Advanced Volatility and associated
topics
Introduction
In this section, we discuss details of volatility. Volatility is one of the most
interesting and important aspects of Quantitative Finance. Volatility is the
key parameter determining the price of an option, yet it is also the hardest to
measure. There are many types of volatility; the precise nature and di¤erence
is very important - it is crucial that we know which volatility we are talking
about. This adds to the di¢ culty of volatility considerations.
In the Black-Scholes model, the SDE for the stock has two parameters, and
but later the drift disappears even though the stock depends on it. Some
…nd this counter-intuitive that the value of a call option does not depend on
whether the underlying stock is more likely to go up than it is to go down.
Recall this is a consequence of hedging. Hence the importance of modelling the
volatility ’correctly’if in the business of derivative pricing. If not things become
increasingly complex! Suppose we are concerned with stock selection,then the
drift comes back in.
The Black-Scholes model is very elegant but it does not perform well in practice.
A basic assumption of the framework is a constant geometric Brownian motion
for the underlying
dS
= dt + dWt
S
and leads to a partial di¤erential equation for which either an analytical solution
exists or can be treated numerically.
Thus far the role of is that of a parameter. It is the most important parameter
when pricing an option, and is also the most di¢ cult to measure. In comparing
the solution to reality, the important question arising is "How plausible is a
constant volatility?". Recapping the model for a Call option C (S; t) ; the
pricing equation and terminal condition in turn
@C 1 2 2 @ 2C
@C
+
S
+
rS
rC = 0
2
@t
2
@S
@S
C (S; T ) = max (S
E; 0) :
The solution is
C (S; t) = SN (d1)
Ee r(T t)N (d2)
where
p
log(S=E ) + (r + 12 2)(T t)
p
d1 =
and d2 = d1
T t:
T t
This pricing formula relates the option price C to six arguments; the variables
S; t and the parameters r; ; E and T .
Quantity Observable?
CM
yes
quoted market option price
S
yes
today’s spot price
t
yes
today’s date
T
yes
expiration
E
yes
strike price
r
yes
today’s interest rate
No
So in derivatives the volatility is the most important parameter.
Drift is not important.
It doesn’t matter if e.g. it is doubling in price or halving in price. It is the level
of noise/randomness that a¤ects the price. But it is very hard to measure volatility cannot be seen/observed. It is how much randomness there is in a
stock price in an instant in time. For these reasons di¤erent types of volatility
needs to be discussed.
Spot Volatility
!
St+ t
: Recall these are popular
St
for many reasons, both theoretic and algorithmic.
De…ne logarithmic returns Rt := log
The mean and variance in turn are
h
mt = E [Rt] ; vt = E (Rt
mt)
The classical Black-Scholes stock price process St; t
is
dSt = Stdt + StdWt;
2
i
0 given by GBM
Using Itô to calculate d(log St) gives the solution as
1 2
t + Wt
2
log St = log S0 +
St
Now look at the mean and variance of the logarithmic return log
S0
"
St
E log
S0
"
V log
!#
=: t :=
St
S0
!#
!
1 2
t
2
=: s2 (t) := 2t
This gives one de…nition of volatility : It is a measure of the variance
of the logarithmic returns, such that the square of the volatility gives the
rate of increase of the log-returns:
d 2
s (t) = 2:
dt
Equivalently, if we denote the quadratic variation of a stochastic process
Yt by [Y ]t ; then we have
[log S ]t = 2t;
so the squared volatility is the rate of change of the quadratic variation of
the log-stock price.
Deterministic Volatility Models
Simplest generalisation of the Black-Scholes constant volatility paradigm is to
allow the volatility to be a deterministic function of time so that the stock
price SDE becomes
dSt = Stdt +
(t) StdWt;
and the modi…ed Black-Scholes equation becomes
@V
1 2
@ 2V
@V
2
+
rV = 0:
+ rS
(t) S
@t
2
@S 2
@S
for the option price V (t; S ) ; where V (T; S ) = h (S ) :
By the Feynman-Kac theorem the option pricing function is given by the riskneutral expectation
h
i
Q
r(T
t)
V (t; S ) = E e
h (ST )j S t = x ;
where under the Q measure, St follows GBM above with
replaced by r :
dSt = Stdt + (t) StdWtQ;
where WtQ is a Q Brownian motion. Under Q; the terminal log-stock price is
now given by
log ST = log St + r (T
t)
Z T
1
2 t
2 (s) ds +
Z T
t
(s) dWsQ
Hence, under Q, given St = x; log ST is normally distributed:
log ST
1 2
2
N log x + r
(T
t) ; 2 (T
t) ;
where 2 is the root mean square (RMS) volatility, given by
2 (T
t) =
Z T
2 (s) ds
Z T
2 (s) ds
t
It follows that one simply prices the option using the Black-Scholes formula
with the volatility replaced by
2
t
=
1
(T
t) t
Thus, in all BS pricing formulas for European, path-independent options, just
replace by t:
For example, the price of a vanilla call at time t is given by
C
2 ; S; t
t
= SN (d1)
Ee r(T t)N (d2)
where
log(S=E ) + (r + 21 2t )(T
p
d1 =
T t
p t
t;
d2 = d1
t T
Z T
1
2 =
2 (s) ds:
t
(T t) t
t)
;
Local Volatility
Further generalisation of Black-Scholes:
dSt = Stdt +
(t; St) StdWt:
The deterministic function
(t; S ) !
(t; St)
is called local volatility.
Option price V (t; St) for terminal payo¤ h (ST ) satis…es the BSE
@V
1 2
@ 2V
@V
2
+
+ rS
rV = 0
(t; S ) S
2
@t
2
@S
@S
V (T; ST ) = h (ST ) :
The model is also referred to as ‘restricted stochastic’ volatility model,
since the volatility path t = (t; St) is stochastic, but the only source
of randomness enters through the state variable S .
Special cases are
i.
= (t) is a function of time alone: There is a term-structure, but
the model fails to predict smiles and skews. As discussed above.
ii.
= (S ) is a function of the stock alone: An important family are
constant elasticity of variance (CEV) models of the form
dSt = Stdt + St dWt
In the previous table we note that CM ; S; t; T; E and r are observables. For
a plain vanilla option we de…ne the implied volatility denoted i to be that
value of the unobservable ; which gives the market price of the option when
substituted into the Black-Scholes option pricing formula. All the other observables are …xed in this process. It is described as the market’s view of the
future actual volatility during the life of the option.
The implied volatility according to the Black-Scholes model should be independent of both strike and expiration; in reality it depends on both.
Consider the following example: A trader can see on their screen that a certain
call option with one year until expiry and a strike of £ 100 is trading at £ 10.45
with the underlying at £ 100 and a short-term interest rate of 5%. Can we use
this information in some way?
We can take invert this relationship between volatility and an option price by
asking “What volatility must I use to get the correct market price?”
This is called the implied volatility. If CBS denotes the Black-Scholes theoretical price then solving
CM (S; t) = CBS (S; T; r; E; i (E; T ))
for i becomes a root-…nding problem. Use of e.g. Newton-Raphson method
will work.
For implied volatility to be a useful concept means there should be a unique
implied vol. This is only true if the options vega ; where
@V
(S; t) =
(S; t; ; E; T ; r; ::::) ;
@
does not change sign for any value of S; t or other parameters, besides ;
involved. While this is true for European calls and puts it is true for all options.
Smiles
According to the classical Black-Scholes analysis
dS
= dt + dW
S
so that is a property of S alone. For a vanilla call or put option the strike E
and the expiry T are properties only of the option. Thus the volatility (which
in the Black-Scholes model should be the same thing as the implied volatility)
should be independent of both strike E and the expiry T or a vanilla put or
call.
In practice, we …nd the implied volatility for a vanilla call (or put) depends on
both the strike and the expiry, the so-called smile. This implies that there is
something wrong with the Black-Scholes model.
The dependence of implied volatility on expiry could imply a term structure
for volatility, b (t) ; rather than a constant volatility : This is not a serious
problem; we know how to deal with time dependent volatilities; we just replace
2 in the Black-Scholes formulae by
1
Z T
b 2 (s) ds:
T t t
The dependence on strike is a serious problem. It is quite inconsistent with the
Black-Scholes analysis.
One way of explaining the smile e¤ect is to assume that the volatility is a
function of both spot price and time;
=
(S; t) :
This is not the only possible explanation.
Volatility Surfaces
One means of implementing a no-arbitrage model is to assume that
(St; t) : The stock price process dynamics follow
dSt
= tdt + (St; t) dWt
St
The Black-Scholes equation then becomes
@V
1 2
@ 2V
@V
2
+
(S; t) S
+ rS
2
@t
2
@S
@S
rV = 0;
and its solution can be written in the form
V (S; t) = e r(T t)
Z 1
0
p S; t; S 0; T V S 0; T dS 0:
=
V S 0; T is the payo¤ and p S; t; S 0; T is the risk-neutral probability density
associated with the Kolmogorov equations
2
@p
1 @2
@
0p ;
2
0
0
=
rS
S
;
t
S
p
@T
2 @S 02
@S 0
2 @ 2p
1 2
@p
@p
=
+ rS :
(S; t) S
2
@t
2
@S
@S
That is p S; t; S 0; T can be viewed in two ways (given that the above analysis
only makes sense if t < T ):
If S and t are …xed (today’s spot price and date) then we can regard p S; t; S 0; T
as the probability density that at time T > t the spot price will be S 0: This
is a conditional probability density for future values, S 0 and T; given that the
present values are S and t < T:
If S 0 and T are …xed (some given value of the spot price and date, say) then
p S; t; S 0; T is the probability density that at time t < T the spot price was
S ; again this is a conditional probability function; the probability that the spot
price was S at time t given that the spot price is S 0 at time T:
Dupire’s method
Suppose now that we want to …nd (S; t) from market data. In fact we shall
…nd (E; T ) : More correctly, we …nd (S; t; E; T ) with the usual notation
of the function arguments.
Using
@p
1 @2
=
@T
2 @S 02
2 S 0 ; t S 02 p
@
0p
rS
@S 0
and that
C (S; t) = e r(T t)
Then
@C
=
@T
Z 1
E
p S; t; S 0; T
rC + e r(T t)
Z 1
@p
E @T
S0
S0
E dS 0:
E dS 0:
Also (from Leibniz)
Z
1
@C
r(T
t)
=
e
p S; t; S 0; T dS 0
@E
E
2
@ C
r(T t) p (S; t; E; T )
=
e
@E 2
Now use the fact that
@p
@2
= 12 @S
02
@T
where as earlier,
2 S 0 ; t S 02 p
@ rS 0 p
@S 0
p S; t; S 0; T dS 0
denotes the risk-neutral probability of a spot price in S 0; S 0 + dS 0 at time
T > t; contingent on the spot price being S at t:
Recall that p is contingent on today’s information, (S; t) ; in general
p S1; t1; S 0; T = p S2; t2; S 0; T
for say today, (S1; t1) ; and tomorrow (S2; t2) :
@p
Substituting for @T
and using integration by parts we arrive, after lengthy
calculation, at
@C
= rC + e r(T t)
@T
Later we will use
Z 1
E
S 0pdS 0 =
(E; T )2 E 2p (S; t; E; T ) + r
Z 1
E
S0
E pdS 0 + E
and from earlier we note that
Z
Z 1
E
Z 1
E
S 0pdS 0 :
pdS 0
1
@C
@ 2C
r(T
t)
0
r(T t) p (S; t; E; T ) ;
=e
pdS ;
=
e
@E
@E 2
E
where the second expression gives
2C
@
:
p (S; t; E; T ) = er(T t)
@E 2
@p
Now use the Kolmogorov equation to express @T
in @C
@T
@C
=
@T
=
Z 1
@p
0
0
rC + e r(T t)
S
E
dS
E @T
rC + Z
1
r(T
t)
2 S 0 ; t S 02 p
1 @2
e
2 @S 02
E
@ rS 0 p
@S 0
S0
The integral
Z 1
E
1 @2
2 @S 02
Z 1
2 S 0 ; t S 02 p
@ rS 0 p
@S 0
2 S 0 ; t S 02 p
1 @2
02
E 2 @S
Z 1
@ rS 0 p S 0
0
E @S
S0
S0
E dS 0
E dS 0
E dS 0 =
E dS 0
In what follows, we assume p decays su¢ ciently fast.
Z 1
1
@2
2 E @S 02
2 S 0 ; T S 02 p
v=
S0
v0 = 1
S0
E
E dS 0 :
2
2
@
= @S 02 2 S 0; T S 0 p
@
2 S 0 ; T S 02 p
u = @S
0
u0
Z 1
=
2 S 0 ; T S 02 p
0
1
@2
S
02
2 E @S
2 S 0 ; T S 02 p
0
1 @2
S
E
2 @S 02
{z
|
=0
=
1 2 (E; T ) E 2 p (S; t; E; T )
2
=
E dS 0
1
E}
Z 1
@
1
2 E @S 0
2 S 0; T
1 2 (E; T ) E 2 er(T t) @ 2 C
2
@E 2
2
0
S p dS 0
Similarly
Z 1
@ rS 0 p
0
E @S
S0
E dS 0 :
v = S0
v0 = 1
E
Z 1
@
=
Now using
Z 1
E
0
E @S
rS 0p S 0
|
S 0pdS 0
rS 0p
{z
=0
=
Z 1
E
@ rS 0 p
u0 = @S
0
u = rS 0p
E
S0
S0
1
E}
E dS 0
r
Z 1
E
E pdS 0
S 0pdS 0
+E
Z 1
E
pdS 0
For the …rst integral term it is the expected payo¤ (i.e. option price
without discount factor), i.e. er(T t)C .
@C =
The second integral term from earlier @E
@C
er(T t) @E
e r(T t)
Z 1
E
pdS 0 gives
Hence
Z 1
@
E @S
0
rS 0p
S0
@C
rEer(T t)
@E
@C
rer(T t) C E
@E
E dS 0 = rer(T t)C
=
Putting everything together
@C
=
@T
=
@ 2 C + rer(T t) C
rC + e r(T t) 12 2 (E; T ) E 2er(T t) @E
2
@C
@ 2 C + rC
rC + 21 2 (E; T ) E 2 @E
rE
2
@E
Hence
@C
1 2
@ 2C
2
=
(E; T ) E
@T
2
@E 2
@C
rE
:
@E
We can now, in principle solve for
2 (E; T )
=
@C + rE @C
@T
@E :
1 E 2 @ 2C
2
@E 2
E
@C
@E
If we now retrace our calculations we …nd that, because the call value C is a
function of today’s spot; S; today’s date; t; the call’s strike E and the call’s
maturity; T; C = C (S; t; E; T ) ; what we have called (E; T ) is actually
2 (S; t; E; T )
=
@C(S;t;E;T )
@C(S;t;E;T )
+
rE
@T
@E
:
2
1 E 2 @ C(S;t;E;T )
2
@E 2
Recall that, in practice, when we compute (S; t; E; T ) ; today’s spot price
S and date t are …xed. We can vary only the strike E and the maturity T:
That is, we have found a local volatility surface (E; T ) ; or more correctly
(S; t; E; T ) ; as it is conditional on today’s spot price S and date t.
Practical problems with this approach
requires continuum of strikes and maturities (interpolation, extrapolation)
numerical di¤erentiation is ill conditioned
the denominator
@ 2C
@E 2
tends to zero for E ! 1
The last problem can be circumvented to some extent by switching from quoted
prices to implied volatilities.
Implied and local volatility
If we use implied volatilities
theorem gives
2 (E; T )
i,
repeated application of the implicit function
=
p
1 + Ed1 T
2
i
+ 2 i (T
2
t @@Ei
t) @@Ti + 2r iE (T
+ i (T
t) E 2
@2 i
@E 2
t) @@Ei
@ i 2p
d1 @E
T
where, as usual,
log(S=E ) + (r + 12 2i )(T
p
d1 =
t
i T
t)
t
Finding Roots
A fundamental problem in numerical analysis consists of obtaining the zero of
a function. Given a function y = f (x) obtain the root of f (x) = 0; i.e. …nd
the value of x = c which satis…es f (c) = 0. e.g.
f ( x) = x
sin x:
Four broad categories of root …nding:
(i) Methods which do not use derivatives of the function
(ii) Methods which do use f 0 (x)
(iii) Methods for polynomials
(iv) Methods which deal with complex roots
Bisection
The simplest method is that of bisection. The following theorem, from calculus
class, insures the success of the method.
Intermediate Value Theorem
Suppose f (x) is continuous on [a; b] then for any y s.t y is between f (a)
and f (b) there 9 c 2 [a; b] s.t f (c) = y:
Example 1
1
The function f (x) = is not continuous at 0: Thus if 0 2 [a; b], we cannot
x
apply the IVT. In particular, if 0 2 [a; b] it happens to be the case that for
every y between f (a) ; f (b) there is no c 2 [a; b] such that f (c) = y:
In particular, the IVT tells us that if f (x) is continuous and we know a; b
such that f (a) ; f (b) have di¤erent sign, then there is some root in [a; b] :
This is a fundamental test we can apply.
Example 2 Show that the function g (x) = x3 + 2x2 + 5x
lying between 0 and 1:
We note f (0) =
s.t. 2 (0; 1) :
1 has a root
1; f (1) = 7: The sign change con…rms that 9 a root
Once location of a root
is established then a reasonable estimate of
is
a+b
c=
. We can check whether f (c) = 0. If this does not hold then one
2
and only one of the two following options holds:
1. f (a) ; f (c) have di¤erent signs.
2. f (c) ; f (b) have di¤erent signs.
We now choose to recursively apply bisection to either [a; c] or [c; b], respectively, depending on which of these two options hold.
Whichever an interval is chosen,the new interval containing the root can be
further subdivided. If the …rst interval is jb aj ; then the second is half the
length and so on.
After n steps of bisection the interval containing the root will be reduced in
size to
jb aj
2n
where in the earlier example the value b = 1 and a = 0:
If the size of the interval becomes smaller than some speci…ed tolerance, t,
then the calculation stops and convergence has been attained.
Theorem 2 (Bisection Method Theorem)
If f (x) is a continuous function on [a; b] such that f (a)f (b) < 0, then after
n steps, the method will return c such that
jc
where
j
is some approximate root of f .
jb
aj
2n
Example Consider f (x) = x
e1=x. There is a root in [1; 2] :
Use the bisection method to show that the root of f (x) = x
interval [1; 2] is 1:763 (correct to 3 decimal places).
f (x0) = f (1) = 1 e1 < 0
f (x1) = f (2) > 0
f (1:5) = 1:5 e2=3 = f (x2) =
f ( x0 ) f ( x 2 ) > 0
0:4477
e1=x in the
x0 + x1
! x2 =
= 1 :5
2
) root in [x2; x1] i.e. in [1:5; 2]
x2 + x1
1 :5 + 2
=
= 1:75
2
2
f (x3) = 1:75 e1=1:75 = 0:0208
f (x3) f (x2) > 0 ) root in [x3; x1] i.e. in [1:75; 2]
x3 =
1:75 + 2
3:75
=
= 1:875
2
2
f (x4) = f (1:875) = 0:1704
f (x3) f (x4) < 0 ) root in [x3; x4] i.e. in [1:75; 1:875]
x4 =
1:75 + 1:875
x3 + x4
=
= 1:8125
x5 =
2
2
f (x5) = f (1:8125) = 0:0763
f ( x3 ) f ( x5 ) < 0
Continuing in this way we have we …nd the root in [x9; x10] i.e. in [1:761718; 1:7636715]
x9 + x10
= 1:76269
2
= 1:763 to 3 decimal places
x11 =
Newton’s Method
Newton’s method is an iterative method for root …nding. That is, starting
from some guess at the root, x0, one iteration of the algorithm produces a
number x1, which is supposed to be closer to a root; guesses x2; x3; :::; xn
follow identically.
We know from Taylor that
f ( x + h ) = f ( x) + f 0 ( x) h + O h 2 :
This approximation is better when f 00(:) is "well-behaved" between x and
x + h. Newton’s method attempts to …nd some h such that
0 = f (x + h) = f (x) + f 0 (x) h:
This is easily solved as
f ( x)
h= 0
f ( x)
An iteration of Newton’s method, then, takes some guess xn and returns xn+1
de…ned by
xn+1 = xn
f ( xn )
:
0
f ( xn )
f (xn )
A
B θ
xn+1
C
xn
AC
f ( xn )
From above we see that tan =
=
BC
(xn xn+1)
But
= f 0 ( xn )
f ( xn )
f 0 ( xn ) =
(xn xn+1)
tan
xn+1 = xn
f ( xn )
f 0 ( xn )
This is the Newton-Raphson Technique.
Example: Solve for roots, the function f (x) = x
cos x:
Start by considering x = cos x: That is draw y = x and y = cos x to obtain
an initial guess for the root(s).
5
4
3
2
1
0
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
-2
-3
-4
-5
Clearly the diagram above shows that there is only one root
2 (0; 1) :
We use the Newton formula
xn+1 = xn
f ( xn )
; n = 0; 1; ::::
f 0 ( xn )
where n = 0 is the initial guess. f (xn) = xn
1 + sin xn:
x
f ( x)
0
1
1 0:75
cos xn
! f 0 ( xn ) =
so numerically we also see that f (0) f (1) < 0 =) 2 (0; 1) : NR formula
for this function becomes
xn cos xn
xn+1 = xn
; x0 = 1
1 + sin xn
x0 cos x0
= 0:75036
1 + sin x0
0:75036 cos 0:75036
x2 = 0:75036
= 0:73911
1 + sin 0:75036
0:73911 cos 0:73911
= 0:73909
x3 = 0:73911
1 + sin 0:73911
00:73909 cos 0:73909
x4 = 0:73909
= 0:73909
1 + sin 0:73909
which gives the root
0:73909:
x1 = x0
Problems
As mentioned above, convergence is dependent on f (x), and the initial estimate x0. A number of conceivable problems might come up. We illustrate
them here.
ln x
,
x
+
with initial estimate x0 = 3: Note that f (x) is continuous on R : It has a
single root at x = 1. Our initial guess is not too far from this root. However,
consider the derivative:
1 ln x
0
f ( x) =
x2
Example Consider Newton’s method applied to the function f (x) =
If x > e1, then 1 ln x < 0, and so f 0 (x) < 0. However, for x > 1; we
know f (x) > 0. Thus taking
f ( xn )
> xn
0
f ( xn )
The estimates will diverge from the root x = 1:
xn+1 = xn
Fourier Transforms
If f = f (x) then consider
Z 1
1
fb ( ) = p
f (x) eix dx:
1
2
If this integral converges, it is called the Fourier Transform of f (x) : Similar
to the case of Laplace Transforms, it is denoted as F (f ) ; i.e.
F (f ) =
Z 1
1
f (x) eix dx = fb ( ) :
The Inverse Fourier Transform is then
F 1 fb ( ) =
Z 1
1
fb ( ) e ix d = f (x) :
The convergent property means that fb ( ) is bounded and we have
Z 1
1
jf (x)j dx < 1:
Functions of this type f (x) 2 L1 ( 1; 1) and are called square integrable.
We know from integration (basic property of Riemann integral) that
Z b
a
Z b
f (x) dx
a
jf (x)j dx:
Hence
fb ( )
=
Z
ZR
R
f (x) eix dx
f (x) eix dx
and Euler’s identity ei = cos +i sin implies that ei
1; therefore
fb ( )
Z
jf (x)j dx < 1:
=
q
cos2 + sin2
R
In addition to the boundedness of fb ( ) ; it is also continuous (requires a
proof).
=
Note: If f (x) represents the probability density of some random variable X
then the Fourier transform is the characteristic function of f (x) ; i.e.
fb ( ) =
E
h
i
i
x
e
:
Example: Obtain the Fourier transform of f (x) = e jxj
fb ( ) = F (f ) =
=
=
=
Z 1
1
Z 0
1
Z 0
1
Z 0
1
Z 1
f (x) eix dx
1
e jxjeix dx
e jxjeix dx +
exeix dx +
Z 1
0
Z 1
0
e jxjeix dx
e xeix dx =
exp [(1 + i ) x] dx +
0
Z 1
0
exp [
(1
i ) x] dx
1
1
exp [(1 + i ) x]
exp [
=
(1 + i )
(1 i )
1
1
2
1
+
=
=
(1 + i ) (1 i )
1+ 2
(1
i ) x]
1
0
Our interest in di¤erential equations continues, hence the reason for introducing
this transform. We now look at obtaining Fourier transforms of derivative
terms. We assume that f (x) is continuous and f (x) ! 0 as x !
Consider
o
n
F f 0 ( x) =
Z
R
1:
f 0 (x) eix dx
which is simpli…ed using integration by parts
1
f (x) eix
1
so
F
n
f 0 ( x)
o
=
i
Z
R
i
Z
R
f (x) eix dx
f (x) eix dx =
i fb ( ) :
We can obtain the Fourier transform for the second derivative by performing
integration by parts (twice) to give
F
n
f 00 (x)
o
= ( i )2 F ff (x)g =
F f 0 ( x) =
F f 00 (x) =
i fb ( )
2 fb (
)
2 fb (
):
Example: Solve the di¤usion equation problem
@u
@ 2u
=
;
@t
@x2
u (x; 0) = e jxj;
1 < x < 1; t > 0
1<x<1
Here u = u (x; t) ; so we begin by de…ning
F fu (x; t)g =
Z 1
1
b ( ; t) :
u (x; t) eix dx = u
Now take Fourier transforms of our PDE, i.e.
F
@u
@t
=F
(
@ 2u
@x2
)
to obtain
b
du
2u
b ( ; t) :
=
dt
We note that the second order PDE has been reduced to a …rst order equation
of type variable separable. This has general solution
b ( ; t) = Ce
u
2t
:
We can …nd the constant C by transforming the initial condition
F fu (x; 0)g = F
b ( ; 0) =
u
n
e jxj
Z 1
1
o
e jxjeix dx =
b ( ; t) gives
Applying this to the solution u
hence
b ( ; 0) = C =
u
b ( ; t) =
u
1+
2
1+
2
2
e
2
2t
;
:
2
1+
2
:
b ( ; t))
We now use the inverse transform to get u (x; t) = F 1 (u
=
Z 1
= 2
= 2
= 2
b ( ; t) e ix d
u
Z1
1
1 e
2
1 (1+ )
Z 1
1
1
Z 1
1+
2
1 e
2
1 (1+ )
2t
e ix d
2t
e
2t
(cos x
i sin x) d
Z 1
cos x d
This now simpli…es nicely because
2i
1
1+
2
e
1 e
2
1 (1+ )
2t
sin x
is an odd function, hence
Z 1
1
1
1+
2
e
2t
sin x d = 0:
2t
sin x d :
Therefore
u (x; t) = 2
Z 1
1
1
1+
2
e
2t
cos x d :
In order to solve this we now need to use Residues (Complex Analysis).
Complex Variables
In the following sections we shall begin our study of analytic functions of a complex variable. Complex variable theory is one of the most beautiful branches
of pure mathematics but it also has important applications in applied mathematics. More excitingly, complex variables are now used in derivative pricing,
when solving the pricing equations via the Fourier Transform approach.
In what follows we shall convey some of the basic ideas of complex analysis
without emphasis on any rigor.
Review of Complex Numbers
A complex number z = x + iy; is a pair (x; y ) of real numbers.
x = real part = Re z ; y = imaginary part = Im z
Operations on complex numbers:
1. Addition: (x1; y1) + (x2; y2) = (x1 + x2; y1 + y2)
2. Multiplication: (x1; y1) (x2; y2) = (x1x2
y 1 y 2 ; x 1 y 2 + x2 y 1 )
The set of all complex numbers de…ned by C is called a …eld, i.e. addition and
multiplication are associative and commutative
(z1 + z2 ) + z3 = z1 + (z2 + z3 )
z1 (z2z3) = (z1z2) z3
distributive
z1 (z2 + z3 ) = z1 z2 + z1 z3
zero:
(0; 0) + (x; y ) = (x; y )
identity:
(1; 0) s.t. (1; 0) (x; y ) = (x; y )
Non-zero complex numbers have inverses, i.e. given (x; y ) 6= (0; 0) 9 x0; y 0
s.t.
(x; y )
x0; y 0 = (1; 0)
In fact
x0 ; y 0
=
x
y
; 2
2
2
x + y x + y2
!
Look at the complex numbers (x; 0) ;
(x1; 0) + (x2; 0) = (x1 + x2; 0)
(x1; 0) (x2; 0) = (x1x2; 0)
So f(x; 0) 2 Cg is a sub…eld of C.
In fact it is the same as R
x 2 R 7 ! (x; 0) 2 C:
Geometrical Representation
There is a 1-1 correspondence between C and R2
z = x + iy = (x; y )
! the point with coordinates (x; y ) :
R2 is called an Argand diagram or the Complex Plane.
x
z = x+iy
r
y
θ
This is polar coordinate form (r; )
x = r cos ; y = r sin
r =
so
sin
cos
q
x2 + y 2
= q
= q
y
x2 + y 2
x
x2 + y 2
= y=r
= x=r
giving us an alternative representation of complex numbers, i.e.
z = x + iy = z = r (cos + i sin )
= rei
The …nal form is Euler’s identity/formula and called the mod-arg form of z:
r = modulus of z = mod z = jzj
= arg z = argument of z:
is only determined by x and y up to the addition of an integer multiple of
2 :
e.g. z =
To …nd
1
i; x =
1=y !r=
2
solve
1=
to get
p
=
3
4
p
2 cos
or
1=
p
2 sin
or 54 :
The values of arg z are
2n ; n 2 Z:
3 ; 5 ; 13 ; 21 ; ::::;
4
4
4
4
The Principal Value of arg z is the argument
11 ;
4
19 ;
4
which satis…es
i.e.
<
3
4
+
:
So we see that the angle is not unique, there are many values for the argument.
The set
f + 2n : n 2 Zg
is written Argz:
Examples: z = 1
i
jzj =
p
2
arg z = arctan
n
Argz = :::;
9 ;
4
o
7 ; 15 ; :::::
;
4 4
4
1
=
1
4
2(
; ]
z=
p
3+i
jzj = 2
Argz =
n
Argz = :::;
:::;
5
arg z =
6
9 ;
4
o
7
15
4 ; 4 ; 4 ; :::::
7 5 17
; ;
; :::::
6 6
6
For any z 2 C; the expression
eiz
e iz
eiz + e iz
sin z
sin z =
, cos z =
and tan z =
2i
2
cos z
de…nes the generalized circular functions, and
ez + e z
sinh z
; cosh z =
and tanh z =
sinh z =
2
2
cosh z
the generalized hyperbolic function.
ez
e z
Using Euler’s formula with positive and negative components we have
ei
e i
= cos + i sin
= cos
i sin
Adding gives
2 cos =
ei
+e i )
ei + e i
cos =
2
and subtracting gives
e i
2i sin =
sin =
:
2i
We can extend these results to consider other functions:
1
1
1
cos ec z =
; sec =
; cot z =
sin z
cos z
tan z
1
1
1
; sec =
; cot z =
cosh ec z =
sinh z
cosh z
tanh z
We can also obtain a relationship between circular and hyperbolic functions:
ei
we know 1=i =
ei
e i )
1
sin (iz ) =
e z
2i
i hence
sin (iz ) =
1
i: e z
2
ez
ez
1 z
= i: e
2
e z
so
sin (iz ) = i sinh z:
Similarly it can be shown that
sinh (iz ) = i sin z
cos (iz ) = cosh z
cosh (iz ) = cos z
Example:
Let z = x + iy be any complex number, …nd all the values for which cosh z =
0:
We use the hyperbolic identity
cosh(a + b) = cosh a cosh b + sinh a sinh b
to give
cosh z = cosh (x + iy ) = cosh x cosh iy + sinh x sinh iy
= cosh x cos y + i sinh x sin y
i.e.
cosh x cos y + i sinh x sin y = 0
so equating real and imaginary parts we have two equations
cosh x cos y = 0
sinh x sin y = 0
From the …rst we know that cosh x 6= 0 so we require cos y = 0 ) y =
+ n 8n 2 Z:
2
Putting this in the second equation gives
sinh x sin (2n + 1)
2
=0
where
sin (2n + 1)
2
= cos n = ( 1)n
so
sinh x = 0
which has the solution x = 0: Therefore the solution to our equation cosh z =
0 is
zn = i (2n + 1) ; n 2 Z
2
De Moivres Theorem
(cos + i sin )n =
ei
n
= ein
= cos n + i sin n
Similarly
(cos + i sin ) n = cos n
It is quite common to write cos + i sin
i sin n :
as cis:
If
z = ei = cos + i sin
then
_
1
= e i = z = cos
z
i sin :
So
cos
sin
Also z n = ein
_
1
1
1
= Re z =
z+z =
z+
2
2
z
_
1
1
1
= Im z =
z z =
z
:
2i
2i
z
!
z n + z n = (cos n + i sin n ) + (cos n
= 2 cos n
)rearranging gives
1 n
1
cos n =
z + n :
2
z
Similarly
1 n
sin n =
z
2
1
zn
i sin n )
Finding Roots of Complex Numbers
Consider a number w; which is an nth root of the complex number z: That is,
if wn = z; and hence we can write
w = z 1=n:
We begin by writing in polar/mod-arg form
z = r (cos + i sin ) :
hence
z 1=n = r1=n (cos + i sin )1=n
and then by DMT we have
+ 2k
+ 2k
=
cos
+ i sin
n
n
Any other values of k would lead to repetition.
z 1=n
r1=n
k = 0; 1; :::::; n
1:
This method is particularly useful for obtaining the n
requires solving the equation
roots of unity. This
z n = 1:
There are only two real solutions here, z = 1; which corresponds to the case
of even values of n: If n is odd, then there exists one real solution, z = 1: Any
other solutions will be complex. Unity can be expressed as
1 = cos 2k + i sin 2k
which is true for all k 2 Z: So z n = 1 becomes
r n (cos n + i sin (n )) = cos 2k + i sin 2k :
The modulus and argument for z = 1 is one and zero, in turn. Equating the
modulus and argument of both sides gives the following equations
rn = 1
and n = 2k
Therefore
2k
2k
+ i sin
=1
n
n
2k i
= exp
k = 0; :::; n
n
z = cos
If we set ! = exp 2kn i then the n
1
roots of unity are 1; !; ! 2; :::::; ! n 1:
These roots can be represented geometrically as the vertices of an n sided
regular polygon which is inscribed in a circle of radius 1 and centred at the
origin. Such a circle which has equation given by jzj = 1 and is called the unit
circle.
More generally the equation
jz
z0 j = R
represents a circle centred at z0 of radius R: If z0 = a + ib; then
jz
z0j = j(x; y ) (a; b)j
= j( x a ) + i ( y
b)j
and
j(x
(x
a ) + i (y
a )2 + ( y
b)j2 = R2
b)2 = R2
which is the Cartesian form for a circle, centred at (a; b) with radius R:
The unit circle is de…ned as
jzj = 1
and the unit disk is jzj
1.
If
jzj < 1 then the disk is the open unit disk
jzj
1 then the disk is the closed unit disk
These are examples of open and closed disks
jz
z0j < ; jz
z1 j
Consider the annulus (ring shaped region)
r < jz
z0j < R:
For the special case r = 0; i.e.
0 < jz
z0j < R;
we call this the punctured disk of radius R around the point z0.
De…nition 1
The open disc centre z0 2 C and radius r > 0 is the set Nr (z0) given by
De…nition 2
Nr (z0) = fz 2 C : jz
z0j < rg
The closed disc centre z0 2 C and radius r > 0 is the set Nr (z0) given by
Nr (z0) = fz 2 C : jz
z0 j
rg
Applications
Example 1
Calculate the inde…nite integral
Z
We begin by expressing cos4
in terms of cos n
cos
24 cos4
cos4
d :
(for di¤erent n).
1
1
1 4
4
4
=
z+
) 2 cos = z +
)
2
z
z
1
1
1
1
4
3
2
= z + 4z + 6z 2 + 4z 3 + 4 using Pascals triangle
z
z
z
z
1
1
= z 4 + 4z 2 + 6 + 4 2 + 4
z
z
1
1
= z4 + 4 + 4 z2 + 2 + 6
z
z
We know
1
2
zn
1
+ n
z
= cos n
24 cos4
1
1
= 2: 12 z 4 + 4 + 4:2: 21 z 2 + 2 + 6
z
z
hence
24 cos4
cos4
= 2 cos 4 + 8 cos 2 + 6
1
=
(cos 4 + 4 cos 2 + 3) )
8
Now integrating
Z
cos4 d
Z
1
=
(cos 4 + 4 cos 2 + 3) d
8
1
3
1
sin 4 + sin 2 +
+K
=
32
4
8
Example 2
As another application , express cos 4
in terms of cosn :
We know from De Moivres theorem that
cos 4 = Re (cos 4 + i sin 4 )
So
cos 4 = Re (cos + i sin )4 ;
and put c
cos ;
is
i sin ; to give
cos 4
= Re c4 + 4c3 (is) + 6c2 (is)2 + 4c (is)3 + (is)4
cos 4
= Re c4 + i4c3s
cos 4
= c4
6 c2 s 2 + s 4
6 c2 s 2
i4cs3 + s4
Now s2 = 1
c2 ; )
cos 4
= c4
6 c2 1
cos 4
= 8 cos4
c2 + 1
c2
2
= 8 c4
8 cos2 + 1:
8c2 + 1 )
Example 3
Find the square roots of 1 ; i.e. solve z 2 =
has a modulus of one and argument ; so
1: The complex number
1 = cos ( + 2k ) + i sin ( + 2k ) :
Hence,
( 1)1=2 = (cos ( + 2k ) + i sin ( + 2k ))1=2
+ 2k
+ 2k
+ i sin
= cos
2
2
1
for k = 0; 1 :
( 1)1=2 = cos
2
+ i sin
2
=0+i
3
3
= cos
+ i sin
=0 i
( 1)
2
2
Therefore the square roots of 1 are z0 = i and z1 = i:
1=2
Example 4
Find the …fth roots of 1 ; i.e. solve z 5 =
a modulus of one and argument ; so
1: The complex number
( 1)1=5 = (cos ( + 2k ) + i sin ( + 2k ))1=5
+ 2k
+ 2k
= cos
+ i sin
5
5
1 has
for k = 0; 1; 2; 3; 4 :
z0 = cos
5
3
z1 = cos
5
+ i sin
5
3
+ i sin
5
z2 = cos ( ) + i sin ( )
z3 = cos
7
5
+ i sin
7
5
z4 = cos
9
5
+ i sin
9
5
Example 5
Find all z 2 C such that z 3 = 1 + i: So we wish to …nd the cube roots of
(1 + i) : The argument of this complex number is = arctan 1 = =4: The
modulus of (1 + i) is r =
as
p
2: We can express (1 + i) compactly in r exp (i )
1+i=
p
2 exp i
4
So
(1 + i)
1=3
=
21=6 exp
(8k + 1)
i
12
for k = 0; 1; 2:
z0 = 21=6 exp i
12
z1 =
9
1=6
2
exp i
z2 =
17
1=6
2
exp i
12
12
!
Example 6: We can apply Euler’s formula to integral problems. Consider the
earlier example
Z
ex cos xdx
which was simpli…ed using the integration by parts method. We know Re ei =
cos ; so the above becomes
Z
ex Re eixdx =
Z
1 e(i+1)x
Re e(i+1)xdx = Re 1+i
ix
1
1 i
= ex Re 1+i
e
eix = ex Re (1+i)(1
i)
= 21 ex Re (1 i) eix = 12 ex Re eix ieix
=
=
1 ex Re (cos x + i sin x
2
1 ex (cos x + sin x)
2
i cos x + sin x)
Exercise: Repeat this method of working for evaluating
Z
ex sin xdx
Functions
Polynomial Functions: A polynomial function of z has the form
f (z ) = a 0 + a 1 z + a 2 z 2 + O z 3 =
1
X
an z n
n=0
and is of degree n: The domain is the set C of all complex numbers: So for
example a 3rd degree polynomial is 2 z + a2z 2 + 3z 3:
Rational Functions: A rational function has the form
R (z ) =
P1 (z )
P2 (z )
where P1; P2 are polynomials. The domain is the set C zeroes of P2 (z ).
For example
2z + 3
2z + 3
f (z ) = 2
=
z
3z + 2
(z 1) (z 2)
and domain is C
f 1; 2g :
Power Series:
exp ( z ) = 1
1
z + z2
2!
1
n
X
1 3
nz
z + ::::: =
( 1)
3!
n!
n=0
1
X
1 3
1 5
z 2n+1
sinh z = z + z + z + ::::: =
3!
5!
n=0 (2n + 1)!
1
X
1 4
1 2
z 2n
cosh z = 1 + z + z + ::::: =
2!
4!
n=0 (2n)!
sin z = z
1 3
1 5
z + z
3!
5!
cos z = 1
1 2
1 4
z + z
2!
4!
1
X
z 2n+1
::::: =
( 1)
(2n + 1)!
n=0
1
2n
X
n z
::::: =
( 1)
(2n)!
n=0
n
Functions of a Complex Variable
A rule which assigns to every complex number
z = x + iy = rei
in some region D; a unique complex number
w = u + iv = ei :
w is called a function of a complex variable.
So w = f (z )
= u (x; y ) + iv (x; y )
So we see that
Re w = u (x; y )
Im w = v (x; y )
e.g.
w = z2
= (x + iy )2
= x2 y 2 + 2xyi
Here
u (x; y ) = x2 y 2
v (x; y ) = 2xy
Note
@u
@v
= 2x =
@x
@y
@u
@v
=
2y =
@y
@x
Exponential Function:
w = f (z ) = ez
= ex+iy = exeiy
Re ez : u (x; y ) = ex cos y
Im ez : v (x; y ) = ex sin y
jexp zj = ex and y is the argument.
Logarithmic Function:
If
ew = z
we say w is a logarithm and we write
w = Logz
which is not unique, for suppose
w = Logz = u + iv
or
eu+iv = z
eueiv = z
eu (cos v + i sin v ) = z
therefore
eu = jzj =) u = ln jzj
and
v = arg z + 2n
Thus we can write
Logz = ln jzj + i (arg z + 2n )
Logz has in…nitely many vales. If we take the principal value of arg z then the
corresponding value of Logz is called the principal value of Logz and written
log z where
log z = ln jzj + i arg z
and log z is now a function.
Example: z =
p
1+i 3
jzj = 2; arg z = arctan
p
3 =
3
=
2
3
hence
2
Logz = ln j2j + i
+ 2n
3
2
log z = ln j2j + i
3
; n2Z
Power Series
We de…ne a power series in (z
a 0 + a 1 (z
a ) + a 2 (z
a) or about z = a as
2
a) + :::::: + an (z
n
a) + ::: =
1
X
a n (z
a )n
n=0
This in…nite series converges for z = a (plus other points).
9 R 2 Q+ s.t.
R:
1
X
a n (z
n=0
The special case jz
a)n converges jz
aj < R and diverges jz
aj = R may or may not converge.
(y)
aj >
z=a
R
Γ
(y) will converge at all points in and diverge everywhere outside : On
do not know (needs additional work).
The special cases R = a and R = 1 correspond in turn to
R = a corresponds to converges at z = a only
R = 1 corresponds to converges 8 …nite values of z
R
radius of convergence
circle of convergence
we
Various Tests
Absolute Convergence
If
1
X
n=1
junj converges then
1
X
un converges
n=1
Comparison Test:
If
1
X
n=1
If
1
X
n=1
jvnj converges and junj
jvnj diverges and junj
or may not converge.
jvnj then
jvnj then
1
X
un converges absolutely.
n=1
1
X
n=1
junj diverges but
1
X
n=1
un may
Ratio Test:
If
un+1
=L
n!1 un
lim
then
1
X
un
n=1
p
series Test:
8
>
<
converges (absolutely) L < 1
diverges
L>1
>
: test fails
L=1
1
X
1
converges for any constant p > 1 and diverges for p
p
n=1 n
1:
Example 1: Show that
1
X
zn
n=1 n (n + 1)
converges absolutely for jzj
If jzj
1:
1 then
zn
n (n + 1)
1
jz nj
=
n (n + 1)
n (n + 1)
1
n2
and by the p series test for p = 2 we know that it converges, hence comparison test implies convergence.
Let’s re-do but using the Ratio test
un+1
n!1 un
lim
where
zn
z n+1
un =
; un+1 =
n (n + 1)
(n + 1) (n + 2)
so
un+1
n!1 un
lim
=
=
=
z n+1
(n+1)(n+2)
lim
zn
n!1
n(n+1)
z n+1 n
lim
n!1 n + 2 z n
n
jzj
n!1 n + 2 |{z}
| {z }
lim
<1
< 1
so we have convergence by the Ratio test.
1
Example 2: Calculate the radius of convergence of
1
X
(z + 2)n 1
3
n
n=1 4 (n + 1)
For z =
2 this converges.
Use the Ratio test with un =
un+1
lim
n!1 un
(z+2)n 1
;
4n(n+1)3
=
=
lim
n!1
un+1 =
(z+2)n
4n+1 (n+2)3
(z+2)n
4n+1 (n+2)3
(z+2)n 1
4n(n+1)3
n+1 3 z
+2
n!1 n + 2
4
8
> < 1 abs cgce
jz + 2j <
= 1 test fails
4 >
: > 1 diverges
lim
Therefore jz + 2j < 4 gives R = 4; which is the region of convergence. Circle
centred at ( 2; 0) and radius 4:
We also see that z =
2 is included in jz + 2j < 4:
What about the boundary of the circle jz + 2j = 4? The Ratio test does not
assist here.
So try the Comparison Test
look at
(z + 2)n 1
4n (n + 1)3
the numerator becomes, using jz + 2j = 4; 4n 1
4n 1
1
=
4n (n + 1)3
4 (n + 1)3
which converges (from comparison test for p = 3):
1
n3
So the series is absolutely convergent for jz + 2j
centre 2 and radius 4; including the boundary.
4; i.e. region of circle
Di¤erentiation in The Complex Plane
Recall that for a real variable
df
f (x + x) f (x)
= lim
:
x !0
dx
x
For functions of complex variables there are an in…nite number of paths along
which z ! 0 and so as many values of
f (z + z )
z !0
z
f 0 (z ) = lim
are possible.
f (z )
If all these limits are the same we say that f (z ) is di¤erentiable and the
derivative is the value of the limit. So in other words if the derivative exists it
must be independent of the way in which z tends to zero.
Another de…nition for f 0 (z ) at the point z0 is
f 0 (z
f (z )
0 ) = z lim
!z0
z
f (z0 )
:
z0
Holomorphic Functions
Consider a region U. If the derivative f 0 (z ) exists at all points in U then
the function is said to be Holomorphic in U. This is a relatively new term,
some of the older books use the synonyms regular and analytic. We then write
f (z ) 2 H (U) :
A function f (z ) is said to be holomorphic at a point z0 if 9 a neighbourhood
jz
at all points of which f 0 (z ) exists.
z0 j <
If a function is holomorphic everywhere we simply say f (z ) is a holomorphic
function, i.e. f (z ) 2 H (C) :
Example: Show that f (z ) = z is not di¤erentiable at any point.
z = x + iy and z = x + i y
then
z + z = (x + x) + i (y + y ) :
f 0 (z )
=
=
=
Now consider the limits.
f (z + z ) f (z )
lim
z !0
z
(z + z ) z
lim
z !0
z
x i y
lim
z !0 x + i y
First let y ! 0 and then x ! 0
x
x i y
= lim
=1
lim
x !0 x
z !0 x + i y
now x ! 0 and then y ! 0
x i y
i y
= lim
=
z !0 x + i y
x !0 i y
lim
1
as the results di¤er f (z ) = z is not di¤erentiable at any point.
A point at which f (z ) is not di¤erentiable is called a singularity ; or a singular
point of f (z ) :
As we cannot test all the paths as z ! 0; this provides us with a way to
establish non-di¤erentiability - by simply …nding two paths which give di¤erent
limits.
Example: Prove (using the de…nition) that
8
< x3(1+i) y 3(1 i)
x2 +y 2
f (z ) =
: 0
z 6= 0
z=0
is not holomorphic at z = 0:
f (z) f (z0 )
; which becomes
Let’s use the de…nition f 0 (z0) = lim
z z0
z !z0
f (z )
z !0
z
f 0 (0) = lim
f (0)
0
lim
x3 (1 + i)
z !0
y 3 (1
i)
x2 + y 2 (x + iy )
Let z ! 0 along the line y = mx and examine the limit
x3 (1 + i)
m3x3 (1
i)
x2 + m2x2 (x + imx)
=
(1 + i)
m3 (1
i)
1 + m2 (1 + im)
hence
lim
x !0
(1 + i)
m3 (1
i)
1 + m2 (1 + im)
has many values depending on m which implies that f 0 (0) does not exist.
If f (z ) is a function of z; e.g. z 2; ez ; sin z then di¤erentiate in the normal/real
way and the various rules (e.g. product/quotient) apply.
Example: If f (z ) = cosecz
f 0 (z )
=
cos ecz cot z =
1
cos z
=
sin z sin z
cos z
sin2 z
which has singularities where sin z = 0 () z = n : n 2 Z:
As an exercise verify these singularities by solving sin z = 0 for z = x + iy:
The Cauchy-Riemann Equations
We need a way of showing that a function is di¤erentiable as the de…nition of
di¤erentiability is really only useful for establishing non-di¤erentiability.
Let
z = x + iy ; f (z ) = u (x; y ) + iv (x; y ) :
f (z + z )
If f (z ) is di¤erentiable at a given point z then the ratio
z
0
f (z ) no matter how z ! 0
f (z )
u(x+ x;y+ y)+iv(x+ x;y+ y) u(x;y) iv(x;y)
:
x+i y
z !0
f 0 (z ) = lim
We consider this in two steps:
!
1. Let z ! 0 horizontally i.e. y = 0 and x ! 0
f 0 (z ) =
=
u(x+ x;y)+iv(x+ x;y) u(x;y) iv(x;y)
x
x !0
u(x+ x;y) u(x;y)
v(x+ x;y) v(x;y)
lim
+
i
x
x
x !0
lim
@v
@u
+i :
=
@x
@x
2. Let z ! 0 vertically i.e. x = 0 and y ! 0
f 0 (z ) =
=
=
u(x;y+ y)+iv(x;y+ y) u(x;y) iv(x;y)
i y
y !0
lim
lim
y !0
1 @u
i @y
1 u(x;y+ y) u(x;y)
i
y
@v
+
=
@y
@u @v
i
+
@y @y
+
v(x;y+ y) v(x;y)
y
If f 0 (z ) exists these two limits must be equal and hence
@u
@v
+i
=
@x
@x
i
@u @v
+
@y @y
equating real and imaginary parts (in turn) gives
@v
@u
=
@x
@y
@u
@v
=
@y
@x
These are the Cauchy-Riemann Equations.
They are necessary for di¤erentiability but not su¢ cient, i.e. if C-R equations
are satis…ed, f (z ) may or may not be di¤erentiable. We can say with certainty
that if the conditions are not satis…ed then the function is non-di¤erentiable.
Example: Show that the functions
x
u= 2
; v=
x + y2
y
x2 + y 2
satisfy the Cauchy-Riemann equations everywhere except at (0; 0) :
This can be done simply by verifying
@v
@u
=
@x
@y
@v
@u
=
@y
@x
for the given u (x; y ) and v (x; y ) :
y2
x2
@v
2xy
=
2
@x
2
2
x +y
@u
=
@x
x2 + y 2
@u
=
@y
@v
y 2 x2
;
=
2
2
@y
x2 + y 2
x2 + y 2
2xy
;
2
so C-R equations are satis…ed. The partial derivatives are continuous everywhere except at (0; 0) ; where they do not exist.
f (z ) = u + iv
x
y
= 2
+i 2
x + y2
x + y2
1
=
(x iy )
2
x2 + y 2
further simpli…cation gives
f (z ) =
1
2
jzj
1
=
z
zz
1
=
z
z
A function
(x; y ) is called harmonic if it satis…es Laplace’s Equation
@2
@2
+
=0
2
2
@x
@y
The real and imaginary parts of a Holomorphic function satisfy Laplace’s Equation.
This is very easy to verify, for if f (z ) = u + iv 2 H (C) then the C-R
equations are satis…ed.
@u
@v
=
@x
@y
@u
@v
=
@y
@x
(1)
(2)
and we can di¤erentiate these partially. Di¤erentiate (1) w.r.t x; and (2) w.r.t
y
@ 2u
@ 2v
=
@x2
@x@y
@ 2u
=
@y 2
@ 2v
@y@x
(3)
(4)
(3) + (4) gives
@ 2u @ 2u
+
=0
2
2
@x
@y
Similarly di¤erentiating (1) and (2) wrt to y and x respectively gives
@ 2v @ 2v
+
=0
@x2 @y 2
So we see that both real and parts of a holomorphic function are harmonic.
They are sometimes called harmonic conjugates. Given one harmonic function
we can use the C-R equations to …nd a conjugate harmonic function.
Consider the following
u (x; y ) = ex
2
y 2 sin 2xy
does this satisfy the PDE above?
2 y2
@u
x
= 2e
(y cos 2xy + x sin 2xy )
@x
2 y2
@ 2u
x
2
2+2 +
=
e
sin
2
xy
4
x
4
y
@x2
2
2
8xyex y cos 2xy
2 y2
@ 2u
x
2 sin 2xy
=
e
4
x
@y 2
so clearly uxx + uyy = 0:
8xy cos 2xy + 4y 2 sin 2xy
2 sin 2xy
Complex Integration
A complex integral is an integral taken along a curve (contour) in the complex
plane. We will denote this by :
We base our de…nition of such an integral on real integrals to avoid unnecessary
work.
Consider …rst the type of curve along which we will integrate. A curve can be
written in parametric form if it can be expressed as
z = z (t) : a
t
b
where t is a real parameter with initial and …nal points z (a) and z (b) ; in turn.
Examples:
1. The circle centre 0; radius r starting and …nishing at the point A (jzj = r) :
Positively described means anti-clockwise,
z = reit
= r (cos t + i sin t) : 0
t
2
2. Semi-circle, centre 0; radius 1 lying in the right hand half of the plane.
The initial point is A (z = i) and …nal point B (z = i)
z = eit :
2
t
2
3. Circle centre z0 = x0 + iy0 of radius r
z = z0 + reit : 0
t
2
4. The positive real axis starting at 0
z=t:0
What about the real axis from
5. The imaginary axis from z =
t<1
3 to 2?
2i to z =
z = it :
2
5i
5
t
6. The line segment from a + ic to b + ic
z = t + ic : a
t
b
There is a useful general method for obtaining this, if we wish to express
the line segment from z1 to z2 :
z = z1 + t ( z2
z1 ) : 0
t
1
7. The line from a to a + ib
z = a + it : 0
8. The line from
1
t
1
is parameterized by
z = z (t) : a
then
b
1 to 1 + 2i
z = t + i (1 + t) :
If
t
t
b
can be expressed as
z = z(
):
b
t
a
The contour is called closed if z (a) = z (b) ; i.e. the starting point and
end point are the same. For example, consider the circle earlier
z = eit : 0
t
2
here we see z (0) = z (2 ) = 1
A closed contour which does not cross itself is called a simple closed contour.
Integration Along a Contour
Let
be a contour de…ned by (t) = z (t) : t 2 [a; b] : Let f (z ) be a
continuous function on and z 0 (t) is also continuous on ; then
Z
Simple Example:
1 to i: Evaluate
First de…ne
Z b
dz
dt
dt
a
is the …rst quadrant of the unit circle, i.e. centre 0 from
f (z ) dz =
Z
f (z (t))
zdz
:
z = eit 0
dz
= ieit
dt
t
=2
therefore
Z
zdz =
=
=
Example 2: Let
Z
=2
0
Z =2
0
ei
eitieitdt
e2itidt
1
2
=
1 2it =2
= e i
2i
0
1
be the straight line from 1 to 2+ i: Evaluate
:
z = t + (t
dz
= 1+i
dt
1) i : 1
t
2
Z
(1 + 2z ) dz
hence
Z
(1 + 2z ) dz =
Z 2
1
1 + 2 (t + ( t
= (1 + i)
= 3 + 5i
If
Z
Z 2
1
(1
1) i) (1 + i) dt
2i + 2 (1 + i)) dt
dz
is continuous except at t = c1; c2; :::::; cn we can de…ne
dt
f (z ) dz =
Z c
1
a
f (z (t)) dz
dt dt +
Z c
2
c1
f (z (t)) dz
dt dt + :::::: +
Z b
cn
f (z (t)) dz
dt dt
Example 1: Let
be the line from
(t)
:
dz
=
dt
Then
Z
(
(
1 to 0 together with the line from 0 to i
z=t
z = it
1
1 t 0
0 t 1
1
i
0
t<0
t 1
zdz =
Z 0
1
=
1
t 1dt +
Z 1
0
it idt
Example 2:
Consider the following parametrization:
(t) =
Evaluate
Z
=
(t
>
: (3
t) i
t 2 [0; 1]
t 2 [1; 2]
t 2 [2; 3]
1) i + (2
t) i) (i
1) i + (2
t) i
Re (z ) dz =
Z 1
0
Z 3
=
8
>
< t
2
Z 1
Re (t) dt +
Re ((3
t dt +
0
1
t2
2 0
+ (i
Z 2
1
Re ((t
t) i) ( i) dt
Z 2
1
(2
1) 2t
t) (i
1) dt +
!
2 2
t
2
1
= i
2
1
Z 3
2
0 dt
1) dt +
Properties of the Integral
Z
As an example consider
Z
(t)
Z
f (z ) dz =
z dz where
:
f (z ) dz
is the straight line from 0 to 1 + i
z = t + it
0
1
t
z0 = 1 + i
Z
z dz =
Z 1
(t + it) (1 + i) dt
t
it
0
= i
Now consider
(t) : z =
1
t
0
Z
z dz =
=
If
Z 0
Z1
( t
it) ( 1
i) dt =
i
z dz
has initial and …nal point z1 and z2; in turn and if f (z ) = dF
dz on
Z
Z
dF
f (z ) dz =
dz
dz
= F ( z2 ) F (z1 )
for suppose : z = z (t) a t b; i.e. z1 = z (a) & z2 = z (b) then
Z
Z
dF
f (z ) dz =
dz
dz
Z b
dF (z (t)) dz
=
dt
dz
dt
a
= F (z (t))jba = F (z (b)) F (z (a))
= F (z2 ) F (z1 )
then
Note:
1. If z2 = z1; i.e.
0
is a closed contour then
Z
f (z ) dz = F (z1) F (z1) =
2. These results mean that if f (z ) can be integrated directly then we do not
need to parameterize :
Example 1:
Then
Z
is the circular contour joining 1 to i:
z3
2
z dz =
3
i
z3
=
3 1
1 (1 + i) :
3
Note that this answer only depends on the initial and …nal points of ; not on
itself.
Example 2: Let
be the line from 1 to 2 + i
Z
(1 + 2z ) dz = z + z 2
= z + z2
2+i
1
= 5i + 3
Cauchy’s Theorem
If
is any closed contour and if f (z ) is di¤erentiable inside and on
Z
then
f (z ) dz = 0
There are many versions of this theorem which make di¤erent assumptions
about the contour :
As an example, a polynomial P (z ) is di¤erentiable everywhere. The exponential, circular and hyperbolic functions are holomorphic on C: Therefore given
a closed contour
Z
P (z ) dz =
Z
ez dz =
Z
sin zdz = ::::::: =
e.g. if C is the unit circle jzj = 1 then
Z
C
z 2 + 6z
3 dz = 0:
Z
cosh zdz = 0
A rational function R (z ) = P1 (z ) =P2 (z ) is holomorphic everywhere except
R
P1 (z ) =P2 (z ) dz = 0 on any closed
at the zeroes of P2 (z ) : Therefore
contour which does not contain or pass through any zero of P2 (z ) :
Corollary to Cauchy’s Theorem
If 1 and 2 are any two contours with the same initial and …nal points and if
f (z ) is di¤erentiable inside and on 1
2 then
Z
f (z ) dz =
1
Z
f (z ) dz
2
i.e. the integral does not depend on the contour, only on the initial and …nal
points.
1
2
is closed now and f (z ) is di¤erentiable inside and on 1
2
(given).
Therefore (by Cauchy)
Z
hence
R
1
f (z ) dz =
Z
f (z ) dz +
Z1
R
2
f (z ) dz
1
f (z ) dz:
f (z ) dz = 0
Z1
2
f (z ) dz = 0
Z
2
f (z ) dz = 0
2
R
dz
Example: Evaluate
where
2
z +2z+2
in the upper 1/2 plane.
is the semi-circle joining
1 and 1
This contour is not closed. How ever by introducing the line segment L which
1
goes from 1 to 1 along the real axis
L is now closed. z 2+2z+2
is
holomorphic except where z 2 + 2z + 2 = 0; i.e. z =
outside
L:
It follows by the Corollary to Cauchy’s Theorem that
Z
dz
z2
+ 2z + 2
=
Z
dz
L z2
+ 2z + 2
So we solve along L by parameterizing:
L : z (t) = t;
1
t
1
1
i; which lies
Z
dz
=
2
z + 2z + 2
=
Z 1
dt
1 t2 + 2t + 2
Z 1
dt
1 (t + 1)2
Use a substitution u = t + 1; which gives
Z 2
du
1u 2
=
tan
0
0 u2 + 1
= tan 1 2
+1
2
Example: By integrating e z around the rectangle with sides y = 0; y = b;
R
dz
x = R; z 2+2z+2
where is the semi-circle joining 1 and 1 in the upper
1/2 plane.
This contour is not closed. How ever by introducing the line segment L which
1
goes from 1 to 1 along the real axis
L is now closed. z 2+2z+2
is
holomorphic except where z 2 + 2z + 2 = 0; i.e. z =
outside
L:
It follows by the Corollary to Cauchy’s Theorem that
Z
dz
z2
+ 2z + 2
=
Z
dz
L z2
+ 2z + 2
So we solve along L by parameterizing:
L : z (t) = t;
1
t
1
1
i; which lies
Z
dz
=
2
z + 2z + 2
=
Z 1
dt
1 t2 + 2t + 2
Z 1
dt
1 (t + 1)2
Use a substitution u = t + 1; which gives
Z 2
du
1u 2
=
tan
0
0 u2 + 1
= tan 1 2
+1
An Extension of Cauchy’s Theorem
When there is a simple type of singularity of f (z ) on C; let f (z ) be regular
in and on C except for a single singularity at z = a which is on C .
With centre z = a and radius draw an arc of a circle to indent the contour
C at a forming a new contour : Since f (z ) is holomorphic in and on so
by Cauchy’s Theorem
Let
!0
Z
Z
C
f (z ) dz = 0
f (z ) dz + lim
Z
!0 indent
f (z ) dz = 0
Cauchy’s Integral Formula
The following important result is due to Cauchy, and is also a Theorem.
Let C be a simple closed contour and suppose that f is holomorphic in and
on C: If z = is a point inside C then
I
1
f (z )
f( )=
dz;
2 i Cz
the integral being taken in the positive (anti-clockwise) sense.
It is possible to deduce from Cauchy’s integral formula that f is di¤erentiable
at and that the derivative of f to all orders n; can be computed by formally
di¤erentiating with respect to z under the integral sign. Thus
I
n!
f (z )
f (n) ( ) =
dz n 2 N
n+1
2 i C (z
)
Now for some examples.
Example 1
I
ez
Evaluate
dz , where
C z
C : z ( ) = ei ; 0
2
by using Cauchy’s integral formula.
Let f (z ) = ez : Then f is a holomorphic function we may apply Cauchy’s
integral formula in the form
I
1
f (z )
f (0) =
dz
2 i Cz 0
It follows that
I
ez
dz = 2 i
C z
Example 2
Evaluate
I
1
f (z )
1 =
2 i Cz 0
I
C (z
ez
1) (z
3)
dz
taken round the circle C given by jzj = 2 in the positive (anti-clockwise)
sense. What is the value of the integral taken around the circle jzj = 1=2 in
the positive sense?
Put
f (z ) = ez = (z
3)
Then f is holomorphic in a domain which contains the circle jzj = 2 and its
interior (but not, of course, the point z = 3). Cauchy’s integral formula is
applicable and we have
I
I
1
1
f (z )
ez = (z 3)
f (1) =
dz =
dz
2 i C (z 1)
2 i C (z 1)
where f (1) =
e=2
We conclude that
I
C (z
ez
1) (z
3)
dz = 2 if (1) =
ei
By Cauchy’s theorem the integral taken round the circle jzj = 1=2 in the
positive sense is zero because the integrand is holomorphic in a domain which
contains the circle and its interior.
Taylor’s Theorem
If f (z ) be holomorphic in the a neighbourhood of z = a then it has a power
series expansion
f (z ) =
1
X
a n (z
a )n
0
with a non-zero radius of convergence R:
Example: Expand
f (z ) =
(z
1
1) (z
2)
about the origin and the point at in…nity.
1. About z = 0: f (z ) has singularities at z = 1; 2: Since z = 0 is a
regular point there is a Taylor expansion about z = 0 of the form
1
X
0
an z n
convergent for jzj < 1
(z
1
1) (z
1
2)
1
=
1
+
1
2 (1 z=2) (1 z )
2) (z 1)
1
1
1
1
1X
z n X
+
+ zn
=
=
2 (1 z=2) 1 z
2 0 2
0
1
X
1
n
=
1
z
2n+1
0
(z
1
X
(n) (0)
f
: We know from above
Or Note f (z ) =
anz n where an =
n
!
0
that f (z ) = (z 1 2)
f 0 (z )
=
f 000 (z ) =
1
(z 1)
1
(z
2
+
4
+
2)
2 3
1
(z
2
1)
3
2
;
4
;
f 00 (z )
=
f (z ) =
(z
2)3
(z
(z 2)
(z 1)
n
n+1
n
!
n!
(
1)
(
1)
(n)
f
+
!
(z ) =
n+1
n+1
(z 2)
(z 1)
f (n) (0)
1
(n)
)
f
(0) = n! 1
=1
2n+1
n!
About z = point at in…nity:
1
1) (z
2
2
1
2n+1
:
2)
z is point at in…nity =) z1 = 0 so let t = 1=z , so
f (t) = 1
t
1
1
1
1
t
2
=
(1
t2
t) (1
2t)
(z
1)3
t = 0 is a regular point therefore we can expand in a Taylor series valid
for jtj < 1=2:
t2
3t=2 1=2
1
f (t) =
= +
2 (1 t) (1 2t)
(1 t) (1 2t)
1
1
1
+
=
2 1 t 2 (1 2t)
1
1
1
1 X
1 X
1X
n
n
=
t +
(2t) = +
2n 1
2
2 0
2
0
0
=
1
X
1
2n 1
1
tn
=
1
X
1
To expand f (z ) about z = 3 put t = z
(about t = 0) :
2n 1
1
1 tn
1
jzj > 2
n
z
3 and expand in powers of t
Laurent’s Theorem
Let f (z ) be holomorphic in the annulus b < jz
series expansion
I
1
X
An (z
aj < c then it has a power
a )n
1
1
f (z )
Here An =
dz: C is any circle jz aj = R where b <
2 i C (z a)n+1
R < c and the expansion is valid for any z in the annulus.
The Nature of Singularities
If f (z ) has an isolated singularity at z = a then by Laurent’s theorem
f (z ) =
b1
(z
a)
+
b2
(z
bn
+ ::::: +
n + a 0 + a 1 (z
2
(z a )
a)
a) + ::::
If there is no singularity at z = a then by Taylor’s Theorem
f (z ) =
1
X
An (z
a )n
0
Hence
1
X
1
br
(z
a)
r
has been produced by the singularity .
We call this part of the expansion the Principal Part (PP) or Laurent Part.
We de…ne the type of singularity according to the shape of the principal part:-
a) If the PP has an in…nite number of terms, we say that f (z ) has an Isolated
Essential Singularity (IES) at z = a.
b) If the PP has a …nite number of terms we say that f (z ) has a pole at
1
z = a. The ORDER of the pole equals the highest power of
which
z a
occurs.
Example:
1. If
f (z ) =
6
(z
|
+
6
2)
(z
3
{z
PP
+
2
(z
2)
f (z ) has a 6th order pole at z = 2:
1
2)
}
+ 4 + 2 (z
2)2 + :::
2.
ez
z2
z3
1
1+z+
+
+ :::
z
z
2!
3!
1
z2
z
=
+ :::
+1+ +
z
2!
3!
|{z}
f (z ) =
=
!
PP
First order pole - also called simple pole at z = 0:
We can also examine the point at in…nity. Put t = 1=z to get
1
1
1
= t 1+ +
+
+ :::
2
3
t 2!t
3!t
1
1
+
+ :::
= t+1+
2
2!t 3!t
PP has an in…nite number of terms there IES at t = 0 i.e. z = point at
in…nity.
te1=t
3.
So put t = z
ez
f (z ) =
at z = 1
z
1 expand in powers of t:
et+1
= e (1 + t) 1 et
t+1
t + t2
= e 1
t3 + :::
1+t+
t2
2!
+ :::
!
= A0 + A1t + A2t2 + :::
If f (z ) has a pole of order n at z = a
f (z ) =
b1
(z
a)
+
b2
(z
bn
+ ::::: +
n + a 0 + a 1 (z
2
(
z
a
)
a)
In the residue theorem (next) we …nd that the coe¢ cient of
a) + ::::
1
i.e.
(z a )
b1 is very important. b1 is called the residue (poles only) of f (z ) at
z = a:
To Find The Residue
1. Use the de…nition of the residue - expand f (z ) in powers of (z
1
pick out the coe¢ cient of
:
z a
(a) For the simple pole
f (z ) =
(z
b1
(z
a)
+ a 0 + a 1 (z
a) f (z ) = b1 + a0 + a1 (z
b1 = lim (z
z!a
a ) f (z )
a) + ::::
a)2 + ::::
a) and
(b) For an nth order pole
bn
b1
+
::::
+
+ a0 + a1 (z a) + ::::
n
(z a )
(z a )
a)n f (z ) = bn + bn 1 (z a) + ::: + b1 (z a)n 1 + a0 (z a)n + :::
f (z ) =
(z
Di¤erentiate (n
dn 1
((z
n
1
dz
Thus
1) times
a)n f (z )) = b1 (n
b1 =
1
(n
1)!+A0 (z
dn 1
lim [(z
n
1
z!a
1)! dz
a)+A1 (z
a)n f (z )]
a)2+::::
Examples
1.
f (z ) =
z=
e1=z
2)2 (z + 1)
(z
1 is a simple pole
residue =
lim (z + 1)
z! 1
e1=z
(z
1
=
2
9e
2) (z + 1)
z = 2 is a double pole
"
d
residue = lim
(z
z!2 dz
2)
e1=z
2
(z
2
2) (z + 1)
#
=
7e1=2
36
Also examine z = 0 by expanding in powers of z
1 1=z
z 2
1
f (z ) = e
1
(1 + z )
4
2
1
1
1
2
=
1+ +
+
:::
1
z
+
z
::: (1 + z::)
2
4
z 2!z
Bn
B1
= ::: n + :::: +
+ A0 + A1z + ::::
2
z
(z a )
Hence there is an IES at z = 0:
2.
f (z ) =
=
=
1
=
2
z sin z
z2
1
z3
1
"
2z 2
1
3z 3
z
3!
4z 4
3! + 5!
2z 2
4z 4
1
1+
3
z
3!
+ :::
5!
+
4z 4
3!
+ :::
#
Hence a 3rd order pole at z = 0 and residue = =6:
The Residue Theorem
If f (z ) is holomorphic in and on a simple closed curve C apart from a
number of poles in C then
Z
C
f (z ) dz = 2 i
sum of residues of f (z ) at all its poles in C .
Example: Show that
Z 1
x+3
1 x2
+9
2
dx =
18
Start by constructing a suitable contour. C consists of the straight line from
R to R and the semi-circular contour of radius R:
Z
z+3
C z2
+9
2
dz
There are singularities at z =
Z 1
x+3
1 x2
+9
Z
dx+ lim
2
R !1 0
R ei + 3
R2e2i + 9
3i: Let R ! 1:
i d = 2 i residue at 3i
i
R
e
2
Now using the earlier result
Z b
a
f (x) dx
Z b
a
jf (x)j dx
we have
Z
Z
0
0
(R + 3) R
R2
9
2
d
R 2 + 3R
= 2 ! 0 as R ! 1
4
2
R + 81 18R
R
=
Z 1
x+3
1 x2
+9
2
dx = 2 i
residue at 3i
Residue at 3i =
2
d 6
lim
4(z
z!3i dz
Z 1
3i)
+9
z+3
z2 + 9
x+3
1 x2
2
2
dx = 2 i
3
1
7
=
25
36i
1
= :
36i
18
Stochastic Volatility Models
An observation when pricing derivatives is the fact that volatility of an asset
price is anything but constant. We have seen in the much celebrated BlackScholes framework that the assumptions do not consider these market features.
Volatility does not behave how the Black–Scholes equation would like it to
behave; it is not constant, it is not predictable, it’s not even directly observable.
Volatility is di¢ cult to forecast - although not impossible.
This makes it a prime candidate for modelling as a random (stochastic) variable.
There are many economic, empirical, and mathematical reasons for choosing
a model with such a form. Empirical studies have shown that an asset’s logreturn distribution is non-Gaussian. It is characterised by heavy tails and high
peaks (leptokurtic). There is also empirical evidence and economic arguments
that suggest that equity returns and implied volatility are negatively correlated
(also termed ‘the leverage e¤ect’).
These reasons has been cited as evidence for non-constant volatility.
Stochastic volatility models were …rst introduced by Hull and White (1987),
Scott (1987) and Wiggins (1987) to overcome the drawbacks of the Black
and Scholes (1973) and Merton (1973) model. So it seems plausible to model
volatility as a stochastic process. The method gives more parameters to …t,
hence popular for calibration purposes.
These are systems of bi-variate SDEs. We continue to assume that S satis…es
GBM
dS = Sdt + SdW1;
but we further assume that volatility
satis…es an arbitrary SDE
d = p(S; ; t)dt + q (S; ; t)dW2:
Here both drift and di¤usion are arbitrary, with q (S; ; t) being volatility of
the volatility (vol of vol).
The two increments dW1 and dW2 have a correlation of
EP [dW1dW2] = dt:
Here P represents the physical measure. The choice of functions p(S; ; t) and
q (S; ; t) is crucial to the evolution of the volatility, and thus to the pricing of
derivatives.
The value of an option with stochastic volatility is a function of three variables,
V (S; ; t).
Let’s do the general theory …rst and then think about speci…c forms for p and
q:
The pricing equation
The new stochastic quantity that we are modelling, the volatility, is not a
traded asset. So as with the spot rate we cannot hold volatility. Thus, when
volatility is stochastic we are faced with the problem of having a source of
randomness that cannot be easily hedged away.
Because we have two sources of randomness we must hedge our option with
two other contracts, one being the underlying asset as usual, but now we also
need another option to hedge the volatility risk.
We therefore must set up a portfolio containing one option, with value denoted
by V (S; ; t), a quantity
of the asset and a quantity
1 of another
option with value V1(S; ; t).
We have
=V
S
1 V1
The change in this portfolio in a time dt is given by
@ 2V
@ 2V
@ 2V
!
@V
+ 21 2S 2 2 + qS
+ 12 q 2 2 dt
@t
@S
@S@
@
!
2
2
2
@ V1
@ V1
@V1 1 2 2 @ V1
2
1
+2 S
+ 2q
+ qS
dt
1
@t
@S 2
@S@
@ 2
@V1
@V
dS
+
1
@S
@S
@V
@V1
+
d :
1
@
@
where a higher dimensional form of Itô has been used on functions of S; and
t:
d
=
To eliminate all randomness from the portfolio we must choose
@V
@S
@V1
1
@S
= 0;
to eliminate the dS terms, which are the sources of randomness, and
@V
@
@V1
= 0;
1
@
to get rid o¤ d terms.
Therefore our choice of delta terms to make the portfolio risk free become
1
=
@V
@
@V1
@
and
@V
=
@S
@V
@ @V1 :
@V1 @S
@
This leaves us with
d
@ 2V
@ 2V
@ 2V
!
@V
1 2 2
1 2
+
S
+
q
+
qS
dt
2
2
@t
2
@S
@S@
2 @
!
2
2
2
1 2 @ V1
@ V1
@V1 1 2 2 @ V1
+
S
+ q
+ qS
dt
1
@t
2
@S 2
@S@
2 @ 2
= r dt = r (V
S
1 V1 ) dt;
=
where we have used arbitrage arguments to set the return on the portfolio
equal to the risk-free rate.
As it stands this is one equation in the two unknowns V and V1:
This contrasts with the earlier Black–Scholes case with one equation in the
one unknown - but presents the same type of problem when deriving the bond
pricing equation.
Collecting all V terms on the left-hand side and all V1 terms on the right-hand
side we …nd that
@V
@t
=
@V1
@t
2
+ 12 2S 2 @@SV2 +
+
1 2 S 2 @ 2 V1
2
@S 2
+
@ 2 V + 1 q 2 @ 2 V + rS @V
qS @S@
2 @ 2
@S
@V
@
@ 2 V1
qS @S@
@V1
@
+
1 q 2 @ 2 V1
2
@ 2
rV
1
+ rS @V
@S
rV1
We are lucky that the left-hand side is a functional of V but not V1 and the
right-hand side is a function of V1 but not V .
Therefore both sides can only be functions of the independent variables, S;
and t. So set both sides equal to
f (S; ; t) :
Thus we have
@V 1 2 2 @ 2V
+
S
+
2
@t 2
@S
@ 2V
1 2 @ 2V
@V
Sq
+ q
+ rS
2
@S@
2 @
@S
rV =
(p
@V
q)
;
@
for some function
(S; ; t) :
Reordering this equation, we usually write
@V 1 2 2 @ 2V
1 2 @ 2V
@V
@ 2V
@V
+
S
+
q
+(
p
q
)
rV = 0:
+
Sq
+
rS
2
2
@t 2
@S
@S@
2 @
@S
@
The function (S; ; t) is called the market price of (volatility ) risk.
The market price of volatility risk
If we can solve the pricing equation on the previous slide then we have found
the value of the option, and the hedge ratios.
@V :
But note that we …nd two hedge ratios, @V
and
@S
@
We have two hedge ratios because we have two sources of randomness that
we must hedge away.
Because one of the modelled quantities, the volatility, is not traded we …nd
that the pricing equation contains a market price of risk term.
What does this term mean?
Let’s see what happens if we only hedge to remove the stock risk.
Suppose we hold one of the option with value V , and satisfying the pricing
equation, delta hedged with the underlying asset only i.e. we have
=V
S:
The change in this portfolio value is
d
=
@ 2V
@ 2V
@ 2V
@V
1 2 2
1 2
+
S
+
qS
+
q
2
@t
2
@S
@S@
2 @ 2
@V
@V
+
dS +
d :
@S
@
!
dt
Because we are delta hedging the coe¢ cient of dS is zero, leaving
d
=
@ 2V
@V
1 2 2
+
S
+
2
@t
2
@S
qS
@ 2V
@ 2V
1
+ q2 2
@S@
2 @
!
dt +
@V
d :
@
Now from the pricing PDE we have
@V 1 2 2 @ 2V
+
S
+
2
@t 2
@S
We …nd that
1 2 @ 2V
@ 2V
+ q
Sq
=
2
@S@
2 @
d
@V
rS
@S
r dt =
@V
(p
q)
+ rV dt + @V
@ d
@
@V
@V
=
(p
q)
dt +
(pdt + qdW2)
@
@
Now simplifying this last term gives
@V
rS
@S
q
(p
@V
q)
+ rV:
@
@V
@V
dt +
qdW2
@
@
r V
@V
S dt
@S
Observe that for every unit of volatility risk, represented by dW2, there are
units of extra return, represented by dt. Hence the name ‘market price of risk.’
The return on this partially hedged portfolio in excess of the risk-free return is
@V
q
( dt + dW2)
@
Returning to the pricing equation
1 2 @ 2V
@V
@V 1 2 2 @ 2V
@ 2V
@V
+
S
+ q
+(p
q)
rV = 0:
+ Sq
+rS
2
2
@t 2
@S
@S@
2 @
@S
@
The quantity p
q is called the risk-neutral drift rate of the volatility.
Recall that the risk-neutral drift of the underlying asset is r and not .
When it comes to pricing derivatives, it is the risk-neutral drift that matters
and not the real drift, whether it is the drift of the asset or of the volatility.
stochastic volatility: an example for particular value of p; q;
The option price is shown for varying stock and volatility.
This is a snapshot at a …xed point in time. We notice it looks like a typical
European option.
Note for larger
we have greater curvature (i.e. larger di¤usion).
In addition to the model for GBM we have SDE for volatility, where v = 2:
The equations look nicer expressed in terms of the variance (important quantity).
Many volatility models are of the form
dv = A (v ) dt + cv dW2;
for some value
and mean reverting drift A (v ) ; where the variance v = 2:
In the presence of a continuous dividend yield, the earlier PDE can be written
as
p
@V 1 2 @ 2V
@ 2V
1 2 @ 2V
+ vS
+ vSq
+ q
+(r
2
2
@t 2
@S
@S@
2 @
D) S
@V
@V
+cv
@S
@
rV = 0:
Popular Models
GARCH - di¤usion: Generalized Autoregressive Conditional Heteroskedasticity.
A commonly used popular discrete time model in econometrics. It can be
turned into the continuous time limit of many GARCH-processes by the following SDE
dv = (a
bv ) dt + cvdW2:
The popularity lies in the ease with which the positive valued parameters a; b
and c can be estimated, hence allowing the pricing of options.
There is a mean reverting drift with speed b and mean rate a=b: c is the vol
of vol which sets the scale for the random nature of volatility. In the case
a = b = 0; the GARCH di¤usion model reduces to the log-normal process
without drift in the Hull and White (1987) model.
Given v (0) > 0;
v (t) = v (0) e
b+ 21 c2 t+cWt
Heston:
+a
Z t
0
e
b+ 12 c2 (s t)+c(Wt Ws )
ds:
He takes
dv =
(m
v ) dt +
p
vdW2
Also called the square root model because of the term in the di¤usion - which
gives a closed form solution, hence the popularity. This means it is easier to
calibrate. Heston takes 6= 0: In this model the the process is proportional to
the square root of its level.
Must be comfortable with Complex Analysis Methods, as it requires the use of
Fourier Transforms.
3/2 model:
Pronounced the three-halves model because of the 3=2 power in the di¤usion.
dv = v (a
bv ) dt + cv 3=2dW2
Again mean reverting - the existence of a Closed-form solution makes it a
popular model. But note the mean reverting and volatility parameters are now
stochastic.
See Alan Lewis’ book on Option valuation under stochastic volatility, where
he presents analytical solutions for this model.
This does a supposedly better job of calibrating than Heston, although Heston
is more popular.
Hull & White
dv
= dt + dW2
v
No mean reversion. They take = 0: Note the lognormal structure hence it
can grow inde…nitely.
Stein & Stein
d =
(
The model allows mean-reversion but
= 0:
m) dt + dW2
can become negative. They take
Ornstein-Uhlenbeck process:
This model is expressed in terms of the log of the variance.
Writing y = log x
dy = (a
by ) dt + cdW2
Already seen the O-U-P interest rate model (looks very similar). This model
matches data well.
This has a steady state distribution which is lognormal.
A closed form solution does not exist so requires numerical treatment.
The Heston Model
In his model the variance follows a mean-reverting square root process, …rst
used by Cox-Ingersoll-Ross in 1985 to capture the dynamics of the spot rate
where the mean reversion rate m > 0; and the speed > 0: The vol of vol
> 0:
p
dS = (
D) Sdt + vSdW1;
dv =
(m
v ) dt +
p
vdW2
Solving problems numerically is simple (FDM or Monte Carlo). In the case of
MC take the stock drift as (r D) :
In order for the mean-reverting square root dynamics for the variance to remain
positive, there are a number of analytical results available. In particular is the
Feller condition, i.e. if
m
2
then the variance process cannot become negative. If this condition is not
satis…ed then the origin is attainable and strongly rejecting so that the variance
process may attain zero in …nite time, without spending time at this point.
In deriving the PDE for Heston, he takes
f (S; v; t) =
(m
v) +
(S; v; t)
p
v
giving the following pricing PDE for the option U (S; v; t)
@U
1 2 @ 2U
+ vS
+
2
@t
2
@S
(m
v)
(S; v; t)
1 2 @ 2U
@ 2U
+
v 2+
vS
@S@v 2
@v
p
v
@U
@U
+ rS
@v
@S
rU = 0
Consider the pricing of a call option subject to the …nal condition C (S; v; T ) =
max (ST E; 0) with the following boundary conditions
C (0; v; t)
@C
lim
(S; v; t)
S!1 @S
@C
@C
@C
+ rS
+ m
@t
@S
@v
lim C (S; v; t)
v!1
= 0
= 1
= rC
= S
How to use Heston
There are four parameters in the model, speed of mean reversion, level of mean
reversion, volatility of volatility, correlation. That is b; a=b; c; respectively.
And also potentially a market price of volatility risk parameter.
The main four parameters can be chosen by matching data or by calibration.
Experience suggests that calibrated parameters are very unstable, and often
unreasonable. (For example, the best …t to market prices might result in a
correlation of exactly 1:)
Consider calibrating. Suppose
Parameters Today Next week
a=
14
p 487
12
b=
29
c=
0:01
1000
=
0 :6
3
so a somewhat exaggerated sarcastic example, but nevertheless shows that
when recalibrating it hasn’t worked - the parameters which were …xed are
totally di¤erent!
The Heston model with jumps
Increasingly popular are stochastic volatility with jumps models (SVJ).
Jump models require a parameter to measure probability of a jump (a Poisson
process) and a distribution for the jumps.
Also have SVJJ - jumps in the stock and jumps in the volatility.
Pros: More parameters allow better …tting. The jump component of the model
has most impact over short time scales.
Therefore use longer-dated options to …t the stochastic volatility parameters
and the shorter-dated options to …t the jump component.
Cons: Mathematics slightly more complicated (and again we must work in the
transform domain).
Hedging is even harder when the underlying stock process is potentially discontinuous.
People also looking at stochastic correlation models.
Whilst there is no such thing as the perfect model, you can always pretend to
have the ideal one by introducing more parameters.
More parameters means more quantities to calibrate.
Case Study: The REGARCH model and its di¤usion limit
REGARCH = Range-based Exponential GARCH
Although a closed form solution does not exist, a fairly nice model which looks
very plausible.
‘Range-based’ refers to the use of the daily range, de…ned as the di¤erence
between the highest and lowest log asset price recorded throughout the day,
rather than simply the closing prices.
‘Exponential’refers to modelling the logarithm of the variance.
Di¤usion limits exist for all GARCH-type of processes. That is, they can be
expressed in continuous time using stochastic di¤erential equations.
(This is achieved via ‘moment matching.’ The statistical properties of the
discrete-time GARCH processes are recreated with the continuous-time SDEs.)
REGARCH is another econometrics discrete time model, but can be turned
into the following three-factor model:
dS =
Sdt + 1SdW0
d (log 1) = a1 (log 2 log 1) dt + b1dW1
d (log 2) = a2 (c2 log 2) dt + b2dW2:
(a)
(b)
(c)
This is a three-factor (higher dimensional) model, with two volatilities.
represents the actual (short term) volatility of the asset returns, which is
stochastic.
1
The 2 represents the (longer term) level to which
stochastic.
1
reverts, and is itself
What are the dynamics of this model?
We have the usual GBM random walk for the stock given by (a) which has
actual volatility 1. This is short term.
Note from (b) that the log of 1 mean reverts to log 2: So rather than
being constant, it is ‡uctuating and 1 is chasing that.
2
From (c) we observe that 2 reverts to a constant mean c2:
For pricing options we must replace these SDEs with the risk-neutral versions:
dS = rSdt + 1SdW0
d (log 1) = a1 (log 2 log 1
b1=a1) dt + b1dW1
d (log 2) = a2 (c2 log 2
b2=a2) dt + b2dW2:
The
terms represent the market prices of risk.
The a and b coe¢ cients and the correlations between the three sources of
randomness give this system seven parameters.
These parameters are related to the parameters of the original REGARCH
model and can be estimated from asset data.
Example: let’s look at some parameters.
a1 = 56:6; b1 = 1:138; a2 = 2:82; b2 = 0:388; c2 =
0 :)
1:25. ( 1 = 2 =
is very rapidly mean reverting to the level of 2. This is a ‘short-term’
volatility. The time scale for mean reversion is about one week.
1
2,
the ‘long-term volatility, reverts more slowly, over a period of about six
months.
a1 is the speed for log 1: The bigger this is, the faster the reversion to log 2:
a1dt is non-dimensional therefore a1 has dimensions of 1=time =) 1=a1 has
dimensions of time.
So a time scale of approximately 1 week, for log 1 to mean revert.
b1
b2; volatility of d (log 1) much greater than d (log 2)
chasing.
which it is
1=a2 is approximately 0:5 years, so it takes log 2 6 months to revert back to
its (long term) mean.
How do you solve these equations?
Monte Carlo: The solutions of the two-factor partial di¤erential equations
you get with stochastic volatility models can still be interpreted as ‘the
present value of the expected payo¤.’ So all you have to do is to simulate
the relevant random walks for the underlying and volatility (risk neutral)
many times, calculate the average payo¤ and then present value it.
Finite di¤erences: The partial di¤erential equations can still be solved by
…nite di¤erences but you will need to work with a three-dimensional grid.
Pros and cons of stochastic volatility models
Pros:
Evidence (and common sense) suggests that volatility changes, possibly
randomly
More parameters means that calibration can be ‘better’
Cons:
As with any incomplete-market model hedging is only possible if you believe in the market price of (volatility) risk
Jump Di¤usion Models
Introduction
Some of the ideal assumptions of the classic Black-Scholes framework continue
to be addressed in this chapter. Brownian motion has been the canonical
random process driving asset price models. A basic property of Brownian
motion is that it has continuous sample paths. It follows a Gaussian distribution
whose thin exponentially decaying tails make large changes in the underlying
less probable than actually observed in the market. This fact that it often …ts
…nancial data very poorly is widely acknowledged.
An observation when pricing derivatives is the fact that the underlying ocassionally jumps. The use of processes with jumps have become increasingly popular.
Their detailed practice has already been seen in modelling credit events (jumps
to default) although given the extreme market moves of 2008, they may well
become more common in other asset classes as well. It is important to note
how ever that large moves are very rare occurrences.
Jump processes have discontinuous sample paths and, therefore, they allow for
large sudden moves in the underlying price process. They can also capture
skewness and excess kurtosis in price returns.
So far, our model for asset prices has been
dS = A (S; t) dt + B (S; t) dW
with the usual properties E [dW ] = 0 and V [dW ] = dt: As dt ! 0; it gives
a continuous realisation of the random walk for S:
We cannot always rely on complete markets.
In complete markets we can hedge derivatives with the underlying in such a
way as to eliminate risk.
Most Quant Finance books deal with the Black Scholes model or Binomial
Model which are examples of Complete Markets Models.
If markets are complete then derivatives are redundant because we can replicate
them using the underlying
=V
S)V =
+
S
The whole purpose of derivatives is that markets are incomplete!
In …xed income we model the spot rate r which is random, but we can’t trade
it so can’t use it to get rid o¤ risk.
This is the idea underpinning derivatives theory, i.e. dynamic/delta hedging.
The presence of jumps means we cannot hedge continuously because we need
a continuous process with which to hedge. We will use the Poisson Process
for modelling jumps.
Discontinuous in practice often refers to a move which is signi…cantly large so
that we can’t hedge our way through it.
The foundations of Mathematical Finance are based upon the idea of continuous hedging - so if stock is not continuous - then we cannot hedge.
Equally if we can’t hedge quickly, the asset path may as well be discontinuous.
To model a discontinuous realization we need a Poisson process or jump
process.
This gives the building block for the jump-di¤usion model for an asset price.
One simple way to represent jumps is using this Poisson process.
This is an example of a counting process.
A random process fq (t)gt 0 is called a counting process if q (t) represents
the total number of occurrences that have taken place in the interval [0; t] :
q (t) is an integer value quantity with q (0) = 0:
We will start to build up the theory using this process in conjunction with
Brownian motion (Itô calculus).
The Poisson process is also used in Credit Modelling.
Usual notation to use is dq (t) with the following de…nition
dq =
(
1
0
with P = dt
with P =1
dt
Thus in each interval either dq (t) stays …xed, or it increases by 1:
So we think of dq as a Poisson counter.
The parameter is called the intensity of the Poisson process. The larger it
is, the greater likelihood there is of a jump.
The scaling of the probability of a jump with the size of the time step dt is
crucial in making the resulting process ‘meaningful,’ i.e. there being a …nite
chance of a jump occurring in a …nite time, with q remaining …nite. This is a
classic Poisson process.
q is the integral of dq
q
3
2
1
t
This is a typical representation of a counting process.
Properties of the Poisson process
A counting process q (t) is called a Poisson process with non-negative intensity (or mean arrival rate) of an event in a time interval dt if
q (0) = 0;
q (t) has independent increments.
The number of jumps in a …nite time horizon t has a Poisson distribution with
parameter t: Then
P [q (t) = n] = exp (
( t)n
; n = 0; 1; 2; :::
t)
n!
E [q (t)] =
V [q (t)] =
t
t
We can propose the following model for S
dS = c (S; t) dq
where c (S; t) itself can be unpredictable so that both the size and timing of
the jumps is random.
However a more sensible and realistic model is to use a jump di¤usion version
of Geometric Brownian Motion, i.e.
dS = a (S; t) dt + b (S; t) dW + c (S; t) dq:
So a model that follows GBM most of the time and every now and again, there
is a jump. Since we are interested in the stock return it makes sense to write
dS
= dt + dW + (J
S
which is the jump di¤usion model.
1) dq
The two basic building blocks of every jump-di¤usion model are the Geometric Brownian motion (the di¤usion part) and the Poisson process (the jump
component).
We assume that the Brownian motion and Poisson process are uncorrelated.
So there are two sources of risk: dW; dq:
J is a random number with property E [J ] = 1:
Most of the time dq = 0; so we have di¤usion.
Occasionally at random intervals there is a contribution from dq when it takes
value one and then there is a jump because it is big.
When dq = 1
S + dS ! S + (J
1) S = JS
So S goes immediately goes to the value JS: Hence
dS = JS
As an example if J = 0:9 then S
S = (J
1) S
! 0:9S; i.e. a 10% fall.
So J is a factor which determines what happens to assets when there is a
jump.
J < 1 =) fall in value
J > 1 =) rise in asset
J = 0 =) stock falls to zero
J can be random with its own distribution. There are a number of parameters
here: ; ; ; J:
J could follow any distribution with its own set of parameters, so plenty of
scope for calibration/data …tting.
A convenient form for J is lognormal, so
1 2
J
E [log J ] = e 2
V [log J ] = 2J
The advantage is that closed form solutions are possible (see Merton’s argument later). Consider the following example.
In anticipation of using Itô calculus, we need a framework for extending to
Poisson.
When dq = 0 we know.
1 2
dt + dW
2
If dq = 1, S ! JS; so in addition to the expression above we have log S
log (JS ) = log S + log J:
d (log S ) =
!
So when dq = 1 we have d (log S ) = usual Itô terms plus log J; which can be
written compactly as
1 2
d (log S ) =
dt + dW + (log J ) dq;
2
when dq = 1; we "switch on" the jumps.
Hedging options when there are jumps
Now start building up a theory of derivatives in the presence of jumps.
Usual construction of a portfolio by holding the option and
(in the usual way):
= V (S; t)
of the asset
S:
Across a time step dt the change is
d
=
@ 2V
!
1 2 2
@V
@V
+
S
dt
+
@t
2
@S 2
@S
+ (V (JS; t) V (S; t)
(J
( Sdt + SdW )
1) S ) dq:
Again, this is a jump-di¤usion version of Itô.
How do we get the second line in the expression above?
Before jump: V (S; t)
S:
After jump: V (JS; t)
JS; because S has jumped to JS:
So jump in portfolio is V (JS; t)
S (1 J ) :
JS V (S; t)+ S = (V (JS; t)
V (S; t))+
That is, the jump in the portfolio equals the jump in option price and jump in
stock.
The risk sources here are dW; dq and potentially J . Yet we only have one
delta term with which to hedge.
Hence Incomplete Markets.
If there is no jump at time t so that dq = 0, then we could have chosen
= @V =@S to eliminate the risk.
If there is a jump and dq = 1 then the portfolio changes in value by an O(1)
amount, that cannot be hedged away.
In that case perhaps we should choose
to minimize the variance of d .
This presents us with a dilemma.
We don’t know whether to hedge the small(ish) di¤usive changes in the underlying which are always present, or the large moves which happen rarely.
Let us pursue both of these possibilities.
Hedging the di¤usion
If we choose
@V
@S
we are following a Black-Scholes type of strategy, hedging away the di¤usive
movements.
=
The change in the portfolio value is then
d
=
@ 2V
1 2 2
@V
+
S
@t
2
@S 2
!
dt +
@V
dq:
@S
The portfolio now evolves in a deterministic fashion, except that every so often
there is a non-deterministic jump in its value.
V (JS; t)
V (S; t)
(J
1) S
Merton’s Approach
One classic approach is Merton’s 1976 model who argued that if the jump
component of the asset price process is uncorrelated with the market as a
whole, then the risk in the discontinuity should not be priced into the option
as it is diversi…able (there is no excess reward for it).
In other words non systematic risk is not rewarded on average, so E [d ] =
r dt:
Recall since there is uncertainty present there should be some compensation
for taking risk.
Merton argued that if the dW is eliminated then there should be no compensation for the dq component:
In other words, we can take expectations of this expression and set that value
equal to the risk-free return from the portfolio
E [d ] = r dt;
where
E [() dq ] = E [()j jump occurs dq ] : dt + E [()j no jump dq ] : (1
@V
1 2 2 @ 2V
@V
+
S
+
rS
@t
2
@S 2
@S
E [V (JS; t)
= 0;
V (S; t)]
dt)
rV +
@V
S E [J
@S
1]
where E [ ] is the expectation taken over the jump size J; which can also be
written
E [X ] =
Z
xp (J ) dJ;
where p (J ) is the pdf for the jump size.
The equation is of the form
LBS (V ) +
Z 1
0
V (JS; t)
V (S; t) p (J ) dJ = 0;
i.e. a PIDE (partial integro-di¤erential equation).
If J is known then just drop the E [ ] : So the original Black Scholes terms plus
a new part.
As an example
E [V (JS; t)] =
Z 1
0
V (JS; t) p (J ) dJ:
V now depends on all stocks when there are jumps between 0 and 1:
Aside: Are we working with real or risk-neutral expectations?
At the moment real (Merton’s argument), but later we’ll look at the concept
of risk neutrality when there are jumps.
This is a pricing equation for an option when there are jumps in the underlying.
The important point to note about this equation that makes it di¤erent from
others we have derived is its non-local nature.
That is, the equation links together option values at distant S values, instead
of just containing local derivatives.
Naturally, the value of an option here and now depends on the prices to which
it can instantaneously jump.
There is a simple closed-form solution of this equation in a special case.
That special case if when J is lognormally distributed. i.e. the logarithm of J
is Normally distributed.
To solve put
S
log
= x
E
J = e y
which gives
PDE +
Z 1
1
V (x
y; t) f (y; t) dy = 0:
Solve this using a Fourier Transform in x:
If the logarithm of J is Normally distributed with standard deviation and
‘mean’k = E[J 1] then the price of a European non-path-dependent option
can be written as
1
X
where
e
0 (T
t)
0 (T
t) n
n!
{z
}
n=0 |
= Probability of getting n jumps
VBS (S; t; n; rn) ;
n 02
n log (1 + k)
= (1 + k) ;
=
+
and rn = r
k+
;
T t
T t
and VBS is the Black-Scholes formula for the option value in the absence of
jumps.
0
2
n
2
So it is a Black-Scholes pricing formula for 0; 1; 2; :::::: jumps.
This formula can be interpreted as the sum of individual Black–Scholes values
each of which assumes that there have been n jumps, and they are weighted
according to the probability that there will have been n jumps before expiry.
There are 3 parameters we could calibrate.
Method 2: Hedging the jumps
In the above we hedged the di¤usive element of the random walk for the
underlying.
Another possibility is to hedge both the di¤usion and jumps ‘together.’
For example, we could choose
to minimize the variance of the hedged portfolio, after all, this is ultimately what hedging is about.
So let’s return to the d
equation.
The change in the value of the portfolio with an arbitrary
(ignoring higher order terms),
d
@V
=
@S
+(
is, to leading order
( Sdt + SdW )
(J
1) S + V (JS; t)
V (S; t)) dq + :::
Square this term and take expectations, then subtract o¤ the square of E [d ] :
The variance in this change, which is a measure of the risk in the portfolio, is
@V
V [d ] =
@S h
+ E (
2
2 S 2 dt +
(J
1) S + V (JS; t)
V (S; t))
2
i
dt + :::
which is to leading order (2 terms) - a di¤usive part and a jump component.
Putting
= @V
@S only eliminates the di¤usive part, not the jumps.
This is minimized by the choice
=
E [(J
This is obtained as follows:
1) (V (JS; t)
h
S E (J
V (S; t))] + 2S @V
@S :
i
1)2 + 2S
@
(V [d ]) = 0
@
which gives
When
hedge.
for the minimum. This choice of
gives the least variance.
= @V
@S ; the usual Black-Scholes
= 0; the expression collapses to
If we value the options as a pure discounted real expectation under this besthedge strategy then we …nd that
@ 2V
2
@V
1 2 2
@V
+
S
+
S
@t
2
@S 2
@S
+ E V (JS; t)
( + k
d
J 1
( + k
d
V (S; t) 1
= 0
where
h
d = E (J
Note how this choice brings in :
1)
2
i
+ 2:
r)
!
r)
rV
Here we are not getting rid of dW; but minimizing risk so we are still left with
so need to measure this term.
Often happens when moving away from Complete Markets.
What about risk neutrality?
Does the concept of risk neutrality have any role when there are jumps?
N.B. The above uses ‘real’expectations.
Let’s see a special case, known jump size, J .
So start with
dS = Sdt + SdW + (J
1) Sdq:
but with J given.
There are now two sources of risk (there were three before), dW and dq . (No
J risk)
Let’s see if we can eliminate risk by having two hedging instruments, the stock
and another option.
(You will recall this from the stochastic interest rate lecture and will see it
again in stochastic volatility modelling.)
Construct a portfolio of the option and
option, V1 :
= V (S; t)
of the asset, and
S
1 V1
1
of another
Doing Itô’s Lemma gives the change in the portfolio as
d
=
+ (V (JS; t)
@ 2V
2
2 S 2 @ V1
@S 2
@V
1 2 2
@V1 1
+
S
+
1
2
@t
2
@S
@t
2
@V1
@V
+
( Sdt + SdW )
1
@S
@S
V (S; t)
(J
1) S
To eliminate dW terms choose
@V
@S
and to eliminate dq terms choose
V (JS; t)
V (S; t)
(J
1 (V1 (JS; t)
1
=
(J
dt
V1 (S; t))) dq
@V1
=0
1
@S
1) S
1 (V1 (JS; t)
V1 (S; t)) = 0:
We obtain the messy expressions
(J
!!
1) S @V
@S
V (JS; t) + V (S; t)
1
1) S @V
@S
V1 (JS; t) + V1 (S; t)
and
=
@V
@S
(J
@V1
@S
(J
1) S @V
@S
V (JS; t) + V (S; t)
1
1) S @V
@S
V1 (JS; t) + V1 (S; t)
:
All risk is now eliminated, so set return on portfolio equal to risk-free rate.
End result:
@V
@t
(J
=
@V1
@t
(J
2
+ 12 2S 2 @@SV2 + rS @V
rV
@S
V (JS; t) + V (S; t)
1) S @V
@S
@ 2 V1
1
2
2
1
+ 2 S @S 2 + rS @V
rV1
@S
:
@V1
1) S @S
V1 (JS; t) + V1 (S; t)
Same functional form on each side.
So one equation in two unknowns . LHS is independent of V1; RHS is inde-
pendent of V:
@V
@t
(J
=
2V
1
@
2
2
+ 2 S @S 2 + rS @V
rV
@S
1) S @V
V (JS; t) + V (S; t)
@S
universal quantity, independent of option type
0:
=
Final equation is
@V
1 2 2 @ 2V
@V
+
S
+ rS
@t
2
@S 2
@S
+
0
V (JS; t)
V (S; t)
rV
(J
@V
1) S
@S
= 0:
This is the same equation as before but with risk-neutral 0 instead of real .
How do you solve these equations?
Monte Carlo: The solutions of the partial integro-di¤erential equations you get
with jump-di¤usion models can still be interpreted as ‘the present value of the
expected payo¤.’ So all you have to do is to simulate the relevant random
walk for the underlying (risk neutral) many times, calculate the average payo¤
and then present value it. As always!
Finite di¤erences: The partial integro-di¤erential equations can still be solved
by …nite di¤erences but the method will no longer be ‘local’since the governing
equation contains integrations over all asset prices.
Pros and cons of jump-di¤usion models
Pros:
Evidence (and common sense) suggests that assets can jump in value
Jump models can capture extreme implied volatility skews (such as seen
close to expiration)
More parameters means that calibration can be ‘better’
Cons:
The foundations are a bit shaky (can’t hedge, hedge di¤usion or minimize
risk, real versus risk neutral)
Fractional Brownian Motion
Fractional Brownian motion written fBm with Hurst exponent H 2 (0; 1) is a
centred Gaussian process (Xt)t 0 such that
8s; t
0 : E [XsXt] =
h
1 s2H
2
+ t2H
js
tj
2H
i
The fBm is a generalisation of a standard Brownian motion that allows its
increments to be correlated with
H = 21 Brownian motion (H = 0:5 below)
H > 21 increments are positively correlated so nice smooth paths (H = 0:8 below)
H < 21 increments are negatively correlated giving something very rough
(H = 0:2 below)
Cheridito (2001) and Dieker (2004) respectively.
Motivated by empirical studies, several authors have studies …nancial models
driven by the fBm
Fractional stochastic vol models
Fractional Black-Scholes model
Fractional stochastic volatility models (see Comte and Renauld (1998) or
Comte, Coutin and Renault (2003)) explain better the long-time behaviour
of the implied volatility. The fBm (and then the volatility) are not Markovian,
and this becomes a strong di¢ culty to study and to put these models into
practice (the usual techniques assume the Markov property).
Models driven by the fBm
Consider the fractional Black-Scholes model for a bond (Bt) and a stock (St)
(H > 1=2) :
Bond dynamics: dBt = rBtdt;
follows the SDE:
B0 = 1 ;
0
t
dSt = Stdt + Std XtH ; S0 = S > 0; 0
T A stock S which
t
T
where ; are constants and
6= 0: Here we introduce the Wick-calculus
universe based on the Wick product, which is denoted by the symbol .
From the market de…nition we have the following explicit solutions
Bt = ert
and
St = S0 exp
XtH
+ t
1 2 2H
t
2
A portfolio (trading strategy) is a pair of progressively-measurable processes
= ( t; t) where t and t in turn represent the
amount invested in the bank account and the shares of stocks. The value
process of such a portfolio is:
Vt ( ) = tBt + tSt
Wick-self …nancing: The value process is assumed to follow
and the prtfolio
Vt ( ) = tBt + t St
= ( t; t) is called self-…nancing if for all t 2 [0; T ]
dVt ( ) =
=
+ t d St
trBtdt +
tStdt +
tdBt
H
tStd Xt
Applying a fractional form of Girsanov with a risk-neutral measure QH
dVt ( ) = rVt ( ) dt +
fH
S
d
X
t
t
t
where
dQH
= exp
dP
and
Z T
0
qt (s; t) dt =
fH = X H +
Note: X
t
t
r
Z T
0
r
qtd XtH
1 2
jqj
2
!
holds for all s 2 [0; T ] :
is a fBM under the measure QH :
Under QH the stock price follows
1 2 2H
t
:
2
Ultimately the price of a European call option at initial time t = 0 was obtained:
St = S0 exp
fH + rt
X
t
V0 = e rT EQH [VT ( )]