Payo Function Decomposition by Fabrice Douglas Rouah

Transcription

1 Payo Fnction Decomposition by Fabrice Doglas Roah For any twice- The following identify is often sed in option pricing. di erentiable fnction f() de ned on R we can write f() = f() + f ()( ) + () f (K)(K ) + dk + f (K)( K) + dk for some threshold. I have seen three proofs of this.. Uses the Dirac delta fnction. Fond in papers by Carr and Madan. 2. Uses the Fndamental Theorem of Calcls. Fond in a paper by John Crosby at Glasgow University. 3. Uses the Taylor eries expansion of f(): Fond in a paper by Attilio Mecci. In this Note these three proofs are each presented in detail. We rst present some fnctions that will be needed. Dirac Delta and Other Fnctions De nition of the Dirac delta fnction (K) (K) = for K = elsewhere and (K)dK = : The sifting property of the Dirac delta fnction is f(a) = a+" a " f(k) (a K) dk for " > : The Heaviside fnction H(K) is de ned as for K > H(K) = for K < = (K>): The derivative of H(K) is easy to chec d H(K) = (K): dk

2 The derivative of (K) + = max(; K) is also easy to chec d dk (K)+ = H(K): The de nitions and derivatives needed for the proofs in this Note are presented in the following table. For ( K) For (K ) ( K) = for K = elsewhere (K ) = for = K elsewhere H( K) = (>K) H(K ) = (K>) d dk H( K) = ( K) d dk H(K ) = (K ) d dk ( K)+ d = (>K) dk (K )+ = (K>) 2 Proof Using the Dirac Delta Fnction Carr and Madan consider > in eqation (), since they se it for an nderlying price : f() = f() + f ()( ) + (2) f (K)(K ) + dk + f (K)( K) + dk: Apply the sifting property of the Dirac delta fnction to f() and choose some threshold > f() = = = = I + I 2 : f(k)( K)dK (3) f(k)( K)dK + f(k)(k )dk + 2. First Integration by Parts f(k)( f(k)( K)dK K)dK Apply integration by parts to each of the two integrals I and I 2. 2

3 2.. Evalating I For I choose = f(k); v = (K ). Hence = f (K); v = (K>). I = 2..2 Evalating I 2 f(k)(k )dk (4) = f(k) (K>) = f() (>) f (K) (K>) dk: f (K) (K>) dk (5) For I 2 choose = f(k); v = ( K) so that = f (K); v = (K>) : I 2 = f(k)( K)dK (6) = f(k) (>K) + f (K) (K<) dk (7) = f() (>) + f (K) (K<) dk: bstitte I and I 2 from eqations (4) and (6) into eqation (3) to obtain f() = f() = f() I 3 + I 4 : f (K) (K>) dk econd Integration by Parts f (K) (K<) dk (8) Apply integration by parts again, to each of the two integrals I 3 and I Evalating I 3 For I 3 choose = f (K); v = (K>) ; = f (K); v = (K ) +. Hence I 3 = f (K) (K>) dk (9) = f (K)(K ) + f (K)(K ) + dk () = f ()( ) + f (K)(K ) + dk: 3

4 2.2.2 Evalating I 4 Finally, for I 4 choose = f (K); v = (>K) ; = f (K); v = ( K) + : I 4 = f (K) (K<) dk () = f (K)( K) + + f (K)( K) + dk (2) = f ()( ) + + f (K)( 2.3 Obtaining the Payo Decomposition K) + dk: bstitte I 3 and I 4 from eqations (9) and () into eqation (8) to btain f() = f() + f () ( ) + ( ) + (3) + f (K)(K ) + dk + f (K)( K) + dk: Eqivalently, since [( ) + ( ) + ] = (which can be veri ed by considering the cases > and < separately), eqation (3) can be written f() = f() + f ()( ) + which is eqation (2). f (K)(K ) + dk + 3 Proof Using the FTC f (K)( K) + This proof ses the Fndamental Theorem of Calcls (FTC) Hence we can write f() as f() f() = f() = f() + f ()d: f ()d: (4) We will focs on the integral R f ()d in eqation (4). Note that the integral can be broen into two parts: when > and when < f ()d = (>) f ()d + (<) f ()d (5) = (>) f ()d 4 (<) f ()d:

5 In R eqation (5) apply the FTC to the rst integral so that f () f () = f (v)dv and to the second integral so that f () f () = R f (v)dv f ()d = (>) f () + f (v)dv d (6) (<) f () f (v)dv d: Re-arrange the integrals to obtain f ()d = (>) f ()d (>) (<) f ()d + (7) f (v)dvd + (<) Consider each of these for integrals in eqation (7) in pairs. f (v)dvd: 3. First Pair of Integrals The rst two integrals in eqation (7) are (>) f ()d (<) f ()d (8) = (>) f ()d + (<) f ()d = f ()d = f () 3.2 econd Pair of Integrals d = f ()( The last two integrals in eqation () can be evalated sing Fbini s theorem changing the order of integration, maing sre to preserve the same domain of integration. The integral has domain of integration <v< << can write the integral (9) as imilarly, the integral f (v)dvd = ): f (v)dvd (9), which is eqivalent to v<< <v< v f (v)ddv = f (v)(. Hence we v)dv: f (v)dvd (2) 5

6 has domain of integration <v< << can write the integral (2) as f (v)dvd = v 3.3 Recombing the Integrals, which is eqivalent to <<v <v< f (v)ddv = f (v)(v. Hence we )dv: We pt these two integrals bac into eqation (7), and sbstitte into eqation (4). We obtain f() = f() + (>) f (v)( v)dv + (<) f (v)(v )dv: (2) ince (>) = only when >, and since (<) = only when <, these can be replaced by the () + fnction on the linear parts of each integrand in eqation (2). Hence we have the desired reslt f() = f() + f ()( ) + which, again, is eqation (). f (v)( v) + dv + 4 Proof Using Taylor eries De ne the fnction g(x) as g(x) = f()+f ()(x )+ f ()(x f (v)(v ) + dv ) + d+ f ()( x) + d: (22) The proof involves demonstrating that g(x) = f(x). This is done by showing that g() = f() and that the derivatives of g and f evalated at any point x are the same, i.e. g (n) (x) = f (n) (x). The Taylor series expansion of f and g will therefore be identical, and conseqently, so will f and g themselves, at all vales of x. First note that for x =, eqation (22) becomes g() = f()+f ()( )+ f ()( ) + d+ f ()( ) + d: (23) In eqation (23), the rst integral is zero since ( ) + = for >, and the second integral is zero since ( ) + = for <. Hence g() = f(). 4. First Derivative Tae the rst derivative of g(x) in eqation (22) g (x) = f () + = f () + f () d dx (x )+ d + f ()H(x 6 )d f () d dx ( f ()H( x)d: x)+ d(24)

7 Consider the second line of eqation (24). In the rst integral, H(x ) = only when < x. Hence the pper limit can be replaced with x. Moreover, H(x ) = only when x >, in which case the integral is R x f ()d. This can be accomplished by inclding the term H(x ) otside the integral. imilarly, for the second integral, H( x) = only when > x so the lower limit can be replaced with x, and H( x) = only when x <, so we pt H( x) otside the integral. This means that eqation () can be written as g (x) = f () + H(x ) x f ()d H( x) x f ()d: (25) To evalate eqation (25) we consider the cases x > and x < separately. 4.. Case For x > we have H( x) = and H(x ) x f ()d = x f ()d = f (x) f () Hence eqation (25) becomes 4..2 Case 2 g (x) = f () + [f (x) f ()] = f (x): For x < we have H(x ) = and H( x) x f ()d = f () f(x): Hence eqation (25) becomes g (x) = f () + [f () f (x)] = f (x): o for both cases we have g (x) = f (x): 4.2 econd Derivative Tae the second derivative of g(x) in eqation (22) g (x) = = f () d2 dx 2 (x )+ d + f ()(x )d + f ()( f () d2 dx 2 ( x)+ d (26) x)d: Consider the second line of eqation (26). The sifting property of the Dirac delta fnction implies that the rst integral is f (x), bt only when x >, 7

8 since this ensres that the domain of integral is s ciently wide for the sifting property to hold. Otherwise, the integral is zero. This can be accomplished by mltiplying the integral by H(x ): imilarly, the second integral is f (x) only when x <, so we mltiply the integral by H( x). This implies that we can write g (x) as g (x) = H(x ) f ()(x )d + H( x) f ()( = H(x )f (x) + H( x)f (x) = f (x) x)d where the second line follows by the sifting property of the Dirac delta fnction. 4.3 Higher Order Derivatives The nth derivative of g(x) is " d n dn g(x) = dxn dx n f() + f ()(x ) + " = dn 2 dx n 2 f ()(x ) + d + f ()( # f () d2 dx 2 (x )+ d + f () d2 dx 2 ( x)+ d : x) + d # From eqation (26) in the preceding section we now that the term inside the sqare bracets is simply d2 dx f(x). Hence we can write 2 d n dx n g(x) = dn 2 d 2 dx n 2 dx 2 f(x) = dn dx n f(x): Apply the Taylor theorem to g(x) g(x) = g() + = f() + = f(x): X n= X n= n! g(n) (x ) n n! f (n) (x ) n ince the fnctions f and g are identical, eqation (22) becomes eqation () and the proof is complete. 8