School of Education, Culture and Communication
Division of Applied Mathematics
MASTER THESIS IN MATHEMATICS / APPLIED MATHEMATICS
Cubature on Wiener Space for the Heath–Jarrow–Morton framework
by
Lutufyo Mwangota
Masterarbete i matematik / tillämpad matematik
DIVISION OF APPLIED MATHEMATICS
MÄLARDALEN UNIVERSITY SE-721 23 VÄSTERÅS, SWEDEN
School of Education, Culture and Communication
Division of Applied Mathematics
Master thesis in mathematics / applied mathematics
Date:
1stFebruary, 2019
Project name:
Cubature on Wiener Space for the Heath–Jarrow–Morton framework
Author: Lutufyo Mwangota Version: 1stFebruary, 2019 Supervisor(s): Anatoliy Malyarenko Reviewer: Lars Hellström Examiner: Ying Ni Comprising: 30 ECTS credits
Abstract
This thesis established the cubature method developed by Gyurkó & Lyons (2010) and Lyons & Victor (2004) for the Heath–Jarrow–Morton (HJM) model. The HJM model was first pro-posed by Heath, Jarrow, and Morton (1992) to model the evolution of interest rates through the dynamics of the forward rate curve. These dynamics are described by an infinite-dimensional stochastic equation with the whole forward rate curve as a state variable. To construct the cubature method, we first discretize the infinite dimensional HJM equation and thereafter ap-ply stochastic Taylor expansion to obtain cubature formulae. We further used their results to construct cubature formulae to degree 3, 5, 7 and 9 in 1-dimensional space. We give, a considerable step by step calculation regarding construction of cubature formulae on Wiener space.
Keywords: Heath–Jarrow–Morton model, stochastic Taylor expansion, Cubature formulae, Brownian signature, forward rate.
Dedication
I dedicated this thesis to my family members, who have made me stronger, better and more fulfilled than I could have ever imagined.
Acknowledgements
Foremost, I would like to express my sincere gratitude to Professor Anatoliy Malyarenko who has been the great ideal thesis supervisor. His valuable advice and patient encouragement aided the writing of this thesis in innumerable ways. Moreover, I would like to thank Lars Hellströmfor taking time to review this thesis. His insightful comments extremely strengthen the quality of this thesis.
Besides my supervisor and reviewer, I gratefully acknowledge the scholarship received from the Swedish Institute (SI) to study Master’s programme in Financial Engineering.
Finally, special gratitude goes to my family members for providing me with unfailing support and continuous encouragement throughout my years of study and throughout the process of writing this thesis. This accomplishment would not have been possible without them.
Contents
1 Introduction 3
1.1 Pricing zero coupon bonds . . . 3
1.2 The content of the thesis . . . 5
1.2.1 Aim and Objectives. . . 5
1.2.2 Overview and Outline . . . 5
2 The formulation of the problem 6 2.1 The HJM model . . . 6
2.2 The discretization problem . . . 9
2.2.1 The Stratonovich integral. . . 9
2.3 Stochastic Taylor expansion . . . 11
3 Cubature formulae 14 3.1 A formulation of the problem in other terms . . . 14
3.2 Examples . . . 21
4 Conclusion 38 A Criteria for a Master thesis 40 B Mathematical tools 42 B.1 Hilbert spaces . . . 42
B.1.1 Scalar or Inner products . . . 42
B.1.2 Orthornormal bases . . . 43
B.2 Linear operators . . . 44
B.2.1 Bounded linear operators . . . 44
B.3 Functions of bounded variation . . . 45
B.4 Fréchet derivatives . . . 46
List of Tables
2.1 A comparison of notation . . . 9
3.1 The nonnegative integer solutions to the inequality 2i1+ i2≤ 3 . . . 22 3.2 The nonnegative integer solutions to the inequality 2i1+ i2+ 4i3≤ 5.. . . 25
Chapter 1
Introduction
1.1
Pricing zero coupon bonds
Financial engineering is a part of applied mathematics which constructs and studies mathem-atical models of financial markets. Constructing a model starts from description of a market in words of a human language.
Definition 1. A zero coupon bond is a financial contract which guarantees the holder one money unit to be paid on a prescribed date in the future.
The next step is to introduce notation. Consider the interval [0, T ], where T is a positive real number. Assume that 0 describes current time, T determines time of payment, or the maturity of the bond, and t ∈ [0, T ] is an arbitrary current time and maturity. Then denote;
• p(t, T ): The market price at time t of a bond with maturity date T and
• f (t, T ): The instantaneous forward rate at time t for a bond with maturity date T . Our next task is to create a mathematical model that describes the above introduced functions p(t, T ) and f (t, T ).
Definition 2. The instantaneous forward rate f (t, T ) with maturity T , contracted at t is f(t, T ) = −∂ ln p(t, T )
∂ T . (1.1)
Note that,
• The function p(t, T ) takes positive values and is differentiable in the second variable, • The today’s (observed at time 0) instantaneous forward rate f (0, T ) is known. Denote it
by f◦(T ).
Our task is to describe a model for the instantaneous forward rate (Equation (1.1)). Most of the existing models belong to the following three families.
1. Short-rate models. 2. Libor market models.
3. The HJM (Heath–Jarrow–Morton) models.
For the first two approaches, we only give the relevant references. Short rate models are described in [19, 27, 35], among others. Libor market models are described in [5,4, 24, 34], among others.
The third approach was introduced in [23]. Heath, Jarrow, and Morton assumed that the
instantaneous forward rate f (t, T ) is a random field. That is, there is a probability space (Ω, F, P) in a function of three variables f (t, T, ω) defined for 0 ≤ t ≤ T < ∞ and ω ∈ Ω such that for any fixed t0≤ T0the function f (t0, T0, ω) is a random variable. They write a stochastic
differential equation for the function f (t, T ), see Equation (2.1).
As we will see in Section2.1, for different times t, the functions f (t, T ) have different domains
of definition. This complication was solved by Musiela [36]. He modified the HJM model in
an appropriate way and obtained another stochastic differential equation, see Theorem2. The Musiela equation is an infinite-dimensional stochastic differential equation. The exact sense of the above sentence will be discussed in Section2.1.
We are interested in calculation of the fair price of an European type contingent claim written on the instantaneous forward rate. As in finite-dimensional case, there are two general methods to perform this calculation: solving the Kolmogorov backward equation and Monte Carlo simulation. We work with a modern version of the second named method.
For Monte Carlo simulation, there exist two approaches: a classical one and a modern one. The classical approach is as follows.
1. The space discretisation of the equation: The original equation is projected on some finite-dimensional subspace H ⊂ Hγ (see Definition 5), and the derivative ∂ x∂ rt(x) is
approximated by some linear operator defined on H.
2. Time discretisation: the stochastic differential equation on H is approximated by a stochastic difference equation using the Euler method or its generalisations.
3. Discretisation on the space of trajectories: Monte Carlo simulation. This approach is described in [18,22,25,40,41], among others.
The modern approach, according to [3], is to solve the problem in the reversed order.
1. Discretisation on the space of trajectories. The d-dimensional Wiener process W(t) is replacing by a set of finitely many trajectories with well-defined probabilities according to a well-defined matching criterium.
2. The resulting deterministic problem is solved by standard numerical methods for de-terministic partial differential equations, e.g., finite-element or finite-difference meth-ods.
This approach was proposed by [30,31] and developed by [3,14,33,37,38,39], among oth-ers. Our task is to describe how the discretisation on the space of trajectories performs.
1.2
The content of the thesis
1.2.1
Aim and Objectives
This thesis aims to describe the cubature formulae developed in [20, 33] for the Heath—
Jarrow—Mortonmodel. To make more clear, we want to prove the cubature method to the
Heath–Jarrow–Morton infinite dimensional stochastic equation.
1.2.2
Overview and Outline
The thesis is structured as follows:
Chapter1has briefly introduced the research topic and noting down the research objectives. We specifically introduced the modern algebraic approach, described in [3] as the method of solving the infinite-dimensional HJM equation. We started formulating the problem in Chapter
2by establishing some theorems and definitions. In the process of discretization, we recalled and apply the stochastic Taylor formula to obtain cubature formulae on the Wiener space.
Chapter 3focused on the algebra expression regarding construction of cubature formulae on
Wiener space. The constructed examples of cubature formulae to degree 3, 5, 7, and 9 in 1–dimensional space, proved that the description of the task is clear and accurate. Finally, in
Chapter 2
The formulation of the problem
2.1
The HJM model
In 1990 David Heath, Bob Jarrow, and Andy Morton (HJM) published a paper describing the non-arbitrage conditions to model the entire forward rate curve as their (infinite dimensional) state of variable. To describe their model, we start as follows,
Define the instantaneous short rate at time t by the values of the instantaneous forward rate on the diagonal:
r(t) = f (t,t). Mathematically, the money account process is
B(t) = exp Z t 0 r(s) ds .
Financially, the money account process is described in a bank with the short rate r(t). Following [23], introduce the following definition.
Definition 3. The Heath–Jarrow–Morton or HJM framework is the set of market models given by the stochastic differential equation
d f (t, T ) = α(t, T ) dt + σσσ>(t, T ) dW(t),
f(0, T ) = f◦(0, T ), (2.1)
where W(t) is the vector-valued Wiener process, α(t, T ) is the drift, and σσσ (t, T ) is the volat-ility vector.
Equation (2.1) holds true under a given risk-neutral measure Q and for every fixed maturity T > 0. The forward rate curve f◦(0, T ) observed at time 0 is used as the initial condition. We
assume that the stochastic processes α(t, T ) and σσσ (t, T ) are adapted to the filtration { Ft: t ≥
0 }, generated by the Wiener process W(t).
To obtain a particular model, one has to specify the drift structure α(t, T ) and the volatility structure σσσ (t, T ).
Theorem 1 (The HJM drift condition [Heath, Jarrow, and Morton, 1992 – [23]). The model
(2.1) is arbitrage-free if and only if
α (t, T ) = σσσ>(t, T )
Z T
t
σ
σσ (t, s) ds.
See appendixB.5for the derivation of the HJM drift term.
Note that for different times t, the functions f (t, T ) have different domains of definition. To overcome this difficulty, introduced the Musiela parametrisation by letting x = T − t be the time to maturity, so that the forward rate r(t, x) is defined as
rt(x) = f (t,t + x),
Note that rt(0) is the instantaneous short rate at time t.
Theorem 2 (The Musiela equation [Musiela, 1993 [36]). If the forward rate dynamics satisfies (2.1) under Q, then the r-dynamics is given by
drt(x) = ∂ ∂ xrt(x) + σσσ > 0(t, x) Z x 0 σσσ0(t, u) du dt + σσσ>0(t, x) dW(t) (2.2) where σ σσ0(t, x) = σσσ (t, t + x).
The Musiela equation is an infinite-dimensional stochastic differential equation. To give the exact sense to this sentence, introduce the following definitions.
Let r(x) : [0, ∞) → R be an integrable function, that is,
Z ∞ 0
|r(x)| dx < ∞.
Recall that the support of a continuous function ϕ(x) : [0, ∞) → R is the closure of the set { x ∈ [0, ∞) : ϕ(x) 6= 0 }.
Definition 4. An integrable function g(x) is called the weak derivative of the function f (x) if for all infinitely differentiable functions ϕ(x) : [0, ∞) → R with closed and bounded support we have Z ∞ 0 f(x)ϕ0(x) dx = − Z ∞ 0 g(x)ϕ(x) dx.
Note that if the classical derivative f0(x) exists and is integrable, then it satisfies the above definition: it becomes the integration by parts formula. This definition is a particular case of [29, Definition 4.3.1].
Definition 5. For a fixed positive real number γ, the Filipovi´c space Hγ is the space of all
differentiable functions f (x) of the nonnegative variable x satisfying the condition k f kγ< ∞,
where the square of the norm is given by k f k2γ= Z ∞ 0 f2(x)e−γxdx + Z ∞ 0 ( f0)2(x)e−γxdx, where f0(x) is the weak derivative of the function r(x).
The above space is named after Filipovi´c [16]. If we replace the weak derivative with the classical one, then the resulting space fails to be complete.
In addition, the volatility function may depend on the function r(t, x). Postulate the volatility function of the form σσσ : [0, ∞) × Hγ× [0, ∞) → Rm. Let rt be the forward rate curve at time t.
The Musiela equation (2.2) takes the form drt(x) = ∂ ∂ x rt(x) + σσσ>(t, rt, x) Z x 0 σσσ (t, rt, u) du dt + σσσ>(t, rt, x) dW(t), r0(x) = r◦(x). (2.3)
The main reference for such equations is [13].
For simplicity, assume that σσσ : Hγ× [0, ∞) → Rm, that is, σσσ does not depend on t. We are
interested in calculation of the fair price EQ[ϕ(rT)] of an European type contingent claim with
payoff ϕ(rT). As in finite-dimensional case, there are two general methods to perform this
calculation.
1. It is proved in [13] that under some conditions the function v(t, rt) = EQ[ϕ(rt)] is a
solution of the initial value problem for Kolmogorov’s backward equation ∂ v(t, rt) ∂ t = 1 2tr ∂2v(t, rt) ∂ (rt)2 σ σ σ (rt, x)σσσ>(rt, x) + ∂ ∂ xrt(x) + σσσ >(t, r t, x) Z x 0 σσσ (t, rt, u) du ,∂ v(t, rt) ∂ rt Hγ , v(0, rt) = ϕ(rt).
2. Use either Monte Carlo simulation or modern algebraic methods.
2.2
The discretization problem
We would like to apply the results of [3] to our model. We note that our Equation (2.3) is a particular case of [3, Equation (1.1)] such that,
dXtx= {AXtx+ α(Xtx)} dt +
d
∑
i=1
βi(Xtx)dβti,
The difference is from notations as shown in the table below.
Notation by [3] Our notation
Xtx rt(x) A ∂ ∂ x α (Xtx) σT(ti, rt, x)R0xσ (t, rt, u)du βi(Xtx) σi(ti, rt, x) βti Wi(t) x r◦(x)
Table 2.1: A comparison of notation Where, the symbols within the Our notation column are defined as,
• rt(x): An instantaneous forward rate,
• ∂
∂ x: The linear operator,
• σi(ti, rt, x): The volatility vector,
• Wi(t): A vector-valued Wiener process,
• r◦(x): Initial condition (an instantaneous forward rate observed at time 0)
2.2.1
The Stratonovich integral
Our approach of Stratonovich integral (developed by Ruslan L. Stratonovich and D. L. Fisk) is found in [26] and [28]. This is a method of representing stochastic integrals and differential equations. The method is equivalent to It ˆomethod but has its own computational techniques which are easier to manipulate. The integrals in Stratonovich function are defined in such way that the chain rule of ordinary calculus holds.
Definition 6. Let {ψ(t)} be a simple adapted stochastic process partitioned in time inter-vals 0 = t0 < t1< . . . .. < tn= T and random variables ξ1, ξ2. . . ξn−1 such that ξi is ℑt
-measurable then, our Stratonovich integral approach can be defined in a manner similar to the Riemann integral, that is a limit of Riemann-Stieltjes sums.
Jn= n−1
∑
i=0
ψ (ti) + ψ(ti+1)
2 {W (ti+1) −W (ti)}, That is limn→∞Jn=
Z T
0
ψ (t) ◦ dW(t),
where W (t) is the Wiener process.
Then we write Equation (2.3) in Stratonovich form, as drt(x) = ∂ ∂ xrt(x) + σσσ > (t, rt, x) Z x 0 σσσ (t, rt, u)du − 1 2Dσσσ (t, rt, u)σσσ > (t, rt, x) dt + σσσ>(t, rt, x) ◦ dW(t), (2.4)
and, following [3], replace “◦dW” with “dωωω ”: drt(x) = ∂ ∂ xrt(x) + σσσ >(t, r t, x) Z x 0 σσσ (t, rt, u)du − 1 2Dσσσ (t, rt, u)σσσ >(t, r t, x) dt + σσσ>(t, rt, x) dωωω ,
which has the form similar to equation (1.5) in [3], i.e. dXtx(ω) = ( AXtx(ω) + α(Xtx(ω)) −1 2 d
∑
i=1 Dβi(Xtx(ω)) · βi(Xtx(ω)) ) dt + d∑
i=1 βi(Xtx(ω))dωi(t), where, DF(x) · v = ∂ ∂ ε ε =0 F(x + εv) denotes the Fréchet derivative of a function or vector field F.For some d-dimensional paths we can re-write our solution in Stratonovich form as, drt(x) = ( ∂ ∂ x rt(x) + σσσ>(t, rt, x) Z x 0 σσσ (t, rt, u)du − 1 2 d
∑
i=1 Dσi(t, rt, x) · σi(t, rt, x) ) dt + d∑
i=1 σi(t, rt, x)◦dWi(t), (2.5)which is exactly the same as [3, equation (3.2)], i.e. dXtx= ( A(x) + α(x) −1 2 d
∑
i=1 Dβi(x) · βi(x) ) dt + d∑
i=1 βi(Xtx)◦dβi(t).2.3
Stochastic Taylor expansion
We assume that our function (2.5) is continuously differentiable in such a way that it could be easy to formulate the multiple Stratonovich stochastic integrals. To form the core of cubature on the Wiener space, we formulate a notation Am, which denote the set of all multi-indices in
d-dimensional paths. Let ηηη be a multi-index η
ηη = (i1, . . . , ik),
where k ≥ 0, and where 0 ≤ ij≤ d for all j with 1 ≤ j ≤ k.
For notational purpose we denote,
kηηη k = k + ]{1 ≤ j ≤ k : ij= 0},
where ] denotes the number of elements in the sets. Introducing our notation Amwe have,
Am= {(i1, . . . , ik) ∈ {0, ...., d}k: kηηη k ≤ m},
Then for the multiple Stratonovich integrals that form the blocks of the stochastic Taylor expansion, we have I(t, ηηη , ◦dWtikk) = Z 0<t1<....<tk<T ◦dWi1(t 1) · · · ◦ dWik(tk).
The main reference for this idea is in [28] and [33].
Define by Cbv([0,t]; Rd) the set of all functions ωωω = (ω0, . . . , ωd) : [0,t] → Rd+1 such that
ω0(s) = s and each ωi(s), 1 ≤ i ≤ d, is a continuous function of bounded variation with
ωi(0) = 0. Let rt(x, ωωω ) be the solution to the equation drt(x, ωωω ) = ∂ ∂ xrt(x) + σσσ >(t, r t, x) Z x 0 σσσ (t, rt, u)du − 1 2Dσσσ (t, rt, u)σσσ >(t, r t, x) dt + σσσ>(t, rt, x) dωωω (t).
which is obtained from Equation (2.4) by replacing dW(t) with dωωω (t).
Definition 7. Let m be a positive integer. Then we propose that our paths ωωω1, . . . , ωωωn ∈
Cbv([0, T ], Rd) and our positive weights λ1...λndefine a cubature formula on Wiener space
of degree ≤ m at time T , if and only if, for all (η1...ηk) ∈ AAAmmm, we have
E[I(t, ηηη , ◦dWtikk)] = n
∑
j=1 λj Z 0<t1<....<tk<T ◦dωi1 j (t1).... ◦ dω ik j (tk). (2.6)With the above definition and then applying the Tchakaloff’s theorem, we could easily see that it is possible to construct cubature formulae on Wiener space for d-dimension paths when t= 1 (see [33] for the proof of the Tchakaloff’s theorem).
Let ddxmmr denote the mth classical derivative of the function r(x). Let D1 be the subset of the
Filipovi´c space Hγ defined by
D1= ( r∈ Hγ: dr dx γ < ∞ ) ,
that is the domain of the operator
A= ∂
∂ x, and let Dmbe the domain of the operator
Am= ∂
m
∂ xm,
that is, the set of all functions r ∈ Hγ are m times differentiable in the classical sense, with
kddxmmrkγ < ∞. Denote by ˜Amthe set
˜
Am= { (i0, . . . , ik) : k(i1, . . . , ik)k ≤ m < k(i0, . . . , ik)k }.
In other words, ˜Amis the set of multi-indices that have norm greater than m but after deleting the first element the norm of the remaining multi-index becomes less than or equal to m. Let f : Hγ→ R be a Fréchet differentiable function. The function σiof Equation (2.4) acts on
f by
(σif)(y) = D f (y) · σi(y) =
∂ ∂ ε ε =0 f(y + εσi(y)), y∈ Hγ.
Theorem 3 (Stochastic Taylor expansion [3,33]). For any m ≥ 1, there exists a positive integer n= bm/2c + 1 such that for all infinitely differentiable functions f : Hγ→ R, for all x ∈ Dn, and for all t∈ (0, 1) we have
f(rt(x)) =
∑
η ηη ∈Am (σi1· · · σikf)(rt(x))I(t, ηηη , ◦dW ηk tk ) +∑
ηηη ∈ ˜Am (σi0· · · σikf)(rt(x))I(t, ηηη , ◦dW ηk tk ). (2.7) Moreover, there exists a constant C> 0 that depends on f such that the variance of the last term is small for small t: E
∑
ηηη ∈ ˜Am (σi0· · · σikf)(rt(x))I(t, ηηη , ◦dW ηk tk ) 2 1/2 ≤ Ct(m+1)/2.Let assume we want to calculate the payoff of the derivative using Equation (2.7) above. Under risk neutral probability, we take expectation on both sides of the equation.
E ( f (rt(x))) = E "
∑
η η η ∈Am (σi1· · · σikf)(x)I(t, ηηη , ◦dW ηk tk ) +∑
ηηη ∈ ˜Am (σi0· · · σikf)(x)I(t, ηηη , ◦dW ηk tk ) =∑
η η η ∈Am (σi1· · · σikf)(x)EI(t, ηηη , ◦dW ηk tk ) +∑
η η η ∈ ˜Am (σi0· · · σikf)(x)EI(t, ηηη , ◦dW ηk tk ) . (2.8)Equation (2.8) tells us that, the quantity of the derivative( f (rt(x))) is obtained by the sum of
the two term. The first term which is known (can be calculated) regarding definition7and
Chapter 3
Cubature formulae
3.1
A formulation of the problem in other terms
Denote C([0, T ], Rd) as the space of Rd-valued continuous functions defined in [0, T ] and Cbv([0, T ], Rd) as it’s subset made of bounded variation paths.
Then to find a cubature formula, we use the following idea. Consider a set, call it T, and a map X : Cbv([0, T ], Rd) → T. The problem (2.6) translates to another problem. We have to
carefully choose T and X in such a way that the new problem is simpler than the problem (2.6).
Fortunately, such a space and a variant of such a map has been already proposed by K.T. Chen in [10]. For any integer i with 0 ≤ i ≤ d, denote by εεεi∈ R1+d the vector whose ith component
is 1 and all other are 0. Denote
Bn= { ηηη : kηηη k = n }, n≥ 0.
Let W1, . . . , Wn be finitely many real finite-dimensional linear spaces with inner products
(·, ·)1, . . . , (·, ·)n.
Definition 8. The tensor product of the empty family of linear spaces is equal to R1. The
tensor product W1⊗ W2⊗ · · · ⊗ Wn is the linear space of all multi-linear forms f : W1× W2×
· · · ×Wn→ R, that is,
f(v1, . . . , αv0i+ β v00i, . . . , vn) = α f (v1, . . . , v0i, . . . , vn) + β f (v1, . . . , v00i, . . . , vn)
for all 1 ≤ i ≤ n and for all α, β ∈ R.
In particular, the tensor product of the one-element family {W1} must be the linear space W1∗
of the linear forms on W1, but we identify W1∗and W1using the isomorphic map acting from W1
Definition 9. Let vi∈ Wi, 1 ≤ i ≤ n. The tensor product of vectors v1⊗ v2⊗ · · · ⊗ vn is the
multi-linear form acting by
v1⊗ v2⊗ · · · ⊗ vn(x1, x2. . . , xn) = (v1, x1)1(v2, x2)2· · · (vn, xn)n.
Let V0 be the space R1 with the basis B0 = {1}, and let Vn be the linear space with the
basis
Bn= { εεεi1⊗ · · · ⊗ εεεik: ηηη = (i1, . . . , ik) ∈ Bn}. (3.1)
For example, when d = 1, then V1is the one-dimensional space generated by εεε1, and V2is the
two-dimensional space generated by εεε0and εεε1⊗ εεε1.
Denote T = ∞ M n=0 Vn, that is, the elements of T are infinite sequences
a = (a0, a1, . . . , an, . . . )
with an∈ Vn. The set T has a rich structure.
1. The sum a + b of two elements a = (a0, a1, . . . , am, . . . ), b = (b0, b1, . . . , bm, . . . ) ∈ T is
defined by
a + b = (a0+ b0, a1+ b1, . . . , am+ bm, . . . ).
The product λ a ∈ T of λ ∈ R and a ∈ T is
λ a = (λ a0, λ a1, . . . , λ am, . . . ).
With this operations, T is a linear space.
2. Note that if a ∈ Vmand b ∈ Vn then a ⊗ b ∈ Vm+n. The tensor product of two arbitrary
elements of T is defined as follows:
a ⊗ b = (c0, c1, . . . , cm, . . . ),
where for each m ≥ 0 we have,
cm= m
∑
i=0
ai⊗ bm−i.
For example, write down the first few terms in the tensor product a ⊗ b. c0= a0⊗ b0,
c1= a0⊗ b1+ a1⊗ b0,
c2= a0⊗ b2+ a1⊗ b1+ a2⊗ b0, and so on.
3. The exponent of an element a ∈ T is defined by the Taylor series exp(a) = ∞
∑
k=0 a⊗k k! .4. Let a ∈ T with a0> 0. Define the logarithm of a by the Taylor series
ln(a) = ln(a0) + ∞
∑
k=1 (−1)k k (a − a0)⊗k ak0 .5. The projection operator πmis defined by
πma = (a0, a1, . . . , am, 0, 0, . . . ).
6. Finally, the Lie bracket of two elements of T is defined by [a, b] = a ⊗ b − b ⊗ a.
Let a = 10 and b = 01 ∈ V1. That is A = (0, a, 0, 0, . . . ) ∈ T and B = (0, b, 0, 0, . . . ) ∈ T.
Then, using item2above, we write down all possible terms in tensor product such that, A ⊗ B = (c0, c1, c2, 0, 0, 0, ....) ∈ T, where, c0= 0
∑
i=0 ai⊗ b0−i= a0⊗ b0= 0 × 0 = 0, c1= 1∑
i=0 ai⊗ b1−i= a0⊗ b1+ a1⊗ b0= 0 × b + a ⊗ 0 = 0, c2= 2∑
i=0 ai⊗ b2−i= a0⊗ b2+ a1⊗ b1+ a2⊗ b0= 0 ⊗ 0 + a ⊗ b + 0 ⊗ 0 = a ⊗ b. B ⊗ A = (d0, d1, d2, 0, 0, 0, ....) ∈ T, where, d0= 0∑
i=0 bi⊗ a0−i= b0⊗ a0= 0 × 0 = 0, d1= 1∑
i=0 bi⊗ a1−i= b0⊗ a1+ b1⊗ a0= 0 × a + b ⊗ 0 = 0, d2= 2∑
i=0 bi⊗ a2−i= b0⊗ a2+ b1⊗ a1+ b2⊗ a0= 0 ⊗ 0 + b ⊗ a + 0 ⊗ 0 = b ⊗ a.Note:If m > 2 , then at least one multiplier in each term is 0, then the sum is 0. Then, c2= a ⊗ b = 0 1 0 0 and d2= b ⊗ a = 0 0 1 0 . Moreover, [a, b] = a ⊗ b − b ⊗ a = 0 1 −1 0 ,
which implies that [a, b] 6= 0.
We construct a sequence { Wn: m ≥ 1 } of subspaces of the space T using mathematical
in-duction.
Induction base W1is the linear space with the basis {εεε0, εεε1, . . . , εεεd}.
Induction hypothesis Assume the space Wm−1 is constructed.
Induction step Define
Wm= [W1, Wm−1],
where the right hand side is the set of all possible finite linear combinations of Lie brackets [ak, bk] with ak∈ W1and bk∈ Wm−1.
Denote
U = ⊕∞
m=1Wm.
An element of the space U is called a Lie series. An element of the space Um= πmU
is called a Lie polynomial.
Chen [10] defined a map Xs,t: Cbv([0, T ], Rd) → T by
Xs,t(ωωω ) = ∞
∑
k=0ηηη ∈B∑
k Z s≤t1≤···≤tk≤t dωi1(t 1) · · · dωik(tk)εεεi1⊗ · · · ⊗ εεεik (3.2)Theorem 4. The map Xs,t(ωωω ) is multiplicative, that is, for any 0 ≤ s < t < u ≤ T we have
Xs,u(ωωω ) = Xs,t(ωωω ) ⊗ Xt,u(ωωω ).
For any ωωω ∈ Cbv([0, T ], Rd), ln(Xs,t(ωωω )) is a Lie series. For any Lie polynomial L ∈ Um, there
exist a path ωωω ∈ Cbv([0, T ], Rd) such that
πm(ln(Xs,t(ωωω ))) = L.
Define a random element Xs,t(◦W) of the space T by the series of Stratonovich iterated
integ-rals similar to (3.2): Xs,t(◦W) = ∞
∑
k=0ηηη ∈B∑
k Z s≤t1≤···≤tk≤t ◦dWi1(t 1) · · · dWik(tk)εεεi1⊗ · · · ⊗ εεεik. Denote X(m)s,t (ωωω ) = m∑
k=0ηηη ∈B∑
k Z s≤t1≤···≤tk≤t dωi1(t 1) · · · dωik(tk)εεεi1⊗ · · · ⊗ εεεik, X(m)s,t (◦W) = m∑
k=0ηηη ∈B∑
k Z s≤t1≤···≤tk≤t ◦dWi1(t 1) · · · dWik(tk)εεεi1⊗ · · · ⊗ εεεik.It is obvious that the paths ωωω1, . . . , ωωωn∈ Cbv([0, 1], Rd) and positive weights λ1, . . . , λndefine
a cubature formula on Wiener space of degree m at time 1 if and only if E[X(m)0,1(◦W)] =
n
∑
i=1
λiX(m)0,1(ωωωi). (3.3)
Assume that the paths ωωω1, . . . , ωωωn∈ Cbv([0, 1], Rd) and positive weights λ1, . . . , λn define
a cubature formula on Wiener space of degree m at time 1. By Theorem 4, the elements
ln(X0,1(ωωω1)), . . . , ln(X0,1(ωωωn)) of the space T are in fact Lie series. Then, the elements
Li= πm(ln(X0,1(ωωωi))), 1 ≤ i ≤ n
are Lie polynomials. Equation (3.3) takes the form E[X(m)0,1(◦W)] =
n
∑
i=1
λiπm(exp(Li)). (3.4)
Conversely, assume that (3.4) holds true. By Theorem4, there exist paths ωωωi∈ Cbv([0, 1], Rd)
such that
πn(ln(X0,1(ωωωi))) = Li, 1 ≤ i ≤ m.
The paths ωωω1, . . . , ωωωn and positive weights λ1, . . . , λndefine a cubature formula on Wiener
It follows that the problem of finding cubature formulae is equivalent to finding solutions to Equation (3.4).
First, we need to calculate the left hand side of (3.4). By [33, Proposition 4.10], we have
E[X0,1(◦W)] = exp εεε0+ 1 2 d
∑
i=1 ε ε εi⊗ εεεi ! .This means the following: to find a cubature formula on Wiener space of a degree m, we need to find Lie polynomials L1, . . . , Ln∈ Umand positive weights λ1, . . . , λnsuch that
πm exp εεε0+ 1 2 d
∑
i=1 εεεi⊗ εεεi !! = n∑
i=1 λiπm(exp(Li)). (3.5)To do that, we introduce a basis in U. Use mathematical induction.
Induction base PutH1= {εεε0, εεε1, . . . , εεεd} and order the above elements as
ε
εε0< εεε1< · · · < εεεd.
PutH2= { [εεεi, εεεj] : 0 ≤ i < j ≤ d } and order the above elements as [εεεi, εεεj] < [εεεk, εεεl] if
and only if either i < k or i = k and j < l, like in a dictionary.
Induction hypothesis Assume that the setsH1, H2, . . . , Hn−1 and their ordering are
con-structed.
Induction step DefineHn= {[a, [b, c]]}, where the elements a, b, and c satisfy the following
requirements:
• [a, [b, c]] contain exactly n symbols from the setH1.
• a, b, c, and [b, c] belong toH1∪ · · · ∪Hn−1.
• b ≤ a < [b, c], b < c
and order the constructed elements like in a dictionary
The union of all Hn is a basis in U called the Philip Hall basis, see [32, Theorem 5.3.1].
Denote the above basis byH = ∪n≥0Hn.
Example 1. Let d = 1. By induction base we haveH1= {εεε0, εεε1} andH2= {[εεε0, εεε1]}. For
H3we have two choices:
a = b = εεε0, c = εεε1
and
a = c = εεε1, b = εεε0.
ForH4we have three choices: a = b = εεε0, c = [εεε0, εεε1], a = εεε1, b = εεε0, c = [εεε0, εεε1], a = b = εεε1, c = [εεε0, εεε1]. It follows that H4= {[εεε0, [εεε0, [εεε0, εεε1]]], [εεε1, [εεε0, [εεε0, εεε1]]], [εεε1, [εεε1, [εεε0, εεε1]]]}.
The space T has a basisB, the union of the bases (3.1) andB0= {1}. Using the definition of
exponent (3), we may expand the left hand side of (3.5) in this basis as follows
πm exp εεε0+ 1 2 d
∑
i=1 εεεi⊗ εεεi !! = bm/2c∑
k=0 1 k! εεε0+ 1 2 d∑
i=1 ε εεi⊗ εεεi !⊗k . (3.6)Following [20], we may expand this expression in another basis. Let n be a positive integer, let Sn be the set of all permutations of order n, and let L1, . . . , Ln be Lie polynomials. The
symmetric productof the family {L1, . . . , Ln} is defined as
(L1, . . . , Ln) = 1
n!
∑
σ ∈Sn
Lσ (1)⊗ · · · ⊗ Lσ (n).
The celebrated Poincaré–Birkhoff–Witt theorem (see [20] and [33]) tells that the set
∞
[
n=0
{ (`1, . . . , `n) : `1, . . . , `n∈H ,`1≤ · · · ≤ `n}
is a basis of the space T.
Expand the left hand side of (3.5) in the Poincaré–Birkhoff–Witt basis. Let SL,mbe the set of
all ` ∈H that participate in this expansion. Assume it contains kmelements. We search the
Lie polynomials Lithat belong to the right hand side of (3.5) in the form
Li= km
∑
j=1
βi j`j, βi j∈ R.
To find a cubature formula on Wiener space of a degree m, we need to find the positive integer n, the real numbers βi j, 1 ≤ i ≤ km, 1 ≤ j ≤ n, and positive weights λ1, . . . , λnsuch that
πm exp εεε0+ 1 2 d
∑
i=1 ε εεi⊗ εεεi !! = n∑
i=1 λiπm exp ∑
`j∈SL,m βi j`j . (3.7)Example 2. We use [20, Appendix A]. Let d = 1. Then we have SL,3= {εεε0, εεε1},
SL,5= {εεε0, εεε1, [εεε1, [εεε0, εεε1]]},
SL,7= {εεε0, εεε1, [εεε0, εεε1], [εεε1, [εεε0, εεε1]], [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]}.
(3.8)
The set SL,9 contains 11 elements, and the set SL,11 contains 24 elements, see [20,
Ap-pendix A].
Let d = 2. The set SL,5 contains 8 elements, the set SL,7contains 26 elements.
When d = 3, the set SL,7contains 91 elements.
By [20, Lemma 4], for each m and for each Lie polynomial
Li=
∑
`j∈SL,m
βi j`j
there is a positive integer K such that
πm(exp(Li)) = K
∑
k=0ηηη =(i1,...,ik) : `
∑
i j∈SL,m,1≤ j≤kmηηη(βii1, βii2, . . . , βiik)(`i1, . . . , `ik),
where each function of k variables mηηη is a monomial. Express the left hand side of (3.7) in
the same basis: πm exp εεε0+ 1 2 d
∑
i=1 εεεi⊗ εεεi !! = K∑
k=0ηηη =(i1,...,ik) : `∑
i j∈SL,m,1≤ j≤k cηηη(`i1, . . . , `ik). (3.9)By equating coefficients of the two expansions, we obtain the system of equations
n
∑
i=1
λimηηη(βii1, βii2, . . . , βiik) = cηηη.
3.2
Examples
Example 3. Put d = 1 and m = 3. By (3.6), π3 exp εεε0+ 1 2εεε1⊗ εεε1 = 1 + εεε0+ 1 2εεε1⊗ εεε1= 1 + εεε0+ 1 2(εεε1, εεε1). (3.10) In other words, the nonzero coefficients in the right hand side of (3.9) are
c∅= c(0)= 1, c(1,1)= 1 2.
We see that the set SL,3indeed has the form given in the first line of (3.8). The Lie polynomials
take the form Li= εεε0+ βi1εεε1.
To expand the expression π3(exp(εεε0+ βi1εεε1)), we may use [33, Proposition 4.3]: for any
positive integer n, for any Lie polynomials `1, . . . , `n, and for any real numbers α1, . . . , αnwe
have exp n
∑
i=1 αi`i ! = ∞∑
k=0i1+···+i∑
n=k α1i1· · · αin n i1! · · · in! (`1, . . . , `1, . . . , `n, . . . , `n), (3.11)where the term `j of the symmetric product appears ij times. We need to take all possible
terms with k = 0 and k = 1, the terms with k = 2 and i1∈ {0, 1}, and the term with k = 3 and
i1= 0.
We see that from the expression π3(exp(εεε0+ βi1εεε1)), the norm of the first term is 2, while that
of the second term is 1. We find all nonnegative integer solution to the inequality 2i1+ i2≤ 3.
The trial and error method gives Table3.1. k i1 i2 2i1+ i2 1i1βi1i2 i1!i2! (`1, . . . , `1, `2, . . . , `2) 0 0 0 0 1 0 βi10 0!0!1 = 1 1 0 1 1 1 0 βi11 0!1! (εεε1) = βi1εεε1 1 1 0 2 1 1 βi10 1!0!(εεε0) = εεε0 2 0 2 2 10βi12 0!2! (εεε1, εεε1) = βi12 2 (εεε1, εεε1) 2 1 1 3 1 1 βi11 1!1!(εεε0, εεε1) = βi1(εεε0, εεε1) 3 0 3 3 1 0 βi13 0!3!(εεε1, εεε1, εεε1) = βi13 6 (εεε1, εεε1, εεε1)
Table 3.1: The nonnegative integer solutions to the inequality 2i1+ i2≤ 3
By Equation (3.11), the expression π3(exp (εεε0+ βi1εεε1)) is the sum of the terms
1i1βi2
i1
i1!i2!
(`1, . . . , `1, `2, . . . , `2),
over all lines of Table3.1, where `1= εεε0 and `2= εεε1. In the above expression, there are i1
copies of the symbol `1and i2copies of the symbol `2. Hence, we obtain
π3(exp(εεε0+ βi1εεε1)) =1 + εεε0+ βi1εεε1+ 1 2β 2 i1(εεε1, εεε1) + βi1(εεε0, εεε1) +1 6β 3 i1(εεε1, εεε1, εεε1).
Then we have n
∑
i=1 λiπ3(exp(Li)) = n∑
i=1 λi(1 + εεε0) + n∑
i=1 λiβi1(εεε1+ (εεε0, εεε1)) +1 2 n∑
i=1 λiβi12(εεε1, εεε1) + 1 6 n∑
i=1 λiβi13(εεε1, εεε1, εεε1). (3.12)Equating the coefficients in the right hand sides of (3.10) and (3.12), we obtain 1, εεε0: n
∑
i=1 λi= 1, εεε1, (εεε0, εεε1) : n∑
i=1 λiβi1= 0, (εεε1, εεε1) : 1 2 n∑
i=1 λiβi12 = 1 2, (εεε1, εεε1, εεε1) : 1 6 n∑
i=1 λiβi13 = 0.The same system of equation is written in [20, Appendix A.1] without proof. By trial and
error, it is easy to find a solution: n = 2, λ1= λ2= 1/2, β11= −1, β21 = 1. Then we have
L1= εεε0− εεε1, L2= εεε0+ εεε1.
Example 4. This time, we check [20, Appendix A.3], where d = 1 and m = 5. By (3.6),
π5 exp ε εε0+ 1 2εεε1⊗ εεε1 = π3 exp ε εε0+ 1 2εεε1⊗ εεε1 + 1 2! εεε0+ 1 2εεε1⊗ εεε1 ⊗ ε εε0+ 1 2εεε1⊗ εεε1 = 1 + εεε0+ 1 2(εεε1, εεε1) + 1 2(εεε0, εεε0) + 1 2(εεε0, εεε1⊗ εεε1) +1 8(εεε1, εεε1, εεε1, εεε1).
The fifth term of the right hand side is not an element of the symmetric Poincaré–Birkhoff– Witt basis. To find its expansion in the above basis, note that it contains one copy of εεε0and
two copies of εεε1. The elements of the Poincaré–Birkhoff–Witt basis with the same property
are:
[εεε1, [εεε0, εεε1]], (εεε1, [εεε0, εεε1]), (εεε0, εεε1, εεε1).
It follows that the desired expansion has the form 1
To find the unknown coefficients, expand all terms in the basis (3.1): 1 2(εεε0, εεε1⊗ εεε1) = 1 4εεε0⊗ εεε1⊗ εεε1+ 1 4εεε1⊗ εεε1⊗ εεε0, α [εεε1, [εεε0, εεε1]] = 2αεεε1⊗ εεε0⊗ εεε1+ αεεε1⊗ εεε1⊗ εεε0− αεεε0⊗ εεε1⊗ εεε1, β (εεε1, [εεε0, εεε1]) = 1 2β εεε0⊗ εεε1⊗ εεε1− 1 2β εεε1⊗ εεε1⊗ εεε0, γ (εεε0, εεε1, εεε1) = 1 3γ εεε0⊗ εεε1⊗ εεε1+ 1 3γ εεε1⊗ εεε0⊗ εεε1 1 3γ εεε1⊗ εεε1⊗ εεε0. By equating coefficients we obtain the following system of linear equations.
εεε0⊗ εεε1⊗ εεε1: − α + 1 2β + 1 3γ = 1 4, εεε1⊗ εεε0⊗ εεε1: 2α + 1 3γ = 0, εεε1⊗ εεε1⊗ εεε0: α − 1 2β + 1 3γ = 1 4.
By adding all three equations, we obtain γ = 1/2. From the second equation, we have α = −121, and then from the first equation β = 0. Then
1 2(εεε0, εεε1⊗ εεε1) = 1 2(εεε0, εεε1, εεε1) − 1 12[εεε1, [εεε0, εεε1]]. Finally, π5 exp εεε0+ 1 2εεε1⊗ εεε1 = 1 + εεε0+ 1 2(εεε1, εεε1) + 1 2(εεε0, εεε0) + 1 2(εεε0, εεε1, εεε1) − 1 12[εεε1, [εεε0, εεε1]] + 1 8(εεε1, εεε1, εεε1, εεε1), (3.13)
which coincides with [33, Proposition 5.4]. We see that the set SL,5indeed has the form given
in the second line in (3.8). Then we have
Li= εεε0+ βi1εεε1+ βi2[εεε1, [εεε0, εεε1]].
Again as in Example3, we can expand the expression
π5(εεε0+ βi1εεε1+ βi2[εεε1, [εεε0, εεε1]]),
using Equation (3.11). We see that the norm of the first term is 2, the norm of the second term is 1, while that of the third term is 4. We have to find all nonnegative integer solution to the inequality
The trial and error method gives Table3.2. k i1 i2 i3 2i1+ i2+ i3 1i1βi1i2βi3 i2 i1!i2!i3! (`1, . . . , `1, `2, . . . , `2, `3, . . . , `3) 0 0 0 0 0 1 0 βi10βi20 0!0!0! 1 = 1 1 0 1 0 1 10βi11βi20 0!1!0! (εεε1) = βi1εεε1 1 1 0 0 2 1 1 βi10βi20 1!0!0! (εεε0) = εεε0 1 0 2 0 2 1 0 βi12βi20 0!2!0! (εεε1, εεε1) = 1 2β 2 i1(εεε1, εεε1) 2 1 1 0 3 1 1 βi11βi20 1!1!0! (εεε0, εεε1) = βi1(εεε0, εεε1) 2 0 3 0 3 10βi13βi20 0!3!0! (εεε1, εεε1, εεε1) = 1 6β 3 i1(εεε1, εεε1, εεε1) 2 0 0 1 4 1 0 βi10βi21 0!0!1! [εεε1, [εεε0, εεε1]] = βi2[εεε1, [εεε0, εεε1]] 2 2 0 0 4 1 2 βi10βi20 2!0!0! (εεε0, εεε0) = 1 2(εεε0, εεε0) 3 1 2 0 4 1 1 βi12βi20 1!2!0! (εεε0, εεε1, εεε1) = 1 2β 2 i1(εεε0, εεε1, εεε1) 3 0 4 0 4 1 0 βi10βi24 0!0!4! (εεε1, εεε1, εεε1) = 1 24β 4 i1(εεε1, εεε1, εεε1, εεε1) 3 0 1 1 5 10βi11βi21 0!1!1! (εεε1, [εεε1, [εεε0, εεε1]]) = βi1βi2(εεε1, [εεε1, [εεε0, εεε1]]) 4 2 1 0 5 1 2 βi11βi20 2!1!0! (εεε0, εεε0, εεε1) = 1 2βi1(εεε0, εεε0, εεε1) 4 1 3 0 5 1 1 βi13βi20 1!3!0! (εεε0, εεε1, εεε1, εεε1) = 1 6β 3 i1(εεε0, εεε1, εεε1, εεε1) 5 0 5 0 5 1 0 βi15βi20 0!5!0! (εεε1, εεε1, εεε1, εεε1, εεε1) = 1 120β 5 i1(εεε1, εεε1, εεε1, εεε1, εεε1)
Table 3.2: The nonnegative integer solutions to the inequality 2i1+ i2+ 4i3≤ 5.
Again, by Equation (3.11), the expression π5(exp (εεε0+ βi1εεε1+ βi2[εεε1, [εεε0, εεε1]])) is the sum
of the terms 1i1β
i2 i1βi2i3
i1!i2!i3! (`1, . . . , `1, `2, . . . , `2, `3, . . . , `3) over all lines of Table3.2, where `1= εεε0,
π5(exp (Li)) = 1 + βi1εεε1+ εεε0+ 1 2β 2 i1(εεε1, εεε1) + βi1(εεε0, εεε1) + 1 6β 3 i1(εεε1, εεε1, εεε1) + βi2[εεε1, [εεε0, εεε1]] + 1 2(εεε0, εεε0) + 1 2β 2 i1(εεε0, εεε1, εεε1) + 1 24β 4 i1(εεε1, εεε1, εεε1, εεε1) + βi1βi2(εεε1, [εεε1, [εεε0, εεε1]]) +1 2βi1(εεε0, εεε0, εεε1) + 1 6β 3 i1(εεε0, εεε1, εεε1, εεε1) + 1 120β 5 i1(εεε1, εεε1, εεε1, εεε1, εεε1).
Collecting all terms with the same numerical coefficients, we have
n
∑
i=1 λiπ5(exp(Li)) = n∑
i=1 λi 1 + εεε0+ 1 2(εεε0, εεε0) +1 2 n∑
i=1 λiβi1 εεε1+ (εεε0, εεε1) + 1 2(εεε0, εεε1, εεε1) +1 2 n∑
i=1 λiβi2[εεε1, [εεε0, εεε1]] + n∑
i=1 λiβi1βi2(εεε1, [εεε1, [εεε0, εεε1]]) +1 2 n∑
i=1 λiβi12((εεε0, εεε1, εεε1) + (εεε1, εεε1)) + 1 6 n∑
i=1 λiβi13((εεε1, εεε1, εεε1) + (εεε0, εεε1, εεε1, εεε1)) + 1 24 n∑
i=1 λiβi14(εεε1, εεε1, εεε1, εεε1) + 1 120 n∑
i=1 λiβi15(εεε1, εεε1, εεε1, εεε1, εεε1) (3.14) Equating the coefficients in the right hand sides of (3.13) and (3.14), we obtain1, εεε0, (εεε0, εεε0) : n
∑
i=1 λi= 1, ε εε1, (εεε0, εεε1), (εεε0, εεε0, εεε1) : n∑
i=1 λiβi1= 0, [εεε1, [εεε0, εεε1]] : n∑
i=1 λiβi2= − 1 12, [εεε1, [εεε1, [εεε0, εεε1]] : n∑
i=1 λiβi1βi2= 0, (εεε0, εεε1, εεε1), (εεε1, εεε1) : n∑
i=1 λiβi12 = 1, (εεε1, εεε1, εεε1), (εεε0, εεε1, εεε1, εεε1) : n∑
i=1 λiβi13 = 1 (εεε1, εεε1, εεε1, εεε1) : n∑
i=1 λiβi14 = 3, (εεε1, εεε1, εεε1, εεε1, εεε1) : n∑
i=1 λiβi15 = 0,A solution to this system is found in [20] and is as follows: n = 3, λ1 = λ3= 16, λ2= 23,
β11= −
√
3, β21= 0, and β31=
√
3, β12= β22= β32= −121. The Lie polynomials Litake the
form L1= εεε0− √ 3εεε1− 1 12[εεε1, [εεε0, εεε1]], L2= εεε0− 1 12[εεε1, [εεε0, εεε1]], L3= εεε0+ √ 3εεε1− 1 12[εεε1, [εεε0, εεε1]]. Example 5. Put d = 1 and m = 7. By (3.6),
π7 exp ε εε0+ 1 2εεε1⊗ εεε1 = π5 exp ε εε0+ 1 2εεε1⊗ εεε1 + 1 3! εεε0+ 1 2εεε1⊗ εεε1 ⊗3 = π5 exp ε εε0+ 1 2εεε1⊗ εεε1 +1 6(εεε0, εεε0, εεε0) + 1 12εεε1⊗ εεε1⊗ εεε0⊗ εεε0+ 1 12εεε0⊗ εεε1⊗ εεε1⊗ εεε0 + 1 24εεε1⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε0+ 1 12εεε0⊗ εεε0⊗ εεε1⊗ εεε1 + 1 24εεε1⊗ εεε1⊗ εεε0⊗ εεε1⊗ εεε1+ 1 24εεε0⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε1 + 1 48(εεε1, εεε1, εεε1, εεε1, εεε1, εεε1).
There are 42 = 6 elements of the basis (3.1) that contain two copies of εεε0 and two copies
of εεε1. By the Poincaré–Birkhoff–Witt theorem, there should be 6 elements of the symmetric
Poincaré–Birkhoff–Witt basis with the same property. They are as follows. • The element ofH4: [εεε1, [εεε0, [εεε0, εεε1]]].
• The symmetric tensor products that include the elements ofH3: (εεε1, [εεε0, [εεε0, εεε1]]) and
(εεε0, [εεε1, [εεε0, εεε1]]).
• The symmetric tensor products that include the elements ofH2: ([εεε0, εεε1], [εεε0, εεε1]) and
(εεε0, εεε1, [εεε0, εεε1]).
• The symmetric tensor product that includes only the elements ofH1: (εεε0, εεε0, εεε1, εεε1).
It follows that 1
12(εεε1⊗ εεε1⊗ εεε0⊗ εεε0+ εεε0⊗ εεε1⊗ εεε1⊗ εεε0+ εεε0⊗ εεε0⊗ εεε1⊗ εεε1) = α[εεε1, [εεε0, [εεε0, εεε1]]] + β (εεε1, [εεε0, [εεε0, εεε1]]) + γ(εεε0, [εεε1, [εεε0, εεε1]])
Expanding the tensors in the right hand side in the basis (3.1) and equating coefficients, we obtain the system of linear equations
−α +1 2β − 1 2γ + 1 6ε + 1 6ζ = 1 12, 2α − β + γ + δ + +1 6ε + 1 6ζ = 0, −γ − δ +1 6ζ = 1 12, β − δ +1 6ζ = 0, −2α − β + γ + δ −1 6ε + 1 6ζ = 0, α +1 2β − 1 2γ − 1 6ε + 1 6ζ = 1 12.
The solution to this system is α = β = ε = 0, γ = −121, δ = −121, ζ = 14. Then 1 12(εεε1⊗ εεε1⊗ εεε0⊗ εεε0+ εεε0⊗ εεε1⊗ εεε1⊗ εεε0+ εεε0⊗ εεε0⊗ εεε1⊗ εεε1) = − 1 12(εεε0, [εεε1, [εεε0, εεε1]]) + 1 24([εεε0, εεε1], [εεε0, εεε1]) + 1 4(εεε0, εεε0, εεε1, εεε1).
Similarly, there are 51 = 5 elements of the basis (3.1) that contain one copy of εεε0 and four
copies of εεε1. By the Poincaré–Birkhoff–Witt theorem, there should be 5 elements of the
symmetric Poincaré–Birkhoff–Witt basis with the same property. They are as follows. • The element ofH5: [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]].
• The symmetric tensor product that includes the elements ofH4: (εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]).
• The symmetric tensor product that includes the elements ofH3: (εεε1, εεε1, [εεε1, [εεε0, εεε1]]).
• The symmetric tensor product that includes the elements ofH2: (εεε1, εεε1, εεε1, [εεε0, εεε1]).
• The symmetric tensor product that includes only the elements ofH1: (εεε0, εεε1, εεε1, εεε1, εεε1).
It follows that 1
24(εεε1⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε0+ εεε1⊗ εεε1⊗ εεε0⊗ εεε1⊗ εεε1+ εεε0⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε1) = α[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] + β (εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]) + γ(εεε1, εεε1, [εεε1, [εεε0, εεε1]])
+ δ (εεε1, εεε1, εεε1, [εεε0, εεε1]) + ε(εεε0, εεε1, εεε1, εεε1, εεε1).
obtain the system of linear equations −α +1 2β − 1 3γ + 1 2δ + 1 5ε = 1 24, 4α − β +1 3γ − 1 4δ + 1 5ε = 0, −6α +1 5ε = 1 24, 4α + β +1 3γ − 1 4δ + 1 5ε = 0, −α −1 2β − 1 3γ + 1 5ε = 1 24. The solution is α = −3601 , β = δ = 0, γ = −241, ε = 18. We have
1 24(εεε1⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε0+ εεε1⊗ εεε1⊗ εεε0⊗ εεε1⊗ εεε1+ εεε0⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε1) = − 1 360[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] − 1 24(εεε1, εεε1, [εεε1, [εεε0, εεε1]]) + 1 8(εεε0, εεε1, εεε1, εεε1, εεε1). The complete expansion of the element π7 exp εεε0+12εεε1⊗ εεε1 in the symmetric Poincaré–
Birkhoff–Witt basis takes the form π7 exp ε ε ε0+ 1 2εεε1⊗ εεε1 = 1 + εεε0+ 1 2(εεε1, εεε1) + 1 2(εεε0, εεε0) + 1 2(εεε0, εεε1, εεε1) − 1 12[εεε1, [εεε0, εεε1]] + 1 8(εεε1, εεε1, εεε1, εεε1) + 1 6(εεε0, εεε0, εεε0) − 1 12(εεε0, [εεε1, [εεε0, εεε1]]) + 1 24([εεε0, εεε1], [εεε0, εεε1]) +1 4(εεε0, εεε0, εεε1, εεε1) − 1 360[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] − 1 24(εεε1, εεε1, [εεε1, [εεε0, εεε1]]) + 1 8(εεε0, εεε1, εεε1, εεε1, εεε1) + 1 48(εεε1, εεε1, εεε1, εεε1, εεε1, εεε1). (3.15)
This confirms the third line in (3.8). Then we have
Li= εεε0+ βi1εεε1+ βi2[εεε0, εεε1] + βi3[εεε1, [εεε0, εεε1]], +βi4[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]].
We see that the norm of the first term is 2, that of the second term is 1, that of the third term is 3, that of the fourth term is 4, and that of the fifth term is 6. There are 40 nonnegative integer solutions to the inequality
2i1+ i2+ 3i3+ 4i4+ 6i5≤ 7. (3.16)
The expression π7(Li) is the sum of the terms
1i1βi2 i1β i3 i2β i4 i3β i5 i4 i1!i2!i3!i4!i5! (`1, . . . , `1, . . . , `5, . . . , `5),
where `1= εεε0, . . . , `5= [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]].
The terms 1, εεε0, 12(εεε0, εεε0), and 16(εεε0, εεε0, εεε0) of the above sum generate the same terms in the
expansion (3.15) and the equation
n
∑
i=1
λi= 1.
The terms12βi12(εεε1, εεε1), 12βi12(εεε0, εεε1, εεε1), and14βi12(εεε0, εεε0, εεε1, εεε1) generate the terms 12(εεε1, εεε1), 1
2(εεε0, εεε1, εεε1), and 1
4(εεε0, εεε0, εεε1, εεε1) of the expansion (3.15) and the equation n
∑
i=1
λiβi12 = 1.
The terms βi3[εεε1, [εεε0, εεε1]] and βi3(εεε0, [εεε1, [εεε0, εεε1]]) generate the terms −121[εεε1, [εεε0, εεε1]] and 1
12(εεε0, [εεε1, [εεε0, εεε1]]) of the expansion (3.15) and the equation n
∑
i=1 λiβi3= − 1 12.The terms241βi14(εεε1, εεε1, εεε1, εεε1) and241βi14(εεε0, εεε1, εεε1, εεε1, εεε1) generate the terms81(εεε1, εεε1, εεε1, εεε1)
and 18(εεε0, εεε1, εεε1, εεε1, εεε1) of the expansion (3.15) and the equation n
∑
i=1
λiβi14 = 3.
The term βi4[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] generates the term −3601 [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] of the
ex-pansion (3.15) and the equation
n
∑
i=1 λiβi4= − 1 360.The term 12βi12βi3(εεε1, εεε1, [εεε1, [εεε0, εεε1]]) generates the term −241(εεε1, εεε1, [εεε1, [εεε0, εεε1]]) of the
ex-pansion (3.15) and the equation
n
∑
i=1 λiβi12βi3= − 1 12.The term12βi22([εεε0, εεε1], [εεε0, εεε1]) generates the term241([εεε0, εεε1], [εεε0, εεε1]) of the expansion (3.15)
and the equation
n
∑
i=1 λiβi22 = 1 12.Finally, the term 7201 βi16 generates the term 481(εεε1, εεε1, εεε1, εεε1, εεε1, εεε1) of the expansion (3.15)
and the equation
n
∑
i=1
The above 8 equations with nonzero right hand side are listed in [20]. The remaining 25 solutions to the inequality (3.16) generate 14 additional equations with zero right hand side not listed in [20]. They are as follows.
The solutions (0, 1, 0, 0, 0), (1, 1, 0, 0, 0), (2, 1, 0, 0, 0), and (3, 1, 0, 0, 0) generate the equation
n
∑
i=1
λiβi1= 0.
The solutions (0, 0, 1, 0, 0), (1, 0, 1, 0, 0), and (2, 0, 1, 0, 0) generate the equation
n
∑
i=1
λiβi2= 0.
The solutions (0, 3, 0, 0, 0), (1, 3, 0, 0, 0), and (2, 3, 0, 0, 0) generate the equation
n
∑
i=1
λiβi13 = 0.
The solutions (0, 1, 1, 0, 0) and (1, 1, 1, 0, 0) generate the equation
n
∑
i=1
λiβi1βi2= 0.
The solutions (0, 2, 1, 0, 0) and (1, 2, 1, 0, 0) generate the equation
n
∑
i=1
λiβi12βi2= 0.
The solutions (0, 5, 0, 0, 0) and (1, 5, 0, 0, 0) generate the equation
n
∑
i=1
λiβi15 = 0. The solution (0, 3, 1, 0, 0) generates the equation
n
∑
i=1
λiβi13βi2= 0.
The solution (0, 1, 0, 0, 1) generates the equation
n
∑
i=1
λiβi1βi4= 0.
The solution (0, 0, 1, 1, 0) generates the equation
n
∑
i=1
The solution (1, 1, 0, 1, 0) generates the equation
n
∑
i=1
λiβi1βi3= 0.
The solution (0, 3, 0, 1, 0) generates the equation
n
∑
i=1
λiβi13βi3= 0.
The solution (0, 1, 2, 0, 0) generates the equation
n
∑
i=1
λiβi1βi22 = 0.
The solution (0, 4, 1, 0, 0) generates the equation
n
∑
i=1
λiβi14βi2= 0.
Finally, the solution (0, 7, 0, 0, 0) generates the equation
n
∑
i=1
λiβi17 = 0.
The solution to the obtained system of equations can be found in [20].
All the previous examples had the following feature. Consider the expressionn!1 εεε0+12εεε1⊗ εεε1
⊗n . By the binomial formula, it contains nk tensor products of n − k copies of εεε0and 2k copies of
ε ε
ε1, 0 ≤ k ≤ n. The total number of such tensor products is n+k2k. By the Poincaré–Birkhoff–
Witt theorem, the number of elements of the symmetric Poincaré–Birkhoff–Witt basis that contain n − k copies of εεε0 and 2k copies of εεε1, is equal to n+k2k . Thus, in order to represent
the sum of nk tensor products of n − k copies of εεε0and 2k copies of εεε1in the above basis, we
have to solve a system of n+k2k linear equations with the same amount of unknowns. Indeed, in Example5, where n = 3, we solved the system of 42 = 6 linear equations to represent the sum
1
12(εεε1⊗ εεε1⊗ εεε0⊗ εεε0+ εεε0⊗ εεε1⊗ εεε1⊗ εεε0+ εεε0⊗ εεε0⊗ εεε1⊗ εεε1) in the above basis and the system of 54 = 5 linear equations to represent the sum
1
24(εεε1⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε0+ εεε1⊗ εεε1⊗ εεε0⊗ εεε1⊗ εεε1+ εεε0⊗ εεε1⊗ εεε1⊗ εεε1⊗ εεε1).
Among n+k2k numbers that solve each system considered before, there were exactly nk
nonzero numbers, the rest of them were zeroes. In other words, the expansions of the expres-sion π2n+1 εεε0+12εεε1⊗ εεε1 in the basis (3.1) and in the symmetric Poincaré–Birkhoff–Witt
basis contain the same amount of terms, when1 ≤ n ≤ 3.
Example 6. Consider the case of d = 1, m = 9. In order to represent the sum of 41 tensor products of 3 copies of εεε0and 2 copies of εεε1in the above basis, we have to solve a system of
5
2 = 10 linear equations with the same amount of unknowns. For the case of 2 copies of εεε0
and 4 copies of εεε1the number of equations is 64 = 15. Instead of solving these equations, we
propose another idea.
The set SL,9contains 11 elements, and the Lie polynomials have the form
Li= εεε0+ βi1εεε1+ βi2[εεε0, εεε1] + βi3[εεε1, [εεε0, εεε1]] + βi4[εεε1, [εεε1, [εεε0, εεε1]]]
+ βi5[εεε1, [εεε0, [εεε0, [εεε0, εεε1]]]] + βi6[εεε1, [εεε1, [εεε0, [εεε0, εεε1]]]] + βi7[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]
+ βi8[[εεε0, εεε1], [εεε0, [εεε0, εεε1]]] + βi9[[εεε0, εεε1], [εεε1, [εεε0, εεε1]]]
+ βi10[εεε1, [εεε1, [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]]].
The system of equations for λi and βi j, 1 ≤ j ≤ 10, with nonzero right hand sides is given
in [20]. Each equation generates some amount of terms to the expansion of the expression
π9 εεε0+12εεε1⊗ εεε1 in the symmetric Poincaré–Birkhoff–Witt basis.
The first equation is
n
∑
i=1
λi= 1.
This equation generates the following terms:
1 + 1 1!εεε0+ 1 2!(εεε0, εεε0) + 1 3!(εεε0, εεε0, εεε0) + 1 4!(εεε0, εεε0, εεε0, εεε0). The next one is
n
∑
i=1
λiβi12 = 1.
It generates the terms 1 2!(εεε1, εεε1) + 1 1!2!(εεε0, εεε1, εεε1) + 1 2!2!(εεε0, εεε0, εεε1, εεε1) + 1 3!2!(εεε0, εεε0, εεε0, εεε1, εεε1). The next one is
n
∑
i=1
λiβi14 = 3.
It generates the terms 3 4!(εεε1, εεε1, εεε1, εεε1) + 3 1!4!(εεε0, εεε1, εεε1, εεε1, εεε1) + 3 2!4!(εεε0, εεε0, εεε1, εεε1, εεε1, εεε1). The next one is
n
∑
i=1
It generates the terms 15
6!(εεε1, εεε1, εεε1, εεε1, εεε1, εεε1) + 15
1!6!(εεε0, εεε1, εεε1, εεε1, εεε1, εεε1, εεε1). The next one is
n
∑
i=1
λiβi18 = 105.
It generates the term
105
8! (εεε1, εεε1, εεε1, εεε1, εεε1, εεε1, εεε1, εεε1). The next equation
n
∑
i=1 λiβi22 = 1 12 generates the terms1
12 · 2!([εεε0, εεε1], [εεε0, εεε1]) + 1
12 · 1! · 2!(εεε0, [εεε0, εεε1], [εεε0, εεε1]). The next equation
n
∑
i=1 λiβi3= − 1 12 generates the terms− 1 12[εεε1, [εεε0, εεε1]] − 1 12 · 1!(εεε0, [εεε1, [εεε0, εεε1]]) − 1 12 · 2!(εεε0, εεε0, [εεε1, [εεε0, εεε1]]). The equation n
∑
i=1 λiβi32 = 1 72 generates the term1 72 · 2!([εεε1, [εεε0, εεε1]], [εεε1, [εεε0, εεε1]]). The equation n
∑
i=1 λiβi5= 1 720 generates the term1
720[εεε1, [εεε0, [εεε0, [εεε0, εεε1]]]]. The next equation is
n
∑
i=1 λiβi7= − 1 360.It generates the terms
− 1
360[εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]] − 1
360 · 1!(εεε0, [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]). The next equation
n
∑
i=1 λiβi8= − 1 720 generates the term− 1 720[[εεε0, εεε1], [εεε0, [εεε0, εεε1]]]. The equation n
∑
i=1 λiβi10= − 1 6720 generates the term− 1 6720[εεε1, [εεε1, [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]]]. The equation n
∑
i=1 λiβi1βi6= 1 360 generates the term1 360(εεε1, [εεε1, [εεε1, [εεε0, [εεε0, εεε1]]]]). The equation n
∑
i=1 λiβi1βi9= − 1 720 generates the term− 1
720(εεε1, [[εεε0, εεε1], [εεε1, [εεε0, εεε1]]]). The next one
n
∑
i=1 λiβi12βi22 = 1 12 generates the term1 12 · 2! · 2!(εεε1, εεε1, [εεε0, εεε1], [εεε0, εεε1]). The equation n
∑
i=1 λiβi12βi3= − 1 12 generates the terms− 1
12 · 2!(εεε1, εεε1, [εεε1, [εεε0, εεε1]]) − 1
The equation n
∑
i=1 λiβi14βi3= − 1 4 generates the term− 1
4 · 4!(εεε1, εεε1, εεε1, εεε1, [εεε1, [εεε0, εεε1]]). Finally, the equation
n
∑
i=1 λiβi2βi4= 1 144 generates the last term1
144([εεε0, εεε1], [εεε1, [εεε1, [εεε0, εεε1]]]).
The expansion of the expression π9 εεε0+12εεε1⊗ εεε1 in the basis (3.1) contains 41 = 4
differ-ent tensor products of 3 copies of εεε0and 2 copies of εεε1. In the symmetric Poincaré–Birkhoff–
Witt basis, there are 5 nonzero terms: 1 12(εεε0, εεε0, εεε0, εεε1, εεε1) + 1 24(εεε0, [εεε0, εεε1], [εεε0, εεε1]) − 1 24(εεε0, εεε0, [εεε1, [εεε0, εεε1]]) + 1 720[εεε1, [εεε0, [εεε0, [εεε0, εεε1]]]] − 1 720[[εεε0, εεε1], [εεε0, [εεε0, εεε1]]].
Similarly, the above expansion contains 42 = 6 different tensor products of 2 copies of εεε0and
4 copies of εεε1. In the symmetric Poincaré–Birkhoff–Witt basis, there are 8 nonzero terms:
1 16(εεε0, εεε0, εεε1, εεε1, εεε1, εεε1) + 1 144([εεε1, [εεε0, εεε1]], [εεε1, [εεε0, εεε1]]) − 1 360(εεε0, [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]) + 1 360(εεε1, [εεε1, [εεε1, [εεε0, [εεε0, εεε1]]]]) − 1 720(εεε1, [[εεε0, εεε1], [εεε1, [εεε0, εεε1]]]) + 1 48(εεε1, εεε1, [εεε0, εεε1], [εεε0, εεε1]) − 1 24(εεε0, εεε1, εεε1, [εεε1, [εεε0, εεε1]]) + 1 144([εεε0, εεε1], [εεε1, [εεε1, [εεε0, εεε1]]]).
Moreover, the above expansion contains 43 = 4 different tensor products of 1 copy of εεε0
and 6 copies of εεε1. In the symmetric Poincaré–Birkhoff–Witt basis, there are only 3 nonzero
terms: 1 48(εεε0, εεε1, εεε1, εεε1, εεε1, εεε1, εεε1)− 1 6720[εεε1, [εεε1, [εεε1, [εεε1, [εεε1, [εεε0, εεε1]]]]]]− 1 96(εεε1, εεε1, εεε1, εεε1, [εεε1, [εεε0, εεε1]]). Assume that we found the weights λ1, . . . , λn and Lie polynomials L1, . . . , Ln. How to find
the paths ωωω1, . . . , ωωωn∈ Cbv([0, 1], Rd) such that
for i = 1, 2, . . . , n? Following [20] and [33], we divide the interval [0, 1] into l subintervals of lengths ϕ0 j, 1 ≤ j ≤ l. Consider a piecewise linear path ωωωi∈ Cbv([0, 1], Rd) with derivatives
equal to ϕ1
0 j(ϕ0 j, ϕ1 j, . . . , ϕd j) in the jth interval. Then we have
X0,1(ωωωi) = exp d
∑
k=0 ϕk1εεεk ! ⊗ · · · ⊗ exp d∑
k=0 ϕklεεεk !by Theorem4. We have to solve the equation
πm ln exp d
∑
k=0 ϕk1εεεk ! ⊗ · · · ⊗ exp d∑
k=0 ϕklεεεk !!! = Li. (3.17)To calculate the logarithm, we use the explicit Baker–Campbell–Hausdorff formula. There are many variants of the above formula in the literature. We choose a variant represented in [8]. Other variants may be found in the original papers [2,6,7,15,21].
Let U be a free Lie algebra generated by the symbols X and Y , and letH be its Philip Hall
basis.
Theorem 5 (Baker–Campbell–Hausdorff). There exist rational numbers { zE: E ∈H } such
that
ln(exp(X ) exp(Y )) =
∑
E∈H
zEE. (3.18)
The first terms of the Baker–Campbell–Hausdorff series (3.18) are as follows:
ln(exp(X ) exp(Y )) = X +Y +1 2[X ,Y ] + 1 12[X , [X ,Y ]] − 1 12[Y, [X ,Y ]] + 1 24[X , [Y, [Y, X ]]] + · · · . Casas and Murua in [8] calculated the coefficients zE for 111013 elements of the Philip Hall
basis containing up to 20 symbols. Of these coefficients, 109697 are not equal to 0. The question of the convergence of the Baker–Campbell–Hausdorff series is delicate and will not be considered here.
To solve Equation (3.17), we apply Theorem5to its left hand side. As a result, we obtain a system of polynomial equations on the coefficients ϕkm, 0 ≤ k ≤ d, 1 ≤ m ≤ l, for each Li. An
Chapter 4
Conclusion
This thesis covered several areas regarding construction of cubature formulae for the
Heath-Jarrow-Morton (HJM) model. Clearly, we proved the cubature method developed in [20,33]
to the infinite dimensional stochastic differential equation–HJM. The chapters that aimed this were organized as follows;
Chapter1has briefly introduced the research topic and noting down the research objectives.
We specifically introduced the modern algebraic approach, described in [3] as the method
of solving the infinite-dimensional HJM equation. We started formulating the problem in
Chapter 2by establishing some theorems and definitions. We presented a discussion on the
stochastic Taylor expansion based on the iterated Stratonovich integrals and assumed that the Stratonovich Equation (2.5) is continuously differentiable in such a way that it could be easy to formulate the multiple Stratonovich stochastic integrals. Then, we proposed that the paths ω
ω
ω1, . . . , ωωωn ∈ Cbv([0, T ], Rd) and positive weights λ1...λn define a cubature formula on
Wiener space of degree ≤ m at time T , if and only if, for all set of multi indices, we have E[I(t, ηηη , ◦dWtikk)] = n
∑
j=1 λj Z 0<t1<....<tk<T ◦dωi1 j (t1).... ◦ dω ik j (tk). (4.1)Applying the Tchakaloff’s theorem (as explained in [33]) to (4.1), one could easily find that it is possible to construct cubature formulae on Wiener space for d-dimension paths when t= 1.
In Chapter 3, we concentrated on the algebra expression regarding construction of cubature
formulae on Wiener space. We closed this chapter by showing some examples regarding construction of cubature formulae for degree 3, 5, 7, and 9 in 1–dimensional space.
Generally, in this study, we have covered the general idea of the cubature formulae on Wiener space. We focused to prove this method to the HJM infinite dimensional stochastic equation. The description of the model was just the theoretical part of the stochastic differential equation and algebraic expression applied to mathematical finance. In connection to this study, we
proposed more investigation on the models that incorporate stochastic volatility and jump processes simply because forward curves are normally distributed with time. On the other hand, regarding the theory of algebraic mathematics, we recommend further study to consider the algebraic model itself to come up with a conclusion.
Appendix A
Criteria for a Master thesis
This section provides a brief declaration of the fulfilment of the objectives set as requirements of the Swedish National Agency for Higher Education to Master theses.
Objective 1: Knowledge and understanding.
The thesis starts with an introduction of different approaches that can be used in Pricing zero coupon bonds. The focus is on the development of the HJM (Heath—Jarrow-–Morton) model using modern algebraic approach. Different research papers have been reviewed in this field to build the content of the thesis.
Objective 2: Methodological knowledge.
In this report, the author shows how a problem is formulated as a mathematical problem. A deeper presentation is in chapter2 and 3. Moreover, the report also included and explained some of the mathematical proofs in the appendices.
Objective 3: Critically and Systematically Integrate Knowledge.
The thesis uses information from different sources to develop the main concept. Many sources were suggested by the thesis supervisor (Anatoliy Malyarenko) to extensively elaborate the particular concept of the project.
Objective 4: Independently and Creatively Identify and Carry out Advanced Tasks. With the help of guidance from the thesis supervisor, the plan was followed thoroughly through-out the research. Moreover, the author has shown a significant ability to identify and formulate questions, in the given time frame.
Objective 5: Present and Discuss Conclusions and Knowledge.
The language and concepts used in the thesis are universal to any reader of financial mathemat-ics and especially to anyone who is familiar with the basic concepts of modern algebraic meth-ods. The thesis will be presented on 1stFebruary, 2019 at the Mälardalen University.
Objective 6: Scientific, Social and Ethical Aspects.
math-ematics in constructing different financial mathematical models. Carefully, ideas from differ-ent authors have used as inputs in this thesis.
