Exercise: Show that. Remarks: (i) Fc(l) is not continuous at l=c. (ii) In general, we have. yn ¾¾. Solution:
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1 Exercie: Show ha Soluio: y ¾ y ¾¾ L c Þ y ¾¾ p c. ¾ L c Þ F y (l Fc (l I[c,(l "l¹c Þ P( y c <e P(ce<y<c+e ³ P(c e <y c+ e TC Remar: (i Fc(l i o coiuou a lc. (ii I geeral, we have y ¾¾ p y Þ y ¾¾ L y. The la exercie hereore eablihe he equivalece o covergece i probabiliy ad covergece i law (diribuio i he cae o covergece o a coa. F (c+ e F (c e y y Fc(c+ e Fc(c e
2 The mea o a ample x,,x rom a AR( proce (xm(xm+u wih mea m, ARparameer aiyig <, iovaio u wih variace >, ad auocovariace g(e(xm(x+m i a coie eimaor o m. We eablih (wea coiecy by uig he ac ha mea quare covergece implie covergece i probabiliy. We have x E( x E Ex m For rog coiecy, almo ure covergece i required raher ha covergece i probabiliy. ad x Var( x E( m E( ( x m ( E(xm + +E(xm +E(xm(xm+ +E(xm(xm M +E(xm(xm +E(xm(xm+ +E(xm(xm M +E(xm(xm ( g ( [g(+(g(+ +g(] g ( g(. TM
3 Now aume ha m ad he iovaio u are a i.i.d. equece o N(, radom variable. Exercie: Show ha E x <. Hi: Ue he MA( repreeaio o he AR( proce a well a (which will be how laer. x u xx+u Euuulum E u 3 Soluio: E x E ( u l m 3 3 ( l m l m Eu 3 ( < l u m u l u m T 3
4 Exercie: Show ha Euuulum E u 3. Soluio: ¹,l,m: Euu ulum EuEu ulum ¹lm: Euuulum E u E u l Lemma: lm: Euuulum E u 3 E xx+hxx+h g (h +g ( +g (+hg (h Proo: For coveiece, we rewrie he MA( repreeaio o he AR( proce a xx+u x y u, where y or ³ ad y or <. E xx+h xx+h E y u y u + h y u y u + h E y u ψ u ( + h ψ u ( + ( ψ u ( + (+ Ey ψ hu ψ ( u ψ ( + h u u y y hy l( y m( + h Eu uulu l m Slm ¼ + SS¹lm ¼ + SSl¹m ¼ + SSm¹l ¼ Slm ¼ + (SS, lm ¼ Slm ¼ + ( ( +( ( ( ( + (ψ ( ψhψ( + h Y ψ + ( ψ ( + h hψ( ψ h ψ Y TG m
5 Exercie: Show ha or ay equece w(,,w, w ( w. S ( I i i ow ha m, he impliied ample covariace ˆ g ( h h h x x + h h h i a coie eimaor o g (h, becaue E( ˆ g ( h h ad h Var( ˆ g ( h E( h ( h h h ( h h h ( x m ( x + h m, h>, m ( + h h E( x x m h x x + h E xx+hxx+h g (h g (h (h g (hg(h [g (h +g ( +g((+hg ((h]g (h ( h h h h ( h (( ( h h h h [g ( +g ((+hg ((h] (( h ( h (g ( +g(+hg(h + h φ σ + φ (( + h φ σ φ h ( Noe ha ½+h½+½h½, becaue ½+h½+½h½ ½(+h+(h½½½. ( h. Remar: We have how above ha he ir ad ecod ample mome o a Gauia AR( proce coverge i probabiliy o heir heoreical couerpar. Such a proce i aid o be ergodic or he ir ad ecod mome. V 5
6 The OLSeimaor o he parameer i give by ˆ x x x The radom variable aiy ad x ( x + u x z u x EzEuxEuEx x u +. x E ( z z, z,... E ( u x ux,u x3,... Eu E x u x,u x,..., ( 3 becaue u i idepede o x ad ux, ux3, Thu, he equece o radom variable z i a marigale dierece equece. O Exercie: Show ha he equece o radom variable z i whie oie. Exercie: Show ha he equece o radom variable (u σ w x i a marigale dierece equece. W D Exercie: Show ha he equece o radom variable w i whie oie. N Remar: Sice he radom variable u are Gauia, he equece o radom variable z ad w, repecively, are o oly wealy aioary bu alo ricly aioary, which mea ha he oi diribuio o ay iie ubequece o radom variable wih idice,, i he ame a ha o he radom variable wih idice +,,+. 6
7 Wea law o large umber ae codiio uder which he ample mea coverge i probabiliy oward he populaio mea. The wea law o large umber or ricly aioary marigale dierece equece e require oly abolue iegrabiliy. Thu, E z ad E w E u x E u E x E( + u E( + x E u σ x Eu Ex + σ Ex Þ z ¾¾ p < Þ w ¾¾ p. The ceral limi heorem or marigale dierece equece ae ha > i Eε, Eε <, e ¾¾ L ε ¾¾ p N(, Eε Eε. < Thu, becaue ad z ( z E z E(u x Eu E x E z E(u x E z u E x u x ( σ ¾¾ p + σ z φ ¾¾ L N(,, 3 E x <, E. >, + L 7
8 8 Now i i raighorward o derive he aympoic diribuio o he OLSeimaor ˆ. The umeraor o he aiic ( ˆ x u x coverge i law o N(, ad i deomiaor coverge i probabiliy o φ σ, hece (ˆ ¾ ¾ L plim x N(, x p lim N(, N(, N(, ø ö ç è æ N(,. A
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