A.5 Calculus, Fourier Transformations and Partial Differential Equations
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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) := √
2π
+∞
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) = √
2π
+∞
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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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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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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