John Geweke a and Gianni Amisano b a Departments of Economics and Statistics, University of Iowa, USA b European Central Bank, Frankfurt, Germany

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1 Herarchcal Markov Normal Mxure models wh Applcaons o Fnancal Asse Reurns Appendx: Proofs of Theorems and Condonal Poseror Dsrbuons John Geweke a and Gann Amsano b a Deparmens of Economcs and Sascs, Unversy of Iowa, USA b European Cenral Bank, Frankfur, Germany and Unversy of Bresca, Bresca, Ialy December, 008 1

2 Proofs of Theorems Proof of Theorem 1 Usng he mehods of Ryden e al. (1998) for he Markov normal mxure model, cov (y ; y s x 1 ; : : : ; x T ) = 0 B s0 = 0 B s (s = 1; ; : : :), (1) where =dag(); B = P e m1 0, whch esablshes su cency. If he egenvalues of P are dsnc hen P s dagonable and has specral decomposon P = Q 1 Q, where he marx =dag ( 1 ; : : : ; m1 ) conans he ordered egenvalues of P, 1 3 : : : m1. The marx Q has orhogonal columns and we may ake Q = [q 1 ; q ; : : : ; q m ] 0 = [; q ; : : : ; q m ] 0, Q 1 = q 1 ; q ; : : : ; q m = e m ; q ; : : : ; q m. If P s also rreducble and aperodc hen 1 = 1 > and we may wre B = Q 1 Q q 1 q 0 1 = Q 1 e Q () where e = dag (0; ; : : : ; m1 ). From (1) absence of seral correlaon s equvalen o 0 Q 1 e s Q =0 (s = 1; ; : : :). The rs elemen of Q s q 0 1 = 0 =0, and so 0 Q 1 s Q =0 (s = 1; ; : : :). (3) De ne he m 1 m 1 marx D = m1 1.. m 1. m 1 1 m 1 m 1 m , m m1 1 whose deermnan s ( ) 6= 0 (Rao (1965), p 8). Le A = D 1 and le ; denoe he Kronecker dela funcon; hen < m 1 s=1 a s s = ; =) m 1 m 1 a s s = I m1 ( = 1; : : : ; m 1 ), s=1

3 and from (3) m 1 m 1 a s 0 Q 1 s Q = 0 = s=1 = 0. Proof of Theorem The nsananeous varance marx 0 s mmedaely aaned by consderng h h 0 0 = E z z = E z z 0 0 h = z z 0 s = 0 = R + m 0 u 0 : The dynamc covarance marces (p > 0) are obaned by condonng on s and s u, explong seral ndependence of observables afer condonng on he saes, and hen by margnalzng ou he saes: u = cov z ; z u = E z z 0 u 0 = E z z 0 u s = ; s u = [P u ] M 0 M 0 = = E z s = E where B u = (P e m 0 ) u = P u e m 0. Proof of Theorem 3 z 0 u s = [P u ] M 0 M 0 0 [P u ] e 0 mm 0 = M B u0 M 0 ; Adop he noaon n he proof of Theorem. From (), B u = P m = u q q 0. Subsung n he expresson for n he saemen of he heorem, where u = u r+1 u M q q 0 M 0 = u A 0 (u = 1; ; 3; : : :) = A 0 = hh M q h q h0 M 0, H = h : q 0 hp = q 0 h; M q h 6= 0.: = 3

4 Observe ha r s he number of dsnc egenvalues of P wh modulus n he open un nerval assocaed wh as leas one column of Q 0 no n he column null space of M. In oher words, r can be less han m 1 because some egenvalues are equal o zero (as n he compound Markov model nerpreed as havng m = m 1 m saes), because some egenvalues are repeaed, or because some egenvalues are assocaed wh columns of Q 0 all n he column null space of M. De ne now a sochasc process v wh auocovarances ~ u = P r+1 = u A 0 (u > 0) and ~ 0 = P r+1 = A0. Then for u > 0, ~ u = u, whle ~ 0 = r+1 A 0 = = 0 0 : Noce ha he marx ~ 0 0 = P m R s posve (sem) de ne, snce each R s a varance marx. Gven ha here are r dsnc egenvalues of P, ; : : : ; r+1, wh modulus n he open un nerval, conrbung o he deermnaon of u = ~ u, here exss a unque se of consans 1 ; : : : ; r such ha r r r = 0 ( = ; : : : ; r + 1) : The coe cens 1 ; : : : ; r deermne a degree r polynomal whose roos are 1 ; : : : ; 1 r. Thus for all u > r, ~ u r r+1 ~ u = u A 0 The auocovarance funcon of = = r+1 n = v u r r+1 u = r u herefore sas es he ule-walker equaons for a VAR(r) process wh coe cen marces I np o A 0 A 0 = 0: ( = 1; : : : ; r). 4

5 Deals of he Markov chan Mone Carlo algorhm Le s 1 = (s 11 ; : : : ; s T 1 ) 0. Then p s 1 = s11 T = p s 1;1 s 1 = s11 p T ; (4) where T s he number of ransons from perssen sae o n s 1. The n n Markov ranson marx P s rreducble and aperodc, and = ( 1 ; : : : ; m1 ) 0 s he unque saonary dsrbuon of fs 1 g. Le s = (s 1 ; : : : ; s T ) 0 denoe all T ransory saes. Then p s s 1 ; = T m s = U : (5) where U s he number of occurrences of s = (; ) ( = 1; : : : ; T ). The observables y depend on he laen saes s and he deermnsc varables x. If s = (; ) hen y = 0 x " ; " s N 0; (h h h ) 1 : (6) Condonal on (x ; s ) ( = 1; : : : ; T ) he y are ndependen. From (6) one expresson for hs dsrbuon s p (y s; ) = () T n= h T= m 1 exp 4 h h m h T n= h m :s =(;) h U n= 3 " =5 ; (7) The uncondonal mean of he ransory saes whn each permanen sae s 0 0, whch s equvalen o = 0 ( = 1; : : : ; m 1 ). Le C be an m (m 1) orhonormal complemen of, de ne he (m 1) 1 vecors e 0 = C 0, and noe ha = C e ( = 1; : : : ; m 1 ). Consruc he m 1 m m 1 (m 1) block dagonal marx C = Blockdag [C 1 ; : : : ; C m1 ] and he m 1 (m 1) 1 vecor e = e 0 1; : : : ; e 0 m 1 0. Then = C e, and subsung n equaon (7) a he end of Secon.1.1, y = 0 x + e 0 C 0 0z 1 + e 0 C 0 z + " : (8) 5

6 Ths expresson has he form y = 0 w + " n whch he (k + m 1 m 1) 1 vecor = 0 ; ~ 0 ; e 0 0 and w 0 = x 0 ; z 10 C 0 ; z 0 C : (9) Thus condonal on he laen saes s (equvalenly z 1 and z ) ( = 1; : : : ; T ), and gven he resrcons on he sae means, (6) s a lnear regresson model wh hghly srucured heeroscedascy. If we ake = h s1 h s, hen " # p (y s; ) = () T= h T= T = () T= h T= T exp " h n= exp n= T h " = # T (y w) 0 = : (10) The kernel of he pror densy s he produc of he followng expressons. h p () / exp 0 H = m (11) p (p ) / p r 1 1 ( = 1; : : : ; m 1 ) (1) p ( ) / r 1 ( = 1; : : : ; m 1 ) (13) p (h) / h ( 1)= exp s h= (14) p (h ) / h ( 1 1)= exp s 1h = ( = 1; : : : ; m 1 ) (15) p (h ) / h ( 1)= exp s h = ( = 1; : : : ; m 1 ; = 1; : : : ; m ) (16) p e h / h (m 1 1)= exp h h e 0 = e 1 1 = h (m 1 1)= exp h h e = p e h ; h / (h h ) (m 1)= exp h h h e 0 e = ( = 1; : : : ; m 1 ) (17) (18) 6

7 / h m 1(m 1)= p e h1 ; : : : ; h m ; h m 1 exp h h h e 0 e = h (m 1)= " = h m 1(m 1)= h (m 1)= exp n exp = h m 1(m 1)= m 1 1 h h h h (m 1)= e = h h e 0 [dag (h 1 ; : : : ; h m1 ) I m 1] e o = # (19) (0) (1) Condonal poseror dsrbuon of h. From (14), (18), (0) and (7), s h s () ; s = s + h e 0 e + h m 1 h e 0 e + T ", = + (m 1 1) + m 1 (m 1) + T: Condonal poseror dsrbuon of he h. From (15), (0), and (7), s h s ( ) ; s = s 1 + h h e 0 e m + h h ", :s =(;) = 1 + m 1 + nt ( = 1; : : : ; m 1 ). Condonal poseror dsrbuon of he h. From (16) and (7), s h s ( ) ; s = s + h h = + U :s =(;) ( = 1; : : : ; m 1 ; = 1; : : : ; m ). Condonal poseror dsrbuon of P. From (1), (7), and (4), T p (P) / s11 p r 1+T 1 exp h " = : 7 ",

8 Use a Meropols whn Gbbs sep for each for each row of P. Draw he canddae p s Bea (r 1 + T 1 ; : : : ; r 1 + T m1 ), and le C 0 be he orhonormal complemen of correspondng o he resulng P. Accoun mus be aken of he fac ha because " = y 0 x s z 10 C 0, e C0 s a funcon of and herefore of P. Le C 0 be he orhonormal complemen of and compue " = y 0 x s z 10 C e 0. The Meropols accepance rao s s 11 exp s11 exp h P T " = h P T " = If he canddae s acceped, hen P s updaed o P, o, and C 0 o C 0. The orhonormal complemen of C 0 of s no unque. As dscussed n Secon.1. nohng subsanve n he model depends on whch C 0 s used. However, f C 0 s no a smooh funcon of hen he canddae wll be reeced more ofen han f s, because C 0 e wll change more. To consruc a unque orhonormal complemen C ha s a smooh funcon of a vecor of probables wh P m = 1, noe ha (0; 1) wh probably 1 ( = 1; : : : ; m). Consruc a marx C as follows. The rs column of C s c 11 =, c 1 = 1, c 1 = 0 ( = 3; : : : ; m). The h column of C s c = ( = 1; : : : ; ), c +1; = P = +1, c = 0 ( = + ; : : : ; m). Consruc C from C by normalzng he columns o each have Eucldan lengh 1. Condonal poseror dsrbuon of R. From (13), (7), and (5), m p / k=1 r +U k 1 k exp h : :s 1 = " = Use a Meropols whn Gbbs sep for each for each row of R. Noe ha n " = y 0 x s z 0 C ~, C s a funcon of whenever s 1 =. Draw he canddae from Bea (r + U 1 ; : : : ; r + U ;m ). Le C be he orhonormal complemen of. For all for whch s 1 =, compue " = y 0 x s z 0 C ~ :The Meropols accepance rao s exp h P :s 1 = " = exp h P : :s 1 = " = The Meropols sep s used only afer he rs 1,000 eraons. Condonal poseror dsrbuon of. Recall ha y = w 0 + ", wh 0 = 0 ; ~ 0 ; ~ 0 and From (11), (17), (1) and (10), w 0 = x 0 ; z 10 C 0 ; z 0 C : () s N ; H 1 ; H = H + h 8. T w w 0

9 where wh he mean s = H 1 c wh H = 4 H H H 3 5, H = h hi m1 1and H = h h Dag (h 1 ; : : : ; h m1 ) ; c = c + h T w y, c 0 = 0 H 0 ; 0 0 : Drawng he sae marx S. The nal sep of he MCMC algorhm s he draw of he T marx of laen saes from s dsrbuon condonal on he parameers and observed and. De ne d = p [y s = (; ) ; x ; ] h = () 1= (hh h ) 1= exp hh h y 0 x = and m d = p (y s 1 = ; x ; ) = d. We draw s sp (s ; y; ) as a wo sep margnal-condonal, s 1 s P (s 1 ; y; ) followed by s s P (s s 1,; y; ). Frs, gven d ( = 1; : : : ; T; = 1; : : : ; m 1 ) and P, he algorhm of Chb (1996) draws s 1 s P (s 1 ; y; ) and provdes p (y ) as a byproduc of he compuaons. Then he ransory saes s are condonally ndependen wh P (s = s 1 = ; y ; x ; ) _ d. References Chb S Calculang poseror dsrbuons and modal esmaes n Markov mxure models. Journal of Economercs 75: Rao CR Lnear Sascal Inference and Is Applcaons. New ork: Wley. Rydén T, Teräsvra T, Åsbrnk S Sylzed facs of daly reurn seres and he hdden Markov model. Journal of Appled Economercs 13: