Homework 3 Solutions

Transcription

1 Math 4506 Sprig 04 Homework 3 Solutios. a The ACF of a MA process has a o-zero value oly at lags, 0, ad. Problem 4.3 from the textbook which you did t do, so I did t expect you to metio this shows that ρ. b The ACF of a MA process has a o-zero value oly at lags,, 0,, ad. c The ACF of a AR process decays expoetially. If φ > 0, all values are positive, otherwise, the values alterate betwee positive ad egative with ρ egative. d If the roots of the characteristic equatio are complex, the autocorrelatio fuctio is a damped sie wave of course oly sampled at iteger values of h. Otherwise, it ca have a umber of differet shapes. e The ACF of a ARMA, process is a expoetially decayig fuctio startig at lag h.. If αz α i z i, βz β j z j, ad we defie W t αbβby t, the W t i Z α i B i β k B k Y t η j Y t j. To express the coefficiets η j i terms of α i ad β k, we eed to solve the followig polyomial equatio: α i z i β k z k η j z j, i Z which ca be re-writte as Note that α i β k z i+k η j z j. i Z i + k j i j k k j i. Usig the first equality i ad the fact that if k is fixed, the j rus through all values of Z if ad oly if i rus through all values of Z, the polyomial equatio becomes α j k β k z j η j z j α j k β k z j η j z j, so that η j α j k β k.

2 Usig the secod equality i, we get implyig that α i β j i z j η j z j, i Z η j i Z α i β j i Puttig everythig together ad chagig the idex i the last sum, we get η j α k β k i β k α j k. 3. Note first that for ay φ ad z such that, Now sice >, <, so. is the sum of a geometric series. More specifically, i0 i i, i0 so that i j. i0 j Note that the geometric series i this problem is actually a series of complex umbers thik about this, but the argumet works just as if it were a real geometric series, sice absolute covergece of a series implies covergece. 4. a Sice the equatio that defies {X t } i terms of {Z t } is autoregressive with φ >, there is o causal solutio to this equatio. b We kow from Problem 3 that c X t φ j B j Z t φ j Z t+j. j Sice j φ j <, sice φ >, we see that writte this way, {X t } is a liear process. j E[W t ] E[X t φ X t ] E[X t ] φ E[X t ] 0, sice by, E[X t ] 0 for all t Z. Note that for h > 0, γ X h Cov φ j Z t+j, φ j Z t+h+j j j φ j+h φ j σ σ φ h φ φ σ φ h φ. j

3 Suppose h > 0. The Also, CovW t, W t+h CovX t φ X t, X t+h φ X t+h γ X h φ γ Xh φ γ Xh + + φ γ Xh σ φ φ h φ φ φ h φ φ h+ + φ φ h σ φ φ φ h φ h φ h+ + φ h+ 0. VarW t γ X 0 φ γ X φ γ X + φ γ X0 γ X 0 + φ φ γ X σ φ + φ φ σ φ. Therefore, so {W t } is white oise. γ W h { σ φ h 0 0 h 0, d Viewed this ew way, {X t } is causal, sice it is a AR process with parameter /φ satisfyig /φ <. I other words, whe defied i coectio to the white oise {W t }, {X t } is a causal time series. e I this problem, we showed that ay o-causal AR time series is actually a causal time series whe cosidered relatively to aother white oise. Note that this is ot oly true for AR time series but for ay o-causal ARMA time series. Similarly, every o-ivertible time series ca be expressed as a ivertible time series

4 5. a We ca use the fact that X i,k X i to see that X i,k X i X j,k X j X i,k Xj + X j,k Xi X i Xj X j X i,k X i X j,k + X i Xj X j X i X i X j + X i Xj X i Xj X i Xj X i Xj. b Usig the fact that CovX i,k, X j,l 0 if k l, we get Cov X i, X j Cov X i,k, X j,l l CovX i,k, X j,k l c We eed to show that E[Q i,j ] CovX i, X j. Usig the fact that ad that CovX i,k, X j,l CovX i, X j CovX i, X j CovX i,k, X j,k E[X i,k X j,k ] E[X i,k ]E[X j,k ] E[X i X j ] E[X i ]E[X j ] Cov X i, X j E[ X i Xj ] E[ X i ]E[ X j ] E[ X i Xj ] E[X i ]E[X j ],

5 we get [ E[Q i,j ] E [ part a E part b ] X i,k X i X j,k X j ] X i Xj E[X i,k X j,k ] E[ X i Xj ] CovX i,k, X j,k + E[X i ]E[X j ] Cov X i, X j + E[X i ]E[X j ] CovX i,k, X j,k Cov X i, X j CovX i,k, X j,k CovX i, X j CovX i, X j CovX i, X j 6. a We ca for istace use the followig code twice to obtai the images below. Of course, your images will probably be differet from these. > Nrorm00 > XN > for i i :00 X[i]N[i]+0.5*N[i-] > plotx,type l X X Idex Idex b The commad > acfx gives the followig two sample ACFs:

6 Series X Series X ACF ACF Lag Lag I both cases, the sample ACF is likely ozero at lag ad possibly zero at all two of the remaiig 9 lags. Sice it is ot highly surprisig that for a 95% cofidece iterval, the true value would be outside of the iterval times out of 9 o average it would be time out of 9, we ca t exclude the possibility that this is the sample ACF of a MA process. c The commads ad output are the followig: > Box.testX,lag3,type L Box-Ljug test data: X X-squared.8359, df 3, p-value 4.369e-05 > Box.testX,lag3,type L Box-Ljug test data: X X-squared , df 3, p-value 9.78e-06 I both cases, the p-value is less tha 5%, so we would reject at the 5% level the hypothesis that the time series is white oise. This is of course ot surprisig sice we geerated the time series ad kow that it is t white oise. 7. We kow that if Y t W t + W t, where W t is white oise, the σy VarY t γ Y 0 σw σ W ad if if Y t W t W t, the σy VarY t γ Y 0 σw σ W. Also, sice for MA processes Y, ρ Y θ + θ, we see that i this problem, ρ Y / +/ 5 ad ρ Y / + / 5. If k, ρ Y k 0, sice both Y s are MA processes.

7 Equatio 3..3 i the textbook tells us that VarȲ σ Y + k ρ Y k, which for our two processes yields a Var X VarȲ σ Y + σ 5 Y σ W σw 9 4 i the first case. Note that we used the fact that sice Y t µ + X t, the processes X ad Y oly differ by a costat, so their autocorrelatios are the same. b Var X VarȲ σ Y + σ 5 Y σ W σw 4 + i the secod case. c I the white oise case, we already kow that Var X σ X σ W. We otice that for the three processes discussed i this problem, if we assume that their variace is the same, the Var X is smallest for the MA process with θ ad largest for the MA process with θ. d First, we geerate a 00 matrix of stadard ormals > Zrorm000 > dimzc00,0 We eed colums, so that we ca use 0 successive pairs of elemets from which to defie our MA process. We also defie a matrix structure i which to put the MA processes: > Xmatrixrow00,col0 The, we ca geerate the 00 realizatios of legth 0 of each of the two MA processes of legth 00 with the commads, respectively, > for i i :00 for j i :0 X[i,j]Z[i,j]+Z[i,j+]/ ad > for i i :00 for j i :0 X[i,j]Z[i,j]-Z[i,j+]/ The we ca get the mea for each of the 00 MA processes: > mmatrixrow00,col > for i i :00 m[i]meax[i,] Fially, we compute the sample variace for the 00 meas: > varm If the very first thig you wrote was set.seed, the the umerical value you obtaied should have bee, respectively, ad If you did t set the seed to be, you probably got somethig very close to these umbers. How does this compare to what we would have expected to see? Usig σw ad 0 i a ad b, we get 0. ad 0.05, respectively.