Modelli Clamfim Lezione 8 13 ottobre 2011 Black Scholes equation

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1 Unverstà d Bologna Modell Clamfm Lezone 8 13 ottobre 2011 Black Scholes equaton professor Danele Rtell 1/12?

2 Exercse Solve the Cauchy problem u t = u xx x R, t > 0 u(x, 0) = sn x x R then use formula (H s ) to nfer the ntegraton formula + e s2 cos(as)ds = π e a2 /4, a R 2/12?

3 Ier abbamo vsto che la soluzone ha la forma u(x, t) = sn x π e s2 cos(2s t)ds = sn x π f(t) pertanto mponendo che detta funzone rsolva l problema s trova che f(t) rsolve l equazone dfferenzale sn x π f (t) = sn x π f(t) da cu s ottene l problema d Cauchy f (t) = f(t) f(0) = e s2 cos(0)ds = π 3/12?

4 Black-Scholes equaton V t σ2 S 2 2 V V + rs S2 S S 0, t [0, T ] rv = 0, (1) s the Black-Scholes equaton where V (S, t) s the value of the opton, S the prce of the underlyng, t the tme, T the expraton date, σ the volatlty of the underlyng and r the rsk-free nterest rate. We suppose r and σ to be constant. 4/12?

5 Françose Coppex Solvng the Black-Scholes equaton: a demystfcaton We wll reduce the Black-Scholes equaton to a general parabolc equatons wth constant coeffcents. 5/12?

6 Françose Coppex Solvng the Black-Scholes equaton: a demystfcaton We wll reduce the Black-Scholes equaton to a general parabolc equatons wth constant coeffcents. Consder the followng change of varables: S = Ke x, (2a) V (S, t) = Kv(x, τ), (2b) τ = (T t)σ 2 /2. (2c) 5/12?

7 The partal dervatves of V (S, t) therefore read V t V S 2 V (2b) = K v τ (2b) = K v x S 2 (3b) = S τ t x S ( K S (2c) = K σ2 2 (2a) = K v x ) v x = K S 2 v x + K S = K S 2 v x + K S 2 2 v x 2. v τ, S ln(s/k) = K S = K v S 2 x + K S ( ) x v S x x S v x v x, (3a) (3b) (9c) 6/12?

8 Insertng Eqs. (3) nto Eq. (1) gves v τ = 2 v x + 2 ( 2r σ 2 1 ) v x 2r σ2v. (4) 7/12?

9 Insertng Eqs. (3) nto Eq. (1) gves We defne v τ = 2 v x + 2 a = ( 2r σ 2 1 2r σ 2 1, ) v b = 2r = (1 + a), σ (5b) 2 therefore Eq. (4) takes the form of Eq. (p): v τ = 2 v x + a v 2 x x 2r σ2v. (4) (5a) + bv. (6) 7/12?

10 General parabolc equatons constant coeffcents We show how a lnear parabolc equaton of the form v t = v xx + av x + bv, (p) can always be reduced to the heat equaton h t = h xx (h) 8/12?

11 General parabolc equatons constant coeffcents We show how a lnear parabolc equaton of the form v t = v xx + av x + bv, (p) can always be reduced to the heat equaton h t = h xx (h) The soluton of Eq. (p) must be of the form v(x, t) = f(t)g(x)h(x, t) (a) 8/12?

12 General parabolc equatons constant coeffcents We show how a lnear parabolc equaton of the form v t = v xx + av x + bv, (p) can always be reduced to the heat equaton h t = h xx (h) The soluton of Eq. (p) must be of the form v(x, t) = f(t)g(x)h(x, t) (a) v(x, τ) = c 1 e (a2 /4+a+1)τ e (a/2)x h(x, τ), c 1 R 8/12?

13 In order to sum up, the BS equaton takes the form of a dffuson equaton wth the followng change of varables: h τ = 2 h x 2, x R, τ [0, σ2 T/2], S = Ke x, τ = (T t)σ 2 /2, (7) V (S, t) = Ke (a2 /4+a+1)τ e (a/2)x h(x, τ), a = 2r/σ /12?

14 The partal dervatves of v(x, t) defned by (a) therefore read v t = f t gh + fgh t (8a) v x = fg x h + fgh x (8b) v xx = fg xx h + 2fg x h x + fgh xx (8c) 10/12?

15 The partal dervatves of v(x, t) defned by (a) therefore read v t = f t gh + fgh t (8a) v x = fg x h + fgh x (8b) v xx = fg xx h + 2fg x h x + fgh xx (8c) Insertng Eqs. (8) nto Eq. (p) gves f t gh + fgh t = fg xx h + 2fg x h x + fgh xx + afg x h + afgh x + bfgh (9) 10/12?

16 The partal dervatves of v(x, t) defned by (a) therefore read v t = f t gh + fgh t (8a) v x = fg x h + fgh x (8b) v xx = fg xx h + 2fg x h x + fgh xx (8c) Insertng Eqs. (8) nto Eq. (p) gves f t gh + fgh t = fg xx h + 2fg x h x + fgh xx + afg x h + afgh x + bfgh (9) Eq. (9) can be satsfed f f and g are of the form f(t) = c 1 exp[ f(t)], g(x) = c 2 exp[ g(x)], (10a) (10b) where c 1 R and c 2 R are two constants. 10/12?

17 Therefore: f t = f f t (11a) g x = g g x (11b) g xx = g g xx + g g 2 x (11c) 11/12?

18 Therefore: f t = f f t (11a) g x = g g x (11b) g xx = g g xx + g g x 2 (11c) Insertng Eqs. (10) and (11) nto Eq. (9) gves [ h t = h xx + h x 2 gx + a ] + h [ f t + g xx + g x 2 + a g x + b ] (12) 11/12?

19 Therefore: Insertng Eqs. (10) and (11) nto Eq. (9) gves f t = f f t (11a) g x = g g x (11b) g xx = g g xx + g g 2 x (11c) h t = h xx + h x [ 2 gx + a ] + h [ f t + g xx + g 2 x + a g x + b ] (12) In order for Eq. (12) to be of the form of Eq. (h), t s requred that: 2 g x + a = 0 = g(x) = ax 2 + c 1, (13a) f t + g xx + g 2 x + a g x + b = 0 = f(t) = (b a2 4 )t + c 2 (13b) 11/12?

20 uttng back Eqs. (13) nto the soluton (a) one gets v(x, t) = c 1 e (b a2 /4)t e (a/2)x h(x, t) c 1 R. (14) 12/12?

21 uttng back Eqs. (13) nto the soluton (a) one gets v(x, t) = c 1 e (b a2 /4)t e (a/2)x h(x, t) c 1 R. (14) To solve (6) we use formula (14) v(x, τ) = c 1 e (a2 /4+a+1)τ e (a/2)x h(x, τ), c 1 R. (15) 12/12?