Contents
1 Introduction 1
2 Barrier Options 4
3 CEV Di¤usion 6
4 Knock Out Options Under The CEV Model 8
4.1 Pricing Rebates. . . 17
5 Laplace Transform 20
6 Numerical Implementation 22
6.1 Numerical Results of the C++ Program . . . 23
7 Conclusion 25
A Appendix - Proof of Proposition 2 26
B Appendix - Closed-Form for Integrals I (K; A; B) and J (K; A; B) 37
C Appendix - Proof of Proposition 5 48
D Appendix - C++ Program 53
D.1 The Gamma Function . . . 54 D.2 Whittaker Functions . . . 61 D.3 Modi…ed Bessel Functions . . . 66
D.4 Function for the solutions and . . . 89
D.5 Function for the Wronskian and (A; B) . . . 96
D.6 Laplace transform of the Double Knock Out Barrier option . . . 98
D.7 Program for Euler Abate and Whitt method . . . 99
D.8 The main funtion . . . 102
Abstract
The aim of this thesis is to …nd a closed-form solution for evaluation of knock-out barrier options under the Constant Elasticity of Variance model. To achieve this goal, the pricing solution of the double knock out barrier option is written in terms of Laplace Transforms, that are inverted using the Abate and Whitt Euller method. Then the analytical solution is implemented using a C++ program.
Key words: Path-dependent options, barrier options, Constant Elasticity of Variance model, Laplace Transforms, Abate and Whitt Euller method.
1
Introduction
The goal of this thesis is to achieve an analytical solution to price barrier options under the Constant Elasticity of Variance (CEV) process.
Option pricing is one of the most important areas of modern …nance. Financial histori-cians attribute the …rst attempts to model stochastic processes to Bachlier. In his PhD thesis entitled The Theory of Speculation, published in 1900, Bachlier discussed the use of the Brownian motion to evaluate stock options. This thesis is historically the …rst paper to use advanced mathematics in the study of …nance (Courtault et al , 2000). But the model that revolutionized the …nancial markets was the Black and Scholes (1973) model. The Black and Scholes (1973) model assumes that the price of the underlying asset follows a geometric Brownian motion, and as a consequence the underlying asset follows a lognormal process. Using the assumptions of Black and Scholes (1973), Merton (1973) was the …rst to derive a closed-form solution for the standard European call and put. Merton (1973) was also the …rst to get a closed-form solution for a down-and-out European call option. This was the …rst step to the development of further studies on the pricing of barrier options. Rubin-stein and Reiner (1991) and Carr (1995) derive analytical solutions for all types of single barrier options. The study of barrier options using a binomial method was presented by Boyle and Lau (1994). Kunimoto and Ikeda (1992) covered the valuation of double barrier options expressing the probability density as a sum of normal density functions. German and Yor (1996) derive expressions for the Laplace Transform of a double barrier option price, that is inverted analytically. Later, Schroder (2000) derives the formulas of Kunimoto and Ikeda (1992) inverting the Laplace Transforms derived in German and Yor (1996). Pelsser (2000) derived analytical expressions to price double barrier options inverting the Laplace Transform using a contour integration method.
One of the assumptions of the Black and Scholes (1973) model is that the volatility of the underlying asset price is constant; however, the market shows that such volatility is not constant, and does not follow a lognormal distribution ( Hull, 2002). Black (1976) and Christie (1982) consider the e¤ect of …nancial leverage on the variance of the stock. A fall in
the stock price increases the debt / equity ratio of the …rm, and so increases the risk and the variance of the stock. Empirical evidences shows that the constant volatility is inapplicable in real market situations. Schmalensee and Trippi (1978) …nd a strong negative relationship between stock prices changes and changes in implied volatility. When the implicit volatility of the asset is analyzed we see that volatility is not constant. The plot of implied volatility, known as volatility smile (see …gure in Hull (2002), p 437) evidences that the volatility of the asset is related to the strike price. This e¤ect of volatility smile is not captured by the Black and Scholes (1973) model, and therefore new models were studied with the intention of capturing this e¤ect evidenced by the market. The Constant Elasticity of Variance (CEV) model is one model that incorporates the volatility smile. The CEV model was developed by Cox (1975) and in this model the volatility of the underlying asset is related to his price level. This fact is consistent with empirical observations in which stock volatility changes with stock prices oscillations. Cox (1975) was the …rst to price vanilla European call and put options under the CEV process. Cox (1975) studied the CEV model for the case 2, and Emanuel and MacBeth (1982) and Schroder (1989) extended this analysis to the case > 2. Originally Cox (1975) has restricted the parameter to the interval [0; 2], but Reiner (1994) and Jackweth and Rubinstein (1998) found out that the values of the CEV elasticity implicit in the post crash S&P500 stock index options prices were as low as = 4 or = 6.
The comparison between the Black and Scholes (1973) model and the CEV model has been made by several authors. Macbeth and Merville (1980) compared the two models and concluded that prices obtained by the CEV model were closer to market quotes than the ones obtained by Black and Scholes (1973), specially in case of < 2. They also concluded that the volatility of the underlying asset decreases when the price increases, and this matches the speci…cations of the CEV model that assume the volatility of the asset as a function of the asset price. Boyle and Tian (1999) concluded that the di¤erence of prices between the Black and Scholes (1973) and the CEV models is greater for path dependent options than for standard options.
The pricing of barrier options under CEV model is also well established. Boyle and Tian (1999) used a trinomial lattice to price single and double barrier options under the CEV
model. Later Davidov and Linetsky (2001) derived closed-form expressions for the Laplace Transforms of single and double barrier options prices in time to maturity. Davidov and Linetsky (2003) developed eigenfunctions expansions for single and double barrier options. These eigenfunctions expansions were used to invert the Laplace transform of Davidov and Linetsky (2001). All the above derivations assume the model parameters as volatility, interest rate and dividend yield are constant. However, the pricing of barrier options was also studied using time dependent parameters: Lo, Yuen and Hui (2000) priced path dependent options under the CEV model using Lie -algebraic techniques. Lo, Tang, Ku and Hui (2009) derived the analytical kernels of the pricing formulae of the CEV knockout options with time-dependent parameters for a parametric class of moving barriers.
This thesis considers the CEV model with time independent parameters, and is based on the results of Davidov and Linetsky (2001): I will derive the Laplace transform of double knock out barrier options (and also of single barrier knock out options). Then, I will imple-ment the computation of the inversion of the Laplace Transform in a C++ program using the Abate and Whitt (1995) algorithm.
This thesis is further organized as follows: In Section 2 the barrier options and contracts will be introduced. In Section 3 the CEV di¤usion and its assumptions will be described. In section 4, starting from the de…nition of …nal payo¤ of a double barrier option, the analytical expression of the Laplace Transform of such payo¤ will be computed. The same thing is done using the de…nition of the …rst hitting or exit time, and the analytical expression of the Lapalce Transform of the rebate price will also be computed. In Section 5 the Abate and Whitt (1995) Euler-type method to invert Laplace Transforms will be presented, and the parameters of the method used to invert the Laplace Transform will be de…ned. In Section 6 the analytical closed-form solution deduced in the other chapters is implemented using a C++ program. This program will then be used to price down-and-out, up-and-out and double knock out call options with di¤erent strikes, and di¤erent betas. In Section 7 the conclusions of this work will be presented.
2
Barrier Options
Option trading has become increasingly popular over these years because of its ‡exibility to ful…ll the needs of investors. Options with more complex payo¤s than standard European or American options are called exotic options. Exotic options where the payo¤ depends on the path followed by the price of the underlying asset are referred as path dependent options. Barrier options are one of the most popular types of path dependent options. Barrier options are options where the …nal payo¤ depends on the values that the underlying asset reaches during the life of the option. As the name suggest, in barrier options we have a pre-de…ned price level, or levels, that in case of being reached will in‡uence the …nal payo¤ of the option. The barrier options are classi…ed in two groups according to the type of the barrier:
Knock Out Barrier Options - these are options where the option becomes inactive if the underlying asset price reaches the value of the barrier. The barrier can be hit at any moment of the option life, and becomes inactive at the moment the barrier is hited. The payo¤ is a standard vanilla option if the barrier never is hited, or nothing (or a rebate value) if the barrier is hited. In knock out options we can …nd:
1. Up-and-out Options: when the barrier is approached from below, that is, the value of the barrier is bigger than the spot price.
2. Down-and-out Options: The value of the barrier is lower than the spot price.
3. Double knock out Options: In this case, there exists up and down barriers that limit the price of the underlying asset. If the price goes out of the interval de…ned by the two barriers, the option becames inactive. The option payo¤ is a standard vanilla option if the barrier is hited, or nothing (or a rebate value) if the barrier is never hited.
Knock In Barrier Options - In this case, the option contract is only activated if the price of the underlying asset reaches the value of the barrier. If that value is never reached during the option life, then the payo¤ is zero. There are three cases of knock in options:
1. Up-and-in Options: The barrier value is lower than the spot price. 2. Down-and-in Options: The barrier value is greater than the spot price.
3. Double knock in Options: There exists up and down barrier levels that limit the price of the underlying asset. If the price touches one of the barriers then the option becomes active.
If knock in options never become active, or if knock out option become inactive, then there is a possibility to pre-establish a rebate value that is paid in these cases. In knock in options the rebate can only be paid at maturity, but in knock out options the price of the underlying asset reaches the barrier at any moment, and so the rebate can be paid at any moment.
There are two ways of monitoring the barrier: continuously or discretely. Continuous monitoring of the barrier happens when the barrier condition is applied at any time prior to maturity. In discrete monitoring we de…ne a discrete time interval, and at the end of this interval we see if the barrier condition is reached, but not at any other time. This thesis is only concerned with the continuous monitoring of the barrier option.
3
CEV Di¤usion
The Black and Scholes (1973) option pricing model assumes that the underlying asset price follows a Geometric Brownian motion
dSt = Stdt + Sdw Q
t ; (1)
where wQt is a standard Brownian motion. This means that the stock price @SStt in the
small interval dt is normally distributed with mean dt and variance 2dt: And it is known that the lognormality assumption does not hold empirically for stock prices. Alternative stochastic processes have been studied and applied to option pricing. Cox (1973) studied a general class of stochastic processes known as the CEV process:
dSt= Stdt + S2dw Q
t ; (2)
where dSt is the change of the stock price S over the short increment of time dt, ; and
are constants, and dwQt is a Wienner process increment.
Note that the di¤erence between equations (1) and (2) consists in the di¤usion term. Notice that the volatility of the percentage change in the underlying asset price in (2) is (S) = S2 1 and is not constant. And it has an inverse relationship with the underlying
asset price S as long as < 2:This is, the volatility is an increasing (decreasing) function of S when > 2 ( < 2).
The CEV model nests, as special cases, other pricing models:
If = 2; then the volatility is the constant ; and so we have the Black and Scholes (1973) model.
If = 1; then we have the Square Root Process that was …rst presented by Cox and Ross (1976) as an alternative di¤usion process for options valuation.
Cox and Ross (1975) de…ne the elasticity factor as , but I will use, along this thesis, + 1 = 2 in (2) for a simpli…cation of the valuation formulas. And so we get:
dSt= Stdt + S +1 t dw Q t ; t 0; S0 = S 0 (3) where:
Q is an equivalent martingale measure (risk neutral probability measure). Under Q the asset price fSt , t 0g is a time-homogeneous, non-negative di¤usion process
fwtQ; t 0g is a standard Brownian motion de…ned on a …ltered probability space
( ; z; fztgt 0; Q):
is a constant ( = r q; where r 0 and q 0 are respectively the risk-free rate and the constant dividend yield).
is a constant elasticity factor. (St) = St
Next some results are presented in order to evaluate knock out barriers options under the CEV model.
4
Knock Out Options Under The CEV Model
All over this thesis the variable t denotes the running time, and it is also assumed that all options are written at time t = 0, and expire at time t = T > 0: The time left behind expiration is denoted by = T t: Suppose the initial asset price is S and the lower and upper barrier levels are L and U , where L < S < U: The …rst hitting and exit times of the barriers can be de…ned as:
First hitting time of the lower barrier:
=L = inff t 0 : St= Lg: (4)
First hitting time of the upper barrier:
=U = infft 0 : St= Ug: (5)
First exit time from an interval between two barriers:
=(L;U )= infft 0 : St2 (L; U)g:= (6)
Using Zhang (1998, p 206-207) the …nal payo¤s of knock out options are:
The …nal payo¤ of a down and out option (PDO) with strike K and no rebate at expiration is (call if = 1, and put if = 1)is equal to
P DO = 11f=L>T g( St K)+: (7)
The …nal payo¤ of a up and out option (PUO) with strike K and no rebate at expiration is (call if = 1, and put if = 1) is equal to
P U O = 11f=U>T g( St K)+: (8)
The …nal payo¤ of a double knock out option (PDKO) with strike K and no rebate at expiration is (call if = 1, and put if = 1)is equal to
P DKO = 11f=(L;U )>T g( St K)+ (9)
=L; =U and =(L;U ) are de…ned in (4), (5) and (6).
11fAg is the indicator function de…ned as 11fAg(x) = 10if x2Aif x =2A x+= max
fx; 0g:
As seeing before in (7), (8) and (9) the …nal payo¤ of a knock out barrier option involves hitting times that can happen at = =L; =U and =(L;U ) with T: From Ingersoll (1987,
p371-372), the value of perpetual claims can help in the valuation of such hitting times. So, the …rst step is to analyze the valuation of three perpetual claims that pay one EUR at times =L; =U or =(L;U ) and have no expiration date. This result will be used later to …nd
an expression for the expected value of …nal payo¤s and also of rebates:
Proposition 1 Suppose the risk neutral asset price process follows the di¤usion (3) and the constant risk free interest rate is r > 0. Then, the prices at t = 0 of the three perpetual claims are (assuming S0 = S):
One EUR paid at =L :
EQ e r=L11
f=L<1g =
r(S) r(L)
; S L (10)
One EUR paid at =U :
EQ[e r=U11
f=U<=0g] =
r(S) r(U )
; S U (11)
One EUR paid at =(L;U ) :
EQ[e r=(L;U )] = r(L; S) + r(S; U )
r(L; U )
; L S0 U; (12)
where (for any 0 < A < B < 1)
r(A; B) := r(A) r(B) r(A) r(B); (13)
and the functions r(S) and r(S) can be characterized as the unique (up a
multiplica-tive constant) solutions of the ordinary di¤erential equation (ODE) 1 2 2 (S)S2@ 2u @S2 + S @u @S ru = 0 ; S 2 (0; 1) (14)
The functions r(S) and r(S) are characterized, and have properties described in
Borodin and Salminen (2002, p.18-19), and r(S) is the increasing solution and r(S) is the decreasing solution.
Proof. For equation (10) and (11):
=L and =U are, respectively, the …rst hitting time of L and U, so using n.10 of Borodin
and Salminen (2002, p. 18) then equations (10) and (11) follow, where r and r can be
characterized as the unique (up to a multiplicative constant) solutions of (14) demanding that
r is increasing and r is decreasing. The boundary conditions of r and r are explained in
Borodin and Salminen (2002, p. 19). Note that =0 = inff t 0 : St= 0g; so this represents
the default. Hence, the de…nition of (11) presume that the upper barrier must be hitted before the default.
For equation (12):
Equation (12) can be rewritten using e r=(L;U ) = e r=L11
f=L<=Ug+ e
r=U11
f=U<=Lg
and from the property of linearity of an expected value (Shreve, 2004, p. 55), EQ[e r=(L;U )] = EQ e r=L11
f=L<=Ug + E
Q e r=U11
f=U<=Lg :
EQ e r=L11
f=L<=Ug satis…es the ODE (14) because it is a Kolmogorov equation. But any
solution of (14) can only be expressed as a combination of the two fundamental solutions of
r and r. Therefore, the solution of (12) will be of the form g(s) = a1 r(s)+ a2 r(s): To
…nd the value of a1 and a2 the boundary conditions in the limits L and U are used. So,
a1 r(L) + a2 r(L) = 1 a1 r(U ) + a2 r(U ) = 0 () a2 = r(U ) r(L) r(U ) r(U ) r(L) a1 = r (U ) r(L) r(U ) r(U ) r(L)
Replacing a1 and a2 in g(S) the solution is:
g(S) = r(U ) r(L) r(U ) r(U ) r(L) r(S) + r(U ) r(L) r(U ) r(U ) r(L) r(S) = r(S) r(U ) r(S) r(U ) r(L) r(U ) r(U ) r(L)
So doing the same for EQ e r=U11
f=U<=Lg , and using the following boundary conditions in
the limits L and U
a1 r(L) + a2 r(L) = 0
a1 r(U ) + a2 r(U ) = 1
we …nd that the solution is
r(L) r(S) r(L) r(S) r(L) r(U ) r(U ) r(L) Consequently, EQ[e r=(L;U )] = r(S) r(U ) r(S) r(U ) r(L) r(U ) r(U ) r(L) + r(L) r(S) r(L) r(S) r(L) r(U ) r(U ) r(L) ; and if r(A; B) = r(A) r(B) r(B) r(A) then:
EQ[e r=(L;U )] = r(S; U ) + r(L; S)
r(L; U )
From Proposition 1 we know that r(S)and r(S) are the fundamental solutions of the
ODE (14), but we still need to …nd the expression of those solutions. The next result gives a closed-form expression for the two fundamental solutions (S)and (S).
Proposition 2 The fundamental solutions (S) (increasing solution) and (S) (decreas-ing solution) of the CEV ODE (14), are (up to a multiplicative constant):
(S) = 8 > > > > > > < > > > > > > : S +12e2x(S)Mk;m(x(S)); < 0; 6= 0 S +12e2x(S)W k;m(x(S)); > 0; 6= 0 S +12I (p2 z(S)); < 0; = 0 S +12K (p2 z(S)); > 0; = 0 ; (15) and (S) = 8 > > > > > > < > > > > > > : S +12e2x(S)Wk;m(x(S)); < 0; 6= 0 S +12e2x(S)M k;m(x(S)); > 0; 6= 0 S +12K (p2 z(S)); < 0; = 0 S +12I (p2 z(S)); > 0; = 0 ; (16)
where Mk;m(x) and Wk;m(x) are the Whittaker functions de…ned in Abramowitz and Stegun
(1972, p 504), and I (p2 z) and K (p2 z) are the modi…ed Bessel functions also de…ned in Abramowitz and Stegun (1972, p 374),with:
x(S) = 2j j j jS 2 (17) z(S) = 1 j jS (18) = sign( ) = 1 if > 0 1 if < 0 (19) m = 1 4j j (20) k = (1 2 + 1 4 ) 2j j (21) = 1 2j j (22)
The Wronskian of the functions and with respect to the scale density of the CEV di¤usion is ! = 2j j (2m+1) 2 (m+k 1 2) ; 6= 0 j j ; = 0 (23)
where (x) is the Euler Gamma function.
Proof. See Appendix A.
Now that the closed-form solutions for (S) and (S) are available, it is possible to price European knock out barrier options. From now on, the focus is on the European double barrier options because the case of the single barrier option (down-and-out and up-and-out) can be obtained from the double barrier option. For instance, the down and out barrier is found if the upper barrier U tends to in…nity, and the up and out barrier option is found if the down barrier L of a double barrier option is zero.
The next results are obtained only for the call case; the put case can be obtained simul-taneously.
To price European double barrier calls we must …nd the expected value of its terminal payo¤. Consequently, from (9) the price is e rtEQ[11
f=(L;U )>T g(ST K)
the expected value of the terminal payo¤ of a knock out double barrier option, its the Laplace Transform that will be obtained, and then inverted. Next result presents the expected value of the terminal payo¤ of a knock out double barrier option in terms of its Laplace Transform:
Proposition 3 The Laplace Transform of EQ[11
f=(L;U )>T g(ST K) +] in time to expiration T is given by: Z 1 0 e TEQ[11f=(L;U )>T g(ST K)+]dT (24) = 1 ! (L; U ) 8 > > > > > > < > > > > > > : (L; S)[ (U )J (K; K; U ) (U )I (K; K; U )], if S K (L; S)[ (U )J (K; S; U ) (U )I (K; S; U )] + (S; U )[ (L)I (K; K; S) (L)J (K; K; S)], if S > K where (A; B) is de…ned in equation (13), ! is the Wronskian as de…ned in Borodin and Salminen (p. 19, 2002), and has the closed-form expression in (23). Moreover,
I (K; A; B) = Z B A (Y K) (Y )m(Y )dY; (25) J (K; A; B) = Z B A (Y K) (Y )m(Y )dY; (26)
where m(Y ) is the speed density of the di¤usion (3).
Proof. Transforming the expected value in to an integral having in mind that 11f=(L;U )>T g; and ST K, and using the Fubbini Theorem:
Z 1 0 e TEQ[11f=(L;U )>T g(ST K)+]dT = Z 1 0 e T Z U K (Y K)p(T ; ST; Y )dY dT = Z U K (Y K) Z 1 0 e Tp(T ; ST; Y )dT dY; (27)
where p(T ; ST; Y ) is the transition density of ST with respect to the speed measure (see
Borodin and Salminen (2002, p.13)).
The Green function de…ned in Borodin and Salminen (2002, p.19) is present in the second integral. So, Z U K (Y K) Z 1 0 e Tp(T ; S; Y )dT dY = Z U K (Y K)G (S; Y )dY (28)
The solution of the Green function in Borodin and Salminen (2002, p.19), with modi…cation introduced by Davidov and Linetsky (2001, p 962) is
G (S; Y ) = m(Y ) W
(S) (Y )if S Y
(Y ) (S)if S > Y (29)
where W is the Wronskian of the function and , which is de…ned by (see Borodin and Salminen (p 19, 2002)):
W = (S) (S) (S) (S); (30)
and (S)and (S)can be expressed only in terms of the fundamental solutions of equation (14) (S)and (S).
From Davidov and Linetsky (2001, p 962) (S) = (S; U ) and (S) = (L; S); where (A; B)is de…ned in (13). And, consequently,
(S) = 0(S) (U ) 0(S) (U ) (31)
(S) = (L) 0(S) (L) 0(S) (32)
Using (30), (31) and (32) the Wronskian - W - is :
W = ( (S) (U ) (S) (U ))( (L) 0(S) (L) 0(S)) ( (L) (S) (L) (S))( 0(S) (U ) 0(S) (U )) = ( (S) (U ) (L) 0(S) (S) (U ) (L) 0(S) (S) (U ) (L) 0(S) + (S) (U ) (L) 0(S) (L) (S) 0(S) (U ) + (L) (S) 0(S) (U ) + (L) (S) 0(S) (U ) (L) (S) 0 (S) (U ) Hence, W = (L) (U )( (S) 0(S) (S) 0 (S)) + (U ) (L)( (S) 0 (S) (S) 0(S)) so, W = ( (S) 0(S) (S) 0(S))( (L) (U ) (L) (U )) (33) = ! (L; U );
where
! = (S) 0(S) (S) 0(S) Then (29) can be rewritten as:
G (S; Y ) = m(Y ) ! (L; U )
(L; S) (Y; U ) if S Y
(L; Y ) (S; U ) if S > Y (34) Going back to (28) and (34):
Z U K (Y K)G (S; Y )dY = = 8 < : RU K(Y K) m(Y ) ! (L;U ) (L; S) (Y; U ) dY if S Y RU K(Y K) m(Y ) ! (L;U ) (L; Y ) (S; U ) dY if S > Y (35)
For the case S K:
Using that if S K and Y 2 [K; U]; then S Y and (35) Z U K (Y K)G (S; Y )dY = Z U K (Y K) m(Y ) ! (L; U ) (L; S) (Y; U )dY passing the factors that do not depend on Y out of the integral, using (13),
Z U K (Y K)G (S; Y )dY = (L; S) ! (L; U ) Z U K (Y K)m(Y )( (Y ) (U ) (Y ) (U ))dY = (L; S) ! (L; U ) (U ) Z U K (Y K)m(Y ) (Y )dY (U ) Z U K (Y K)m(Y ) (Y ))dY
and using the de…nitions (26) and (25)
= 1
! (L; U ) (L; S) [ (U )J (K; K; U ) (Y )I (K; K; U )] (36) For the case S> K :
As S 2 [L; U] (by de…nition of double barrier option), and S > K; using the integral properties in Theorem 1.17 of Apostol (1994, p 98)
Z U K (Y K)G (S; Y )dY = Z S K (Y K)G (S; Y )dY + Z U S (Y K)G (S; Y )dY
As Y 2 [K; S] ) S > Y , and Y 2 [S; U] ) S Y: Applying this to the …rst and second integral above, using (34)
Z S K (Y K)G (S; Y )dY + Z U S (Y K)G (S; Y )dY = Z S K (Y K) m(Y ) ! (L; U ) (L; Y ) (S; U )dY + Z U S (Y K) m(Y ) ! (L; U ) (L; S) (Y; U )dY = (S; U ) ! (L; U ) Z S K (Y K)m(Y ) (L; Y )dY + (L; S) ! (L; U ) Z U S
(Y K)m(Y ) (Y; U )dY
= 1 ! (L; U ) (S; U ) Z S K (Y K)m(Y )( (L) (Y ) (L) (Y ))dY + (L; S) Z U S (Y K)m(Y )( (Y ) (U ) (Y ) (U ))dY = 1 ! (L; U ) (S; U ) (L) Z S K (Y K)m(Y ) (Y )dY (L) Z S K (Y K)m(Y ) (Y )dY + (L; S) (U ) Z U S (Y K)m(Y ) (Y )dY (U ) Z U S (Y K)m(Y ) (Y ))dY and using the de…nitions (25) and (26) we have
= 1
! (L; U )[ (S; U ) [ (L)I (K; K; S) (L)J (K; K; S)] + (L; S) [ (U )J (K; S; U ) (U )I (K; S; U )]]
Therefore, equation (24) arises immediately from this last equation and from equation (36):
The derivation of a closed-form expression for integrals I (K; A; B) and J (K; A; B) is given in Appendix B. Note that the close-form expression for integrals I (K; A; B) and J (K; A; B) deduced in this thesis are di¤erent from the solutions in Davidov and Linetsky (2001, p 964) for the case = 0 and > 0:
Now, to …nd the value in time t=0 of an European double barrier option. To do that we only have to invert the Laplace Transform (24).
4.1
Pricing Rebates
As seen before, rebates are …xed cash amounts that can be paid if the barrier is hited (for knock out options). Many barriers options have a rebate implicit that is paid in case of knock out. To value the cash rebates included in knock-out options it is necessary to value claims that pay one EUR at times =L; =U and =(L;U ): (as seen in Proposition 1) This means that
expectations of the form EQ[11f T ge r ]must be evaluated. To calculate these expectations the Laplace Transforms will be used, like for the terminal payo¤ in Proposition 3 .
Proposition 4 For any > 0; the Laplace transform of the rebate price in time to expiration
T is: Z
1 0
e TEQ[11f T ge r ]dT = 1EQ[e (r+ ) ] (37)
Proof. The Laplace transform of the rebate price is Z 1
0
e TEQ[11f T ge r ]dT: Using the Fubbini theorem:
Z 1 0 e TEQ[11f T ge r ]dT = EQ[e r Z 1 0 e T11f T g]dT = EQ[e r Z 1 e TdT ]; and, using Barrows formula,
= EQ[e r ( 1(0 e )]dT = 1EQ[e (r+ ) ]
The next result gives a closed-form solution for the rebates of knock-out barrier options.
Proposition 5 The close-form solution (R) of the Laplace Transform in time to expiration T of the unit rebate price for knock out barrier option is:
5.1. Down-and-out Barrier Option: R= 8 > > > > > > > > > > > > > > < > > > > > > > > > > > > > > : 1 U (2 ;1+ 1 2 ;x(S)) U (2 ;1+21;x(L)); if < 0; < 0 1 e x(S)M (21 2 ;1+ 1 2 ;x(S)) e x(L)M ( 1 2 2 ;1+ 1 2 ;x(L)) ; if < 0; > 0 1 e x(S)U (1+21 2 ;1+ 1 2 ;x(S)) e x(L)U (1+1 2 2 ;1+ 1 2 ;x(L)) ; if > 0; < 0 1 M (2 ;1+ 1 2 ;x(S)) M (2 ;1+21 ;x(L)); if > 0; > 0 1 S12K (p2 z(S)) L12K (p2 z(L)); if = 0; < 0 1 S12I (p2 z(S)) L12I (p2 z(L)); if = 0; < 0 ; (38)
where L is the lower barrier. 5.2. Up-and-out barrier option:
R= 8 > > > > > > > > > > > > > > < > > > > > > > > > > > > > > : 1 M (1+21 2 ;1+ 1 2 ; x(S)) M (1+1 2 2 ;1+ 1 2 ; x(U )) ; if < 0; < 0 1 e x(S)U (1+21 2 ;1+ 1 2 ;x(S)) e x(U )U (1+1 2 2 ;1+ 1 2 ;x(U )) ; if < 0; > 0 1 S M (1 2 ;1 1 2 ;x(S)) U M (1 2 ;1 1 2 ;x(U )) ; if > 0; < 0 1 U (2 ;1+ 1 2 ;x(S)) U (2 ;1+21;x(U )); if > 0; > 0 1 S 1 2I (p2 z(S)) U12I (p2 z(U )); if = 0; < 0 1 S12K (p2 z(S)) U12K (p2 z(U )); if = 0; > 0 ; (39)
where U is the upper barrier. 5.3. Double barrier option:
R = 1 r(L) r(S) r(L) r(S) + r(S) r(U ) r(S) r(U )
r(L) r(U ) r(L) r(U )
: (40)
M(a,b,x) and U(a,b,x) are the Kummer’s function de…ned in 13.1.2 and 13.1.3 of Abramowitz and Stegun (1972, p 504), K (z) and I (z) are the modi…ed Bessel function and r(S) and
r(S) are de…ned in (16) and (15). And
= r + : (41)
Then, to …nd the present value of the rebate it is only necessary to invert the Laplace transforms (38), (39)or (40).
5
Laplace Transform
The Laplace Transform method is very useful to solve linear di¤erential equations. Thus, Laplace transformation is one of the most popular methods for solving di¤usion equations. On the other hand, the invertion of the Laplace Transform is the most di¢ cult step in applying the Lapalce Transform technique. However, there are algorithms to provide the inverse of the Laplace Transforms numerically, and they are powerful tools that can be used to extend the applicability of the Laplace Transform technique. As a result, in recent years, numerical transform inversion has been recognized as an important technique for calculating probability distributions in stochastic models. Over the years many di¤erent algorithms have been proposed, and studied, for numerically inverting Laplace Transforms. See, for instance, Stehfest (1970), Abate and Whitt (1991), Abate and Whitt (1995) and Fu et al (1999).
In this work the Euler Abate and Whitt method mentioned in Abate and Whitt (1995, p. 37-39) will be used. This method is called "Euler Abate and Whitt" because it employs the Euler summation, and is based on the Bromwich contour inversion integral, which can be expressed as the integral of a real valued function of a real variable by choosing a speci…c contour. The Abate and Whitt (1995) Euler method (see the proof in Abate and Whitt (1995, p 37-39)) calculates values of a real-valued function f (t) of a positive real variable t for di¤erent t from the Laplace transform
F (s) = Z 1
0
e stf (t)dt; where s is a complex variable with nonnegative real part, and
f (t) = E(m; n; t) where E(m; n; t) is the Euler sum.
The Euler sum is de…ned as,
E(m; n; t) = m X k=0 m k 2 mS n+k(t); (42)
with Sn(t) = eA2 2t Re(F ( A 2t)) + eA2 t n X k=0 ( 1)kak(t); (43) where ak(t) = Re(F ( A + 2k i 2t )) (44)
In a Euler sum the Euler summation is applied to m terms after an initial n. It is applied as an acceleration technique that can be described as the weighted average of the last m partial sums by a binomial probability distribution with parameters m and p = 12.
The parameter A describes the discretization error. To produce a 10 k accuracys A =
k log(10): In Abate and Whitt (1995), the authors suggest the values of K = 8, m = 11 and n = 15.
In this work the Abate and Whitt Euler method is used to invert the Laplace Transforms (24), (38), (39) and (40). However the parameters to use are K = 10, m = 12 and n = 35; as suggested in Skachkov (2002), to achieve a better precision in the computational application.
6
Numerical Implementation
In Proposition 3 we found the expression of the Laplace Transform for a double knock out barrier option. To …nd the value of a double knock out barrier call option it is necessary invert that Laplace Transform. In this section the C++ program "Pricing_Knock_Out_Barrier_ Call_Under_CEV_Di¤usion.cpp" (see Appendix D) is implemented and will invert the Lapalce Transform (24). As a result, it will allow us to …nd the value of the double knock out barrier call option. In this program the inversion of the Laplace Transform is made using Euler Abate and Whitt method mentioned in Abate and Whitt (1995, p. 37-39).
The program "Pricing_Knock_Out_Barrier_Call_Under_CEV_Di¤usion.cpp" is com-posed by functions that will implement equations (19), (15), (16), (25), (26), (23) and (24). The program also includes a function named "dfAbateWhittEu()" that inverts the Laplace Transform using the Euler Abate and Whitt method. This function was based in Skachkov (2002). Some changes have been made to the original function, specially one that allows the inversion of functions with more than one variable. The program also includes the functions Gamma, modi…ed Bessel I (x) and K (x) and the Whittaker functions:
Gamma(); cgamma(), cbessIv() and cbessKv() were adapted from the similar functions programed in Fortran in Zhang and Gin (1996, p. 49, 51, 230-233).
The Whittaker functions were programed using the de…nitions 1.9.7 and 1.9.11 of Slater (1960, p. 14) and 1.3.5 of Slater (1960, p. 5). The Kummer M(a,b,x) function present in those de…nitions was programed in the function KummerM_sum_max(). This function was adapted from the greens_specfunc.cpp function in Koval (2004). The function originally computes the Kummer M(a,b,x) with …rst parameter and argument complex, and second argument real, and it has been adapted to compute the Kummer M(a,b,x) function with …rst parameter complex and the second parameter and argument as real numbers.
The "Pricing Knock Out Barrier Call Under CEV Di¤usion.cpp" program was compiled using the freeware Dev-C++ compiler.
6.1
Numerical Results of the C++ Program
The program "Pricing_Knock_Out_Barrier_Call_Under_CEV_Di¤usion.cpp" to price knock out barrier options without rebate was implemented using similar parameters to the ones used by Davidov and Linetsky (2001, p 957), with some changes in order to include some positive values for and the dividend yield as 10% per annum. So, the initial asset price is S0 = 100, the instantaneous volatility at this price is 0 = 25% per annum (so,
= 0S0 ). The risk free rate and the dividend yield are 10% per annum, and all options
have six months to expiration (T=0.5). And = 3; 2; 1; 0:5; 0:5; 1; 2; 3:The results of the program are shown in the next table:
CEV Barrier Options Beta U L K -3 -2 -1 -0.5 0.5 1 2 3 % Dif. N/A 90 95 7.408 7.453 7.499 7.522 7.570 7.594 7.496 7.386 3% N/A 90 100 5.406 5.512 5.624 5.683 5.806 5.870 5.861 4.853 21% N/A 90 105 3.627 3.807 3.999 4.101 4.317 4.433 4.536 3.664 25% 120 N/A 95 4.861 3.946 3.191 2.869 2.325 2.097 1.718 1.423 242% 120 N/A 100 2.840 2.230 1.733 1.524 1.176 1.034 0.800 0.623 356% 120 N/A 105 1.359 1.029 0.766 0.658 0.483 0.414 0.303 0.222 512% 120 90 95 2.974 2.388 1.904 1.699 1.360 1.222 0.998 0.832 257% 120 90 100 1.848 1.437 1.102 0.963 0.735 0.643 0.497 0.389 375% 120 90 105 0.933 0.697 0.511 0.436 0.316 0.269 0.195 0.144 548%
In the previous table the prices for knock out call options are presented. The …rst three lines, where the upper barrier is N/A,contain the prices of down-and-out call options. These values are found using large values for U, this is, making U ! 1. In the same way, making L ! 0; the values for up-and-out call options can be found. These values are in the next three columns where L is N/A. The last three lines have the values of double barrier call options. For each type of knock out option three di¤erent strikes (K) were considered: in the money (ITM), out of the money (OTM) and at the money (ATM). The last column has the maximum di¤erence, in percentage, between the minimum price and the maximum
price for each strike, and each type of knock out barrier option, for di¤erent betas. As it can be seen from the last column the beta value has a great in‡uence in the price of the option. The impact of the parameter beta is bigger for the up-and-out and double knock out barrier options, and the maximum di¤erence happens for OTM options. This result of greater impact of parameter beta for up-and-out and double knock out barrier options, in case of the equal risk free rate and the dividend yield, con…rms the results of Boyle and Tian (1999) and Davidov and Linetsky (2001) for the cases where no dividends are paid.
7
Conclusion
On this thesis a closed-form solution for the Laplace Transform of the knock out barrier option is obtained. After that, this Laplace Transform is inverted and this allows the evalua-tion of the knock out barrier opevalua-tions under the CEV di¤usion. The soluevalua-tion obtained di¤ers from the expression achieved by Davidov and Linetsky (2001) for integrals I (K; A; B) and integral J (K; A; B) for the case = 0 and > 0. The closed-form solution found was implemented computationally using the C++ language, and a freeware compiler. The ana-lytical formulae obtained allows a fast and accucurate calculation of prices under the CEV di¤usion on a normal PC.
From the results obtained by the implementation of the computational program, the im-pact of the beta value in the pricing of options is evident, and this imim-pact is much more clear for up-and-out and double knock out barrier options. These results con…rm the weaknesses appointed in literature to the Black and Scholes model, and they attest the importance that models for pricing options should incorporate the volatility smile, or incorporate volatility as a time dependent parameter.
A
Appendix - Proof of Proposition 2
Has seen before in CEV model 2(S) = S :So replacing in (14), then it becomes
1 2 2S2 +2@2u @S2 + S @u @S u = 0: (A-1)
Proof of equations (15) and (16)
Next all possible cases are considered:
1. Case 6= 0
First, we make the change of variable,
u(S) = S12+ e2x(S)w(x(S)); (A-2)
where " and x(S) are de…ned by (19) and (17) respectively.
Before replacing u(S) in the ODE ( A-1), @S@u and @S@2u2 will be calculated as
@u @S = ( 1 2 + )S 1 2e2x(S)w(x(S)) + S 1 2+ e2x(S)" 2 j j j j 2 ( 2 ) S 2 1 w(x(S)) (A-3) +S12+ e2x(S)@w @x j j j j 2 ( 2 ) S 2 1 = 1 2+ S 1 2 j j j j 2S 1 2 w(x(S)) 2j j j j 2S 1 2@w @x e2 x(S)
@2u @S2 = 1 2+ 1 2 + S 3 2 j j j j 2 1 2 S 3 2 e2x(S)w(x(S)) (A-4) + 1 2 + S 1 2 j j j j 2S 1 2 e2x(S)" 2 j j j j 2 ( 2 ) S 2 1w(x(S)) + 1 2 + S 1 2 j j j j 2S 1 2 e2x(S)@w @x j j j j 2 ( 2 ) S 2 1 2j j j j 2 1 2 S 3 2e2x(S)@w @x 2 j j j j 2S 1 2e2x(S)" 2 j j j j 2 ( 2 ) S 2 1@w @x 2j j j j 2S 1 2e2x(S)@ 2w @x2 j j j j 2 ( 2 ) S 2 1 = (" 1 2 + 1 2 + S 3 2 + 2 j j2 2 j j2 4 S 3 3 2 # w(x(S)) +4 j j 2 2 j j2 4 S 3 32@w @x + 4 j j2 2 j j2 4S 3 32@2w @x2 ) e2x(S)
Now replacing (A-2), (A-3) and (A-4) in (A-1): As x(S) = j j j j 2S 2 ) S = j jj j2x(S) 1 2 1 2 2S2 +2 (" 1 2 + 1 2 + S 3 2 + 2 j j2 2 j j2 4 S 3 32 # w(x(S))+ 4 j j 2 2 j j2 4 S 3 32@w @x + 4 j j2 2 j j2 4S 3 32@2w @x2 ) e2x(S) + S 1 2 + S 1 2 j j j j 2 S 1 2 w(x(S)) 2j j j j 2S 1 2@w @x e 2x(S) S 1 2+ e2x(S)w(x(S)) = 0: Since e2x(S) 6= 0 and S 1 2 6= 0, 2j j 2 2 j j2 2S @2w @x2 + " 2 j j 2 2 j j2 2 S 2 j j j j 2 S # @w @x + 1 2 2 1 2+ 1 2 + S 3 +1 2 2 j j2 2 j j2 2 S + 1 2 + S j j j j 2 S S # w(x(S)) = 0;
Hence, @2w @x2 + ( 1 4 + ( + 1 2) 2 j j2 2j j2 2 j j 2 j j x + 4 j j2 2j j2 2 2 1 4 j j2 4 j j2x(S)2 ) w(x(S)) = 0; and …nally, @2w @x2 + 1 4+ 1 2j j j j(1 + 1 2 ) 2j j 1 x+ ( 1 4 1 42 2) 1 x(S)2 w(x(S)) = 0: (A-5)
In equation (A-5) j jj j = 11 >0<0, wich is " as de…ned in (19). So, j jj j can be replaced by : Considering k = 12 +41 2j j as in (21); and m = 1 4j j as in (20), equation (A-5) becomes: @2w @x2 + " 1 4 + k x + 1 4 m 2 x2 # w(x(S)) = 0 (A-6)
Equation (A-6) is the Whittaker form of the con‡uent hypergeometric equation - see equation 13.1.31 of Abramowitz and Stegun (1972, p 505). The complete solution of the ODE (A-6) is:
w(x(S)) = A Mk;m(x(S)) + B Wk;m(x(S)); (A-7)
where Mk;m(x) and Wk;m(x) are de…ned respectively in equations 13.1.32 and 13.1.33 of
Abramowitz and Stegun (1972, p 505). So, the complete solutions of the ODE (A-1) are u(S) = S12+ e2x(S)[A M
k;m(x(S)) + B Wk;m(x(S))] : (A-8)
Using equations 1.9.7 and 4.1.7 in Slater (1960, p 14 and 60) for Mk;m(x); and
equa-tions 1.9.11 and 4.1.12 in Slater (1960, p 14 and 60) for Wk;m(x) we arrive at the following
boundaries conditions: lim x!1Mk;m(x) = limx!1 x 1 2+me 1 2x1F1[1 2+ m k; 1 + 2m; x] =1; lim x!0Mk;m(x) = limx!0 x 1 2+me 1 2x1F 1[ 1 2 + m k; 1 + 2m; x] = 0;
lim x!1Wk;m(x) = limx!1 x 1 2+me 1 2xU [1 2+ m k; 1 + 2m; x] = 0; lim x!0Wk;m(x) = limx!0 x 1 2+me 1 2xU [1 2+ m k; 1 + 2m; x] =1;
where 1F1[12 + m k; 1 + 2m; x] and U [12 + m k; 1 + 2m; x] are solutions of the Kummer’s
equation as de…ned in 1.1.7 and 1.3.5 of Slater (1960, p 2 and p 5) respectively. We need to consider two cases:
1.1. Case > 0
If S ! 1; then x(S) ! 0; and when S ! 0; then x(S) ! 1: So, using the limit properties of the Whittakers function described above, Mk;m(x(S))is the decreasing solution,
and Wk;m(x(S)) is the increasing solution (x(S) de…ned in (17)).
1.2. Case < 0
If S ! 1; then x(S) ! 1; and when S ! 0 then x(S) ! 0: So, using the limit properties of the Whittakers function described above, Wk;m(x(S))is the decreasing solution,
and Mk;m(x(S)) is the increasing solution (x(S) de…ned in (17)).
Consequently, using the boundaries conditions of and explained in Borodin and Salminen (2002, p. 19), (S) is the increasing and (S)is the decreasing solution of (A-1). Then, (up a multiplicative constant):
(S) = S +12e2x(S)M k;m(x(S)); < 0; 6= 0 S +12e2x(S)Wk;m(x(S)); > 0; 6= 0 ; (S) = S +12e2x(S)W k;m(x(S)); < 0; 6= 0 S +12e2x(S)M k;m(x(S)); > 0; 6= 0 : 2. Case = 0
Let us make the change of variable,
where z(S) = j j1 S : Consider y =p2 z(S) ) z = py 2 :
Before substituting u(S) in the ODE (A-1), we …rst need to compute @S@u and @@S2u2 :
@u @S = 1 2S 1 2g(p2 z(S)) + S 1 2@g @z @z @y p 2 j j( )S 1 = 1 2S 1 2g(p2 z(S)) + S 1 2@g @z 1 p 2 p 2 j j( )S 1; i.e. @u @S = S 1 2 1 2g( p 2 z(S)) j jS @g @z (A-10) …nally, @2u @S2 = S 3 2 1 4g( p 2 z(S)) + 2 j jS @g @z + 2 2 j j2S 2 @ 2g @z2 (A-11)
Now replacing (A-9), (A-10) and (A-11) in (A-1) but with = 0: 1 2 2 S2 +2S 32 1 4g( p 2 z(S)) + 2 j jS @g @z + 2 2 j j2S 2 @2g @z2 S 1 2g(p2 z(S)) = 0; that is, S12 1 2 2 j j2 @2g @z2 + 1 2 2 j jS @g @z 1 8 2S2 g(p2 z(S)) g(p2 z(S)) = 0
Multiplying both members for 2 1
2 2S 2 with attention to S 1 2 6= 0 : 1 2 j j2S 2 @2g @z2 + 1 j jS @g @z 2 2 2S 2 + 1 4 2 g( p 2 z(S)) = 0 (A-12)
But z(S) = j j1 S ; and we can make = 1
2j j ( like in (22)). So, replacing in equation
(A-12): z2@ 2g @z2 + z @g @z 2 z 2+ 2 g(p2 z(S)) = 0 (A-13)
The di¤erential equation (A-13) is the modi…ed Bessel ODE - see equation 9.6.1 of Abramowitz and Stegun (1972, p 374). This ODE has two independent solutions I (p2 z)and K (p2 z) that are de…ned, respectively, in equations 9.6.10 and 9.6.2 of Abramowitz and Stegun (1972, p. 375).
Consequently, the complete solutions of the ODE (A-1) with = 0 is
Using equations 9.6.7 and 9.7.1 in Abramowitz and Stegun (1972, p 375 and p 377) for I (z) and equations 9.6.9 and 9.7.2 in Abramowitz and Stegun (1972, p 375 and p 378) for for K (z) we get the boundaries conditions:
lim z!1I (z) =1 lim z!0I (z) = 0 lim z!1K (z) = 0 lim z!0K (z) =1
We need to considerer two cases:
2.1. Case > 0
If S ! 1; then z(S) ! 0; and when S ! 0; then z(S) ! 1:So, using the limit properties of the Modi…ed Bessel function described above, I (p2 z(S)) is the decreasing solution, and K (p2 z(S)) is the increasing solution (z(S) de…ned in (18)).
2.2. Case < 0
If S ! 1; then z(S) ! 1; and when S ! 0; then z(S) ! 0:Consequently, using the limit properties of the Modi…ed Bessel function described above K (p2 z)is the decreasing solution, and I (p2 z(S)) is the increasing solution (z(S) de…ned in (18)).
Thus, using the boundaries conditions of and explained in Borodin and Salminen (2002, p. 19), r(S) is the increasing and r(S) is the decreasing solution of (A-1). Then,
(up to a multiplicative constant):
(S) = S +12I (p2 z(S)); < 0; = 0 S +12K (p2 z(S)); > 0; = 0 ; and (S) = S +12K (p2 z(S)); < 0; = 0 S +12I (p2 z(S)); > 0; = 0 :
Proof of the wronskian expression (23):
From Borodin and Salminen (2002, p.17) the speed density m(S) is
m(S) = 2 2S 2 2e S R 0 2 2Y 2 2 Y dY (A-15) Computing S R 0 2 2Y 2 2 Y dY : S Z 0 2 2Y 2 2 Y dY = 2 2 S Z 0 Y 2 1dY = 2 2 Y 2 2 S 0 = 2S 2 = j j j j 2S 2 (A-16) So replacing the last result in (A-15):
m(S) = 2 2 S2 +2e
x(S); (A-17)
where x(S) and are respectivately de…ned in (17) and (19). To compute the sclale density it can be used (A-16):
%(S) is the scale density de…ned in Borodin and Salminen (2002, p.17) as
%(S) = exp( S Z 0 2 2Y 2 2 Y dY ): (A-18) Using (A-16) %(S) = e x(S) (A-19)
where " and x(S) are respectivately de…ned in (19),and(17).
From Borodin and Salminen (2002, p.19), the Wronskian (wro) with respect to scale density - %(S) - is
wro %(S) = 0 (S) (S) (S) 0(S) (A-20)
where %(S) is de…ned in (A-19)
Using (15) and (16),
(S) = S +12e2x(S)Wk;m(x(S));
(S) = S +12e2x(S)Mk;m(x(S));
where x(S) and are de…ned in (17) and (19). Consequently, x(S) = j j2 S 2 ) x0(S) = 2j j2 S 2 1 0(S) = ( +1 2)S 1 2e2x(S)M k;m(x(S))+S 1 2e2x(S) "j j 2 Mk;m(x(S)) + 2j j 2 Mk;m0 (x(S)) 0 (S) = ( +1 2)S 1 2e2x(S)Wk;m(x(S)) + S 1 2e2x(S) "j j 2 Wk;m(x(S)) + 2j j 2 Wk;m0 (x(S)) So, 0(S) (S) = ( + 1 2)S 2 ex(S)M k;m(x(S))Wk;m(x(S)) + ex(S) "j j 2 Mk;m(x(S))Wk;m(A-21)(x(S)) +ex(S)2j j2 Mk;m0 (x(S))Wk;m(x(S)) (S) 0(S) = ( + 1 2)S 2 ex(S)M k;m(x(S))Wk;m(x(S)) + ex(S) "j j 2 Mk;m(x(S))Wk;m(A-22)(x(S)) +ex(S)2j j2 Mk;m(x(S))Wk;m0 (x(S))
Replacing (A-21) and (A-22) in equation (A-20), and eliminating the symmetric expressions, wro %(S) = ex(S)2j j2 Mk;m0 (x(S))Wk;m(x(S)) ex(S)
2j j
2 Mk;m(x(S))Wk;m0 (x(S))(A-23)
= ex(S)2j j2 (Mk;m(x(S))Wk;m0 (x(S)) Wk;m(x(S))Mk;m0 (x(S)))
Using the result 2.4.27 of Slater (1960, p 2 and p 26)
wro %(S) = e x(S)2j j2 (1 + 2m)
(12 + m k) (A-24)
where (x)is the Euler Gamma function. From equation (A-19), %(S) = e x(S); where x(S)
is de…ned in (17): Replacing it in equation (A-24) we obtain wro = 2j j2 (1 + 2m)
2. If > 0 and 6= 0
Using (15) and (16),
(S) = S +12e2x(S)M
k;m(x(S));
(S) = S +12e2x(S)Wk;m(x(S));
where x(S) and are de…ned in (17) and (19). Consequently, x(S) = j j2 S 2 ) x0(S) = 2j j2 S 2 1 0(S) = ( +1 2)S 1 2e2x(S)Wk;m(x(S))+S 1 2e2x(S) "j j 2 Wk;m(x(S)) 2j j 2 Wk;m0 (x(S)) 0 (S) = ( +1 2)S 1 2e2x(S)Mk;m(x(S))+S 1 2e2x(S) "j j 2 Mk;m(x(S)) 2j j 2 Wk;m0 (x(S)) So, 0(S) (S) = ( + 1 2)S 2 ex(S)W k;m(x(S))Mk;m(x(S)) ex(S) "j j 2 Wk;m(x(S))Mk;m(A-26)(x(S)) e x(S)2j j2 Wk;m0 (x(S))Mk;m(x(S)) (S) 0(S) = ( + 1 2)S 2 ex(S)W k;m(x(S))Mk;m(x(S)) ex(S) "j j 2 Wk;m(x(S))Mk;m(A-27)(x(S)) e x(S)2j j2 Wk;m(x(S))Mk;m0 (x(S))
Replacing (A-26) and (A-27) in equation (A-20), and eliminating the symmetric expressions, wro %(S) = ex(S)2j j2 Wk;m0 (x(S))Mk;m(x(S)) + ex(S)
2j j
2 Wk;m(x(S))Mk;m0 (x(S))(A-28)
= ex(S)2j j2 (Mk;m(x(S))Wk;m0 (x(S)) Wk;m(x(S))Mk;m0 (x(S)))
Using the result 2.4.27 of Slater (1960, p 2 and p 26)
wro %(S) = e x(S)2j j2 (1 + 2m) (1
2 + m k)
(A-29) From equation (A-19), %(S) = ex(S); where x(S) is de…ned in (17). Replacing (A-19) in
(A-29) it follows that
wro = 2j j2 (1 + 2m)
3. If < 0 and = 0
Using (15) and (16),
(S) = S12K (p2 z(S));
(S) = S12I (p2 z(S));
where z is de…ned in (18) as z(S) = 1S :Making p2 z(S) = w
) w = p2 S , then @w @S = p 2 S 1: Hence, (S) = S12K (p2 z(S)); and (S) = S12I (p2 z(S)): So, 0(S) = 1 2S 1 2I (w) + S 1 2 p 2 I0(w) 0(S) = 1 2S 1 2K (w) + S 1 2 p 2 K0(w) and, 0(S) (S) = 1 2I (w)K (w) + S p 2 I0(w)K (w) (A-31) (S) 0(S) = 1 2I (w)K (w) + S p 2 I (w)K0(w) (A-32)
Replacing (A-31) and (A-32) in equation (A-20), and eliminating the symmetric expressions,
wro %(S) = S
p 2
[K (w)I0(w) K0(w)I (w)] Using the result 9.6.15 of Abramowitz and Stegun (1972, p. 375),
wro %(S) = S
p 2 1
w (A-33)
Since = 0, then %(S) = 1: Replacing %(S) and w in (A-33) it follows that
wro = S p 2 p2 S ! 1 = (A-34) 4. If > 0 and = 0
Using (15) and (16),
(S) = S12I (p2 z(S));
(S) = S12K (p2 z(S));
where z(S) is de…ned in (18) as z(S) = 1 S :Making p2 z(S) = w) w = p2 S , then
@w @S = p 2 S 1: Hence, 0(S) = 1 2S 1 2K (w) S 1 2 p 2 K0(w) 0(S) = 1 2S 1 2I (w) S 1 2 p 2 I0(w) So, 0(S) (S) = 1 2K (w)I (w) S p 2 K0(w)I (w) (A-35) (S) 0(S) = 1 2K (w)I (w) S p 2 K (w)I0(w) (A-36)
Replacing (A-35) and (A-36) in equation (A-20), and eliminating the symmetric expressions,
wro %(S) = S
p 2
[K (w)I0(w) I (w)K0(w)] : Using the result 9.6.15 of Abramowitz and Stegun (1972, p 375), then
wro %(S) = S
p 2 1
w (A-37)
Since = 0, then %(S) = 1: Replacing %(S) and w in (A-37) it follows that
wro = S p 2 p2 S ! 1 = (A-38)
As a result, from (A-25), (A-30), (A-34) and (A-38), we conclude that
wro = 2j j 2 (1+2m) (12+m k) if 6= 0 j j if = 0
B
Appendix - Closed-Form for Integrals
I (K; A; B)
and
J (K; A; B)
A. I (K; A; B) =RAB(Y K) m(Y )dY 1. Case 6= 0
m(Y ) was calculated in equation (A-17) of Appendix A.
A - 1. I (K; A; B) =RAB(Y K) (Y )m(Y )dY, with 6= 0 A - 1.1 If > 0 and < 0 In this case " = 1, m(Y ) = 2 2Y 2 2ex(Y ), (Y ) = Y +12e 12x(Y )M k;m(x(Y )), x(Y ) = ( ) 2Y 2 ; m = 1
4 ;where ; m(Y ); x(Y ); k and m are de…ned in (19), (A-17), (17), (21) and (20),
respectively. Hence, I (K; A; B) = Z B A (Y K)Y +12e 1
2x(Y )Mk;m(x(Y ))2 2Y 2 2ex(Y )dY (B-1)
= 2 2 Z B A Y 12e x(Y ) 2 Mk;m(x(Y ))dY 2K 2 Z B A Y 32e x(Y ) 2 Mk;m(x(Y ))dY
Let us consider the …rst integral 2 2hRABY 12e x(Y )
2 Mk;m(x(Y ))dY
i
: Making the change of variable, x(Y ) = ( ) 2Y 2 ) Y = 2 x(Y ) 1 2 ;
@Y @x = 2 2 2 x 1 2 1 and then, 2 2 Z B A Y 12e x(Y ) 2 M k;m(x(Y ))dY = 2 2 8 < : Z x(B) x(A) " 2 x 1 2 # 1 2 ex2Mk;m(x) 2 2 2 x 1 2 1 dx 9 = ; = 1 2 41 1 2 Z x(B) x(A) xm 12e x 2Mk;m(x)dx
Applying the result of equation 2.4.7 of Slater (1960, p 24), the last equation become: 2 2 Z B A Y 12e x(Y ) 2 Mk;m(x(Y ))dY = 1 2 41 1 2 1 2m + 1e x 2xmM k+1 2;m+ 1 2(x) x(B) x(A) = 1 2 41 1 2 " 1 2m + 1e x 2 ( ) 2Y 2 1 4 Mk+1 2;m+ 1 2(x) #x(B) x(A) = p1 j j " Y 12 2m + 1e x 2M k+12;m+12(x) #x(B) x(A)
but as x(Y ) is a function of Y , the last equation can be rewritten as
2 2 Z B A Y 12e x(Y ) 2 M k;m(x(Y ))dY = 1 p j j " Y 12 2m + 1e x(Y ) 2 M k+12;m+12(x(Y )) #Y =B Y =A (B-2) Now we consider the second integral 2K 2hRABY 32e
x(Y )
2 Mk;m(x(Y ))dY
i : Making the same change of variable that it was done for the …rst integral:
2K 2 Z B A Y 32e x(Y ) 2 Mk;m(x(Y )dY = 2K 2 Z x(B) x(A) " 2 x 1 2 # 3 2 ex2M k;m(x) 2 2 2 x 1 2 1 dx = K 2 41 1 2 Z x(B) x(A) x m 12e x 2M k;m(x)dx
Applying equation 2.4.8 of Slater (1960, p 24) to the last equation: 2K 2 Z B A Y 32e x(Y ) 2 Mk;m(x(Y ))dY (B-3) = K 2 41 1 2 2m m k 12e x 2x mM k+1 2;m 1 2(x) x(B) x(A) = 1 2 41 1 2 " 2mK m k 1 2 ex2 ( ) 2Y 2 1 4 Mk+1 2;m 1 2(x) #x(B) x(A) = p1 j j " 2mKY 12 m k 12e x 2M k+12;m 12(x) #Y =B Y =A Replacing (B-2) and (B-3) in (B-1): I (K; A; B) (B-4) = p1 j j " Y 12 2m + 1e x(Y ) 2 M k+12;m+12(x(Y )) 2mKY 12 m k 12e x(Y ) 2 M k+12;m 12(x(Y )) #Y =B Y =A A - 1.2 If < 0 and < 0 In this case " = 1, m(Y ) = 2 2Y 2 2e x(Y ), (Y ) = Y +12e 1 2x(Y )Mk;m(x(Y )), x(Y ) = 2Y 2 ;
m = 41 ; where ; m(Y ); x; k and m are de…ned in (19), (A-17), (17), (21) and (20), respectively. Hence, I (K; A; B) = Z B A (Y K)Y +12e 1
2x(Y )Mk;m(x(Y ))2 2Y 2 2e x(Y )dY (B-5)
= 2 2 Z B A Y 12e x(Y ) 2 Mk;m(x(Y ))dY 2K 2 Z B A Y 32e x(Y ) 2 M k;m(x(Y ))dY (B-6)
Consider the …rst integral 2 2hRABY 12e x(Y )
2 Mk;m(x(Y ))dY
i :
Doing the same change of variable that was done in case A - 1.1 for the …rst integral, and then applying equation 2.4.9 of Slater (1960, p 24):
2 2 Z B A Y 12e x(Y ) 2 Mk;m(x(Y ))dY = p1 j j " Y 12 2m + 1e x 2M k 12;m+12(x) #Y =B Y =A (B-7)
For the second integral 2K 2hRABY 32e x(Y )
2 Mk;m(x(Y ))dY
i
we can apply the same change of variable that was done in case A - 1.1 for the second integral, and next applying equation 2.4.10 of Slater (1960, p 24): 2K 2 Z B A Y 32e x(Y ) 2 M k;m(x(Y ))dY (B-8) = p1 j j " 2mKY 12 m + k 12e x(Y ) 2 M k 12;m 12(x(Y )) #Y =B Y =A Replacing (B-7) and (B-8) in (B-5): I (K; A; B) = p1 j j " Y 12 2m + 1e x(Y ) 2 M k 1 2;m+ 1 2(x(Y )) (B-9) 2mKY 12 m + k 12e x(Y ) 2 M k 12;m 12(x(Y )) #Y =B Y =A A - 1.3 If > 0 and > 0 In this case " = 1, m(Y ) = 2 2Y 2 2e x(Y ), (Y ) = Y +12e12x(Y )W k;m(x(Y )), x(Y ) = 2Y 2 ;
m = 41 ; where ; m(Y ); x; k and m are de…ned in (19), (A-17), (17), (21) and (20), respectively. Hence, I (K; A; B) = Z B A (Y K)Y +12e 1
2x(Y )Wk;m(x(Y ))2 2Y 2 2e x(Y )dY (B-10)
= 2 2 Z B A Y 12e x(Y ) 2 Wk;m(x(Y ))dY 2K 2 Z B A Y 32e x(Y ) 2 W k;m(x(Y ))dY
Making the same changes of variables before and applying equations 2.4.22 and 2.4.23 of Slater (1960, p 25) to the …rst and second integrals respectively , in equation (B-10), we …nd:
I (K; A; B) = p1 j j h Y 12e x(Y ) 2 W k 12;m 12(x(Y )) (B-11) +KY 12e x(Y ) 2 W k 12;m+12(x(Y )) iY =B Y =A A - 1.4 If < 0 and > 0 In this case " = 1, m(Y ) = 2 2Y 2 2ex(Y ), (Y ) = Y +12e 12x(Y )W k;m(x(Y )), x(Y ) = 2Y 2
m = 41 where ; m(Y ); x; k and m are de…ned in (19), (A-17), (17), (21) and (20), respectively. Hence, I (K; A; B) = Z B A (Y K)Y +12e 1 2x(Y )W k;m(x(Y ))2 2Y 2 2ex(Y )dY
Making the same change of variables as before, and applying equations 2.4.19 and 2.4.20 of Slater (1960, p 25) to the …rst and second integrals, respectively, we …nd:
I (K; A; B) = p1 j j " Y12 k m + 12e x(Y ) 2 W k+12;m 12(x(Y )) (B-12) KY 12 k + m + 1 2 ex(Y )2 W k+1 2;m+ 1 2(x(Y )) #Y =B Y =A (B-13)
B.
J (K; A; B) = Z B
A
(Y K) (Y )m(Y )dY B - 1. J (K; A; B) =RAB(Y K) (Y )m(Y )dY, with 6= 0
The closed-form solution for the integral J (K; A; B) is obtained in the same way as for I (K; A; B); with 6= 0 B - 1.1 If > 0 and < 0 J (K; A; B) (B-14) = p1 j j " Y 12 k + m + 12e x(Y ) 2 W k+12;m+12(x(Y )) KY 12 k m + 12e x(Y ) 2 W k+12;m 12(x(Y )) #Y =B Y =A B - 1.2 If < 0 and < 0 J (K; A; B) (B-15) = p1 j j h Y 12e x(Y ) 2 W k 1 2;m+ 1 2(x(Y )) + KY 1 2e x(Y ) 2 W k 1 2;m 1 2(x(Y )) iY =B Y =A B - 1.3 If > 0 and > 0 J (K; A; B) (B-16) = p1 j j " 2mY 12 k + m 12e x(Y ) 2 W k 12;m 12(x(Y )) + KY 12 2m + 1e x(Y ) 2 W k 12;m+12(x(Y )) #Y =B Y =A B -1.4 If < 0 and > 0 J (K; A; B) (B-17) = p1 j j " 2mY 12 k m + 12e x(Y ) 2 W k+12;m 12(x(Y )) + KY 12 2m + 1e x(Y ) 2 W k+12;m+12(x(Y )) #Y =B Y =A 2. Case = 0
Considering = 0 in equation (A-18), then %(Y ) = exp(0) = 1. So, for this case,
A - 2. I (K; A; B) =RAB(Y K) (Y )m(Y )dY, with = 0 A - 2.1 If < 0 m(Y ) = 2 2Y 2 2, (Y ) = Y12I (p2 z(Y )), = 21 z(Y ) = 1 Y ;
m = 41 ; where m(Y ); ; z; m and are de…ned in (B-18), (22), (18), (20), and (15), respectively. So, I (K; A; B) = Z B A (Y K)Y 12I (p2 z(Y ))2 2Y 2 2dY (B-19) = 2 2 Z B A Y 2 12I (p2 z(Y ))dY 2 2K Z B A Y 2 32I (p2 z(Y ))dY
Let us consider the …rst integral 2 2RABY 2 12I (p2 z(Y ))dY: Making the change
of variable w =p2 z(Y )) w =p2 ( 1 Y ); then, Y = ( p 2 w) 1 ; and @Y @w = p2 ( p2 w) 1 1 : So, 2 2 Z B A Y 2 12I (p2 z(Y ))dY = 2 2 Z w(B) w(A) " p 2 w 1# 2 12 I (w)p 2 ( p2 w) 1 1 dw = p2 2 ( p2 ) 1 21 Z w(B) w(A) w1+ I (w)dw
Now, applying equation 5 of Prudnikov et al (1986, p 46) to the last equation, then: 2 2 Z B A Y 2 12I (p2 z(Y ))dY (B-20) = p2 2 ( p2 ) 1 1 2 " p 2 1 Y 1 21 I +1(w) #w(B) w(A) = p2 2 h Y +12I +1( p 2 z(Y ))i Y =B Y =A
For the second integral 2 2KRABY 2 32I (p2 z(Y ))dY;and making the same change
of variable as in the …rst integral: 2 2K Z B A Y 2 32I (p2 z(Y ))dY = 2 2K Z w(B) w(A) " p 2 w 1# 2 32 I (w)p 2 ( p2 w) 1 1 dw = p2K 2 p2 1+21 Z w(B) w(A) w1 I (w)dw
Applying equation 5 of Prudnikov et al (1986, p 46) to the last equation: 2 2K Z B A Y 2 32I (p2 z(Y ))dY = (B-21) 2 p 2 K( p2 ) 1+21 w1 I 1(w) w(B) w(A) = 2 p 2 h KY 12I 1(p2 z(Y )) iY =B Y =A Replacing (B-20) and (B-21) in (B-19): I (K; A; B) = p2 2 h Y +12I +1(p2 z(Y )) KY 1 2I 1(p2 z(Y )) iY =B Y =A (B-22) A - 2.2 If > 0 m(Y ) = 2 2Y 2 2, (Y ) = Y12K (p2 z(Y )), = 21 ; z(Y ) = 1 Y ;
m = 41 ;where m(Y ); ; z; m; and are de…ned in (B-18), (22), (18), (20) and (15), respectively. So, I (K; A; B) = Z B A (Y K)Y 12K (p2 z(Y ))2 2Y 2 2dY (B-23) = 2 2 Z B A Y 2 12K (p2 z(Y ))dY 2 2K Z B A Y 2 32K (p2 z(Y ))dY
Let us consider the …rst integral 2 2RABY 2 12K (p2 z(Y ))dY:Making the change
of variable w =p2 z(Y )) w =p2 ( 1 Y ); then, Y = (p 2 w) 1 and, @Y @w = p2 (p2 w) 1 1 : So, 2 2 Z B A Y 2 12K (p2 z(Y ))dY = 2 2 Z w(B) w(A) " p 2 w 1# 2 12 K (w) p 2 p2 w 1 1 dw = p2 2 (p2 ) 1 21 Z w(B) w(A) w1 K (w)dw
Applying equation 5 of Prudnikov et al (1986, p 47) to the last equation: 2 2 Z B A Y 2 12K (p2 z(Y ))dY = p2 2 (p2 ) 1 21 w1 K1 (w) w(B) w(A) = 2 p 2 h Y +12K 1 ( p 2 z(Y ))i Y =B Y =A
Using, in last equation, the result 9.6.6 from Abramowitz and Stegun (1972, p 375), then 2 2 Z B A Y 2 12K (p2 z(Y ))dY = p2 2 h Y +12K 1(p2 z(Y )) iY =B Y =A (B-24)
For the second integral 2 2KRABY 2 32K (p2 z(Y ))dY, and making the same change
of variable as in the …rst integral, then 2 2K Z B A Y 2 32K (p2 z(Y ))dY = 2 2K Z w(B) w(A) " p 2 w 1# 2 32 K (w) p 2 (p2 w) 1 1 dw = p2K 2 p2 1+21 Z w(B) w(A) w1+ K (w)dw
Applying equation 5 of Prudnikov et al (1986, p 47) to the last equation: 2 2K Z B A Y 2 32K (p2 z(Y ))dY (B-25) = p2 2 (p2 ) 1+21 w1+ K1+ (w) w(B) w(A) = 2 p 2 h KY 12K +1(p2 z(Y )) iY =B Y =A Replacing (B-24) and (B-25) in (B-23): I (K; A; B) = p2 2 h Y +12K 1(p2 z(Y )) KY 1 2K +1(p2 z(Y )) iY =B Y =A (B-26) B - 2. J (K; A; B) =RAB(Y K) (Y )m(Y )dY, with = 0
The expressions for integrals J (K; A; B); with = 0; are obtained in the same way as for I (K; A; B); with = 0: Hence, we just summarise the formulae:
B - 2.1 If < 0 J (K; A; B) = p2 2 h Y +12K +1(p2 z(Y )) + KY 1 2K 1(p2 z(Y )) iY =B Y =A (B-27) B - 2.2 If > 0 J (K; A; B) = p2 2 h Y +12I 1( p 2 z(Y )) + KY 12I +1( p 2 z(Y ))i Y =B Y =A (B-28)
All the various solutions can be compiled into I (K; A; B) = 8 > > > > > > > > > > > > < > > > > > > > > > > > > : (B-4) if < 0 ^ > 0 (B-9) if < 0 ^ < 0 (B-11) if > 0 ^ > 0 (B-12) if > 0 ^ < 0 (B-22) if > 0 ^ = 0 (B-26) if < 0 ^ = 0 (B-29) J (K; A; B) = 8 > > > > > > > > > > > > < > > > > > > > > > > > > : (B-14) if < 0 ^ > 0 (B-15) if < 0 ^ < 0 (B-16) if > 0 ^ > 0 (B-17) if > 0 ^ < 0 (B-27) if > 0 ^ = 0 (B-28) if < 0 ^ = 0 (B-30)
Note that these close-form solutions for integrals I (K; A; B) and J (K; A; B) are di¤erent from the solutions in Davidov and Linetsky (2001, p 964) for the case = 0 and > 0: This di¤erence is in the order of the Bessel function for the cases referred.
C
Appendix - Proof of Proposition 5
For equation (38): Using (10) 1 EQ e (r+ )=L = 1 (S) (L); (C-1) with de…ned in (41).Now the di¤erent cases must be considered:
1. If < 0; < 0
In this case: = 1;
x(S) = 2S 2 ;
m = 41 ,
where ; x(S) and m are de…ned respectively in (19), (17) and (20). Using (16), 1 (S) (L) = 1 S +12e12x(S)W k ;m(x(S)) L +12e 1 2x(L)Wk ;m(x(L)) ; where k = 1 2 + 1 4 2j j; (C-2) and is de…ned in (41).
Using equation 13.1.33 of Abramowitz and Stegun (1972, p 505), and (17) the last equa-tion is rewritten as
1 (S) (L) = 1 S +12e 1 2x(S)e 1 2x(S)x(S)m+ 1 2U m k + 1 2; 1 + 2m; x(S) L +12e 1 2x(L)e 1 2x(L)x(L)m+ 1 2U m k +1 2; 1 + 2m; x(L) = 1 S +12 2S 2 1 2 1 4 U m k + 12; 1 + 2m; x(S) L +12 2L 2 1 2 1 4 U m k + 12; 1 + 2m; x(L) = 1 S U m k + 1 2; 1 + 2m; x(S) L U m k +12; 1 + 2m; x(L) :
Applying equation 13.1.29 of Abramowitz and Stegun (1972, p 505), and using (20), (17) and (C-2), then 1 (S) (L) = 1 S x(S)1 1+42 U 1 + m k +1 2 1 2m; 2 1 2m; x(S) L x(L)1 1+42 U 1 + m k +1 2 1 2m; 2 1 2m; x(S) = 1 U 2 ; 1 + 21 ; x(S) U 2 ; 1 + 21 ; x(L) 2. If < 0; > 0 In this case: = 1; x(S) = 2S 2 ; m = 41 ,
where ; x(S) and m are de…ned respetivately in (19), (17) and (20). Using (16), 1 (S) (L) = 1 S +12e 12x(S)M k ;m(x(S)) L +12e 1 2x(L)Mk ;m(x(L))
where k and are de…ned in (C-2) and (41). Using equation 13.1.32 of Abramowitz and Stegun (1972, p 505), as well as (17) and (C-2), the last equation becomes
1 (S) (L) = 1 S +12e x(S) 2 e x(S) 2 x(S) 1 2+mM (m k +1 2; 1 + 2m; x(S) L +12e x(L)2 e x(L)2 x(L)12+mM (m k + 1 2; 1 + 2m; x(L) = 1 e x(S)M ( 1 4 1 2 + 1 4 2 + 1 2; 1 + 2 1 4 ; x(S) e x(L)M ( 1 4 1 2 + 1 4 2 + 1 2; 1 + 2 1 4 ; x(L) = 1 e x(S)M 1 2 2 + 1 2; 1 + 1 2 ; x(S) e x(L)M 1 2 2 + 1 2; 1 + 1 2 ; x(L) 3. If > 0; < 0 In this case: = 1; x(S) = 2S 2 ; m = 1 4 ,
where ; x(S) and m are de…ned in (19), (17) and (20). Using (16) and (21): 1 (S) (L) = e r S +12e 1 2x(S)Wk ;m(x(S)) L +12e 12x(L)W k ;m(x(L))
where k and are de…ned in (C-2) and (41). Using equation 13.1.33 of Abramowitz and Stegun (1972, p. 505), and (17), the last equation becomes
1 (S) (L) = 1 S +12e 1 2x(S)e 1 2x(S)x(S)m+ 1 2U m k +1 2; 1 + 2m; x(S) L +12e 1 2x(L)e 1 2x(L)x(L)m+ 1 2U m k + 1 2; 1 + 2m; x(L) = 1 S +12 2S 2 1 2 1 4 U m k +12; 1 + 2m; x(S) L +12 2L 2 1 2 1 4 U m k +12; 1 + 2m; x(L) = 1 e x(S)S U m k +1 2; 1 + 2m; x(S) e x(L)L U m k +1 2; 1 + 2m; x(L) :
Applying equation 13.1.29 of Abramowitz and Stegun (1972, p 505), and using (20) and (C-2), then 1 (S) (L) = 1 e x(S)S 2S 2 1 2 U 1 + 21 2 ; 1 + 21 ; x(S) e x(L)L 2L 2 1 2 U 1 +21 2 ; 1 + 21 ; x(S) = 1 e x(S)U 1 + 1 2 2 ; 1 + 1 2 ; x(S) e x(L)U 1 + 1 2 2 ; 1 + 1 2 ; x(L) 4. If > 0; > 0 In this case: = 1; x(S) = 2S 2 ; m = 41 ,
where ; x(S) and m are de…ned respetivately in (19), (17) and (20). Using (16), 1 (S) (L) = 1 S +12e12x(S)M k ;m(x(S)) L +12e 1 2x(L)Mk ;m(x(L))
where k and are de…ned in (C-2) and (41).
Using equation 13.1.32 of Abramowitz and Stegun (1972, p 505), as well as (17) and (20) the last equation becomes
1 (S) (L) = 1 S +12e x(S) 2 e x(S) 2 x(S) 1 2+mM (m k + 1 2; 1 + 2m; x(S) L +12e x(L) 2 e x(L) 2 x(L) 1 2+mM (m k +1 2; 1 + 2m; x(L) = 1 S +12 2S 2 1 2+ 1 4 M 1 4 1 2 1 4 + 2 + 1 2; 1 + 1 2 ; x(S) L +12 2L 2 1 2+ 1 4 M 41 12 41 + 2 + 12; 1 + 21 ; x(L) = 1 M 2 ; 1 + 21 ; x(S) M 2 ; 1 + 21 ; x(S) :
5. If = 0; < 0 Using (16), 1 (S) (L) = 1 S 1 2K p2 Z(S) L12K p2 Z(L)
where and Z(S) are de…ned in (22) and (18).
6. If = 0; > 0 Using (16), 1 (S) (L) = 1 S 1 2I p2 Z(S) L12I p2 Z(L)
where and Z(S) is de…ned in (22) and (18).
D
Appendix - C++ Program
//Pricing Knock Out Barrier Call Under CEV Di¤usion #include <iostream> #include <iomanip> #include <math.h> # include <vector> # include <complex> # include <algorithm> # include <fstream> using namespace std;
typedef long double ldouble; typedef vector<ldouble> fvector; typedef complex<ldouble> ldcomplex; typedef complex<ldouble> dcomplex; typedef vector<ldouble> fvector; typedef ldouble Z;
const ldouble PI=3.141592653589793238462643; const ldouble machineepsilon = 5E-16;
const long double ld1=1.0; //De…nition of sign function ldouble sign(ldouble a, ldouble b)
{ ldouble ab; if (a*b>=0) ab=1.0; else ab=(-1.0); return ab; }
D.1
The Gamma Function
// Gamma function for reals ldouble gamma(ldouble x) { int i,k,m; ldouble ga,gr,r,z; static ldouble g[] = { 1.0, 0.5772156649015329, -0.6558780715202538, -0.420026350340952e-1, 0.1665386113822915, -0.421977345555443e-1, -0.9621971527877e-2, 0.7218943246663e-2,
-0.11651675918591e-2, -0.2152416741149e-3, 0.1280502823882e-3, -0.201348547807e-4, -0.12504934821e-5, 0.1133027232e-5, -0.2056338417e-6, 0.6116095e-8, 0.50020075e-8, -0.11812746e-8, 0.1043427e-9, 0.77823e-11, -0.36968e-11, 0.51e-12, -0.206e-13, -0.54e-14, 0.14e-14};
if (x > 171.0) return 1e308; // This value is an over‡ow ‡ag. if (x == (int)x) {
if (x > 0.0) {
ga = 1.0; // use factorial for (i=2;i<x;i++) {
ga *= i; } } else ga = 1e308; } else { if (fabs(x) > 1.0) { z = fabs(x); m = (int)z; r = 1.0; for (k=1;k<=m;k++) { r *= (z-k); } z -= m; } else z = x; gr = g[24]; for (k=23;k>=0;k–) { gr = gr*z+g[k]; }
