Ritelli, Daniele

English

Almae Matris Studiorum Campus

Solving the Black-Scholes equation: a demystificationFranc¸ois Coppex, ∗Our objective is to show all the details of the derivation of the solution to the Black-Scholesequation without any prior prerequisit. We first show how to transform the Black-Scholes equationinto a diffusion equation by means of change of variables. Applying the Fourier transform methodwe then find the general solution of the diffusion equation for arbitrary initial conditions. Finally, asa particular case of the general solution we establish the explicit closed-form solution for Europeanand binary options.ContentsI. Introduction 2II. Reduction of the Black-Scholes equation to a diffusion equation 2A. Reduction to a general parabolic equation 2B. Reduction to a diffusion equation 3III. Solution of the diffusion equation 4A. Fourier transform of the diffusion equation 4B. Inverse Fourier transform of the diffusion equation 4IV. Application to European options 4A. Initial conditions: transformation of the payoff functions 4B. Calculation of the integrals 5C. Inverse change of variables 5V. Application to cash-or-nothing binary options 6A. Some properties of the Fourier transform 71. Definition 72. The Fourier transform of derivatives 73. The convolution theorem 74. A useful inverse Fourier transform 8B. Direct verification of the general solution of the diffusion equation 9C. A useful property of the Dirac distribution 10D. Calculation of a class of Gaussian integrals 10∗URL: http://www.francoiscoppex.com2I. INTRODUCTIONWhile the derivation of the Black-Scholes (BS) equation can be found in many textbooks it may be harder to comeacross with a detailed presentation of all steps of the derivation of its solution. This article therefore aims at providinga self-contained set of explanations with all necessary details in order to establish the solution of the BS equation forvanilla options. A great deal of attention is put into chosing a route that does not require any prior knowledge ofmathematical theorems or results. However a good ease with mathematics still remains a prerequisit.The hypothesis, motivations and other considerations behind the BS equation are supposed known to the readerand will not be discussed here. Moreover we do not discuss any aspect that is not strictly related to the process ofestablishing the solution. Our starting point will therefore be the BS equation which is taken for granted:∂V∂t+12σ2S2∂2V∂S2+ rS∂V∂S− rV = 0, S ≥ 0, t ∈ [0, T ], (1)where V (S, t) is the value of the option, S the price of the underlying, t the time, T the expiration date, σ the volatilityof the underlying and r the risk-free interest rate. The BS Eq. (1) is a linear parabolic equation of the form∂v∂t=∂2v∂x2+ a∂v∂x+ bv, (2)which can always be reduced to a diffusion equation∂h∂t=∂2h∂x2. (3)Then an integral form of the solution of the diffusion Eq. (3) can be obtained for any initial condition using forexample Fourier transform methods.II. REDUCTION OF THE BLACK-SCHOLES EQUATION TO A DIFFUSION EQUATIONA. Reduction to a general parabolic equationWe suppose r and σ to be constant and consider the following change of variables:S = Kex, (4a)V (S, t) = Kv(x, τ), (4b)τ = (T − t)σ2/2. (4c)The partial derivatives of V (S, t) therefore read∂V∂t(4b)= K∂v∂τ∂τ∂t(4c)= −Kσ22∂v∂τ, (5a)∂V∂S(4b)= K∂v∂x∂x∂S(4a)= K∂v∂x∂∂Sln(S/K) =KS∂v∂x, (5b)∂2V∂S2(5b)=∂∂S(KS∂v∂x)= −KS2∂v∂x+KS∂∂S∂v∂x= −KS2∂v∂x+KS(∂x∂S∂∂x)∂v∂x= −KS2∂v∂x+KS2∂2v∂x2. (5c)Inserting Eqs. (5) into Eq. (1) gives∂v∂τ=∂2v∂x2+(2rσ2− 1)∂v∂x− 2rσ2v. (6)We definea =2rσ2− 1, (7a)b = − 2rσ2= −(1 + a), (7b)therefore Eq. (6) takes the form of Eq. (2):∂v∂τ=∂2v∂x2+ a∂v∂x+ bv. (8)3B. Reduction to a diffusion equationThe solution of Eq. (2) must be of the formv(x, τ) = f(τ)g(x)h(x, τ). (9)The partial derivatives of v(x, τ) therefore read∂v∂τ= (∂τf)gh+ fg(∂τh), (10a)∂v∂x= f(∂xg)h+ fg(∂xh), (10b)∂2v∂x2= f(∂2xg)h+ 2f(∂xg)(∂xh) + fg(∂2xh), (10c)where we have made use of the shorthand notation ∂nx f = ∂nf/∂xn and have omitted explicit functional dependencein order to lighten the notation. Inserting Eqs. (10) into Eq. (8) gives(∂τf)gh+ fg(∂τh) = f(∂2xg)h+ 2f(∂xg)(∂xh) + fg(∂2xh) + af(∂xg)h+ afg(∂xh) + bfgh. (11)Eq. (11) can be satisfied if f and g are of the formf(τ) = c1 exp[fˆ(τ)], (12a)g(x) = c2 exp[gˆ(x)], (12b)where c1 ∈ R and c2 ∈ R are two constants. Therefore:∂τf(τ) = f(∂τ fˆ), (13a)∂xg(x) = g(∂xgˆ), (13b)∂2xg(x) = g(∂2xgˆ) + g(∂xgˆ)2. (13c)Inserting Eqs. (12) and (13) into Eq. (11) gives(∂τh) = (∂2xh) + (∂xh)[2(∂g gˆ) + a]+ h[− (∂τ fˆ) + (∂2xgˆ) + (∂xgˆ)2 + a(∂xgˆ) + b]. (14)In order for Eq. (14) to be of the form of Eq. (3), it is required that:2(∂xgˆ) + a = 0 =⇒ gˆ(x) = −ax/2 + c1, c1 ∈ R, (15a)−(∂τ fˆ) + (∂2xgˆ) + (∂xgˆ)2 + a(∂xgˆ) + b = 0 =⇒ fˆ(τ) = (b− a2/4)τ + c2, c2 ∈ R. (15b)Putting back Eqs. (15) into the solution (9) and expressing fˆ(τ) in terms of a only with Eq. (7b), one getsv(x, τ) = c1e−(a2/4+a+1)τe−(a/2)xh(x, τ), c1 ∈ R. (16)In order to sum up, the BS equation takes the form of a diffusion equation with the following change of variables:∂h∂τ=∂2h∂x2, x ∈ R, τ ∈ [0, σ2T/2],S = Kex,τ = (T − t)σ2/2,V (S, t) = Ke−(a2/4+a+1)τe−(a/2)xh(x, τ),a = 2r/σ2 − 1.(17)4III. SOLUTION OF THE DIFFUSION EQUATIONA. Fourier transform of the diffusion equationThe diffusion equation (17) with initial condition h(x, 0) can be solved in a very straightforward way using Fouriertransforms. The reader may refer to App. A for a reminder of the Fourier transform properties used in this article. Ifwe note F (h) = h˜(k) the Fourier transform of function f by respect to variable x, then thanks to the property (A3)the Fourier transform of the diffusion equation is∂h˜∂τ= −k2h˜. (18)The solution of this equation ish˜(k, τ) = h˜(k, 0)e−k2τ , (19)where h˜(k, 0) is the Fourier transform of the initial condition for h. Translated into the original variables this initialcondition corresponds to the terminal condition at expiry t = T of the option, i.e. to the payoff profile at expiry aswe shall see in Sects. IV and V.B. Inverse Fourier transform of the diffusion equationIn order to find the solution for h(x, τ) it is required to apply the inverse Fourier transform of Eq. (19). If we defineF (h1) = h˜1 = e−k2τ , (20a)F (h2) = h˜2 = h˜(k, 0), (20b)then Eq. (19) can be written ash˜(k, τ) = h˜1(k, τ)h˜2(k, τ). (21)Applying the convolution theorem as reminded in App. A 3, the inverse Fourier transform of Eq. (21) givesh(x, τ) = (h1 ∗ h2)(x, τ) (A6)= 1√2pi∫Rdξ h1(x− ξ, τ)h2(ξ, τ). (22)The explicit calculation of the inverse Fourier transform of Eq. (20a) is shown in App. A 4:F−1(h˜1) = h1 =1√2τe−x2/(4τ), (23a)F−1(h˜2) = h2 = h(x, 0). (23b)Inserting Eqs. (23) in Eq. (22) finally gives the general solution:h(x, τ) =1√4piτ∫Rdξ exp[− (x− ξ)24τ]h(ξ, 0). (24)As a reminder the change of variables is summarized by Eqs. (17). As shown in App. B it is easy to explicitely verifythat the solution (24) satifies the diffusion Eq. (3).IV. APPLICATION TO EUROPEAN OPTIONSA. Initial conditions: transformation of the payoff functionsAccording to the change of variables (17), the initial payoff condition at τ = 0 corresponds to the payoff at expiryt = T . The payoff at expiry for European options is given byV (S, T ) = max[ε(S −K), 0], (25)5where ε = 1 for a call option and ε = −1 for a put option. From Eqs. (17) we can express Eq. (25) in terms of thenew variables:h(x, 0)(17)=1Ke(a/2)xV (Kex, T )(25)=1Ke(a/2)xmax [ε(Kex −K), 0]= max{ε[e(a/2+1)x − e(a/2)x], 0}. (26)B. Calculation of the integralsReplacing Eq. (26) into the general solution (24) givesh(x, τ) =1√4piτ∫Rdξ exp[− (x− ξ)24τ]max{ε[e(a/2+1)ξ − e(a/2)ξ], 0}. (27)Since e(α+1)x − eαx > 0 if and only if x > 0 for all α ∈ R then the integration domain of Eq. (27) can equivalently berewritten as [0,∞[. Changing variables to η = ξ/ε, dη = dξ/ε givesh(x, τ) = ε1√4piτ∫ ∞0dη exp[− (x− εη)24τ] [eε(a/2+1)η − eε(a/2)η](28)= εIa/2+1 − εIa/2, (29)where the integrals Iα are calculated in App. D. Eq. (29) therefore becomesh(x, τ)(D1)= εe(a/2+1)(x+aτ/2+τ)Φ(εx+ τa+ 2τ√2τ)− εe(a/2)(x+aτ/2)Φ(εx+ τa√2τ)(30)where Φ denotes the cumulative standard normal distribution function given by Eq. (D2).C. Inverse change of variablesGoing back to the initial variables with Eq. (17) one getsV (S, t)(17)= εKe−(a2/4+a+1)τe−(a/2)x[e(a/2+1)(x+aτ/2+τ)Φ(εx+ τa+ 2τ√2τ)− e(a/2)(x+aτ/2)Φ(εx+ τa√2τ)]= εKexΦ(εx+ τa+ 2τ√2τ)− εKe−(a+1)τΦ(εx+ τa√2τ)= εKSKΦ(εln(S/K) + (T − t)(r − σ2/2)σ√T − t︸ ︷︷ ︸+d1)− εKe−r(T−t)Φ(εln(S/K) + (T − t)(r + σ2/2)σ√T − t︸ ︷︷ ︸+d2)= εSΦ(εd1)− εKe−r(T−t)Φ(εd2). (31)In order to sum-up, the solution of the Black-Scholes equation for European options is given byV (S, t) = εSΦ(εd1)− εKe−r(T−t)Φ(εd2),d1 =ln(S/K) + (T − t)(r − σ2/2)σ√T − t ,d2 =ln(S/K) + (T − t)(r + σ2/2)σ√T − t ,Φ(ζ) =1√2pi∫ ζ−∞dη e−η2/2,ε ={1 for a call,−1 for a put.(32)6V. APPLICATION TO CASH-OR-NOTHING BINARY OPTIONSAs a further example we consider a binary option whose payoff at expiry is equal to 1 if it is in the money and 0otherwise. The initial condition therefore readsV (S, T ) ={Θ(S −K) for a call,1−Θ(S −K) for a put, (33)where Θ(ζ) is the Heaviside distribution which is equal to 0 if ζ < 0 and is equal to 1 if ζ > 0. From Eqs. (17) wecan express Eq. (33) in terms of the new variables. We first note thatΘ(S −K) (17)= Θ(Kex −K)= Θ(ex − 1)= Θ(x), (34)where we have made use of the properties Θ(αζ) = Θ(ζ) for α ∈ R+ and ex− 1 > 0 if and only if x > 0. If we furthernote that 1−Θ(x) = Θ(−x), the initial condition (33) in the new variables readsh(x, 0) =1Ke(a/2)xΘ(εx), (35)where ε = 1 for a call and ε = −1 for a put. Replacing Eq. (35) into the general solution (24) givesh(x, τ) =1K1√4piτ∫Rdξ exp[− (x− ξ)24τ]e(a/2)ξΘ(εξ)η=ξ/ε=1K1√4piτ∫Rdη exp[− (x− εη)24τ]e(a/2)εηΘ(η)=1K1√4piτ∫ ∞0dη exp[− (x− εη)24τ]e(a/2)εη︸ ︷︷ ︸(D1)= Ia/2(D1)=1Ke(a/2)(x+aτ/2)Φ(εx+ τa√2τ). (36)Going back to the initial variables with Eq. (17) one finally getsV (S, t) = Ke−(a2/4+a+1)τe−(a/2)xh(x, τ)(36)= Ke−(a2/4+a+1)τe−(a/2)x1Ke(a/2)(x+aτ/2)Φ(εd2)= e−r(T−t)Φ(εd2). (37)Note that for a call (put) option with ε = 1 (with ε = −1) one gets as expected the discounted risk neutral probabilitythat the stock price S is above (below) K at time T .7APPENDIX A: SOME PROPERTIES OF THE FOURIER TRANSFORM1. DefinitionThe one-dimensional Fourier transform F (f(x))(k) of a function f(x) such that∫R dx|f(x)|2 <∞ is defined byF (f(x))(k) = f˜(k) =1√2pi∫Rdx e−ikxf(x), (A1)where i =√−1. The inverse Fourier transform is defined byF−1(f˜(k))(x) = f(x) =1√2pi∫Rdk eikxf˜(k). (A2)2. The Fourier transform of derivativesThe Fourier transform of the n-th derivative of a function f is given byF(f (n)(x))(k) = (ik)nF (f(x))(k). (A3)This relation can be proven by successive integrations by parts. As a reminder of integration by parts, for twofunctions f ang g with the appropriate regularity properties we have∫fg′ = fg −∫f ′g (A4)where f ′ denotes the derivative of f . From the definition (A1) and the regularity condition∫R dx|f(x)|2 < ∞ itfollows thatF(f (n)(x))(k)(A1)=1√2pi∫Rdx e−ikxf (n)(x)(A4)= − 1√2pi∫Rdx (−ik)e−ikxf (n−1)(x)︸ ︷︷ ︸=(ik)F(f(n−1))(A4)= (ik) (−1) 1√2pi∫Rdx (−ik)e−ikxf (n−2)(x)︸ ︷︷ ︸=(ik)F(f(n−2))= . . .= (ik)nF (f(x))(k) (A5)which establishes the property (A3).3. The convolution theoremA useful property of the Fourier transform is that the Fourier transform of the convolution product of two functionsf and g is equal to the product of the Fourier transforms of f and g. If we denote by f ∗ g the convolution productof f and g:(f ∗ g)(x) = 1√2pi∫Rdy f(x− y)g(y), (A6)then the Fourier transform of the convolution product isF (f ∗ g) = F (f)F (g). (A7)8The convolution theorem (A7) is easily verified from the definition (A1) of the Fourier transform. The RHS of Eq. (A7)readsF (f)F (g)(A1)=1√2pi∫Rdu e−ikuf(u)1√2pi∫Rdv e−ikvg(v)=12pi∫R2du dv e−ik(u+v)f(u)g(v). (A8)Changing variables t = u+ v, s = u, dudv = |J |dsdt where the Jacobian of the transformation is|J | = det(1 01 1)= 1 (A9)then Eq. (A8) becomesF (f)F (g) =1√2pi∫Rdt e−ikt1√2pi∫Rds f(t− s)g(s)︸ ︷︷ ︸(A6)= (f∗g)(t)(A1)= F (f ∗ g) (A10)which establishes the convolution theorem (A6).4. A useful inverse Fourier transformThis section establishes the resultF−1(e−k2τ)=1√2τe−x2/(4τ). (A11)Proof: according to the definition of the inverse Fourier transform (A2) we haveF−1(e−k2τ)=1√2pi∫Rdk eikxe−k2τ . (A12)By completing the squares we require that−k2τ + ikx = A(k +B)2 + C, (A13)where A, B and C have to be determined. Equating the coefficients of the successive powers in k gives the folowingset of equations for A, B and C:−τ = A, (A14a)ix = 2AB, (A14b)0 = AB2 + C. (A14c)The solution of this set of equations isA = −τ, (A15a)B = −ix/(2τ), (A15b)C = −x2/(4τ). (A15c)Replacing Eqs. (A15) into Eq. (A12) givesF−1(e−k2τ)=1√2pie−x2/(4τ)∫Rdk exp[−τ(k − i x2τ)2]. (A16)9We further change variables to ξ =√τ [k − ix/(2τ)], dξ = dk√τ , therefore Eq. (A16) becomesF−1(e−k2τ)=1√2pie−x2/(4τ) 1√τ∫Rdξ e−ξ2=1√2piτe−x2/(4τ)∫Rdξ e−ξ2︸ ︷︷ ︸=√pi=1√2τe−x2/(4τ) (A17)which establishes Eq. (A11).Note that we have made use of the result ∫Rdx e−x2=√pi (A18)which may be established as follows. If we note x = (x1, x2) ∈ R2 then on one hand we have∫R2dx e−x2=∫Rdx1∫Rdx2 e−x21−x22=∫Rdx1e−x21∫Rdx2 e−x22=(∫Rdx e−x2)2. (A19)On the other hand, using polar coordinates x = (x1, x2) = r(cos θ, sin θ), dx1dx2 = rdrdθ, r ∈ [0,∞[, θ ∈ [0, 2pi[ wehave ∫R2dx e−x2=∫ ∞0dr r∫ 2pi0dθ e−r2= 2pi∫ ∞0dr re−r2= 2pi∫ ∞0drddr(−12e−r2)= pi. (A20)Equating the RHS of Eqs. (A19) and (A20) finally gives the result (A18).APPENDIX B: DIRECT VERIFICATION OF THE GENERAL SOLUTION OF THE DIFFUSIONEQUATIONIt is easy to verify that the general solution (24) satifies the diffusion Eq. (3) with initial condition h(x, 0). Wehave:∂h∂τ(24)= − 12τh+1√4piτ∫Rdξ(x− ξ)24τ2exp[− (x− ξ)24τ]h0(ξ) = − 12τh+14τ2〈(x− ξ)2〉 (B1a)∂h∂x(24)=1√4piτ∫Rdξ−(x− ξ)2τexp[− (x− ξ)24τ]h0(ξ) = − 12τ〈x− ξ〉 , (B1b)∂2h∂x2(24)(B1b)= − 12τh− 12τ(− 12τ)〈(x− ξ)2〉 = − 12τh+14τ2〈(x− ξ)2〉 , (B1c)where we have made use of the shorthand notation 〈A〉 for the average of A with the measure defined by Eq. (24).It is now straightforward to see that the set of Eqs. (B1a) and (B1c) satisfies the diffusion Eq. (3).10It remains to verify the initial condition h(x, 0). From Eq. (24):limτ→0h(x, τ) =∫Rdξ limτ→01√4piτ{exp[− (x− ξ)24τ]}︸ ︷︷ ︸=δ(ξ−x)h(ξ, 0) (B2)= h(x, 0), (B3)where δ(ξ − x) is the Dirac distribution. The reader may refer to Appendix C for a proof of property (B2).APPENDIX C: A USEFUL PROPERTY OF THE DIRAC DISTRIBUTIONThe Dirac distribution δ(x) (or ”delta function”) is defined by∫Rdxϕ(x)δ(x− x0) = ϕ(x0) (C1)for each continuous function ϕ(x). A Dirac distribution may be obtained from the following process: for all f(x) suchthat ∫Rdx f(x) = 1 (C2)thenlimε→01εf(x− x0ε)= δ(x− x0). (C3)We made use of this property in Eq. (B3) with ε = τ1/2. In order to establish property (C3) we first consider thechange of variables y = (x− x0)/ε, dy = dx/ε, thereforelimε→0∫Rdxϕ(x)1εf(x− x0ε)= limε→0∫Rdy ϕ(yε+ x0)f(y)=∫Rdy limε→0ϕ(yε+ x0)f(y)=∫Rdy ϕ(x0)f(y)= ϕ(x0)∫Rdy f(y)︸ ︷︷ ︸(C2)= 1= ϕ(x0) (C4)where we are allowed by the dominated convergence theorem to intervene the limit and integration. This establishesEq. (C3).APPENDIX D: CALCULATION OF A CLASS OF GAUSSIAN INTEGRALSWe establish the following resultIα =1√4piτ∫ ∞0dξ exp[− (x− εξ)24τ+ αεξ]= eα(x+ατ)Φ(εx+ 2τα√2τ), (D1)where ε = ±1 and Φ denotes the cumulative standard normal distribution functionΦ(ζ) =1√2pi∫ ζ−∞dη e−η2/2. (D2)11Proof: by completing the squares of the exponential, we require that:− (x− εξ)24τ+ αεξ = −c1(c2 − εξ)2 + c3, (D3)where ci, i = 1, . . . , 3 are constants by respect to the integration variable. Expanding Eq. (D3) gives− 14τ(εξ)2 + (εξ)(α+x2τ)− 14τx2 = −c1ξ2 + ξ2c1c2 − c1c22 + c3. (D4)Equating powers in εξ in Eq. (D4) givesc1 = 1/(4τ), (D5a)c2 = x+ 2τα, (D5b)c3 = α(x+ ατ). (D5c)Eq. (D1) therefore now readsIα = ec31√4piτ∫ ∞0dξ e−c1(c2−εξ)2. (D6)Changing variables to η = (c2 − εξ)√2c1, dη = −dξε√2c1, Eq. (D6) becomesIα =ec3√2c11√4piτ(−ε∫ −ε∞c2√2c1dη e−η2/2). (D7)If ε = 1 Eq. (D7) becomesIα = ec31√2pi∫ c2√2c1−∞dη e−η2/2= ec3Φ(c2√2c1). (D8)If ε = −1 Eq. (D7) becomesIα = ec31√2pi∫ ∞c2√2c1dη e−η2/2. (D9)Using the change of variables ζ = −η, dζ = −dη, Eq. (D9) becomesIα = ec31√2pi∫ −c2√2c1−∞dζ e−ζ2/2= ec3Φ(−c2√2c1) . (D10)Combining Eqs. (D8) and (D10) givesIα = ec3Φ(εc2√2c1), (D11)whereec3(D5)= eα(x+ατ), (D12a)c2√2c1(D5)=x+ 2τα√2τ. (D12b)Replacing Eqs. (D12) in Eq. (D11) gives the result (D1).

Solving Black Scholes equation di Fancois Coppex

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Solving Black Scholes equation di Fancois Coppex

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