∗2
on H , and it is easy to see that if H < H0 then we have k∗2 < k0
2, and for
H > H0 k
∗
2 > k0
2. Hence H0 is a unique point of intersection of these graphs.
II. For k1 > 0, k2 ≥ k0
2 we have
a) if H > H0 then k∗2 > k0
2, hence
min
k1>0,k2>k0
2
f(T, k1, k2) = min
k1∈R,k2∈R
f(T, k1, k2) = f(T, k∗1, k
∗
2) =
= T 2α+1 − T 2α+1c2H
(
B2(1 − α, α) − 2B(1−α,α)
α
+ (α+1)2
α2(2α+1)
)
.
b) if H < H0 then k∗2 < k0
2, hence for the minimum we have:
min
k1≥0,k2≥k0
2
max
t∈[0,T ]
f(t, k1, k2) = min
k1≥0,k2≥k0
2
f(T, k1, k2)
= f(T )|
k1=
�
c1(H)− k0
2
α+1
�
Tα, k2=k0
2
= −
(
c1(H) − k0
2
α+1
)2
· T 2α+1 + T 2α+1 − 2k0
2
cH
2α+1
T 2α+1 + (k0
2)
2 · T 2α+1
2α+1
.
4 OKSANA BANNA AND YULIYA MISHURA
III. Let k1 > 0, k2 < k0
2. Since the function f(t, k1, k2) is increasing on
[0, t1], [t2,+∞) and decreasing on [t1, t2], we have:
a) if t1 > T then f(t, k1, k2) is increasing on [0, T ], and max
t∈[0,T ]
f(t, k1, k2) =
f(T, k1, k2);
b) if t1 < T then max
t∈[0,T ]
f(t, k1, k2) = max {f(T, k1, k2), f(t1, k1, k2)};
c) if t1 = T then max
t∈[0,T ]
f(t, k1, k2) = f(T, k1, k2) = f(t1, k1, k2).
Hence,
min
k1≥0,k2<k0
2
max
t∈[0,T ]
f(t, k1, k2)
= min
k1≥0,k2<k0
2
{
max{f(T, k1, k2), f(t1, k1, k2)}, t1 < T
f(T, k1, k2), t1 ≥ T
.
However, as mentioned above, in [3] it is proved that for H > H0 and
k∗2 ≥ k0
2 function max
t∈[0,T ]
f(t, k1, k2) has its minimum at k1 = k∗1, k2 = k∗2,
which is equal to
T 2α+1 − T 2α+1c2H
(
B2(1 − α, α) − 2B(1−α,α)
α
+ (α+1)2
α2(2α+1)
)
.
Hence for H > H0 one may ignore this case.
The remaining case, when H < H0 and k∗2 < k0
2, is considered in this
work. As H0 < 1, we call these values of Hurst index separated from 1. We
will call the point (k∗1, k
∗
2) minimal one. The paper is organized as follows:
in Section 2 we consider some auxiliary results demonstrating that it is
impossible to achieve the required minimum for H < H0 in the minimal
point or in some connected point. In Section 3 we demonstrate analytically
and with the help of the graph how to achieve minimum in our case.
2. Some approximation results for Hurst index, separated
from 1, for minimal point and for connected points
Consider the following expression:
f(t1(k1, k2), k1, k2)
= k
2H
α
1 (c1(k2))
2H a(k2) − 2k1
k
α+1
α
1 (c1(k2))
α+1
α + 1
b(k2) + k2
1k
1
α
1 c1(k2)
=: k
2H
α
1 ϕ(k2).
Here a(k2) = 1−k2 · cHH +
k2
2
2H
> 0, b(k2) = c3(H)−k2, ϕ(k2) > 0, k2 < k0
2 < 0.
APPROXIMATION OF FBM 5
Now we present a number of “negative” propositions concerned minimal
point and connected points.
Lemma 2.1. If k∗2 < k0
2, then the minimum reached in case III is less than
the minimum in cases II b) and I.
Proof. First we show that the minimum reached in case II b) is less than
the one reached in case I for k∗2 < k0
2.
Since cH > 0 > k0
2 then in the neighborhood of (k1, k2) = (0, cH) func-
tion f(t, k1, k2) is increasing with respect to t. Hence in this neighborhood
max[0,T ] f(t, k1, k2) = f(T, k1, k2). In [3] it is shown that the point (0, cH)
is not a point of local minimum of f(T, k1, k2) (although it is a point of
minimum of this function on the half-plane {k1 ≤ 0}). Therefore this point
is not the point of local minimum for max[0,T ] f(t, k1, k2) and thus there
exists a point (k1, k2), k1 > 0, k2 > cH , such that max[0,T ] f(t, k1, k2) <
max[0,T ] f(t, 0, cH).
Now we show that the minimum reached in case III, is less than the
minimum in case II b) (for k∗2 < k0
2).
Since in case II b) the minimum is reached at k1 =
(
c1(H) − k0
2
α+1
)
T α >
0, k2 = k0
2, and for k2 = k0
2 we have t1 = t2, we denote
t0 = t1
((
c1(H) − k0
2
α+1
)
T α, k0
2
)
= t2
((
c1(H) − k0
2
α+1
)
T α, k0
2
)
.
We show that t0 < T . Indeed,
t1(k1, k2) =
(
|k1|
�
c3(H)−k2−
√
c23(H)−2H−2k2(c3(H)−cH)
�
2H−2k2cH+k2
2
) 1
α
.
As k2 = k0
2, the discriminant (2) is equal to zero, hence c23(H) − 2H −
2k2(c3(H) − cH) = 0. To simplify the denominator of t1(k1, k2) we add the
discriminant (2) to the denominator. Note that for k2 = k0
2 this discriminant
is equal to zero. We have
2H − 2k2cH + k2
2 + c23(H) − 2H − 2k2(c3(H) − cH)
= c23(H) − 2k2c3(H) + k2
2 = (c3(H) − k2)
2.
The expression for t1(k1, k
0
2) is reduced to :
t1(k1, k
0
2) =
(
k1(c3(H) − k0
2)
(c3(H) − k0
2)
2
) 1
α
=
(
k1
c3(H) − k0
2
) 1
α
.
Then
t0 =
(
c1(H) − k0
2
α+1
c3(H) − k0
2
) 1
α
· T =
(
1
α + 1
) 1
α
· T < T.
6 OKSANA BANNA AND YULIYA MISHURA
As t0 < T , f
(
t0,
(
c1(H) − k0
2
α+1
)
T α, k0
2
)
< f
(
T,
(
c1(H) − k0
2
α+1
)
T α, k0
2
)
,
hence
f(t1, k1, k2) < f(T, k1, k2) (4)
in some “one-side” neighborhood:
{(k1, k2)|
∣∣∣k1 −
(
c1(H) − k0
2
α+1
)
T α
∣∣∣ < r, k0
2 − r < k2 ≤ k0
2}
of the point
((
c1(H) − k0
2
α+1
)
T α, k0
2
)
. Since t1 is continuously dependent
on k1 and k2, f(t1, k1, k2) is also continuously dependent on k1 and k2, and
there exists r > 0 such that in the defined above “one side” neighborhood
inequality (4) holds. In this neighborhood
max[0,T ] f(t, k1, k2) = f(T, k1, k2). Taking into account that for k2 ≥ k0
2
function f(t, k1, k2) is increasing with respect to t, we get max[0,T ] f(t, k1, k2) =
f(T, k1, k2) in a “normal” neighborhood
∣∣∣k1 −
(
c1(H) − k0
2
α+1
)
T α
∣∣∣ < r, |k2 −
k0
2| < r. However point
((
c1(H) − k0
2
α+1
)
T α, k0
2
)
is not a point of local
minimum of f(T, k1, k2). Hence it is not a point of local minimum of
max[0,T ] f(t, k1, k2). �
Lemma 2.2. If k∗2 < k0
2, k1 > 0 then the following inequality holds
t1(k
∗
1, k
∗
2) < T.
Proof. Let k∗2 < k0
2. Since
t1(k1, k2) =
(
|k1|
�
c3(H)−k2−
√
c23(H)−2H−2k2(c3(H)−cH)
�
2H−2k2cH+k2
2
) 1
α
, we can multiply
and divide this fraction by the conjugate expression to enumerator c3(H)−
k2 +
√
c23(H) − 2H − 2k2 (c3(H) − cH), and get
t−α1 (k1, k2) =
c3(H)−k2+
√
c23(H)−2H−2k2(c3(H)−cH)
k1
. In the same vein we get
t−α2 (k1, k2) =
c3(H)−k2−
√
c23(H)−2H−2k2(c3(H)−cH)
k1
, moreover
t−α1 (k1, k2) > t−α2 (k1, k2). (5)
Add t−α1 (k1, k2) to both left and right hand sides of (5), to get 2
tα1 (k1, k2)
>
1
tα1 (k1, k2)
+ 1
tα2 (k1, k2)
= 2(c3(H)−k2)
k1
.
Substitute k1 = k∗1, k2 = k∗2, and derive
2
tα1 (k∗1 , k
∗
2)
>
2(c3(H)−k∗2)
k∗1
=
2
�
c3(H)− (α+1)2cH−(2α+1)c3(H)
α2
�
Tα
�
(α+1)(c3(H)−cH)
α2
� =
= 2
Tα
(α+1)2c3(H)−(α+1)2cH
(α+1)(c3(H)−cH)
= 2
Tα (α + 1) > 2
Tα .
APPROXIMATION OF FBM 7
From here tα1 (k∗1, k
∗
2) < T α, that is, t1(k
∗
1, k
∗
2) < T for k∗2 < k0
2. �
So we are looking for min
k1≥0,k2<k0
2
max{f(T, k1, k2), f(t1, k1, k2)}. We now
strengthen this result.
Lemma 2.3. If k∗2 < k0
2, k1 > 0 then the following inequality holds
t2(k
∗
1, k
∗
2) < T .
Proof. Since f(T, k1, k2) = T 2α+1a(k2) − 2k1
Tα+1
α+1
b(k2) + Tk2
1, and (k∗1, k
∗
2)
is the point of minimum of f(T, k1, k2), we have k∗1 = Tα
α+1
b(k∗2).
Note that f(T, k1, k2) > 0 for all k1, k2 therefore the discriminant of
the quadratic function f(T, k1, k2) with the positive highest coefficient is
negative:
D
4
= T 2(α+1)
(
b2(k2)
(α + 1)2
− a(k2)
)
< 0. (6)
Find the derivative of f(t, k1, k2) with respect to t:
∂f(t,k1,k2)
∂t
= (2α + 1)t2αa(k2) − 2k1t
αb(k2) + k2
1.
Since t1(k1, k2) and t2(k1, k2) are solutions of equation ∂f(t,k1,k2)
∂t
= 0, we
have tα1,2(k
∗
1, k
∗
2) =: x1,2 are roots of quadratic polynomial
x2(2α+ 1)a(k∗2) − 2k∗1xb(k
∗
2) + (k∗1)
2. (7)
We substitute T α instead of x and show that the quadratic polynomial is
positive at T α. Indeed,
∂f(T, k∗1, k
∗
2)
∂T
= T 2α(2α + 1)a(k∗2) − 2k∗1T
αb(k∗2) + (k∗1)
2
= T 2α(2α+ 1)a(k∗2) − 2
T αb(k∗2)
α + 1
T αb(k∗2) +
(
T αb(k∗2)
α + 1
)2
= T 2α
(
(2α + 1)a(k∗2) +
(−(2α + 1)
(α + 1)2
)
b2(k∗2)
)
= T 2α(2α+ 1)
(
a(k∗2) −
b2(k∗2)
(α + 1)2
)
> 0,
as according to (6) the expression in the brackets is positive.
Since the quadratic polynomial with positive highest coefficient (7) has
a positive value at T α, point T α is not between its roots, that is T α /∈
[tα1 (k∗1, k
∗
2), tα2 (k∗1, k
∗
2)]. In Particular by lemma 2.1 since T > t1(k
∗
1, k
∗
2), we
get T > t2(k
∗
1, k
∗
2). �
8 OKSANA BANNA AND YULIYA MISHURA
Lemmas 2.2 and 2.3 demonstrate that it is impossible to compare in a
simple way the values at the points t1(k1, k2) and T because point of min-
imum t2(k1, k2) is situated between these two points. We have just shown
that one should look for min
k1≥0,k2<k0
2
max{f(T, k1, k2), f(t1, k1, k2)}. The case
would be simplified if at least one of the inequalities f(T, k1, k2) ≤ f(t1, k1, k2)
or f(T, k1, k2) ≥ f(t1, k1, k2) hold for all k1, k2. However we are dealing with
a more complicated case as shows the following
Theorem 2.1. For any k2 < k0
2 the equation f(t1(k1, k2), k1, k2) =
f(T, k1, k2) (with unknown k1) has two solutions k̃1 and k1, 0 < k̃1 < k1,
moreover
∂
∂k1
(f(t1(k1, k2), k1, k2) − f(T, k1, k2)) |k1=�k1 > 0,
∂
∂k1
(f(t1(k1, k2), k1, k2) − f(T, k1, k2)) |k1=k1 = 0.
If k1 < k
(1)
1 then f(t1(k1, k2), k1, k2) < f(T, k1, k2) and vice versa.
Proof. Equate f(t1(k1, k2), k1, k2) with f(T, k1, k2) to get
k
2H
α
1 ϕ(k2) = T 2Ha(k2) − 2k1
T α+1
α + 1
b(k2) + k2
1T. (8)
Multiplying the left and right hand sides of (8) by 2H
αk1
we get:
2H
α
k
2H
α
− 1
1 ϕ(k2) =
2H
α
· T
2Ha(k2)
k1
− 4
H · T α+1
α(α + 1)
b(k2) +
2H
α
k1T. (9)
Now differentiate (8) with respect to k1:
2H
α
k
2H
α
− 1
1 ϕ(k2) = −2T α+1
α + 1
b(k2) + 2k1T. (10)
Since in (9) and (10) left hand sides are equal, we may equate their right
hand sides to get:
2H
α
· T
2Ha(k2)
k1
− 4
H · T α+1
α(α + 1)
b(k2) +
2H
α
k1T +
2T α+1
α + 1
b(k2) − 2k1T = 0.
This implies
2H
α
· T
2Ha(k2)
k1
+
T α+1
α + 1
b(k2)
(
2 − 4
H
α
)
+
(
2H
α
− 2
)
k1T = 0. (11)
APPROXIMATION OF FBM 9
Multiply (11) by α k1:
2H · T 2Ha(k2) − 2k1T
α+1b(k2) + k2
1T = 0. (12)
Find discriminant D1 of the quadratic in k1 polynomial f(T, k1, k2) :
D1
4
=
T 2α+2
(α + 1)2
b2(k2) − T 2H+1a(k2) < 0.
Find discriminant D2 of the quadratic in k1 polynomial (12), that is, of the
derivative of the difference between f(t1(k1, k2), k1, k2) and f(T, k1, k2) in
the points of intersection:
D2
4
= T 2α+2b2(k2) − 2HT 2α+2a(k2).
By identical transformation we get:
D2
4
= T 2α+2
(
(c3(H) − k2)
2 − (2H − 2k2 · cH + k2
2)
)
= T 2α+2
(
c23(H) − 2k2c3(H) − 2H + 2k2cH
)
= T 2α+2
(
c23(H) − 2H + 2k2(cH − c3(H))
)
.
Since
k0
2 =
c23(H)−2H
2(c3(H)−cH)
,
inequality k2 < k0
2 holds if and only if D2 > 0.
In this case (12) has roots
k0 = T α
(
b(k2) −
√
b2(k2) − 2Ha(k2)
)
,
and
k1 = T α
(
b(k2) +
√
b2(k2) − 2Ha(k2)
)
.
Now, t1(k1, k2) = k
1
α
1 · c1(k2), where c1(k2) =
(
b(k2)−
√
b2(k2)−2Ha(k2)
2Ha(k2)
) 1
α
.
We want to find when t1(k1, k2) = T . We have
k1 =
(
T
c1(k2)
)α
=
T α · 2Ha(k2)
b(k2) −
√
b2(k2) − 2Ha(k2)
= T α
(
b(k2) +
√
b2(k2) − 2Ha(k2)
)
= k1,
and therefore we checked that if k1 = k1 then the following equality
f(t1(k1, k2), k1, k2) = f(T, k1, k2) holds. Hence point k1 is the intersection
10 OKSANA BANNA AND YULIYA MISHURA
point of f(t1(k1, k2), k1, k2) and f(T, k1, k2), moreover the derivative at
this point ∂(f(t1)−f(T ))
∂k1
= 0. There is another point of intersection less than
k2 and such that the derivative ∂(f(t1)−f(T ))
∂k1
at this point is positive. To prove
this fact, consider at first the value k2 = k0
2. Denote g(k1, k2) = k
2H
α
1 ϕ(k2)−
T 2Ha(k2)+2k1
Tα+1
α+1
b(k2)−k2
1T. It is very easy to see that the second deriva-
tive of this function equals g′′(k1,k2)(k1, k2) = 2H
α
(
2H
α
− 1
)
k
1
α
1 ϕ(k2) − 2T ,
that is equivalent to the equality k1ϕ
α(k2) = Tα2α
Hα(H+1)α . So, this deriva-
tive has the unique zero point of the form k̂1 =
(
Tα2
H(H+1)ϕ(k2)
)α
. It means
that the first derivative decreases for k1 < k̂1 and it increases for k1 > k̂1.
At the same time for k2 = k0
2 it is very easy to see that g(k̂1, k
0
2) = 0,
g′k1(k̂1, k
0
2) = 0. It means that k̂1 is the unique zero point of both the
functions g(k1, k
0
2) and g′k1. If we calculate the value of the first derivative
g′k1(k̂1, k2) it will be equal, up to constant multiplier, to bϕαHα(α+1)α−α2α
ϕαHα(α+1)α .
The first derivative ϕ′(k2) = c1(k2)
2H k−cH
H
+ 2
cα+1
1
α+1
, therefore we can cal-
culate that (ϕαb)′(k2) > 0 for c3(H)cH−2H
c3(H)−cH < k2 < k0
2, (ϕαb)′(k2) < 0 for
k2 <
c3(H)cH−2H
c3(H)−cH and (ϕαb)′(k2) → 0 as k2 → −∞. It means that the first
derivative g′k1(k̂1, k2) is negative for all k2 < k0
2. In tu
rn, it means that the function g(k̂1, k2) is positive, at the point k1 > k̂1
it equals zero and at zero point it is negative. It means that there exists one
more point of intersection of the functions f(T, k1, k2) and f(t1(k1, k2), k1, k2).
Denote this point k̃1 and we complete the proof. �
Now we pass to the investigation of the minimization problem concerned
with minimal point. At first, state an auxiliary result.
Lemma 2.4. Function
ζ(t) :=
(T 2H − t2H)(T − t)
(T α+1 − tα+1)2
(13)
is strictly monotonically decreasing on (0, T ) with respect to t.
Proof. It is sufficient to check that the derivative of the right hand side
of (13) with respect to t is negative. We prove this using the following
transformations:
APPROXIMATION OF FBM 11
∂ζ(t)
∂t
= −(2α + 1)t2αT α+2 − T 3α+2 + (2α+ 2)t2α+1T α+1
+ (2α + 1)t3α+1T + T 2α+1tα+1 − (2α+ 2)t3α+2 + 2(α+ 1)tαT 2α+2
− 2(α + 1)t3α+1T − 2(α+ 1)tα+1T 2α+1 + 2(α+ 1)t3α+2
= T (T α − tα)
(
(2α+ 1)tαT α(T − t) − T 2α+1 + t2α+1
)
.
Since T (T α − tα) > 0, it is sufficient to prove that
(2α + 1)tαT α(T − t) − T 2α+1 + t2α+1 < 0. (14)
Divide (14) by T 2α+1. By substitution t
T
=: x we transform this inequal-
ity to:
x2α+1 − (2α + 1)xα+1 + (2α + 1)xα − 1 < 0, x ∈ (0, 1). (15)
Since at point x = 0 expression (15) is equal to −1, and for x = 1 it
is equal to zero, it is sufficient to check that the function is increasing on
[0, 1], that is, that its derivative is positive. Indeed, take the derivative of
(15) with respect to x, and divide it by 2α + 1 to get
x2α − (α + 1)xα + αxα−1, x ∈ (0, 1). (16)
Divide (16) by xα−1 ∈ (0, 1):
xα+1 − (α + 1)x+ α, x ∈ (0, 1). (17)
Since at x = 0 the value of (17) is equal to α, and for x = 1 it is equal
to zero it is sufficient to prove that the function is decreasing on (0, 1),
namely, that its derivative is negative. Differentiate (17) with respect to x:
(α + 1)xα − (α+ 1) < 0 for xα < 1, and the lemma is complete. �
Now we prove that it is impossible to achieve the required minimum at
the point connected with T and minimal point (k∗1, k
∗
2).
Lemma 2.5. Let a(s) be a function of type a(s) = k1 + k2 · sα, k1, k2 ∈
R, s ∈ [0, T ]. If H < H0 so that k∗2 < k0
2 then the inequality holds:
f(T, k∗1, k
∗
2) < f(t1(k
∗
1, k
∗
2), k
∗
1, k
∗
2).
Proof. Consider quadratic polynomials on k1 > 0 for any fixed 0 < t < T :
k2
1T − 2k1
α+1
b(k∗2)T
α+1 + T 2Ha(k∗2) = f(T, k1, k
∗
2),
k2
1t− 2k1
α+1
b(k∗2)t
α+1 + t2Ha(k∗2) = f(t, k1, k
∗
2).
12 OKSANA BANNA AND YULIYA MISHURA
Here a(k2) = 1 − k2 · cHH +
k2
2
2H
> 0, b(k2) = c3(H) − k2, k2 < k0
2 < 0.
Consider the difference
Δ = Δf(T, t, k1, k
∗
2) := f(T, k1, k
∗
2) − f(t, k1, k
∗
2)
= k2
1(T − t) − 2k1
α+ 1
b(k∗2)(T
α+1 − tα+1) + a(k∗2)(T
2H − t2H).
Divide Δ by (T − t), and get a quadratic polynomial
Δ1 = k2
1 −
2k1
α+ 1
b(k∗2)
T α+1 − tα+1
T − t
+ a(k∗2)
T 2H − t2H
T − t
.
Its discriminant is equal to
D =
b2(k∗2)
(α + 1)2
(
T α+1 − tα+1
T − t
)2
− a(k∗2)
T 2H − t2H
T − t
.
Consider possible cases:
1) D < 0. Then f(T, k1, k
∗
2) > f(t, k1, k
∗
2) for all k1 > 0, in particular
f(T, k∗1, k
∗
2) > f(t, k∗1, k
∗
2).
2) D > 0. Then there exist two roots k1,2
1 =
b(k∗2)
α+1
Tα+1−tα+1
T−t ±√
D,
moreover f(T, k1, k
∗
2) < f(t, k1, k
∗
2) for k1
1 < k1 < k2
1 and an opposite
inequality holds for k1 < k1
1, k1 > k2
1.
3)D = 0. Then f(T, k1, k
∗
2) > f(t, k1, k
∗
2) for all k1 but k1 =
b(k∗2)
α+1
Tα+1−tα+1
T−t .
For this value of k1 we have an equality.
Transform D so that we could easier investigate its sign. Instead of D,
first consider
D3 =
b2(k∗2)
a(k∗2 )(α+1)2
−ϕ(T,t)
ψ(T,t)
, where ϕ(T, t) = T 2H−t2H
T−t , ψ(T, t) =
(
Tα+1−tα+1
T−t
)2
.
It is easy to see ϕ(T,t)
ψ(T,t)
= ζ(t) (see Lemma 2.4).
If t = 0 then ζ(t) = 1, and since a(k∗2) >
b2(k∗2)
(α+1)2
, we have D3 < 0.
If t→ T then
lim
t↑T
ζ(t) = lim
t↑T
(T 2H − t2H)(T − t)
(T α+1 − tα+1)2
= lim
t↑T
−2Ht2H−1T − T 2H + (2H + 1)t2H
−2(α + 1)tαT α+1 + 2(α+ 1)t2α+1
= lim
t↑T
−2H(2H − 1)t2H−2T + (2H + 1)2Ht2H−1
−2(α + 1)α tα−1T α+1 + 2(α + 1)(2α+ 1)t2α
=
−4H2 + 2H + 4H2 + 2H
−2(H + 1
2
)(H − 1
2
) + 2(H + 1
2
)2H
=
4H
(H + 1
2
)(4H − 2H + 1)
=
2H
(H + 1
2
)2
,
APPROXIMATION OF FBM 13
and b2(k∗2) − 2Ha(k∗2) > 0 hence, D3 > 0.
Generally D3 is strictly monotonically increasing with respect to t, as
by Lemma 2.4 ζ(t) is strictly monotonically decreasing with respect to t.
f(t1(k1, k2), k1, k2) = f(T, k1, k2)
Hence there exists a unique t0 such that for 0 < t < t0, D3 < 0, and for
t0 < t ≤ T , D3 > 0. Moreover, if 0 < t < t0, then at k∗1
Δ = Δf(T, t, k∗1, k
∗
2) := f(T, k∗1, k
∗
2) − f(t, k∗1, k
∗
2)
= (k∗1)
2 T − 2k∗1
α + 1
b(k∗2)T
α+1 + T 2α+1a(k∗2)
> (k∗1)
2 t− 2k∗1
α + 1
b(k∗2)t
α+1 + t2α+1a(k∗2).
We do not know whether D3 > 0 or D3 < 0 for t = t1(k
∗
1, k
∗
2). Assume
that we verified one of the two inequalities: either 0 < t1(k
∗
1, k
∗
2) < t0 or
t1(k
∗
1, k
∗
2) > t0 but k∗1 < k1
1. The latter inequality is equivalent to:
b(k∗2)
α + 1
T α <
b(k∗2)(T
α+1 − (t1(k
∗
1, k
∗
2))
α+1)
(α + 1)(T − t1(k∗1, k
∗
2))
−
√
D,
or
D <
b2(k∗2)
(α + 1)2
(t1(k
∗
1, k
∗
2))
2 (T α − (t1(k
∗
1, k
∗
2))
α)2
(T − t1(k
∗
1, k
∗
2))
2
.
0.6 0.7 0.8 0.9
0.05
0.1
0.15
0.2
0.25 a
b
Fig.1
Then in both cases we would have inequality f(T, k∗1, k
∗
2) >
f(t1(k
∗
1, k
∗
2), k
∗
1, k
∗
2). Since both of these statements are very complicated to
check them analytically we suggest to verify them graphically with graphs
(a) and (b) presented in Figure 1 (we used Mathematica tool to plot the
graphs). Graph (a) (firm line) shows D at point t = t1(k
∗
1, k
∗
2) against
H , whereas graph (b) (dotted line) plots
b2(k∗2)
(α+1)2
(t1(k
∗
1, k
∗
2))
2 (Tα−(t1(k∗1 ,k
∗
2))α)2
(T−t1(k∗1 ,k
∗
2))2
14 OKSANA BANNA AND YULIYA MISHURA
against H. Graph (a) shows that D > 0 at point t = t1(k
∗
1, k
∗
2) consequently
D3 > 0 at the same point since they are connected via positive multiplier.
Therefore t1(k
∗
1, k
∗
2) > t0 and we cannot give positive conclusion by this way.
Also, from graph (b) it is clear that
D >
b2(k∗2)
(α+ 1)2
(t1(k
∗
1, k
∗
2))
2 (T α − (t1(k
∗
1, k
∗
2))
α)2
(T − t1(k∗1, k
∗
2))
2
for any H < H0 that is k∗1 < k1
1 so the second possible positive conclusion
does not hold.
The lemma is complete. �
Corollary 2.1. For the solution of minimization problem we must calculate
the value
min
k1≥0,k2<k0
2
max{f(T, k1, k2), f(t1, k1, k2)}.
3. Approximation results for fBm with Hurst index,
separated from 1, and the integrals with integrands of the
form a(s) = k1 + k2 · sα, k1, k2 ∈ R
At last we state the “positive” result concerning the minimal point and
the minimal value
min
k1≥0,k2<k0
2
max{f(T, k1, k2), f(t1, k1, k2)}.
Theorem 3.1. The minimal value min
k1≥0,k2<k0
2
max{f(T, k1, k2), f(t1, k1, k2)}
is achieved at the point k0
1 = b(�k2)Tα
α+1
, k̂2 = cH(H+1/2)2−2Hc3(H)
α2 . This value
equals T 2H
(
a(k̂2) − b2(�k2)
(α+1)2
)
.
Proof. Consider the function f(T, k1, k2) = T 2Ha(k2)− 2k1
Tα+1
α+1
b(k2) + k2
1T.
Evidently, the minimal value of this function for fixed k2 is achieved at
the point k1 = b(k2)Tα
α+1
and equals T 2H
(
a(k2) − b2(k2)
(α+1)2
)
. In turn, this for-
mula gives a quadratic polynomial function of k2 and achieves its mini-
mum at the point k̂2. It is evident from Figure 2 that for H < H0 we
have the inequality k̂2 < k0
2. Moreover, we can see from Figure 3 that
f(T, k0
1, k̂2) = f(t1(k
0
1, k̂2), k
0
1, k̂2) = (k0
1)
2H/αϕ(k̂2). Without any doubt, the
value f(T, k0
1, k̂2) is a required minimal value. It completes the proof. �
APPROXIMATION OF FBM 15
0.5 0.6 0.7 0.8 0.9
-6
-5
-4
-3
-2
-1
a
b
Fig.2
0.6 0.7 0.8 0.9
0.5
1
1.5
2
2.5
a
b
Fig.3
Here are values of the functions shown on the Figure 3.
H a b
0.51 0.00106525 0.00106525
0.55 0.0313579 0.0313579
0.59 0.117903 0.117903
0.63 0.28104 0.28104
0.67 0.539574 0.539574
0.71 0.905731 0.905731
0.75 1.37669 1.37669
0.79 1.9206 1.9206
0.83 2.45322 2.45322
0.87 2.797 2.797
0.91 2.59085 2.59085
0.95 0.981965 0.981965
16 OKSANA BANNA AND YULIYA MISHURA
4. Conclusion
It was shown in [3] that in the case when a(s) is a function of type a(s) =
k1 + k2 · sα, k1, k2 ∈ R, s ∈ [0, T ], and H > H0 is such that k∗2 > k0
2, then
the minimum of the function max
t∈[0,T ]
f(t, k1, k2) is reached at k1 = k∗1, k2 = k∗2
and is equal to T 2α+1 − T 2α+1c2H
(
B2(1 − α, α) − 2B(1−α,α)
α
+ (α+1)2
α2(2α+1)
)
. In
the present paper we obtain the approximation results for the same type of
function, for H < H0 and k∗2 < k0
2. We demonstrate that in this case, in
contrary to the case H > H0, it is impossible to achieve minimum at the
minimal point k1 = k∗1, k2 = k∗2 or in some connected point, and demonstrate
analytically and with the help of the graph where the minimum “is situated”
and how to calculate it. So, this paper completes the problem of minimiza-
tion of the distance between fBm and Wiener integrals with the integrands
of the form a(s) = k1 + k2 · sα, k1, k2 ∈ R in the space L∞([0, T ];L2(Ω)).
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Department of Probability Theory and Mathematical Statistics,
Kyiv National Taras Shevchenko University, Kyiv, Ukraine
E-mail address: bannaya@mail.univ.kiev.ua, myus@univ.kiev.ua
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