Calculation of ACVF for ARMA Process: I consider causal ARMA(p, q) defined by • 2 �(B)X = ✓(B)Z , Z WN(0, � ) t t t { } ⇠ want to determine ACVF �(h) for this process, which can • { } be done using four complementary methods first method is based on MA( ) representation • 1 1 X = Z = (B)Z , t j t j t � j=0 X where (B) = ✓(B)/�(B) have noted (overhead VII–5) that ACVF can be expressed as • 1 2 �(h) = � j j+ h | | j=0 X BD–88, CC–56, SS–27 IX–1 Calculation of ACVF for ARMA Process: II can use recursive scheme to compute ’s (overhead VIII-12): j • p = � + ✓ , j = 0, 1, 2, . . . , j k j k j � k=1 X but in general need ’s for infinite number of integers j j since 0 as j , could compute ’s out to, say, j j • ! ! 1 j = J + h and use | | J 1 2 2 �(h) = � � , j j+ h j j+ h ⇡ | | | | j=0 j=0 X X with approximation getting better with increasing J if we have a manageable expression for ’s (true for some j • processes), can get analytic expression for �(h) BD–88, CC–79, SS–101 IX–2 Example – ARMA(1,1) Process: I for an ARMA(1,1) process, homework exercise says that • 1 1 j 1 X = Z + (� + ✓) � Z Z , t t � t j j t j ⌘ � � j=1 j=0 X X j 1 so = 1 and = (� + ✓)� for j 1 0 j � � j 1 armed with x = (valid for x < 1), away we go: 1 j=0 1 x • | | � �(0) P 1 1 2 2 2j 2 = = 1 + (� + ✓) � j � 2 � j=0 j=1 X X 2 (� + ✓) 1 2 2j = 1 + (� + ✓) � = 1 + 2 1 � j=0 � X BD–89, CC–78, SS–104 IX–3 Example – ARMA(1,1) Process: II j 1 for h > 0, with = (� + ✓)� for j 1, have j � • � �(h) 1 1 = = + j j+h h j j+h 2 � j=0 j=1 X X 1 h 1 2 2j+h 2 = (� + ✓)� + (� + ✓) � � � j=1 X h 2 � (� + ✓) h 1 = (� + ✓)� + � 2 1 � � 2 �(� + ✓) h 1 = � � + ✓ + � 2 1 � ! � note: �(h) = ��(h 1) for h 2 • � � BD–89, CC–78, SS–104 IX–4 Example – ARMA(1,1) Process: III overheads that follow show ACVFs (circles) for ARMA(1,1) • processes X with t { } 2 � = 1 � AR parameter � = 0.9 � MA parameter ✓ ranging from 0.99 down to 0.99 � � have � (h) = �� (h 1) for h 2, but not for h = 1 X X • � � h casual AR(1) process Y has ACVF � (h) = � � (0) t Y | | Y • { } have � (h) = �� (h 1) for h 1 Y Y • � � overheads also show ACVFs (asterisks) for Y with � = 0.9 t • { } and with � (0) set such that � (1) = � (1) Y Y X note: for some ✓, not possible to do! � IX–5 ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.99 * 0 2 * * 5 * 1 * * F V * C 0 1 * A * * * * * 5 * * * * * * * * 0 0 5 10 15 20 h (lag) IX–6 ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.5 2 1 * * 0 1 * * 8 * * F V * 6 C * A * * 4 * * * * * * * 2 * * * * 0 0 5 10 15 20 h (lag) IX–7 ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0 * 5 * * 4 * * * 3 F V * C * A * 2 * * * * * * * 1 * * * * * 0 0 5 10 15 20 h (lag) IX–8 ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.25 � 0 . 3 * * 5 . 2 * * 0 . 2 * F * V C 5 * A .1 * * * 0 * . 1 * * * * 5 * * . * 0 * * * 0 . 0 0 5 10 15 20 h (lag) IX–9 ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.5 � 5 . 1 * * F 0 * . V 1 * C A * * * * 5 * . * 0 * * * * * * * * * * * 0 . 0 0 5 10 15 20 h (lag) IX–10