[PDF] Séries temporelles Solution succincte de l'exercice

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année population

1800185019001950

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temps température

1920 1925 1930 1935 1940

30 40 50 60

année passagers

195019521954195619581960

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`1 X

N(0;1) Y=XU=1XU=0ɍU

ĕ1/2 X

XY ā

(X;Y) = 0 XY

P(Y2I) =P(X2I)1

2+P(X2I)1

2=P(X2I)X=X

(X;Y) =E(XY) =E(X2)E(U=1)E(X2)E(U=0) = 0 (YjX=x) =1

2(x+x)6=(Y)XY

X0=XXt="tXt6= 0 ɍ ("t)t6=0 P("t=

1) = 1/2 X (Xt) = 0(Xt) = 1 t

s6=t s6= 0t6= 0(Xs;Xt) =E(X2)E("s)E("t) = 0

X= (Xt)t2N

Xt=+Xt1+Zt t1 ɍX0= 0 ɍ(Zt)t2N

XX X (Xt)t2N st X(t;t+h) =E((Xtt)(Xt+h(t+h))) =E(St(St+S0h)) =E(S2t) =2t

Xt=XtXt1=+Zt

Z= (Zt)t2Z Zt=Xt+Yt t2Z

X Y

Z(t;t+h) =

X(t;t+h)+

Y(t;t+h)+(Xt;Yt+h)+(Yt;Xt+h) =

x(h)+ Y(h) t Z

Xt=Ztt Xt=Zt+ 1t (Zt)t2Z

Xt=Z1++Ztɍ(Zt)t2Z

Xt=Zt+Zt1 ɍ(Zt)t2Z 2R

Xt=ZtZt1ɍ(Zt)t2Z

Yt= (1)tZtXt=Yt+Ztɍ(Zt)t2Z

X

X(t;t+h) =E((Z1++Zt)2) =2t t

(1)X= 0

X(t;t+h) =2(1 +2)h=0+2jhj=1

X= 0

X(t;t+h) =4h=0

X= (Xt)t2Z

t2ZXt=A(t)+B(t)ɍAB

2 ɍ2R X

X(t) =E(A)(t) +E(B)(t) = 0

X(t;t+h) =E(A2)(t)((t+h)) +E(B2)(t)((t+h)) +

=0z}|{ E(AB) =2Re(iti(t+h)) =2Re(ih) =2(h) t ĕ (0) = 1 (h) =jhj= 1 (h) = 0 (h) = 1h= 0 (h) = 1/hh6= 0 (h) = 1 +(h/2) (h) = (1)jhj (h) =(Xh;X0) =(X0;Xh) = (h) n1v2Rn (hv; nvi=)X 1j;kn vj (kj)vk=X 1j;kn (vjXj;vkXk) = 0 nX j=1 vjXj 1 A0; n0 n= ( (jk))1j;kn= 0 BBBB@ 1 1 1

CCCCA:

jj 1/2

X(h) =2(1 +2)h=0+2h=1

jj 1/2 ĕ n v2Rn hv; nvi<0 >1/2 hv; nvi=X 1j;kn (1)j+k n(j;k) =n2(n1)= (n1) n n12 <0jj>1/2ĕ n >2

21 <1/2 v= (1;1;1;:::)

2= 1 1 ;:::;n= 0 BBBB@ 1 1 1 ::: 1 1

CCCCA;:::

n n t=I+AɍA n= (1)In+Jn Jn= (1)1j;kn

1 0 n1(Jn) =n

1 n 1

n11+n 1 n2 2[0;1]

A2 Mn(C) :(A)Sn

k=1D(Ak;k;P j6=kjAk;jj) (X0)(X1) = 1 t t2 (1;:::;1)t 0 B@ 1 1 1

CA=t(1 + (t1))0

01

X(0;1)

X= (1)h=0 Y

X (Y0) =Yt=Y0 t2Z

X+Y= (1)h=0+

Yt=t+st+Utɍ2R ɍst 4

ɍU= (Ut)t2Z

(Yt)t2Z

Z= (1B4)Y

U t+stā = 0 s6= 0s 4 (1B4)Y(t;t+h) = (1B4)U(t;t+h) =(UtUt4;Ut+hUt+h4) = U(h) U(h4)

U(h+ 4) +

U(h) (1B4)U U

FUɍh=h=0h=4

Xt= kX i=0 aiti+Ut;

ɍ ai R(Ut)

(1B)p(Xt)p2N

ĕ Yt=Xt+StɍSt d

(Yt) (1B)Xt=Pk ti(t1)i=ti iX j=0 tj(1)iji j i1X j=0 tj(1)1+iji j i1 (1B)X k1 (1B)pX kp (1B)p < p (1Bd)Xt=XtXtd+StStd=XtXtd: (Xt)t2Z P i2Zjaij<1 Yt=n;m!1Pm i=naiXti

Yt2L1 t2Z Yt

Yt2L2 t2Z

(Yt)

E(Y) =Y=X

X i2Z ai;

Y(h) =X

i2Z X j2Z aiaj

X(h+ji):

IZ YIt=P

i2IaiXti

EYItYJt

i2ZE(jXij)X i2(I[J)nI\J jaij: " >0 n2N fn;:::;ng I\JP i2(I[J)nI\Jjaij " ĕ P i2Zjaij<1 L1 Yt=P i2ZaiXti=P

I!ZYIt L1

ś L1

ĕ Ĝ [0;1]

E X i2Z jaijjXtij t2ZE(jXtj)X i2Z jaij E(X20)kak1<1; P i2ZjaijjXtij<1 P i2ZaiXti

L1 Y

Yt ĕ

Ĝ E[PjaiXtij] =PjaijE[jXtij]Pjaij

q 2X+ X(0)<

1 ĕ Ĝ

Yt:=PaiXti

aiXtiZ

Ŀŀ Lp1

YItYJt

pX i2(I[J)nI\J kaiXikp i2ZkXikp X i2(I[J)nI\J jaij: E

XaiXti

2 X i;j jaiajjE[jXtiXtjj] Xjaij 2(2X+

X(0))<1

t2Z L2

Y(t) =h1;YtiL2

1;I!ZYIt

L2 =I!Z 1;X i2I aiXti L2 =I!Z X i2I aiE(Xti) =X X i2Z ai

Y(t;t+h) +2Y=

I!ZYIt;J!ZYJt+h

L2 =I;J!Z X i2I;j2J aiajE(XtiXt+hj) =I;J!Z X i2I;j2J aiaj(

X(h+ij) +2X)

=2Y+X i;j2Z aiaj

X(h+ij)

=2Y+X i;j2Z aiaj

X(h+ji)

t Y (Zt)t2Z0;2 t2N Xt= tX i=0 i(ZtiZti1):

2R (Xt)t2N

(XtYt)L2!t!10: (Xt)t2N

E(Xt) = 0 t2N

Xt ā

(Xt) =E[X2t]E[Xt] = 0; =E 2 4 tX i=0 i(ZtiZti1) !23 5 tX i;j=0 i+jE[(ZtiZti1)(ZtjZtj1)] =2 tX i;j=0 i+j(2i=ji=j+1i=j1) =2 tX i=0 22i
t1X i=0 2i+1 tX i=1 2i1 1X i=0 0X i=1 :::= 0 = 22 tX i=0 2i t1X i=0 2iquotesdbs_dbs48.pdfusesText_48

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