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A.5 Calculus, Fourier Transformations and Partial Differential Equations

A.5 Calculus, Fourier Transformations and Partial Differential Equations

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348



A Mathematics



∂h(x, y)/∂x. This means that we cut out a thin slice of the hills along the

x-direction and look at the derivative of the elevation on this slice (which

is just a function in one variable). Of course we can generalize this idea to

arbitrarily many dimensions. An example for a vector composed of partial

derivatives is the gradient of a function h : (x1 , . . . , xn ) → h(x1 , . . . , xn ) ∈ R

which is defined by







∇h(x) := ⎜





∂h(x)

∂x1



..

.



∂h(x)

∂xn





⎟,





where x = (x1 , . . . , xn ) and sometimes Dh is written instead of ∇h.

In the following we need to extend the real numbers R to complex numbers

C that can be written as a linear combination of a real and an imaginary

number, i.e., z = x + iy ∈ C with x, y ∈ R and i2 = −1. For complex numbers

particularly the Euler formula holds:

eiφ = cos φ + i sin φ.

We can now define a Fourier transformation: it maps a function f : R → R

to its Fourier transform F f : R → C and is defined by

1

(F f )(ξ) := √





+∞



e−iξt f (t) dt.



−∞



The Fourier transformation describes in a certain way the frequency distribution of f . The most natural application is in acoustics where f describes the

oscillation (e.g., of the air) and its Fourier transform corresponds to the distribution of frequencies, i.e., its sound. A sine function would lead to a Dirac

distribution as Fourier transform: in other words there is only one frequency,

the sound is a pure tone. Such a tone would sound very harsh and artificial,

since natural tones (like the sound of a piano) are composed of many different tones, i.e., they correspond to a weighted sum of sine functions. Their

Fourier transform is therefore a weighted sum of Dirac distributions. Another

example is a normal distribution: their Fourier transform is again a normal

distribution.

The Fourier transformation has a couple of interesting properties. We collect the most important in the following lemma:

Lemma A.12 (Properties of the Fourier transformation). Let F be the

Fourier transformation, then:

(i) F is a linear map.

(ii) The inverse of the Fourier transformation is given by

1

(F −1 f )(x) = √





+∞



eixt fˆ(t) dt.

−∞



A.5 Calculus, Fourier Transformations and Partial Differential Equations



349



(iii) The Fourier transform of a derivative is a polynomial, more precisely:

F



∂n

f (x) (ξ) = (iξ)n (Ff )(ξ).

∂xn



(iv) It is possible to define an appropriate space (of functions or distributions)

such that F is a bijective map on this space.

Property (iii) can be summarized by saying that the Fourier transformation turns derivatives into a product – although this has nothing to do with

the marketing of options... The property is often the main reason to use the

Fourier transformation and could even be used as an alternative definition

of differentiation. While this is certainly a complicated way for reaching this

goal, it enables us to generalize the definition of derivatives: if we take the

Fourier transform of a multiplication of f with, e.g., i|ξ|1/2 , we get something

like a “half” derivative. If that sounds esoteric to you, then be ensured that it

is not: very much to the contrary it is surprisingly useful. The concept leads

to the definition of pseudo differential operators and they are needed, e.g.,

in solving certain asset pricing problems when the underlying process is of

a more complicated form. We mention this point and generally use Fourier

transformations when we discuss L´evy processes in Section 8.8.

A partial differential equation (short: PDE ) is an equation that contains

different partial derivatives of an unknown function and is used to determine

this function. In contrast, an ordinary differential equation (short: ODE ) only

involves one kind of derivative (e.g., only derivatives with respect to t). As an

example for a simple PDE we consider the heat equation 4 which is needed to

solve the Black-Scholes equation (see Section 8):

∂ 2 u(x, t)

∂u(x, t)

=

,

∂t

∂x2

where t ∈ [0, T ] and x ∈ R. (In the original physics model for heat transport,

t is the time and x the space variable, whereas u(x, t) is the temperature at

time t and position x.) The above equation therefore means that the partial

derivative of u with respect to t equals its second partial derivative with

respect to x.

Typically, PDEs can only be solved uniquely when we have additional

conditions. In the case above this would be an initial condition (specifying u

at t = 0) plus some boundary conditions (e.g., specifying the behavior of u

for x → ±∞).

PDEs are a central modeling tool in all scientific disciplines that rely on

sophisticated mathematical models (like physics, chemistry, biology, engineering – and some areas in finance). Therefore their analytical and numerical

4



This PDE was originally used to describe the transport of heat in a material.

There are, however, various other applications for this equation, therefore it is

sometimes also called, e.g., diffusion equation.



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investigation is very important. But how can we solve such a PDE? First,

we need to stress that there is no general method that works for all kinds of

PDEs. Very much to the contrary, specific methods need to be developed for

different situations, and there is a whole research area in mathematics dedicated to this. In the case of the simple linear heat equation given above, things

look better, of course: there are in fact several methods that can be applied.

In the following we sketch one particularly simple method (the separation of

variables). Other methods that could be used are the Fourier transformation5 ,

variational methods or the finite element method. The latter is the standard

way for numerical computations and works for a large class of PDEs. We refer

the reader to [Eva98] and [RR04] for in-depth introductions to PDEs.

The key idea of the separation of variables is to look not for all possible

solutions, but only for solutions of a special form, namely u(x, t) = a(x) · b(t).

Once we have found such a solution, we only need to prove uniqueness, and

we know that the solution we have found is not any solution, but the only

solution. In fact, the uniqueness proof will be omitted here, but can be found

in most mathematical textbooks on PDEs.

Using the ansatz u(x, t) = a(x) · b(t) we can rewrite the PDE as follows:

a(x)b (t) = a (x)b(t).

Sorting terms (and assuming that non of them vanishes, which is another

point that would have to be justified later) we obtain

a (x)

b (t)

=

.

b(t)

a(x)

The central observation is now the following: the left side only depends on t

and the right side only on x. Since both sides agree for all x and t, both terms

have to be constant. Let us call this constant −λ, then we get two ordinary

differential equations:

b (t) = −λb(t),

a (x) = −λa(x).

The first of these equations can be solved by an exponential function:

b(t) = b(0)e−λt ,

the second can be solved by a combination of sine and cosine functions, e.g.

5



Here we can exploit that the Fourier transform of a derivative is a simple multiplication. Thus after taking the Fourier transform of a PDE some of the derivatives

become multiplications (compare Lemma A.12). The result is typically an ODE

that can be solved much easier, either analytically or numerically. Finally the

solution needs to be Fourier transformed again to return to the original formulation.



A.6 General Axioms for Expected Utility Theory



351





a(x) = sin(x/ λ).

To determine the precise form of a and b we need to take into account the

initial and boundary conditions of the heat equation: we superimpose solutions

for a such that a(0) corresponds to the initial condition . For instance, if

the boundary condition is given by u(0, t) = u(1, t) = 0, then any ak (x) =

sin(kx/π) for k ∈ N satisfies the boundary condition, since sin(kπ) = 0 for

all k ∈ N. Denote uk (x) = ak (x) · bk (x), where bk (x) = b(0)e−kt/π . Then a

weighted sum of the uk can be constructed to fit the initial condition. This

sum still solves the heat equation and the boundary and initial condition, since

the heat equation is linear, i.e., weighted sums of solutions are also solutions.



A.6 General Axioms for Expected Utility Theory

There are many different ways to obtain a general characterization of Expected

Utility Theory for arbitrary probability measures. In the following we sketch

a rather new approach by Chatterjee and Krishna [CK]. The details of this

derivation can be found in their article.

Let Z be a compact metric space. Let P(Z) be the set of all probability

measures on Z.

The Independence Axiom can essentially be stated as in the finite case:

Axiom A.13 (Independence). Let p, q, r ∈ P(Z). Let p

then λp + (1 − λr) λq + (1 − λ)r.



q and λ ∈ (0, 1],



To state the Continuity Axiom we need to generalize the notion of continuity via the concept of open sets (compare App. A.3): we say that a function

f : X → Y is continuous if, for all open sets U ⊂ Y , the set f −1 (U ) is open.

Open sets in P(Z) can be defined via weak- convergence (compare

Sec. 2.4.5): first, define closed sets as sets which contain the limit of any

converging sequence (compare App. A.3). Second, define open sets as complement to these closed sets. We can do more and construct even a metric

d that measures the distance between two probability measures and reflects

the same convergence. Such a metric is given by the so-called “Wasserstein

metric”, compare, e.g., [AGS05].

The Continuity Axiom then becomes:

Axiom A.14 (Continuity). The sets {q ∈ P(Z) | q

q} are open.



p} and {q ∈ P(Z) | p



Theorem A.15. Let Z be a compact metric space (e.g. a bounded and closed

interval in R). Let

be a preference relation, i.e., a complete and transitive

relation on P(Z), satisfying the Continuity and Independence Axioms, then

can be represented by a von Neumann-Morgenstern Expected Utility function

u : Z → R.



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To prove Thm. A.15, Chatterjee and Krishna use intermediate steps: they

prove that the Independence Axiom together with the Continuity Axiom

implies a new axiom, the Translation Invariance Axiom. Together with the

Continuity Axiom, this new axiom implies the existence of a EUT function,

representing . Translation Invariance can be stated as follows:

Axiom A.16 (Translation Invariance). Let r be a signed measure on Z with

average r(Z) = 0, in other words, let r be the difference of two probability

measures on Z. Let p, q ∈ P(Z). Assume moreover that p + r, q + r ∈ P(Z).

Then p q implies p + r q + r.

The intuition behind the translation invariance is that adding a signed

measure r to a lottery does not change the preference relation. This means

that making certain outcomes more likely, others less likely in the same way

for p and q, does not change the original preference between p and q. This

is morally the same as the Independence Axiom and mathematically at least

close enough to show the equivalence of both axioms (under the condition of

continuity) relatively easy.

How can we now use the Independence Axiom to construct an EUT function?

First, one can prove that the indifference sets under the preference relation

are “thin”. This means: for any q ∈ P(Z) and any ε > 0 there are p, r ∈ P(Z)

which are “close” to q, i.e., d(p, q) < ε and d(q, r) < ε, and that p q r.

Second, one can show that, for any p ∈ P(Z), the contour sets {q ∈

P(Z) | q p}, {q ∈ P(Z) | q ∼ p} and {q ∈ P(Z) | q ≺ p} are all convex.

P(Z) is a convex subset of a vector space. We can pick a measure o ∈ P(Z)

and some δz ∈ P(Z) with δz

o. Let us choose moreover some q ∈ P(Z),

q = δz , q = o. The structure of the indifference sets derived above allows

us to find a continuous affine functional f : P(Z) → R such that f (o) = 0,

f (δz ) = 1 and such that the indifference set of q is a contour set of f , i.e., for

all p ∈ P(Z) with q p we have f (q) > f (p).6

Using the translation invariance, one can show that f reflects the preferences on all of P(Z). With a translation, we can also assume that f is not

only affine, but linear.

In the final step, we define u : Z → R by u(z) := f (δz ). We have to show

that this definition is correct, i.e., that

U (p) :=



u(z) dp(z) = f (p).

Z



This is easy to see for measures with finite support: let p =

by linearity of f ,

6



n

i=1



pi δzi , then



More precisely, we first restrict ourselves to a finite dimensional subset, such

that the existence of the affine functional f can be deduced from the Separating

Hyperplane Theorem (see App. A.1). Later one can show that f is independent

of the choice of this finite dimensional subset.



A.6 General Axioms for Expected Utility Theory

n



u(z) dp(z) =

Z



n



pi u(zi ) =

i=1



n



pi f (zi ) = f

i=1



353



p i δz i



= f (p).



i=1



We can approximate any measure p ∈ P(Z) by measures with finite support.

Since f is continuous, this proves that U (p) = f (p) for all p ∈ P(Z) and thus

u is an expected utility function representing the preference relation .



B

Solutions to Tests and Exercises

“Teachers open the door. You enter it by yourself.”

Chinese proverb



The tests are meant to provide an immediate feed back when studying by

yourself, hence we give solutions to all questions. Although some of the questions are tricky and require some thinking about the context of the chapter,

the student should be able of answering most questions correctly after working

through a chapter. If this is not the case, we would recommend to the reader,

to study the chapter a little bit more in detail. A good result, however, can

only ensure that the basic concepts have been understood and memorized.

The exercises then serve as a way to apply and train the ideas and methods

of the chapter. We only give solutions to some of the exercises. This may be

inconvenient for the self-learning student, but it allows to use some of the

exercises for homework assignments.



Solutions to Tests

Chapter 2

Exercise: 1 2 3 4 5 6 7 8 9 10 11 12 13 14

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Answers:

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×

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××

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×

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× ×

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× × × ×



×



×

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356



B Solutions to Tests and Exercises



Chapter 3

Exercise: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Answers: ×



×



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Chapter 4



Exercise: 1 2 3 4 5 6 7 8 9 10

Answers: × × ×



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Chapter 5



Exercise: 1 2 3 4 5 6

Answers:



×



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××××



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Solutions to Exercises

Solutions to the exercises are provided on the web page to this book. See

http:\\www.financial-economics.de

or the publisher’s web page for details.



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