Stefanica D. A Primer for the Mathematics of Financial Engineering
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204
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
The
derivative
of
f
with
respect
to
the
variable t is given
by
the
following
chain
rule formula:
d
8f 8f
dt
(f(g(t),
h(t))) = g'(t)
8x
(g(t), h(t)) + h'(t)
8y
(g(t), h(t)),
which
can
also
be
written
as
df
dt
8f
8x
8f
8y
--+--
8x
8t
8y
8t'
Example:
Let
f (x, y) =
X2
+ y + xy3,
with
x = e
2t
and
y = t
2
.
Compute
-it(f(x,
y)).
Answer:
d
dt
(f(x,
y))
(2x +
y3)
. (2e
2t
) +
(1
+ 3xy2) . (2t)
2(2e
2t
+ t
6
)
e
2t
+ 2t(1 +
3e
2t
t4). D
Let
f(x,
y)
be
a differentiable function.
Assume
that
x
and
yare
differ-
entiable
functions
of
two
other
variables
sand
t, i.e.,
x = ¢(s, t)
and
y =
'IjJ(s,
t).
Then
f
(x,
y)
can
be
regarded as a function
of
sand
t, i.e.,
f(x,y)
= f(¢(s,t),'IjJ(s,t)).
The
partial
derivatives
of
f
with
respect
to
the
variables
sand
t
are
given
by
the
following chain rule formulas:
8f
8f
8x
8s
8x
8s
8f
8¢
8x
8s
8f
8f
8x
8t
8x
8t
8f
8¢
8x
8t
Functions
of
n Variables
+
8f
8y
8y
8s
+
8f
8'IjJ
8y
8s
+
8f
8y
8y
8t
+
8f
8'IjJ
8y
8t
(7.1)
(7.2)
Let
f(Xl,
X2,
.
..
, x
n
)
be
a function of n variables. Assume
that
each variable
Xi,
i = 1 :
n,
is a differentiable
function
of
m
other
variables
denoted
by
tl,
7.2.
CHANGE
OF
VARIABLES
FOR DOUBLE
INTEGRALS
205
t2,
...
, t
m
.
The
derivative
of
the
function f
with
respect
to
the
variable
tj
is
computed
according
to
the
following chain rule formula:
8 f =
~
8 f 8Xi
\j
j = 1 : m.
8t
.
L.,;
8x·
8t·
J
i=l
I J
(7.3)
Example:
Let
f(Xl,
X2,
X3)
=
xI
+
X1X2
+
X1X3
+
2x~,
with
xl(h,
t2)
=
tt
-
t~
+
1,
X2(tl,
t2)
=
t~
+
tl
+
1,
and
X3(tl'
t2)
=
-tt
-
l.
Compute
~£.
Answer:
It
is
easy
to
see
that
:1
1
=
2Xl
+
X2
+
X3,
:1
2
=
Xl,
:13
=
Xl
+
4X3,
and
~~~
= 2tl,
~~:
=
1,
~~~
=
-2tl'
From
(7.3), we
obtain
that
8f
8
X
l
8f
8X2
8f
8X3
--+--+--
8Xl
8t
l
8X2
8t
l
8X3
8t
l
(2Xl
+
X2
+
X3)
(2tl) +
Xl
+
(Xl
+
4X3)
(-2tl)
(tt -
t~
+ tl +
2)
(2tl) + (tt -
t~
+
1)
+
(-3tt
-
t~
-
3)
(-2tl)
8t~
+ 3tt + 10tl -
t~
+
l.
Alternatively,
by
direct
computation,
Then,
f(tl,
t2)
= (tt -
t~
+
1?
+ (tt -
t~
+
1)(t~
+ tl +
1)
+ (tt -
t~
+
1)(-tt
-1)
+
2(-tt
_1)2
2tt +
t~
-
tlt~
+ 5tt -
t~
+ tl +
3.
7.2
Change
of
variables
for
double
integrals
The One Variable Case
Let
f (
x)
be
a continuous function,
and
let g (
s)
be
a continuously differen-
tiable
and
invertible function
1
mapping
an
interval
[c,
d]
into
the
interval
[a,
b],
i.e., g :
[c,
d]
-7
[a,
b]
with
sE[c,d]
----+
x=g(s)E[a,b].
1 Note
that
if
g(
s)
is continuous
and
invertiblc,
thcn
g(
s)
must
be
either
strictly increasing
or
strictly decreasing.
206
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
The
integral
of
f(x)
with
respect
to
the
variable x over
the
interval
[a,
b]
can
be
written
as
an
integral
with
respect
to
the
variable s
by
using
the
substitution
x =
g(
s)
as follows:
(b
f(x)
dx
=
(g-l(b)
f(g(s)) g'(s)
ds;
Ja
Jg-l(a)
(7.4)
see
the
integration
by
substitution
formula (1.17)2.
We
note
that
formula (7.4)
can
also
be
interpreted
as
x = g ( s ) , dx =
g'
(
s)
ds,
x = a
----+
s = g
-1
( a ) ; x = b
----+
S = g
-1
(b)
.
Functions
of
Two Variables
Let
f
(x,
y)
be
a continuous function
of
two variables
to
be
integrated
over
a
domain
D C
]R2.
Let
n c
]R2
be
another
domain
such
that
there
exists
an
one-to-one
and
onto
function (¢,
'ljJ)
with
continuous first
order
derivatives
mapping
n
into
D,
i.e., (¢,
'ljJ)
: n
-7
D
with
(s,
t) E n
----+
(x,
y)
= (¢(s, t),
'ljJ(s,
t)) E
D.
(7.5)
The
integral
of
f(x,
y)
with
respect
to
the
variables x
and
y over D
can
be
written
as
an
integral
with
respect
to
the
variables
sand
t over
the
domain
n
by
using
the
substitution
(x,
y)
= (¢(s, t),
'ljJ(s,
t)) as follows:
J k f(x, y) dxdy = J
1,
h(s, t) dsdt.
(7.6)
To find
the
appropriate
function h(s, t) from (7.6),
note
that
the
function f
can
be
written
in
terms
of
the
variables
sand
t, i.e.,
f(x,
y)
= f(¢(s, t),
'ljJ(s,
t)).
To
estimate
dxdy
in
terms
of
dsdt, recall
that
(x,
y)
= (¢(s, t),
'ljJ(s,
t)),
and
the
gradient
of
(¢,
'ljJ)
is
the
following 2 x 2
square
matrix:
(
8</>
8</»
8s
at
D(¢(s, t),
'ljJ(s,
t)) = ;
8'¢
8'¢
8s
at
(7.7)
cf. (1.41).
Then,
dxdy =
1_8¢_8'ljJ
-
_8¢_8'ljJ1
dsdt.
8s
at
at
8s
(7.8)
----------------------
2If g(8) is increasing,
then
g-l(a) = c
and
g-l(b) =
d.
If
g(8)
is decreasing,
then
g-l(a) = d
and
g-l(b) =
c.
7.2.
CHANGE
OF
VARIABLES
FOR
DOUBLE
INTEGRALS
207
The
factor multiplying dsdt
in
(7.8) is
the
absolute value
of
the
determinant
of
the
matrix
D(¢(s, t),
'ljJ(s,
t)) from (7.7),
and
is called
the
Jacobian
of
the
mapping
(¢,
'ljJ)
: n
-7
D.
Then,
the
two dimensional change
of
variables is given
by
the
formula
J
kf(x,y)
dxdy = J
1,
f(qI(s,t),,p(s,t))
I~~~~
-
:~~I
dsdt.
(7.9)
7.2.1
Change
of
Variables
to
Polar
Coordinates
A commonly used change
of
variables is
the
polar
coordinates
transformation.
In
two dimensions, given (x,
y)
E
]R2,
we
can
find r E
[0,
00)
and
e E
[0,
21f)
uniquely
determined
3
by
the
following conditions:
x =
rcose
and
y = r sin
e.
Using
the
notation
from (7.5), we define ¢(r,
e)
and
'ljJ(r,
e)
as
x = ¢(r,
e)
= r cos
e,
and
y =
'ljJ(r,
e)
= r sin
e.
(7.10)
From (7.8), we find
that
the
Jacobian
of
the
two dimensional
polar
coor-
dinates
change
of
variables is
1
8¢
8'ljJ
8¢
8'ljJ1
. .
8r
8e - 8e 8r = I cos e . r cos e -
(-r
SIn
8) .
SIn
e I
= I r(cos
2
e + sin
2
e)
I =
r,
since cos
2
e + sin
2
e = 1 for
any
8.
From (7.9),
it
follows
that
J
lJ(x,y)dxd
Y
= 1
00
12'r
J(r
cosO,
rsinO)
dOdr.
(7.11)
Similarly,
if
we
integrate
the
function f(x,
y)
over
the
disk
of
radius R cen-
tered
at
the
origin, i.e., if
(x,
y)
E D =
D(O,
R),
then
the
polar
coordi-
nates
change
of
variables is (x,
y)
=
(r
cos
e,
r
sin
e), where
(r,
e)
E n =
[0,
R]
x
[0,
21f),
and
thus
J
(
f(x,y)
dxdy =
{R
(21f
r
f(rcose,rsine)
d8dr.
(7.12)
J
D(O,R)
Jo
J
o
Example:
Let
D =
D(0,2)
be
the
disk of
center
°
and
radius 2,
and
let
f(x,
y)
= 1 x
2
-
y2.
Compute
J
JD
f·
fact
that
the
point
(0,0) in
the
(x,
y)
space is
not
uniquely
mapped
in
the
('1',
B)
space is
just
a technical
matter.
208
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
Answer:
This
integral was
already
computed
in
section 2.1, using
the
def-
inition
of
a double integral. We now use
the
polar
coordinates change
of
variables
(x,y)
=
(rcos8,rsin8),
where (r,8)
En
= [0,2] X
[0,
21f),
to
com-
pute
the
integral in a much easier fashion.
From
(7.12), we
obtain
that
J
(
f(x,
y)
dxdy =
JD(O,2)
{2
{27r
Jo Jo
r
(1
- (r cos
8)2
- (r
sin
8)2) d8dr
{2
(27r
Jo
Jo
r(l
-
r2)
d8dr
21f
12
r(l
-
r2)
dr
-41f. 0
7.3
Finding
relative
extrema
for
multivariable
functions
The One Variable Case
Let
f :
(a,
b)
---+
JR
be
a twice differentiable function such
that
f"
(x) is
continuous. A local
minimum
point
for f (x) is a
point
where f (x) achieves
its
smallest value over
an
interval
around
that
point.
Similarly, a local
maximum
point
for f (x) is a
point
where f (x) achieves
its
largest value over
an
interval
around
that
point.
In
mathematical
terms,
Xo
is a local
minimum
for f (x)
if
and
only if
:3
E > 0 such
that
f(xo) S
f(x),
\;f
x E
(xo
-
E,
Xo
+
E);
Xo
is a local
maximum
for f ( x )
if
and
only
if
:3
E > 0 such
that
f(xo)
~
f(x),
\;f
x E
(xo
-
E,
Xo
+
E).
Any
local
extremum
point
(i.e., a local
minimum
point
or
a local maxi-
mum
point)
of
f(x)
must
be
a critical
point
for
f(x),
i.e., a
point
Xo
where
f'
(
xo)
=
O.
The
question
whether
a critical
point
is a local
extremum
point
is answered
by
the
second derivative
test:
If
f"(xo) > 0,
then
the
critical
point
Xo
is local
minimum
point;
If
f"(xo) <
0,
then
the
critical
point
Xo
is local
maximum
point;
If
f"(xo) = 0,
anything
can
happen:
the
critical
point
Xo
could
be
a a local
maximum
point,
a local
minimum
point,
or
not
a local
extremum
point.
7.3.
RELATIVE
EXTREMA
OF
MULTIVARIABLE
FUNCTIONS
209
Example:
The
point
Xo
= 0 is a critical
point
for
the
following
three
functions
i1(x) = x
3
;
h(x)
=
x4;
f:J(x)
=
_x4,
and
has
the
property
that
the
second derivative
of
each
function
at
the
point
Xo
= 0 is
equal
to
0 as well:
It
is easy
to
see
that
f{(O)
f~(O)
fHo)
f{'
(0)
f~'
(0)
f~'(O)
O·
,
O·
,
o.
Xo
= 0 is
neither
a local minimum,
nor
a local
maximum
point
for
f1
(x);
Xo
= 0 is a local (actually, a global)
minimum
point
for
h(x);
Xo
= 0 is a local (actually, a global)
maximum
point
for f:J(x). 0
Functions
of
Two Variables
Let
U C
JR2
be
an
open
set, e.g.,
the
product
of two one dimensional
open
intervals,
or
the
entire
set
JR
2
.
Let
f : U
---+
JR
be
a function
with
continuous
second
order
partial
derivatives. A local
minimum
point
(or a local
maximum
point)
fot
the
function f (
x,
y)
is a
point
where
the
function achieves
its
smallest (or largest) value
in
a neighborhood of
that
point.
Definition
7.1. The point (xo,
Yo)
E U is a local
minimum
point for the
function f (x,
y)
if
and only
if
there exist E > 0 such that
f(xo,
Yo)
S
f(x,
y),
\;f
(x,
y)
E
(xo
-
E,
Xo
+
E)
X
(Yo
-
E,
Yo
+
E)
C
U.
Definition
7.2. The point (xo,
Yo)
is a
local
maximum
point for the function
f ( x,
y)
if
and only
if
there exist E > 0 such that
f(xo,
Yo)
~
f(x,
y),
\;f
(x,
y)
E
(xo
-
E,
Xo
+
E)
X
(Yo
-
E,
Yo
+
E)
C
U.
As
in
the
one dimensional case,
any
local
extremum
point
is a critical
point. To define
and
classify
the
critical
points
of
f(x,
y),
recall from (1.39)
that
the
gradient
D f
(x,
y)
of f
(x,
y)
is
the
following 1 X 2 row vector:
D f (x,
y)
= (
~~
(x,
y)
~~
(x,
y))
,
and, from (1.40),
that
the
Hessian
D2
f(x,
y)
of
f(x,
y)
is
the
2 X 2
square
matrix
given
by
D
2
f(
) _
ax2
(x,
y)
ayax(x,
y)
(
(Pj
(Pj
)
x, y -
a2
j
a2
j .
aXay(x,y)
a
y
2(X,y)
210
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
Definition
7.3.
The point (xo,
Yo)
is a critical
point
for
the
function
f (
x,
y)
if
and only
if
D
f(xo,
Yo)
= 0,
which
is
equivalent to
8f
and
8y
(xo,
Yo)
=
o.
Lemma
7.1.
Every local
extremum
point
for
the
function
f (x, y) is a critical
point.
Critical
points
are classified using
the
following two dimensional version
of
the
second derivative test:
Theorem
7.1.
Let
(xo,
Yo)
be
a critical
point
for
the
function
f (
x,
y) and
assume
that
all second order partial derivatives
of
f(x,
y) exist and are con-
tinuous.
If
the
matrix
D2
f(xo,
Yo)
is positive definite, i.e.,
if
both eigenvalues
of
the
matrix
D2f(xo,
Yo)
are strictly positive,
then
the critical point
Xo
is a local
minimum
point;
If
the
matrix
D2
f(xo,
Yo)
is negative definite, i.e.,
if
both eigenvalues
of
the
matrix
D2
f(xo,
Yo)
are strictly negative,
then
the critical point
Xo
is a local
maximum
point;
If
the two eigenvalues
of
the
matrix
D2f(xo,
Yo)
are nonzero and
of
oppo-
site signs,
then
the critical
point
(xo,
Yo)
is a saddle point, and
it
is
not
an
extremum
point;
If
one eigenvalues
of
the
matrix
D2
f(xo,
Yo)
is equal to 0 and the other eigen-
value is strictly positive,
then
the critical point
Xo
is local
minimum
point.
If
the
other
eigenvalue is strictly negative,
then
the critical point
Xo
is local
maximum
point.
If
both eigenvalues are 0, i.e.,
if
the
matrix
D2
f(xo,
Yo)
is
itself
equal to 0, anything could happen, i.e., the critical point (xo,
Yo)
could
be
a local
maximum
point, a local
minimum
point,
or
a saddle point.
While
the
eigenvalues
of
a
matrix
with
real entries are,
in
general, complex
numbers,
the
eigenvalues
of
a
symmetric
matrix
are
real numbers.
Note
that
D
2 f ( ) . t .
..
EP
f - fj2
f·
f)2 f d f)2 f
xo,
Yo
IS
a
symme
rIC
matrIX, l.e.,
f)yf)x
-
f)xf)y'
smce
f)yf)x
an
f)xf)y
are
continuous; cf.
Theorem
1.9. Therefore,
the
eigenvalues of
D2
f(xo,
Yo)
are
real
numbers
and
it
makes sense
to
talk
about
their
signs.
The
results
of
Theorem
7.1
are
often
stated
in
terms
of
det(D2f(xo,
Yo)),
the
determinant
of
the
Hessian
of
f
evaluated
at
(xo,
Yo).
Theorem
7.2.
Let
(xo,
Yo)
be
a critical
point
for
the
function
f(x,
y) and
assume
that
all second order derivatives
of
f(x,
y) exist and are continuous.
7.3.
RELATIVE
EXTREMA
OF
MULTIVARIABLE
FUNCTIONS
211
If
det(D2
f(xo,
Yo))
> 0 and
~:{
(xo,
Yo)
>
0)
then the critical point (xo,
Yo)
is
a local
minimum
point;
If
det(D2
f(xo,
Yo))
> 0 and
~:{
(xo,
Yo)
<
0)
then
the critical point
(xo,
Yo)
is
a local
maximum
point;
If
det(D2
f(xo,
Yo))
<
0,
the critical
point
(xo,
Yo)
is called a saddle point, and
it
is
not
an
extremum
point;
If
det(D2
f(xo,
Yo))
=
0,
anything could happen, i.e.,
the
critical point
(xo,
Yo)
could
be
a a local
maximum
point, a local
minimum
point,
or
a saddle point.
We
note
that
Theorem
7.2
can
be
regarded as a corollary
of
Theorem
7.1.
This
can
be
seen from
the
following result:
Lemma
7.2.
Let
A
be
a 2 x 2
symmetric
matrix
as follows:
The eigenvalues
of
A have the
same
sign
(and
are nonzero)
if
and only
if
the
determinant
of
A is strictly positive.
If
det(A) > 0
and
a >
0)
then both eigenvalues
of
A are strictly positive.
If
det(A) > 0 and a < 0, then both eigenvalues
of
A are strictly negative.
Proof.
Let
Al
and
A2
be
the
eigenvalues of
A.
We
do
not
assume
that
Al
and
A2
are necessarily different.
The
determinant
of
any
matrix
is equal
to
the
product
of
its
eigenvalues, i.e.,
Therefore, Al
and
A2
have
the
same
sign
and
are
nonzero if
and
only if
det(A)
is
strictly
positive, i.e.,
det(A)
>
O.
We recall
that,
by
definition,
det(A)
=
ad
- b
2
.
Then,
if
det(A)
> 0, we find
that
a
i=
0,
since otherwise
det(A)
=
-b
2
<
O.
Proving
that
if
det(A)
> 0
and
a > 0
then
both
eigenvalues
of
A
are
positive is
more
subtle. A
short
and
insightful
proof
requires knowledge of
the
following result:
There exists an upper triangular
matrix
U with all entries on the
main
diag-
onal strictly positive such that A =
UtU
if
and only
if
all the eigenvalues
of
the
matrix
A are strictly positive
4
•
4The decomposition UtU = A
is
called
the
Cholesky decomposition
of
the
matrix
A.
212
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
Assume
that
a > °
and
det(A) = ad - b
2
>
0.
Then,
the
matrix
u =
(~
va~~
)
is
upper
triangular
and
has
all entries
on
the
main
diagonal strictly positive.
By
matrix
multiplication, we
obtain
that
t
(Va
° )
U U =
~
Vad-b2
Va
Va
(
Va
~
)
o
Va~b'
= A,
and
conclude
that
both
eigenvalues
of
A
are
strictly
positive in this case.
To complete
the
proof, we need
to
show
that,
if a < °
and
det(A)
ad - b
2
>
0,
then
both
eigenvalues of A are
strictly
negative.
Let
Al
=
-A.
Then
det(A
I
)
= det(A)
and
the
upper
leftmost
entry
of
the
matrix
Al
is
-a
>
0.
From
the
result proved above, we find
that
both
eigenvalues of
Al
are
strictly
positive. Therefore,
the
eigenvalues of A, which
are
the
same
in
absolute value
as
the
eigenvalues of
AI,
but
have opposite
signs,
are
strictly
negative. 0
Example:
The
point (0,0) is a critical
point
for
the
following functions:
h (x) x
2
+
xy
+
y2
h(x)
_x
2
xy
_
y2
h(x)
x
2
+ 3xy +
y2
The
Hessians
of
these functions (which are
constant
matrices) evaluated
at
the
critical
point
(0,0),
and
their
determinants,
are
as follows:
])2
MO,
0)
=
(i
~);
det(])2
ft(O,
0)) =
3;
])212(0,0)
(=i
=~);
det(])2 12(0,0)) =
3;
])2/:.(0,0) =
(~
~);
det(])2h(0,O)) = -
5.
From
Theorem
7.2,
it
follows
that
the
point
(0,0) is a local
minimum
for
h(x,
y), a local maximum for
h(x,
y),
and
a saddle
point
for
h(x,
y). 0
While we do
not
give formal proofs for
Lemma
7.1
and
Theorem
7.1, we
will provide
the
intuition behind
these
results.
7.3.
RELATIVE
EXTREMA
OF
MULTIVARIABLE
FUNCTIONS
213
Recall
the
matrix
form of
the
quadratic
Taylor expansion (5.41) of a
function of two variables.
By
writing this expansion for
the
function
f(x,
y)
around
the
point
(a,
b)
=
(xo,
yo),
we find
that
(
X -
Xo
)
f(x,
y) - f(xo,
Yo)
= D f(xo,
yo)
y -
Yo
1 ( x -
Xo
) t 2 ( X -
Xo
)
+"2
y -
Yo
D f (xo,
Yo)
y -
Yo
+ 0 (Ix -
xol
3
)
+ 0 (Iy -
YOI3)
, (7.13)
where
D f(xo,
Yo)
and
D2
f(xo,
Yo)
are
the
gradient
and
Hessian of
f(x,
y)
evaluated
at
the
point
(xo,
Yo),
respectively.
Assume
that
(xo,
Yo)
E U is a local
extremum
point
for
f(x,
y), e.g., a
local
minimum
point.
Then,
by
definition,
there
exists
an
interval I
around
(xo,
Yo)
such
that
f(x,
y)
2::
f(xo,
Yo)
for all
points
(x, y) E I. From (7.13),
we find
that,
for all
(x,
y)
E
I,
(
X -
Xo
) 1 ( x -
Xo
) t 2 ( X -
Xo
)
Df(xo,yo)
y-yo
+"2
y-yo
D f(xo,yo)
y-yo
+ 0 (Ix -
xol
3
)
+ 0 (Iy -
yol3)
2::
0. (7.14)
If
we let (x, y) E I
be
close
to
the
point (xo,
Yo),
i.e.,
if
Ix
-
xol
and
Iy
-
yol
are very small,
then
the
dominant
term
in (7.14) is
the
linear
term
(
X -
Xo
)
D f(xo,
Yo)
y _
Yo
(7.15)
If D f(xo,
Yo)
l=
0,
the
inequality (7.14)
cannot
hold, since we can always
choose
x
and
y close
to
Xo
and
Yo,
respectively, such
that
the
dominant (and
nonzero)
term
(7.15)
has
negative sign.
The
entire left
hand
side of (7.14)
would therefore
be
negative, which contradicts (7.14),
and
therefore
the
point
(xo,
Yo)
cannot
be
a local
minimum
point.
We conclude
that,
if
(xo,
Yo)
is a local
minimum
point for
f(x,
y),
then
D f(xo,
Yo)
= 0,
and
therefore
(xo,
Yo)
is a critical
point
for
the
function
f
(x,
y),
as
stated
by
Lemma
7.1.
The
intuition
for
the
local maximum case
follows along
the
exact
same
lines of
thought.
We showed
that,
if
(xo,
Yo)
is a critical
point
for
f(x,
y),
then
D f(xo,
yo)
=
0. Formula (7.13) becomes
1 ( x -
Xo
) t 2 ( X -
Xo
)
f(x,
y)
= f(xo,
Yo)
+"2
y _
Yo
D f(xo,
yo)
y -
Yo
+ 0 (Ix -
xol
3
)
+ 0
(Iy
-
YOI3)
. (7.16)
214
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
We now show
that,
if
the
matrix
D2
f(xo,
Yo)
is positive definite,
then
the
critical
point
(xo, yo) is a local
minimum
point.
If
Ix
- xol
and
Iy
-
yol
are
very small,
then
the
terms
0 (Ix - xol
3
)
and
0 (Iy -
YOI3)
from (7.16)
are
dominated
by
the
term
~
(x
- xo ) t
D2f(x
)
(x
- x
o
) >
O'
2 y
Yo
0,
Yo
Y -
Yo
'
(7.17)
the
inequality from (7.17) follows from
the
fact
that
D2
f(xo,
Yo)
is a positive
definite
matrix
5
•
From (7.16),
and
using (7.17), we
can
conclude
that,
for
any
point
(x,
y)
close
to
the
point
(xo, Yo),
1
(x
-
Xo
) t 2 ( X -
Xo
)
f(x,
y)
-
f(xo,
Yo)
~
-2
D
f(xo,
Yo)
>
O.
y -
Yo
Y -
Yo
Therefore,
the
point (xo,
Yo)
is a local
minimum
point
for
the
function
f(x,
y).
The
intuition
for all
the
other
cases from
Theorem
7.1 follow from a similar
analysis
of
the
Taylor expansion (7.16).
Example:
Find
and
classify
the
local
extrema
of
the
function f :
JR2
--t
JR
given
by
f
(x,
y)
= x
2
+
2xy
+ 5y2 +
2x
+ lOy - 5.
Answer:
The
function
f(x,
y)
has
only one critical point,
(0,
-1),
which is
the
unique solution of
the
linear
system
{
~~
=
2x
+
2y
+ 2 = 0;
~[
=
2x
+ lOy + 10 =
O.
Note
that
[)21(0,
-1)
=
(~
1
2
0)
and
det(D21(0,
-1))
=
16
>
O.
Since
the
upper
leftmost
entry
of
the
Hessian D2 f
(0,
-1)
is equal
to
2 >
0,
we conclude
from
Theorem
7.2
that
the
point
(0,
-1)
is a local
minimum
point
for
the
function
f(x,
y).
0
Functions
of
n variables
The
results for
extrema
of functions
of
two variables
obtained
above
extend
naturally
to
the
case of functions of n variables.
Let
U c
JRn
be
an
open set
6
. Let f : U
--t
JR
be
a twice differentiable
function,
with
continuous second
order
partial
derivatives.
5Recall
that
A is a positive definite
matrix
if
and
only if
vtAv
>
0,
for
any
v
i=
O.
6The
proper
definition
of
an
open
set
U c Rn is as follows: For
any
x E U,
there
exists
an
interval I
around
x such
that
the
entire interval I is contained
in
U.
Examples of
open
sets
are
n-dimensional
open
intervals, i.e., U = TIi=l:n(ai,b
i
),
and
the
set
of
n-dimensional
vectors
with
positive entries, i.e., U = TIi=l:n(O, (0).
The
space Rn is also
an
open
set.
7.3.
RELATIVE
EXTREMA
OF
MULTIVARIABLE
FUNCTIONS 215
A point
Xo
E U is a local
minimum
for
the
function f (x) if
and
only if
there
exists a neighborhood
Ie
U of
Xo
such
that
f(xo)
:::;
f(x),
for all x E
I.
A point
Xo
E U is a local
maximum
for
the
function f (
x)
if
and
only if
there
exists a neighborhood I of
Xo
such
that
f (xo)
;:::
f
(x),
for all x E
I.
Recall
that
the
gradient D f (
x)
and
Hessian D2 f (
x)
of
the
function f (
x)
are given by (1.36)
and
(1.37), respectively.
Lemma
7.3.
Every
local
extremum
point
Xo
for
the
function
f(x)
must
be
a
critical point,
i.e.,
D
f(xo)
=
O.
We recall
that
if A is symmetric
and
has
real entries,
then
all
its
the
eigenvalues of A
are
real numbers. (In general,
an
n x n
matrix
with
real
entries
has
n eigenvalues, counted
with
their
multiplies,
but
these eigenvalues
may
very well
be
complex numbers.)
The
second derivative
test
for
n-dimensional
functions is given below:
Theorem
7.3.
Let
f : U
--t
JR
be
a twice differentiable
function,
with
con-
tinuous
second
order
partial derivatives,
and
let
Xo
be
a critical
point
for
the
function
f(x).
Then,
the
Hessian
matrix
D2
f(xo)
of
f evaluated
at
Xo,
is a
symmetric
matrix.
If
all eigenvalues
of
D2
f(xo)
are greater
than
or
equal
to
0,
and
at
least one
in
nonzero,
then
Xo
is a local
minimum
point.
If
all eigenvalues
of
D2
f(xo)
are
smaller
than
or
equal
to
0,
and
at
least one
in
nonzero,
then
Xo
is
a local
maximum
point.
If
the
matrix
D2
f(xo)
has
both positive
and
negative eigenvalues,
then
the
point
Xo
is
a
saddle
point,
i.e.,
neither
a local
minimum,
nor
a local
maximum.
FINANCIAL
APPLICATIONS
The
Theta
of a derivative security.
Show
that
the
density function of
the
standard
normal
variable has
unit
integral over
JR.
The
Box-Muller
method
for generating samples
of
standard
normal variables.
Reducing
the
Black-Scholes
PDE
to
the
heat
equation.
Barrier options.
Arbitrage
arguments
for
the
up-and-in
up-and-out parity.
On
the
optimality
of early exercise for American Call options.
216
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
7.4
The
Theta
of
a
derivative
security
The
Theta
of
a derivative security is
the
rate
of change of
the
value V
of
the
security
with
respect
to
time
t, i.e.,
8(V)
8V
8t
.
(7.18)
Recall from (3.72)
that
the
Theta
of
a
plain
vanilla
European
call
option
is
8ae-
q
(T-t)
dI
--;:::==::===e
-2
2V27r(T - t)
8(C)
=
(7.19)
+
q8e-
q
(T-t)
N(
d
1
) - r K e-r(T-t)
N(
d
2
),
where d
1
and
d
2
given by (3.55)
and
(3.56), respectively, i.e.,
d
1
=
In
(~)
+
(r
- q +
~2)(T
- t) .
~
and
d
2
= d
1
-
av
T - t. (7.20)
avT-t
Of
all
the
Greeks,
Theta
causes
most
confusions, mainly
generated
by
us-
ing
a simplified version
of
the
Black-Scholes formula when
computing
Theta.
The
Black-Scholes formula (3.57) is
often
written
for
time
t = 0, i.e.,
C =
8e-
qT
N(d
1
) -
Ke-
rT
N(d
2
) ,
where d
1
and
d
2
are given
by
(7.20) for
time
t = 0, i.e.,
d
_
In
(~)
+
(r
- q +
~
)T
c;:,
1 -
.Im
and
d
2
= d
1
-
avT.
avT
(7.21)
The
Theta
of
the
call
option
is defined as
the
rate
of
change
of
C
with
respect
to
the
passage
of
time
and
given as
(7.22)
this
formula is correct,
and
corresponds
to
(7.19)
with
t =
0.
We
note
that,
since
8(
C)
is defined as
8(C)
=
a;:
,
this
would
imply
that
(7.22) should
be
obtained
by
differentiating
the
formula
(7.21)
with
respect
to
t.
However,
this
would give
8(C)
= 0, since (7.21)
does
not
depend
on
the
variable
t.
7.4.
THE
THETA
OF
A
DERIVATIVE
SECURITY
217
The
explanation
is
that
(7.21) should
be
regarded as
the
formula for
C(8,
t)
evaluated
at
time
t = 0, where
C(8,
t) is given
by
the
Black-Scholes
formula (3.53), i.e.,
C(8,
t) =
8e-
q
(T-t)
N(d
1
) -
Ke-r(T-t)
N(d
2
),
(7.23)
with
d
1
and
d
2
given
by
(7.20).
Then,
8(C)
at
time
t = ° is given by
the
partial
derivative
of
C(8,
t) from (7.23)
with
respect
to
t, evaluated
at
t =
0,
i.e.,
8(C)
=
8C(8,
t) I .
8t
t=O
This
yields formula (7.19) for
8(C)
evaluated
at
t = 0, which is
the
same as
(7.22).
Another
point
that
requires
attention
is
that
8 (C) is sometimes given as
8(C)
= _ 8C
8T'
(7.24)
This
formula is
mathematically
correct, as
it
can
be
seen
by
differentiating
(7.21)
and
(7.23)
with
respect
to
T
to
obtain
(7.22)
and
(7.19), respectively.
While formula (7.24) also holds for
the
case t = 0,
it
would
be
confusing
to
interpret
it
as
Theta
being
the
derivative
of
the
price
of
the
option
with
respect
to
T,
the
maturity
of
the
option.
The
correct
interpretation
is
that
8(
C)
is
the
rate
of
change
of
the
value
of
the
call
option
with
respect
to
time
left
to
maturity
T -
t.
Mathematically,
this
corresponds
to
8C
8(
C)
= -
a(T
-
tf
(7.25)
By
differentiating formula (7.23)
with
respect
to
the
variable T - t,
it
is easy
to
see
that
(7.25) yields
the
formula (7.19) for
8(C).
A direct
and
more
insightful
proof
of
the
fact
that
(7.25) is correct
can
be
given using chain rule
as follows:
8C(8,
t)
8t
8C(8,
t)
8(T
- t)
8(T
- t) .
at
8C(8,
t)
8(T
-
tf
Here, T is a fixed
constant
and
t is considered
to
be
a variable.
As for
the
mathematical
correctness
of
formula (7.24),
note
that,
in
(7.24),
C is regarded as a
function
of
T,
and
not
as a
function
of
t, as was
the
case
in
(7.23).
Then,
considering t a fixed
constant,
we
apply
chain rule
and
find
that
8C
aT
8C
8(T-t)
---
8(T
- t)
aT
8C
(7.26)
8(T
-
tf
218
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
Using (7.26)
and
the
definition (7.25) for
8(
e), we conclude
that
formula
(7.24) is
mathematically
correct:
Be
8(e)
= - B(T - t)
We emphasize again
that
while formula (7.24) holds
true,
the
financially
insightful formula
to
use is (7.25), which shows
that
the
Theta
of
the
option
is equal
to
the
negative
rate
of change of
the
value of
the
option
with
respect
to
the
time
left until maturity.
7.5
Integrating
the
density
function
of
the
standard
normal
variable
Recall from section 3.3
that
the
probability
density function of
the
standard
normal
variable is
1 t
2
f(t) =
v'2ir
e-
T
.
We
want
to
show
that
f
(t)
is indeed a density function.
It
is clear
that
f(t)
~
0 for
any
t E:JR. According
to
(3.33), we also have
to
prove
that
1
00
1 1
00
t
2
f(t)
dt
= .
fCC
e-
T
dt
=
1.
-00
v
2K
-00
(7.27)
We use
the
substitution
t =
V2
x.
Then
dt
= V2dx,
and
t =
-00
and
t =
00
are
mapped
into
x =
-00
and
x =
00,
respectively. Therefore,
1
00
1 1
00
2
-00
f(t)
dt
=
Vi
-00
e-
x
dx.
Thus,
in
order
to
prove (7.27), we only need
to
show
that
1:
e
-x'
dx
=
ViC·
(7.28)
Rather
surprisingly,
the
polar coordinates change of variables
can
be used
to
prove (7.28). Let
1 = 1
00
e-
x2
dx.
-00
We
want
to
show
that
1
7.5.
INTEGRATING
THE
DENSITY
FUNCTION
OF
Z
219
Since
x is
just
an
integrating
variable, we
can
also
write
the
integral 1 in
terms
of
another
integrating
variable, denoted
by
y,
as
follows:
Then,
1 = 1
00
e-
y2
dy.
-00
(1:
e-
X
'
dx
) .
U:
e-
Y
'
d
Y
)
1:1:
e-x'e-
Y
'
d.'lJdy
J
1,
e -(x'+y') dxdy.
(Note
that,
while
the
second equality
may
be
intuitively clear, a result similar
to
Theorem
2.1 is required for a rigorous derivation.)
We use
the
polar
coordinates
transformation
(7.10)
to
evaluate
the
last
integral. We change
the
variables (x,y) E
:JR2
to
(r,B) E
[0,
(0) X
[0,
2K)
given
by
x = rcosB
and
y =
rsinB.
From
(7.11), we find
that
K.
Thus, we proved
that
12
= K
and
therefore
that
1 =
V1f,
i.e.,
1
00
e-x2
dx = 1 = Vi.
-00
220
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
As shown above,
this
is equivalent
to
showing
that
which completes
the
proof of
the
fact
that
f
(t)
function.
7.6
The
Box-Muller
method
The
Box-Muller
method
is
an
algorithm
for
generating
samples of
the
stan-
dard
normal
variable, which are needed for
Monte
Carlo simulations.
In
the
Box-Muller
method,
two
independent
standard
normal
samples
are
gener-
ated,
starting
with
two
independent
samples from
the
uniform distribution.
The
method
is based
on
the
following fact:
If
Zl
and
Z2
are
independent
standard
normal
variables,
then
R =
Zi
+
Z~
is
an
exponential
random
variable
with
mean
2,
and,
given
R,
the
point
(Zl'
Z2)
is uniformly
distributed
on
the
circle
of
center
°
and
radius
-JR.
Explaining
the
Box-Muller
method
and
the
Marsaglia
polar
method
for
an
efficient
implementation
of
the
Box-Muller
algorithm
is beyond
the
scope
of
this
book. We resume
the
discussion
to
applying
the
polar
coordinates
change
of
variable
to
explain
the
following
part
of
the
Box-Muller
method:
Lemma
7.4. Let
Zl
and
Z2
be
independent standard normal variables) and
let R
= zi +
Z~.
Then R is
an
exponential random variable with mean 2.
Proof. Recall
that
the
cumulative density
function
of
an
exponential
random
variable X
with
parameter
a > ° is
F(x)
=
{1
-
e-
ax
, if x ?
0;
0, otherWise,
and
its
expected
value is E[X] =
i.
Let
R = zi +
Z~.
To prove
that
R is
an
exponential
random
variable
with
mean
E[R] = 2, i.e.,
with
parameter
a =
~,
we will show
that
the
cumulative
density
of
R is
F(x)
=
P(R
< x) =
{1
-
e-'&,
if
x ?
0;
- 0, otherwIse.
(7.29)
Let
x <
0.
Since R =
Zi
+
Z~
~
0, we find
that
P(R::::;
x) =
0.
7.7.
THE
BLACK-SCHOLES
PDE
AND
THE
HEAT
EQUATION
221
Let x
~
0.
Let
h(XI)
=
vk
exp
(_~i)
and
!2(X2) =
vk
exp
(-¥)
be
the
density functions
of
Zl
and
Z2,
respectively, where exp(t) = e
t
.
Since
Zl
and
Z2
are
independent,
the
joint
density
function
of
Zl
and
Z2
is
the
product
function h(XI)!2(X2); cf.
Lemma
4.5. We
note
that
zi +
Z~
::::;
x
if
and
only if
the
point
(Zl'
Z2)
is
in
the
disk
D(O,
Fx)
of
center °
and
radius
Fx,
i.e.,
if
(Zl, Z2) E
D(O,
Fx).
Then,
P(
R
::::;
x)
P
(Zr
+
Z~
::::;
x)
J
(
h(XI)!2(X2)
dXI
dX2
JD(O,y'x)
1 J { exp (
xr
+
X~)
dXldx2.
27r
J D(O,y'x) 2
The
polar
coordinates
change
of
variables
Xl
= r cos e
and
X2
= r sin e
maps
any
point
(Xl,
X2)
from
the
disk
D(O,
Fx)
into
a
point
(r,
e)
from
the
domain
[0,
Vx] x
[0,
27r).
From
(7.12), we find
that
P(R::::;
x) =
-2
1
J { exp (
xr;
X~)
dXldx2
7r
JD(O,y'x)
2~
lv'·
l2~
r exp (
(r
cos
0)2
;
(r
sin
0)2)
dOdr
2~
lv'X
[~
rcxp
( -
~)
dOdr
~
27r
(y'x
r
e-r;
dr
27r
Jo
(-e-r;)
,:
We proved
that
the
cumulative density
of
the
random
variable R is equal
to
1 -
e-'&,
if x
~
0,
and
to
° otherwise.
This
is equivalent
to
showing
that
R =
Zi
+
Z~
is
an
exponential
random
variable
with
mean
2;
cf. (7.29). 0
7.7
Reducing
the
Black-Scholes
PDE
to
the
heat
equation
Let
V(S, t)
be
the
value
at
time
t
of
a
European
call
or
put
option
with
maturity
T,
on
a lognormally
distributed
underlying asset
with
spot
price
222
CHAPTER
7.
MULTIVARIABLE
CALCULUS.
S
and
volatility
(),
paying dividends continuously
at
rate
q.
Recall from
section 6.4
that
V(S, t) satisfies
the
Black-Scholes
PDE
av
+ !(}2S2a
2
V +
(r_q)Sav
_
rV
= 0,
\j
S > 0,
\j
0 < t < T. (7.30)
at 2
aS
2
as
The
risk-free interest
rate
r is assumed
to
be
constant
over
the
lifetime of
the
option.
The
boundary
condition
at
maturity
is V(S, T) =
max(S
- K, 0), for
the
call
option
and
V(S,
T)
= max(K -
S,
0), for
the
put
option.
One
way
to
solve
the
Black-Scholes
PDE
(7.30) is
to
use a change of
variables
to
reduce
it
to
a
boundary
value
problem
for
the
heat
equation
since a closed form solution
to
this
problem is known.
The
change of variables is as follows:
where
V(S, t) =
exp(-ax
-
bT)U(X,
T),
x
T =
In
(!) ,
(T -
t)(}2
2
and
the
constants
a
and
b are given
by
r-q
1
a
(}2
2'
b
(
~+!)2
(}2 2
(7.31 )
(7.32)
(7.33)
(7.34)
(7.35)
(7.36)
A brief explanation of how
the
change of variables (7.32) is chosen is
in
order.
In
general, changing from
the
variables space
(S,
t)
to
the
space
(x,
T)
means
choosing functions
¢(S,
t)
and
'ljJ(S,
t) such
that
x =
¢(S,
t)
and
T =
'ljJ(S,
t). A simpler version of such change
of
variables is
to
have x
and
T
depend
on
only one variable each, i.e., x = ¢(S)
and
T =
'ljJ(t).
Note
that
the
Black-Scholes
PDE
(7.30)
has
nonconstant coefficients,
which makes
it
more challenging
to
solve. However,
the
Black-Scholes
PDE
is homogeneous, in
the
sense
that
all
the
terms
with
nonconstant coefficients
are
of
the
form
and
7.7.
THE
BLACK-SCHOLES
PDE
AND
THE
HEAT
EQUATION
223
The
classical change of variable for
ODEs
with
homogeneous nonconstant
coefficients involves
the
logarithmic function.
In
our
case,
we
choose
the
change of variables (7.33), i.e.,
From a
mathematical
standpoint,
the
change of variables (7.33) is equivalent
to
x = In(S).
The
term
f is introduced for financial reasons:
taking
the
logarithm of a dollar
amount,
as would
be
the
case for In(S), does
not
make
sense. However,
f is a
non-denomination
quantity, i.e., a number,
and
its
logarithm, In
(f),
is well defined.
The
Black-Scholes
PDE
(7.30) is backward
in
time, i.e.,
the
boundary
data
is given
at
time
T
and
the
solution is required
at
time
O.
We choose
the
change
of
variables for T
to
obtain
a forward
PDE
in
T,
i.e.,
with
boundary
data
given
at
T = 0
and
solution required a fixed
time
Tfinal
>
o.
The
change
of
variables (7.34), i.e.,
(T -
t)(}2
T =
2 '
accomplishes this,
and
will also cancel
out
the
term
~2
from
the
coefficient of
the
second order
partial
derivative.
We now prove
that,
if V(S, t) satisfies
the
Black-Scholes
PDE
(7.30),
then
the
function u(x,
T)
given by (7.32) satisfies
the
heat
equation
(7.31).
We do
this
in
two stages.
First,
we make
the
change of variables
V(S, t)
w(x,
T),
where
In(!)
and
(T
-
t)(}2
x =
T
2
It
is easy
to
see
that
ax
1
ax
o·
(7.37)
as
S'
at
,
aT
aT
(}2
(7.38)
O·
-
as
,
at
2
Using chain rule repeatedly, see (7.1)
and
(7.2), as well as (7.37)
and
(7.38)
we find
that
aw aw
ax
aw
aT
as
ax
as
+
aT
as
a
(av)
a
(1
aw)
as
as
=
as
S ax
1
aw
s
ax