Extension of stochastic volatility models with Hull-White interest rate ...

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Extension of stochastic volatility equity models 103Downloaded by [Lech A. Grzelak] at 10:55 24 January 2012For this, we start with the integral for (k),Z sZ sðkÞdk ¼ ð x, iu þ 2 2 B v Þdk00¼ x, iu þ d s2g ð d Þðg 1Þþ log esd g2dg 1 g ¼ C 1 s þ C 2 log esd g, ðA13Þ1 gwhere C 1 ¼ { x, iu þ [( pd )/2g]}, C 2 ¼ [( d )( g 1)]/2dg, ¼ 2( x,v iu), d ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 8 2 and g ¼ ( d )/( þ d ). After substitution of these quantities, we find thatC 1 ¼ D/2 and C 2 ¼ 1.Next, we need to calculate the exponent of the integralof ,Z s exp ðkÞdk ¼ exp C 1 s þ C 2 log esd g01 g¼ exp sd 1 g2 e sd , ðA14Þgand we can include in the integral,Z Z sðsÞ exp ðkÞdk ds0Z 0Z 0¼ ð2B v þ x,r B x B r þ 2 r,v B r B v Þ0 d esd exp2 s gds:ðA15Þ1 gThis integral is split into three parts. The first part can besolved analytically,Z 2B v e ðd=2Þs e sd gds01 gZ 1 e sd ¼ 2b1 e sd e ðd=2Þs e sd gdsg 1 g¼ 2b1 g0e sd=2 e sd¼ 16b sinh 2 ðd=4Þð1 gÞd1 ds f 11 g : ðA16ÞThe second part can also be solved analytically,Z x,r B x B r e ðd=2Þs e sd gds01 gZ 1¼ x,r0 iuðiu 1Þð1 e s Þe sd=2 e sd gds1 g¼ Zx,riuðiu 1Þ e sd=2 ð1 e s Þðe sd gÞdsð1 gÞ 0¼ x,riuðiu 1Þð f 2 f 3 Þ, ðA17Þð1 gÞwheref 2 ¼ 2 d ðed=21Þþ 2gd ðe d=2 1Þ ðA18Þf 3 ¼ 2ðeð=2Þðd 2Þ 1Þ 2gð1 e ð=2Þðdþ2Þ Þ, ðA19Þd 2d þ 2and the third part readsZ 2 r,v B r B v e ðd=2Þs e sd gds01 g¼ 2 Z r,vB r B v e ðd=2Þs ðe sd gÞds1 g 0¼ 2 Zr,vðiu 1Þb e ð1=2Þsðdþ2Þ ðe sd 1Þðe s 1Þdsð1 gÞ 0¼ 2 r,vðiu 1Þbð f 4 þ f 5 Þ,ðA20Þð1 gÞwheref 4 ¼ 2d 24d þ 2d þ 2 ,f 5 ¼ðe ð1=2Þðdþ2Þ Þ 2e ð1 þ e d ÞdSo, finally, we haveZ Z B ðu, Þ ¼exp ðsÞds ðsÞ exp002e dd 2Z s0ðA21Þ2:d þ 2ðA22ÞðkÞdk ds¼ f 0 f 1 þ 1 x,riuðiu 1Þð f 2 f 3 Þþ 1 2 r,vbðiu 1Þð f 4 þ f 5 ÞÞ, ðA23Þwith f 0 ¼ e (d/2) /(e d g), f 2 and f 3 from (A18) and (A19),respectively, f 4 from (A21) and f 5 from (A22).Now we solve the ODE for A(u, ),dd A ¼ 2 B v þ B þ 1 2 2 B 2 r þ 1 2 2 B 2 þ r,B B r ,with solutionZ ðA24ÞZ Aðu,Þ Aðu,0Þ¼ 2 B v ds þ B ds þ 1 Z 00 2 2 Br 2 ds0þ 1 Z Z 2 2 B 2 ds þ r, B B r ds,00ðA25ÞorZ Aðu, Þ ¼ 2 B v þ 1 0 2 2 Br2 ds|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}A 1 ðu, ÞZ þ B þ 1 0 2 2 B þ r, B r ds : ðA26Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}ðu, Þ
104 L. A. Grzelak et al.Downloaded by [Lech A. Grzelak] at 10:55 24 January 2012In order to find A(u, ) we have to evaluate the integralsA 1 (u, ) and (u, ). Integral (u, ) involves a hypergeometricfunction (called the 2 F 1 function or simply theGaussian function), which is computed numerically here.For integral A 1 (u, ) we have two representations,orA 1 ðu, Þ ¼whereA 1 ðu, Þ ¼12 2 log g e sd 1g 11 esd glog221 gþ f 612 3 f 7,þ f 612 3 f 7,ðA27ÞðA28Þf 6 ¼ 1 ð d Þ, ðA29Þ42f 7 ¼ðiu 1Þ 2 ð3 þ e 2 4e 2Þ: ðA30ÞSince a complex-valued logarithm appears in A 1 (u, ), itshould be treated with some care. It turns out that thesecond formulation gives rise to discontinuities that maycause inaccuracies. According to Lord and Kahl (2006),an easy way to avoid any errors due to complex-valueddiscontinuities is to apply numerical integration.We know that the price of a zero coupon bondcan be obtained from the characteristic function SZHW (u, X t , t, T ) by setting u ¼ [0, 0, 0, 0] T . Therefore,Pðt,TÞ¼ð0,X t ,Þ¼ expZ Ttþ B v ð0,Þv t þ B ð0,Þ t Þ:Since er 0 ¼ 0, we havePð0, T Þ¼exps ds expðAð0,ÞþB x ð0,Þx t þ B r ð0,Þer tZ T0ðA31Þs ds expðAð0, ÞþB x ð0, Þx 0þ B v ð0, Þv 0 þ B ð0, Þ 0 Þ,and it is easy to check that B x (0, T ) ¼ 0, B v (0, T ) ¼ 0,B (0, T ) ¼ 0 andAð0, T Þ¼ 1 Z T2 2 B r ð0, sÞ 2 ds0¼ 2 3 12 3 2 2 e 2T þ 2e T þ T : ðA32ÞR TTherefore, Pð0, T Þ¼expð0 s ds þ Að0, T ÞÞ orlog ðPð0, T ÞÞ ¼R T0 s ds þ Að0, T Þ, which finally givesT ¼@@T logPð0,TÞþ @2Að0,TÞ¼fð0,TÞþ@T 2 ð1 e T Þ 2 ,2ðA33Þsince 0 ¼ f(0, 0) r 0 , where r 0 is the initial value of theinterest rate process r t . With u ¼ [u, 0,0,0] T , we find SZHW ðu, X t , t, T Þ¼expðeAðu, ÞþB x ðu, Þx t þ B r ðu, Þ~r tþ B v ðu, Þv t þ B ðu, Þ t Þ, ðA34ÞwitheAðu, Þ¼¼ðiu¼ð1Z TtZ Tt1ÞiuÞð1Z Ts ds þ iuZ TtZ Tts ds þ Aðu, Þtf ð0, sÞþ 2 2 ð1 e s Þ 2 ds þ Aðu, Þ2dðlogðPð0, sÞÞÞ þ ð1 iuÞ 2e s Þ 2 ds þ Aðu, Þ2 2¼ð1 iuÞ log Pð0, T Þ þð1 iuÞ 2Pð0, tÞ2 2 ðT tÞþ 2 ðe T e t 1 2TÞ ðe e2þ Aðu, Þ,2t ÞðA35Þand A(u, ) as in (A26). Now, by setting (x) ¼ exp(x/2)the discounted CF for the Scho¨bel–Zhu–Hull–White hybrid process is determined and the proof isfinished.œA.3. Proof of lemma 2.4Proof: As in the case of the SZHW hybrid model weneed to find the solution ofdd Aðu, Þ ¼ r 0 þ B T a 0 þ 1 2 BT c 0 B, ðA36Þdd Bðu, Þ ¼ r 1 þ a T 1 B þ 1 2 BT c 1 B: ðA37ÞFor the space vector X t ¼½ex t,er t , t Š T we haveanda 0 ¼r 0 ¼ 0, 1T2 S,r 2 ,0, , a 1 ¼2 306 7r 1 ¼ 4 1 5,0:¼ ðX t ÞðX t Þ T ¼2310 12674 0 0 5,0 0 23 t þ S,r 2 S,r x, t4 2 0 5: 2 tThis leads to23S,r 2 S,r 0c 0 ¼ 6 S,r 2 0 745 ,0 0 02ð0, 0, 1Þ ð0, 0, 0Þ3ð0, 0, x, Þ6c 1 ¼ 4 ð0, 0, 0Þ ð0, 0, 0Þ ð0, 0, 0Þ75:ð0, 0, x, Þ ð0, 0, 0Þ ð0, 0, 2 Þ
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