【正文】 )] = = ?2 = E[Xt E(Xt )][Xt2 E(Xt2 )] = E[Xt ][Xt2 ] = E[(ut +??1ut1+?2ut2 )(ut2 +?1ut3+?2ut4 )] = E[( )] = Solution (cont’d) 2 2212 11 ?? ? tt uu ???22121 ????? ?2211 )( ???? ?2 22 ?tu?22??課件 513 ?3 = E[Xt ][Xt3 ] = E[(ut +?1ut1+?2ut2 )(ut3 +?1ut4+?2ut5 )] = 0 So ?s = 0 for s 2. now calculate the autocorrelations: Solution (cont’d) ? ??0 00 1? ?? ??s s s? ? ? ?0 0 2)1()()1()(2221211222212211011 ????????????????????????)1()1()(222122222122022 ??????????????????課件 514 (iii) For ?1 = and ?2 = , substituting these into the formulae above gives ?1 = , ?2 = . Thus the acf plot will appear as follows: ACF Plot 0 . 60 . 40 . 200 . 20 . 40 . 60 . 811 . 20 1 2 3 4 5 6sacf課件 515 ? An autoregressive model of order p, AR(p) can be expressed as Or using the lag operator notation: Lyt = yt1 Liyt = yti ? or or where 4 Autoregressive Processes ? ? ? ?( ) ( . . . )L L L Lp p? ? ? ?1 1 2 2tptpttt uyyyy ?????? ??? ???? . ..2211??? ???pititit uyy1???????pittiit uyLy1??tt uyL ?? ?? )(課件 516 ? 平穩(wěn)性使 AR模型具有一些很好的性質(zhì) 。 如前期誤差項(xiàng)對(duì)當(dāng)前值的影響隨時(shí)間遞減 。 ? The condition for stationarity of a general AR(p) model is that the roots of 特征方程 all lie outside the unit circle. ? Example 1: Is yt = yt1 + ut stationary? The characteristic root is 1, so it is a unit root process (so nonstationary) ? Example 2: p241 ? A stationary AR(p) model is required for it to have an MA(?) representation. The Stationary Condition for an AR Model 1 01 2 2? ? ? ? ?? ? ?z z zp p. . .課件 517 ? States that any stationary series can be deposed into the sum of two unrelated processes, a purely deterministic part and a purely stochastic part, which will be an MA(?). ? For the AR(p) model, , ignoring the intercept, the Wold deposition is where, 可以證明 ， 算子多項(xiàng)式 R(L)的集合與代數(shù)多項(xiàng)式 R(z)的集合是同結(jié)構(gòu)的 ， 因此可以對(duì)算子 L做加 、 減 、 乘和比率運(yùn)算 。 Wold’s Deposition Theorem tt uyL ?)(?tt uLy )(????????????? ?3322112211)1()(LLLLLLL pp???????課件 518 ? The moments of an autoregressive process are as follows. The mean is given by* ? The autocovariances and autocorrelation functions can be obtained by solving what are known as the YuleWalker equations: * ? If the AR model is stationary, the autocorrelation function will decay exponentially to zero. The Moments of an Autoregressive Process pppppppp??????????????????????????????????. . .. . .. . .22112211212111???ptyE ????????? ?211)(課件 519 ? Consider the following simple AR(1) model (i) Calculate the (unconditional) mean of yt. For the remainder of the question, set ?=0 for simplicity. (ii) Calculate the (unconditional) variance of yt. (iii) Derive the autocorrelation function for yt. Sample AR Problem ttt uyy ??? ? 11??課件 520 (i) E(yt)= E(?+?1yt1) =? +?1E(yt1) But also E(yt)= ? +?1 (? +?1E(yt2)) = ? +?1 ? +?12 E(yt2)) = ? +?1 ? +?12 (? +?1E(yt3)) = ? +?1 ? +?12 ? +?13 E(yt3) An infinite number of such substitutions would give E(yt)= ? (1+?1+?12 +...) + ?1? y0 So long as the model is stationary, . , then ?1? = 0. So E(yt)= ? (1+?1+?12 +...) = Solution 1211 ??? ??? ttt uyy ??11 ???11 ??課件 521 (ii) Calculating the variance of yt :* From Wold’s deposition theorem: So long as , this will converge. Solution (cont’d) ttt uyy ?? ?11?tt uLy ?? )1( 1?tt uLy11 )1(??? ?tt uLLy ...)1(2211 ???? ??11 ??. ..22111 ???? ?? tttt uuuy ??課件 522 Var(yt) = E[ytE(yt)][ytE(yt)] but E(yt) = 0, since we are setting ? = 0. Var(yt) = E[(yt)(yt)] *有簡(jiǎn)便方法 = E[ ] = E[ = E[ = = = Solution (cont’d) ? ?? ?.... 2211122111 ?????? ???? tttttt uuuuuu ????)].. .( 224121212 p r o d u c t sc r o s suuu ttt ????? ?? ??.. .) ]( 224121212 ??? ?? ttt uuu ??.. .2412212 ??? uuu ?????. .. )1( 41212 ??? ??? u)1( 212???u課件 523 (iii) Turning now to calculating the acf, first calculate the autocovariances: （ *用簡(jiǎn)便方法 ） ?1 = Cov(yt, yt1) = E[ytE(yt)][yt1E(yt1)] ?1 = E[ytyt1] ?1 = E[ ] = E[ = = Solution (cont’d) ...)( 22111 ??? ?? ttt uuu ?? ... )( 321211 ??? ??? ttt uuu ??]...2231211 p r o du c t sc r o s suu tt ???? ?? ??...25123121 ??? ??????)1( 2121????課件 524 Solution (cont’d) For the second autocorrelation coefficient, ?2 = Cov(yt, yt2) = E[ytE(yt)][yt2E(yt2)] Using the same rules as applied above for the lag 1 covariance ?2 = E[yt yt2] = E[ ] = E[ = = = ...)( 22111 ??? ?? ttt uuu ?? ...)( 421312 ??? ??? ttt uuu ??]...23412221 p r o du c t sc r o s suu tt ???? ?? ??...241221 ?? ????. . .)1( 4121221 ??? ????)1( 21221????課件 525 Solution (cont’d) ? If these steps were repeated for ?3, the following expression would be obtained ?3 = and for any lag s, the autocovariance would be given by ?s = The acf can now be obtained by dividing the covariances by the variance: )1( 21231????)1( 2121????s課件 526 Solution (cont’d) ?0 = ?1 =