Part III - Pricing A Down-And-Out Call Option

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1 Part III - Pricing A Down-And-Out Call Option Gary Schurman MBE, CFA March 202 In Part I we examined the reflection principle and a scaled random walk in discrete time and then extended the reflection principle to a continuous time Brownian motion. In Part II we deried the joint distribution function for the alue of a Brownian motion at time T and its maximum and minimum oer the time interal [0, T ]. In Part III we will use the joint distribution deried in Part II to price a down-and-out call option. Imagine that we hae a call option that can be exercised at time T proided that it doesn t knock-out prior to time T. The knock-out occurs when a barrier is crossed and in our case this barrier is below the current stock price (i.e. is a down-and-out call). Another complication is that the option is currently in-the-money at time zero as the current stock price is greater than the option s exercise price. We will use the following parameters to price the call option... Option Pricing Parameters: * S Stock price at time zero $20.00 K Call option exercise price $8.00 B Down and out barrier price $5.00 T Option expiration (in years) 2.00 σ Annual olatility 30.00% r Annual risk-free rate 5.00% Question: What is the time zero alue of this down-and-out call option? * Gien that S > B and K > B the option qualifies as a alid down-and-out option. The Valuation Plan Since we will be working directly with the Brownian motion we need to know the point at which the option is at-the-money (i.e. stock price equals exercise price) and the point at which the barrier is crossed (i.e. stock price equals the barrier price). We will therefore make the following definitions... x ln ( K S )...and... y ln We will need the joint density function for the alue of a Brownian motion at time T and its minimum alue during the time interal [0, T ]. The equation for this joint distribution function from Part II is... f(m, w) e αw ( B S ) () 2 2 (2m w)2 (2) We also need to define the drift of the Brownian motion (α) and its ariance () which are... α (r 2 σ2 ) T...and... σ 2 T (3) The Fundamental Theorom of Finance states that in a complete market all asset prices discounted at the risk-free rate are martingales under the unique risk-neutral probability measure. The alue of our call option at time t is therefore the expected call payoff at time T under the risk-neutral probability measure discounted at the risk-free rate. The mathematical equialent of this relationship is... { }] C t E [e Q r(t t) Max S T K, 0 (4)

2 Using risk-neutral expectations as outlined in Equation (4) aboe the equation for the alue of our down-and-out call option at time zero (C) is... C 0 w0 mw wx my We can rewrite Equation (5) aboe as... Where... ( ) Se w K e rt f(m, w) δm δw + w m0 w0 my ( ) Se w K e rt f(m, w) δm δw (5) C 0 Se rt I Ke rt I 2 + Se rt I 3 Ke rt I 4 (6) I I 2 I 3 I 4 w0 mw wx my w0 mw wx my w m0 w0 my w m0 w0 my e w f(m, w) δm δw (7) f(m, w) δm δw (8) e w f(m, w) δm δw (9) f(m, w) δm δw (0) How do we know that Equation (5) aboe represents the true alue of the call? In that equation we use the ariable w to represent the alue of the Brownian motion at time T and the ariable m to represent the minimum alue of the Brownian motion during the time interal [0, T ]. Lets take a look at the first double integral... w0 mw wx my ( ) Se w K e rt f(m, w) δm δw () The bounds of integration for the first integral is w x to w 0. Here we want to weight possible stock prices at time T as stock price goes from the exercise price (point x in the Brownian motion) to stock price at time zero (point 0 in the Brownian motion). Gien that we want to know the probabilities that stock price will not cross the minimum barrier and the fact that the minimum of the Brownian motion must be less than stock price the bounds of integration for the second integral are m y, which represents the barrier, to w 0, which represents current stock price. The probability of the w and m combination is f(m, w) δm δw. Lets now take a look at the second double integral... w m0 ( ) Se w K e rt f(m, w) δm δw (2) w0 my For this integral we want to know the w and m combinations where stock price at time t is greater than stock price at time zero. Solution To The First Integral I What follows in this section is the solution to the integral I as defined by Equation (0) aboe. After replacing the joint density function f(m, w) with Equation (2) the integral to be soled becomes... I w0 mw wx my w0 mw wx my w e e αw 2 2 (2m w)2 δm δw e w+ αw 2 2 (2m w)2 δm δw (3) 2

3 If we define the function g(α, w, ) in Appendix Equation (57) to be w + αw (3) becomes... I w0 wx w0 wx w0 wx w0 wx w0 wx e 2 +α w0 { mw} e w+ αw 2 2 (2m w)2 δw my 2 { e w+ αw 2 2 (2w w)2 e w+ αw 2 2 }δw (2y w)2 w0 e w+ αw 2 2 w2 δw e 2 (w2 2w 2αw+α 2) δw wx w0 e 2 (w α )2 e 2 (2 +2α) δw wx e w+ αw 2 2 (w2 4wy+4y 2) δw wx w0 e 2 (w2 2w 2αw+α 2 4wy+4y 2) δw wx then the integral in Equation e 2 (w α 2y)2 e 2 (2 +4αy+4y+2α) δw w0 e 2 (w α )2 δw e 2αy 2 +α+ +2y e 2 (w α 2y)2 δw (4) wx Note that to sole the integral aboe we made the following definitions... (w α ) 2 w 2 2w 2αw + α α (5) (w α 2y) 2 w 2 2w 2αw + α 2 4wy + 4y yα + 4y + 2α (6) z w α Then the solution to the first integral in Equation (4) is... δw...and... δw (7) w0 wx w 0 α e 2 (w α )2 δw e 2 z2 w x α w α w x α N e [ α ] N 2 z2 x α (8) z w α 2y Then the solution to the second integral in Equation (4) is... δw...and... δw (9) w0 wx w 0 α 2y e 2 (w α 2y)2 δw e 2 z2 w x α 2y N w α 2y w x α 2y e [ α 2y ] N 2 z2 x α 2y (20) If we combine Equations (8) and (20) then the solution to the integral I becomes... } } I e α x α 2 {N +α N e 2αy 2 +α+ α 2y x α 2y {N +2y N (2) 3

4 Solution To The Second Integral I2 What follows in this section is the solution to the integral I 2 as defined by Equation (8) aboe. After replacing the joint density function f(m, w) with Equation (2) the integral to be soled becomes... wx w0 I 2 w0 mw wx my If we define the function g(α, w, ) in Appendix Equation (57) to be αw becomes... w0 { mw} I 2 e αw 2 2 (2m w)2 δw wx w0 wx w0 wx w0 wx w0 wx my e αw 2 2 (2m w)2 δm δw (22) 2 { e αw 2 2 (2w w)2 e αw 2 2 }δw (2y w)2 w0 e αw 2 2 w2 δw e 2 (w2 2αw+α 2) δw w0 e 2 (w α)2 δw e 2 (w α)2 δw e 2αy wx w0 e αw 2 2 (w2 4wy+4y 2) δw wx wx w0 then the integral in Equation (22) e 2 (w2 2αw+α 2 4wy+4y 2) δw e 2 (w α 2y)2 e 2 ( 4αy) δw wx Note that to sole the integral aboe we make the following definitions... e 2 (w α 2y)2 δw (23) (w α) 2 w 2 2αw + α 2 (24) (w α 2y) 2 w 2 2αw + α 2 4wy + 4y 2 + 4αy (25) z w α Then the solution to the first integral in Equation (23) is... w0 wx e 2 (w α)2 δw z w α 2y δw...and... δw (26) w 0 α w x α w α N Then the solution to the second integral in Equation (23) is... w0 wx e 2 (w α 2y)2 δw w x α e 2 z2 e [ α ] N 2 z2 x α (27) δw...and... δw (28) w 0 α 2y w x α 2y w α 2y N w x α 2y e 2 z2 e [ α 2y ] N 2 z2 x α 2y (29) If we combine Equations (27) and (29) then the solution to the integral I 2 becomes... { } α x α I 2 N N e 2αy α 2y x α 2y N N (30) 4

5 Solution To The First Integral I3 What follows in this section is the solution to the integral I 3 as defined by Equation (9) aboe. After replacing the joint density function f(m, w) with Equation (2) the integral to be soled becomes... I 3 w m0 w0 my w m0 w0 my w e e αw 2 2 (2m w)2 δm δw e w+ αw If we define the function g(α, w, ) in Appendix Equation (57) to be w + αw (3) becomes... I 3 w w0 w w0 w w0 w w0 w w0 e 2 +α w { m0} e w+ αw 2 2 (2m w)2 δw my { e w+ αw 2 2 (0 w)2 e w+ αw 2 2 }δw (2y w)2 w e w+ αw 2 2 w2 δw e 2 (w2 2w 2αw+α 2) δw w0 w e 2 (w α )2 e 2 (2 +2α) δw w0 e 2 (w α )2 δw e (2m w)2 δm δw (3) 2 e w+ αw 2 2 (w2 4wy+4y 2) δw w0 w w0 e 2 (w2 2w 2αw+α 2 4wy+4y 2) δw w 2αy +α+ +2y then the integral in Equation e 2 (w α 2y)2 e 2 (2 +4αy+4y+2α) δw w0 Note that to sole the integral aboe we make the following definitions... e 2 (w α 2y)2 δw (32) (w α ) 2 w 2 2w 2αw + α α (33) (w α 2y) 2 w 2 2w 2αw + α 2 4wy + 4y yα + 4y + 2α (34) z w α Then the solution to the first integral in Equation (32) is... δw...and... δw (35) w w0 w α e 2 (w α )2 δw w 0 α w w α N N N e 2 z2 e α 2 z2 α (36) z w α 2y δw...and... δw (37) 5

6 Then the solution to the second integral in Equation (32) is... w w0 w α 2y e 2 (w α 2y)2 δw w 0 α 2y w w α 2y N N N e 2 z2 e 2 z2 α 2y α 2y (38) If we combine Equations (36) and (38) then the solution to the integral I 3 becomes... } } I 3 e α 2 { +α N e 2αy 2 +α+ α 2y { +2y N (39) Solution To The Fourth Integral I4 What follows in this section is the solution to the integral I 4 as defined by Equation (??) aboe. After replacing the joint density function f(m, w) with Equation (2) the integral to be soled becomes... I 4 w m0 w0 my e αw 2 2 (2m w)2 δm δw (40) If we define the function g(α, w, ) in Appendix Equation (57) to be αw becomes... w { m0} I 4 e αw 2 2 (2m w)2 δw w0 w w0 w w0 w w0 w w0 w w0 my 2 { e αw 2 2 ( w)2 e αw 2 2 }δw (2y w)2 { e αw 2 2 w2 } e αw 2 2 (w2 4wy+4y 2 ) δw w0 e 2 (w2 2αw+α 2) δw w0 e 2 (w α)2 δw e 2 (w α)2 δw e 2αy wx wx w0 then the integral in Equation (40) e 2 (w2 2αw+α 2 4wy+4y 2) δw e 2 (w α 2y)2 e 2 ( 4αy) δw wx Note that to sole the integral aboe we make the following definitions... e 2 (w α 2y)2 δw (4) (w α) 2 w 2 2αw + α 2 (42) (w α 2y) 2 w 2 2αw + α 2 4wy + 4y 2 + 4αy (43) z w α δw...and... δw (44) 6

7 Then the solution to the first integral in Equation (4) is... w w0 w α e 2 (w α)2 δw w 0 α w w α N N e 2 z2 e α N α 2 z2 (45) z w α 2y Then the solution to the second integral in Equation (4) is... w w0 w α 2y e 2 (w α 2y)2 δw δw...and... δw (46) w 0 α 2y w w α 2y N N N e 2 z2 e 2 z2 (47) α 2y α 2y (48) If we combine Equations (45) and (47) then the solution to the integral I 4 becomes... { } α I 4 N e 2αy α 2y N (49) The Answer To The Pricing Question The alues of x, y, α and per Equations () and (3) aboe are... x ln(k S) ln( ) y ln(b S) ln( ) α (r 0.50 σ 2 )T ( ) σ 2 T The alues of the integrals per Equations (2), (30), (39) and (49) are... I f(x, y, α, ) I 2 f(x, y, α, ) I 3 f(x, y, α, ) I 4 f(x, y, α, ) The present alue of current stock price and exercise price are... PV Stock price S e rt exp( ) 8.0 PV Exercise price K e rt 8.00 exp( ) 6.29 The alue of the down-and-out call option per Equation (6) aboe is... C 0 Se rt I Ke rt I 2 + Se rt I 3 Ke rt I 4 (8.0)( ) (6.29)( ) + (8.0)( ) (6.29)( ) 4.20 (50) 7

8 Appendix We need the anti-deriatie of the following equation with respect to m... A(m, w) Define the ariable lambda and its deriatie with respect to m... e g(α,w,) 2 (2m w)2 (5) λ 2m w Define the ariable theta and its deriatie with respect to λ... δλ δm 2 (52) θ 2 (2m w)2 2 λ2 The deriatie of theta with respect to m is... δθ δλ λ 2m w (53) δθ δm δθ δλ δλ δm Define the function G and its deriatie with respect to m... (54) B(m, w) e g(α,w,) e θ δb(m, w) δθ e αw 2 +w e θ (55) We will now proe that G is the anti-deriatie to F with respect to m... Note that the function G can also be written as... δb(m, w) A(m, w) δm δb(m, w) δθ { δθ δm } { } 2(2m w) e g(α,w,) e θ e g(α,w,) e θ e g(α,w,) 2 (2m w)2 (56) B(m, w) e g(α,w,) 2 (2m w)2 (57) 8