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CH s 11 - B1 - H KTQD Chuyn Photocopy - nh my - In Lun vn, Tiu lun (: 6.280.688

A.T VN .

Trong nhng nm gn y, cng vi s si ng ca th trng chng khon, th vng v du m l hai mt hng rt c gii u t quan tm. Do c tnh vn c ca mnh, vng tr thnh cng c ct tr an ton trong nhng trng hp th trng bin ng. Mt khc, gi vng lin tc bin i, nhiu nh u t a vng vo danh mc u t ca mnh a dng ha danh mc v phng h ri ro. Tuy nhin gi vng vn hng ngy bin ng v bin ng ht sc phc tp khng th d on trc c, do rt kh khn cho cc nh u t trong vic nh gi ri ro ca gi vng. Chnh v vy, qua qu trnh nghin cu v c s hng dn ca PGS.TS Nguyn Quang Dong em la chn ti S dng m hnh ARIMA v m hnh GARCH trong phn tch gi vng nhm c lng v ri ro ca gi vng.

Do hn ch v nhn thc v thi gian nghin cu nn bi vit ca em cn rt nhiu thiu st. Em rt mong nhn c s hng dn ca thy gio bi vit ca em hon thnh hn.

Em xin chn thnh cm n PGS.TS. Nguyn Quang Dong gip em hon thnh ti ny.

B. NI DUNG.

I. L thuyt v m hnh ARIMA v m hnh GARCH.

Trong th trng ti chnh c bit l trong th trng chng khon, vn ri ro v qun l ri ro l mt vn ht sc thit yu. Khi xt phng sai ca mt ti sn ti chnh th phng sai ny c trng cho ri ro ca ti sn.

Vic p dng cc m hnh kinh t lng vo phn tch phng sai ca cc ti sn ti chnh gip ta tr li cho cu hi mc dao ng trong li sut khc nhau liu c ph thuc vo s thay i li sut trong qu kh v mc dao ng ca s thay i ny hay khng?

Vi cc m hnh m t phng sai c iu kin ca sai s thay i bao gi cng gm hai phn.

Phn 1: M t li sut trung bnh.

Phn 2: M t c ch thay i ca phng sai.

t = E(Rt / Ft-1).

2t =Var(Rt / Ft-1).

Ft-1 l tt c cc thng tin c ti thi k (t-1).

Rt = t + ut.

Rt c m t bng qu trnh ARMA(p,q).

Rt = o + i Rt-i + ut + o ut-j

t = o + i Rt-i - o ut-j.1. M hnh ARIMA.

M hnh ARIMA(p,d,q) trong : p l bc t hi quy, d l s ln ly sai phn chui Yt c mt chui dng, q l bc trung bnh trt. p v q l bc tng ng ca chui dng.

Gi s chui quan st l chui lin kt bc 1 (I(1)) th ta c m hnh ARIMA(p,1,q) c biu din nh sau:

Yt = 0+i Yt-i + q ut-q.

2. M hnh ARCH.

M hnh ARCH c dng:

Trong : Rt l li sut ca ti sn ti thi im t. . l bin ngu nhin c lp c cng phn b v c E() = 0, Var( ) =1.

Phng sai di hn

3. M hnh GARCH.

M hnh GARCH(m,s) c dng:

Trong : ,

Phng sai di hn

II.S dng m hnh ARIMA v m hnh GARCH trong phn tch gi vng.

1.S liu v ngun gc s liu.

S liu s dng trong bi l gi vng ca th trng London c quan st theo thng, t thng 1 nm 1968 n thng 8 nm 2007.

Bn cht s liu l s liu chui thi gian.

S dng phn mm Eviews 4.0 v th gi vng t thng 1 nm 1968 n thng 8 nm 2007 ta c kt qu sau:

T hnh v, ta thy tc tng trng ca gi vng trc nm 1980 nhanh v c phn tng vt khng n nh. T sau nm 1980 gi vng lc tng lc gim nhng theo xu hng i ln v vi bin nh hn trc . T nm 2006, gi vng ang c xu hng tng ln. C rt nhiu nguyn nhn lm cho gi vng tng nhanh. Th nht l do nhu cu v vng ngy cng tng, trong khi lng cung cp li hn hp. Th hai l s leo thang ca gi du trong nhng nm gn y. Ngoi ra cn c cc nguyn nhn khch quan nh tnh hnh chnh tr bt n, s ln gi ca ng la hay nhng nguy c v lm pht, v bin ng trong li sut...Cng t hnh v ta thy chui gi vng l chui khng dng.

2. Kim nh tnh dng ca chui li sut ca vng.

{St} l chui gi vng. Li sut ca vng c tnh theo cng thc ghp li lin tc:

Rt = ln

a. V th.

S dng phn mm Eviews 4.0 v th chui li sut gi vng t thng 1 nm 1968 n thng 8 nm 2007 ta c kt qu sau:

Hnh v cho thy li sut Rt dao ng trong khong . thi gian u li sut bin ng rt mnh, nht l trong giai on thp nin 70, 80. Nguyn nhn c l l do cuc suy thoi kinh t trong giai on ny tc ng lm cho gi vng thay i tht thng. n nhng giai on sau th li sut bin ng u n hn.T hnh v, ta thy Rt l chui dng v khng c h s chn.

Thng k m t i vi chui li sut vng.

b.Kim nh tnh dng ca chui li sut vng

Gi thit: Ho: Chui khng dng.

H1 : Chui dng.

S dng Eviews 4.0 vi kim nh Dickey _ Fuller cho chui li sut vng ta c kt qu sau:

T bng trn ta c |qs | = 9.163057 > , vi mi mc ngha =1%, = 5%, = 10% ta kt lun chui li sut ca vng l chui dng.

Kt qu c lng: DW = 1.999820 cho bit ut khng t tng quan.

3. c lng cc tham s ca m hnh ARIMA.

nh dng m hnh ARIMA i vi li sut vng bng lc tng quan.

T lc tng quan ta thy p=1, p=2 v q=1 do ta c lng m hnh ARIMA(2,0,1) nh sau:

T kt qu c lng ta thy h s ca AR(1) v AR(2) bng 0. Ta c cc kim nh sau:

Ho: c(2) =0.

H1: c(2) 0.

T kt qu kim nh ta thy kim nh F c Pvalue > 0.05 v kim nh 2 c Pvalue > 0.05 nn chp nhn gi thit H0 hay h s ca AR(1) bng 0 c ngha thng k.Ho: c(3) =0.

H1: c(3) 0.

T kt qu kim nh ta thy kim nh F c Pvalue > 0.05 v kim nh 2 c Pvalue > 0.05 nn chp nhn gi thit H0 hay h s ca AR(2) bng 0 c ngha thng k.

Trc ht ta b bin AR(2) v c lng m hnh ARIMA(1,0,1) ta c kt qu c lng nh sau:

Kt qu c lng cho thy h s ca AR(1) bng 0. Ta c kim nh sau:

Ho: c(2) =0.

H1: c(2) 0.

T kt qu kim nh ta thy kim nh F c Pvalue > 0.05 v kim nh 2 c Pvalue > 0.05 nn chp nhn gi thit H0 hay h s ca AR(1) bng 0 c ngha thng k.

Ta b bin AR(1) trong m hnh v c lng m hnh ARIMA(0,0,1) nh sau:

Kt qu c lng cho thy cc h s u khc 0 c ngha thng k do cc mc xc sut Pvalue u nh hn mc ngha = 5%.

Thng k DW = 2.059221 nn ut khng t tng quan.

M hnh c dng: rt = 0.006181 + 0.378496*ut-1.

4.c lng m hnh ARCH, GARCH.

4.1.M hnh ARCH(p)

a.c lng tham s p.

T phng trnh ARIMA(0,0,1) c lng trn ta ghi li phn d, k hiu l e. Kim nh tnh dng ca chui phn d ny.

V th:

T th ta thy chui phn d dng v c h s chn. Kim nh tnh dng bng kim nh DF:

Kt qu kim nh cho thy |qs | = 9.410290 > , vi mi mc ngha =1%, = 5%, = 10% ta kt lun chui phn d l chui dng hay phn d l nhiu trng.Kt qu c lng: DW = 1.997882 cho bit ut khng t tng quan.

Vy m hnh ARIMA(0,0,1) c lng trn tn ti.

Xc nh p.To bin e2 l bnh phng ca phn d thu c m hnh ARIMA(0,0,1) c lng trn, s dng lc tng quan vi chui ny xc nh h s p ca m hnh ARCH.

T lc tng quan ta thy p=1, p=4, q=1, q=2, q=3, q=4, q=5 do ta c lng m hnh ARCH nh sau:

Nhn vo kt qu c lng ta thy h s chn, h s ca AR(1), MA(1), MA(2) bng 0 (do xc sut Pvalue > 0.05) v thng k DW =1.885166 do ut t tng quan.

Ta tin hnh hiu chnh m hnh bng cch ln lt b i tng bin m h s ca chng trong m hnh bng 0 v hiu chnh ut khng t tng quan. Cui cng ta thu c m hnh nh sau:

Kt qu c lng cho thy cc h s u khc 0, thng k DW=2.033377 nn ut khng t tong quan.

Nh vy c s sai khc trong dao ng li sut trung bnh trong cc phin.

H s ARCH(1) dng thc s nn dao ng ca li sut trung bnh chu nh hng dng ca li sut vng.

Ta c th kim nh cc gi thit ca m hnh ARMA(1)

Ho: c(2)=0

H1: c(2)>0

T kt qu kim nh ta thy kim nh F c Pvalue < 0.05 v kim nh 2 c Pvalue < 0.05 nn chp nhn gi thit H1 do h s chn > 0.

Ghi li phn d ca m hnh, t tn l e3 v kim nh tnh dng bng kim nh DF.

Kt qu kim nh cho thy |qs | = 9.408831 > , vi mi mc ngha =1%, = 5%, = 10% ta kt lun chui phn d l chui dng hay phn d l nhiu trng.

Kt qu c lng: DW = 1.997923 cho bit ut khng t tng quan.

Vy m hnh tha mn mi gi thit ca m hnh l thuyt.

4.2. M hnh GARCH.

Kt qu c lng cho thy cc h s u khc khng c ngha thng k. Thng k DW=1.931554 nn ut t tng quan.Cc kim nh:Ghi li phn d ca m hnh t tn l e4 ri kim nh tnh dng.Ho: t l nhiu trng

H1: t khng phi l nhiu trng

Kt qu kim nh cho thy |qs | = 9.383637 > , vi mi mc ngha =1%, = 5%, = 10% ta kt lun chui phn d l chui dng hay phn d l nhiu trng.

Thng k DW = 1.998452 nn ut khng t tng quan.

Ho: c2=0

H1: c2>0

Kt qu kim nh cho thy kim nh F c Pvalue < 0.05 v kim nh 2 c Pvalue < 0.05 nn chp nhn gi thit H1 do h s chn > 0.

Ho: c3=0

H1: c3>0

Kt qu kim nh cho thy kim nh F c Pvalue < 0.05 v kim nh 2 c Pvalue < 0.05 nn chp nhn gi thit H1 do h s ca ARCH(1) > 0.

Ho: c4=0

H1: c4>0.

Kt qu kim nh cho thy kim nh F c Pvalue < 0.05 v kim nh 2 c Pvalue < 0.05 nn chp nhn gi thit H1 do h s ca GARCH(1) >0.

Ho: c3+c4=1

H1: c3+c40.05 v kim nh2 c Pvalue >0.05 nn chp nhn gi thit Ho.

Nh vy th m hnh c lng trn khng c phng sai khng iu kin nn s bin ng ca ri ro l ht sc tht thng.

By gi ta s xem xt s nh hng ca cc c sc i vi li sut ca gi vng. Cc m hnh c lng trn c nhc im l khng xem xt c s nh hng ca c sc m, c sc dng i vi li sut. xem xt s nh hng ny bng m hnh TARCH.

Ta thm vo bin gi:

khi ut-1 < 0

khi ut-1 0

Phng sai di hn:

iu ny a ra s tc ng bt cn xng ca cc c sc ut-1.

-1 nu= 1

-1 nu= 0

Ta c kt qu c lng sau:

Ho: c4 = 0

H1: c4 0.

Kt qu kim nh cho thy rng kim nh F c Pvalue