Lecture #35: The Characteristic Function for Heston s Model

Transcription

1 Statistics 441 (all 014) November 8, 014 Prof Michael Kozdron Lecture #35: The Characteristic unction for Heston s Model As we saw last lecture, it is sometimes ossible to determine the characteristic function of a random variable defined via a stochastic di erential equation without actually solving the S The comutation involves the eynman-kac reresentation theorem, but it does require the solution of a artial di erential equation In certain cases where an exlicit solution does not exist for the S, comuting the characteristic function might still be ossible as long as the resulting P is solvable Recall that the Heston model assumes that the asset rice rocess {S t,t S ds t v t S t db (1) t + µs t dt where the variance rocess {v t,t 0} satisfies dv t v t db () t + a(b v t )dt 0} satisfies the and the two driving Brownian motions {B (1) t,t 0} and {B () t,t 0} are correlated with rate, ie, dhb (1),B () i t dt In order to analyze the Heston model, it is easier to work with X t log(s t ) instead Itô s formula imlies that {X t,t dx t dlogs t ds t S t dhsi t S t 0} satisfies the S v t db (1) t + µ We will now determine the characteristic function of X T for any T version of Itô s formula (Theorem 04) imlies that v t dt 0 The multidimensional df(t, X t,v t ) f(t, X t,v t )dt + f 1 (t, X t,v t )dx t + 1 f 11(t, X t,v t )dhxi t + f (t, X t,v t )dv t + 1 f (t, X t,v t )dhvi t + f 1 (t, X t,v t )dhx, vi t f(t, vt X t,v t )dt + f 1 (t, X t,v t ) db (1) t + µ dt + 1 f 11(t, X t,v t )v t dt + f (t, X t,v t ) vt db () t + a(b v t )dt + 1 f (t, X t,v t ) v t dt + f 1 (t, X t,v t ) v t dt f 1 (t, X t,v t ) v t db (1) t + f (t, X t,v t ) v t db () t +(Af)(t, X t,v t )dt v t 1

2 where the di erential oerator A is defined as (Af)(t, x, y) f(t, x, y)+ µ + y f 1 (t, x, y)+ y f 11(t, x, y)+a(b y)f (t, x, y) y f (t, x, y)+ yf 1 (t, x, y) If we now let u(x) e i x,thenthe(multidimensionalformofthe)eynman-kacreresentation theorem imlies f(t, x, y) [u(x T ) X t x, v t y] [e i X T X t x, v t y] is the unique bounded solution of the artial di erential equation subject to the terminal condition (Af)(t, x, y) 0, 0 ale t ale T, x R, y R, (1) f(t,x,y)e i x, x R, y R Note that f(0,x,y)[e i X T X 0 x, v 0 y] ' XT ( ) isthecharacteristicfunctionofx T Guided by the form of the terminal condition and by our exerience with the Ornstein- Uhlenbeck characteristic function, we guess that f(t, x, y) can be written as f(t, x, y) ex{ (t)y + (t)} ex{i x} () for some functions (t) and (t) oft only satisfying (T )0and (T )0 i erentiating we find f(t, x, y) [ 0 (t)y + 0 (t)]f(t, x, y), f 1 (t, x, y) i f(t, x, y), f 11 (t, x, y) f(t, x, y), f (t, x, y) (t)f(t, x, y), f (t, x, y) (t)f(t, x, y), f 1 (t, x, y) i (t)f(t, x, y), so that substituting into the exlicit form of (Af)(t, x, y) 0andfactoringoutthecommon f(t, x, y) gives [ 0 (t)y + 0 (t)] + i µ or equivalently, ale 0 (t)+(i y a) (t)+ y + a (t)(b y)+ (t) (t) i y + i (t)y 0, y + 0 (t)+i µ + ab (t) 0 Since this equation must be true for all 0 ale t ale T, x R, andy R, theonlywaythatis ossible is if the coe cient of y is zero and the constant term is 0 Thus, we must have 0 (t)+(i a) (t)+ (t) i 0 and 0 (t)+i µ + ab (t) 0 (3)

3 The first equation in (3) involves (t) onlyandisoftheform with 0 (t) A (t)+b (t)+c A a i, B, C i + (4) This ordinary di erential equation can be solved by integration; see xercise 1 below The solution is given by (t) + tan(t+ G) where A r C B, B A 4B, B B r C B and G is an arbitrary constant The terminal condition (T )0imlies 0 + tan(t + G) so that G arctan T which gives (t) + tan arctan A 4B, (5) (6) xercise 1 Suose that a, b, andc are non-zero real constants Comute Z dx ax + bx + c Hint: Comlete the square in the denominator The resulting function is an antiderivative of an arctangent function In order to simlify the exression for (t) givenby(6)above,webeginbynotingthat cos arctan and sin arctan + + (7) Using the sum of angles identity for cosine therefore gives cos arctan cos arctan cos ( ) + sin arctan sin ( (T + cos ( ) sin ( (T + t)) cos ( ) sin ( ) (8) + t)) 3

4 Similarly, the sum of angles identity for sine yields cos ( ) sin ( ) sin arctan (9) + Writing tan(z) sin(z) and using (8) and (9) imlies cos(z) cos ( ) sin ( ) tan arctan cos ( ) sin ( ) cot ( ) cot ( ) so that substituting the above exression into (6) for (t) gives ale cot ( ) (t) + cot ( ) ( + ) cot ( ) The next ste is to substitute back for,, and in terms of the original arameters It turns out, however, that it is useful to write them in terms of Thus, substituting (4) into (5) gives Since we conclude that a (t) ( + i )+(a i ) (10) i, i, and i (11) + i cot The final simlification is to note that i + i + i (a i ) cos( iz) cosh(z) and sin( iz) i sinh(z) so that which gives cot(iz) cos(iz) sin(iz) cosh(z) i sinh(z) i coth(z) (t) i coth i + (a i ) coth i + +(a i ) inally, we find 8 < ex{ (t)y} ex : coth (i + )y +(a 9 i ) ; (1) 4

5 Having determined (t), we can now consider the second equation in (3) involving 0 (t) It is easier, however, to maniulate this exression using (t) in the form (6) Thus, the exression for 0 (t) nowbecomes 0 (t) ab i µ ab tan arctan which can be solved by integrating from 0 to t Recallthat Z tan(z)dz log(sec(z)) log(cos(z)) and so (t) (0) abt i µt ab (0) abt i µt Z t 0 tan arctan ab log cos(arctan cos(arctan (T s) ds! T) ) The terminal condition (T )0imliesthat (0) abt + i µt + ab log + cos(arctan! T) using (7), and so we now have (t) ab+i µ+ ab log + cos(arctan! ) As in the calculation of (t), we can simlify this further using (8) so that log cos ( ) (t) ab+i µ+ ab sin ( (T t)) which imlies ex{ (t)} ex{ab+i µ} cos ( (T t)) sin ( (T t)) ab (13) Substituting the exressions given by (11) for,, and in terms of the original arameters into (13) gives n o ex ab(a i ) + i µ ex{ (t)} ab cos i a i i sin i 5

6 As in the calculation of (t), the final simlification is to note that cos( sin( iz) i sinh(z) sothat n o ex ab(a i ) + i µ ex{ (t)} cosh + a i sinh iz) cosh(z) and ab (14) We can now substitute our exression for ex{ (t)y} given by (1) and our exression for ex{ (t)} given by (14) into our guess for f(t, x, y) givenby()toconclude f(t, x, y) ex{ (t)y + ex i x (t)} ex{i x} coth( cosh (i + )y )+(a i ) + ab(a i ) + i µ + a i sinh (T ab t) Taking t 0gives and we are done! ' XT ( ) f(0,x,y) n ex i x (i + )y abt (a i ) + coth T +(a i ) cosh T + a i sinh T o + i µt ab 6