1
\u6642\u9593\u5e8f\u5217\u4f5c\u696d 5 08/04/14 z z100 5.2 The following observations represent the zvalues , form ,..., a series 91 92 \ zu et a fitted by the model t
01.1 0a t \ue001 3\ue000 \ u a t \ue001 e 0 0.28 1
0
2
1
\
u
e
0
0
1
166,172,172,169,164,168,171,167,168,172. z\u02c6 l l (a) Generate the forecasts 100( ) for
\ue002 1, 2,...,12
and draw a graph of the series
a 90 \ue002 values and forecasts (assume 0,a 91 \ue002 0). 2
\u02c6 \u02c6 \ue002 \ue000 l (b) With\ue000 1.103, calculate the estimated standard deviations a
( ) of the
forecast errors and use them calculate to 80% probability limits for the forecasts. Insert these probsbility limits on the graph, on either side of the forecasts. a) p\ue000 d q \ue006 \ue009 \u02c6 \ue002 \ u e 0 1e 0 0\ 1 u e 0 0 1 \ue000 \ue001 z\u02c6 l E z l j a ( ) ( ) \ue005 \ue005 \ue007 00 0 0 \ u t j t j t\ lu e j \ue00a 1 1 j \ue002 j \ue002 \ue008 \ue00b 0 at0 \ue000 4 \at\ue001 u e 0 0 2 \ u e 0 0 1 \ue000 model :\ ztu ae 1.1 0.28 \ue001 1 2 (ie: ARIMA(0,1,2)) t
\u02c6 \ue002 \ 1)u \ue001 e 0\ue001 1 00. 00\10 \ u \ue000 z\u02c6 E \ue000 a 020 0 t\\ u u e 0e 0 1 0 1zt ( l 1at\ lu1e 0\ue001 t ( l) lu 2ee\ue001
\ue003 the forecasts are following: \ (1) \ue000 0 a990\ue002 29 \ 0.28 167.7 z\u02c6 z100u 1.1ae 100 100 (2) \ue002\ue000 0.28a100 \ue002 168.83 z\u02c6 z\u02c6 100(1) 100 \ue002\ e 0\ue002 0 2 \ ...u z\u02c6 168.83. z\u02c6 z\u02c6 100(2) 100(3) 100(12)
u u
0
e e
1
0
0
0
0
1
2
b)
\ue001 1( \ue000 note 1: 1\ue001 probaility limits ) given by zt\ue000 \ue000 l ( ) andzt\ue000 1/2
l \ue001 1 \ue008 \ue00b 2 \u02c6 \ u e 0 \ue006\ue000 0 6 \ u e 0 s0a . 3 ( ) ( ) 1 zt\ue000 z l u \ue003 \ue007 \ue009 l t j \ue00c \ue000 /2 j \ue003 1 \ue00a \ue00d
U note 2: Suppose
N ~
(0,1).
\ue003\ue000\ eud0 00 0for e1 e0 00 00 3 note 3:\ue003 0\0 0u ju\ue004 \ u1 ej0 \u \2uee 02000... 1 3u\ u \ e 3pu de0\ue001 00\ue002 \ue001 0 1 \ \1 eu j j \ue001 p\ue000 j \e j0 10 1 \ue003 \ue003\ue004 with\ue003\ue003 1, j \ue003\ue002 0 forj 0 and\ue002 0 forj q. j 0 2 0 at 3 \ue001 u\ue000 e at\ue001 02 ) 0 1.1\at\ue001 0.28 zt \ (1u B e B 0 ...)( 1
\ue005
\ at u 0.1eat\ue001 0 0 3 \ u e \ue000 0.18a \ue007 t\ue001 j 1 j \ue003 2
0
0 0
\
u
1
80% probability limits for forecasts j
zj(\u00b1)
j
zj(\u00b1)
101 (166.446, 169.134) 107 (167.379, 170.290) 102 (167.484, 170.186)
108 (167.360, 170.310)
103 (167.462, 170.207)
109 (167.340, 170.330)
104 (167.441, 170.228)
110 (167.320, 170.349)
105 (167.420, 170.249)
111 (167.301, 170.369)
106 (167.400, 170.270)
112 (167.282, 170.387)
e
0
0
0
\
u
e
0
0
0
2
\u6642\u9593\u5e8f\u5217\u4f5c\u696d 5 08/04/14
5.3 Suppose that the data of Exercise 5.2 represent monthly sales. (a) Calculate the minimum mean square error forecasts for quarterly sales for t \ue 000 1, 2, 3, 4quarters ahead, using the data up to 100.
(b) Calculate 80% probability limits for these forecasts. a)
forecasts
Data
z97, z98, z99, z100
z90, z91, z92, z93, z94, z95, z96 Residual of z_97 ~ z_100
z\u02c6 z96 \ue001 \ue002 1.1a 96 \ue000 0.28a 95 \ue002 168.124 (1) 96 z\u02c6 z\u02c6 \ue002 \ue000 0.28a 96 \ue002 167.683 96 (2) 96 (1)
4
z\u02c6 z\u02c6 z\u02c6 \ue002 \ue002 96 (2) 96 (3) 96 (4).
2
e96 (1) \ue002 z97 \ue001 z\u02c6 \ue002 2.876 96 (1) e96 (2) \ue002 z98 \ue001 z\u02c6 \ue 002\ue001 0.683 96 (2)
6 9 _ e
e96 (3) \ue002 z99 \ue001 z\u02c6 \ue002 0.318 96 (3) e96 (4) \ue002 z100 \ue001 z\u02c6 \ue002 4.318 96 (4) M
1 2 MSE \ue002 el \ue002 \ue003 6.869 \ue004 M
l \ue002 1
0
2 -
-4
167.80
b) 0(1) 0 \ue001 11.28 \ u Var e 0[e 0(1)] 0 \ue000 z97 ( \ )u ez\u02c6 (1.876, 96 96
167.90
168.00 x_96
3.876)
0(2) 0 \ue001 11.28 \ u Var e 0[e 0 (2)] 0 \ue000 z98( \ )u ez\u02c6 (-1.817, 96 96
0.203)
0(3) 0 \ue001 11.28 \ u Var e 0[e 0 (3)] 0 \ue000 z99 ( \ )u ez\u02c6 (-0.849, 96 96
1.236)
0(4) 0 \ue001 11.28 \ u Var e 0[e 0 (4)] 0 \ue000 z100( \ )u ez\u02c6 (3.119, 96 96
5.268)
168.10
3
\u6642\u9593\u5e8f\u5217\u4f5c\u696d 5 08/04/14 z101 \ue001 5.4 Using the data of Exercise 5.2, and given the further observation 174 z\u02c6 l l (a) Calculate the forecasts( 101 ) for
1, 2,...,11 using the updating formula
\ue001
z\u02c6 l \ue001 z\u02c6 l \ue000 a 1) \ue000 \ue000 t 1( ) t( l t 1 \ue000
\ue000
(b) Verify these f orecasts using the difference equation directly. a) e 01.10 4\ue000 \ ua \ue001 e 0 a t \ue001 a 0.28 t t 2 1
\ zu t
0
2
\
u
e
0
0
0
0
1
a101 \ue002\ue001\ue000 z101 z100 1.1a100 \ue001 0.28a 99 \ue002 6.21 e a \ue000 0 0 z\u02c6 l \ue002\ z\u02c6 l 1)u \ue000 t \ue000 t( l t 1 1( )
0
\
u
e
0
\ue002 \ue000 \ue002 \ue002 z\u02c6 z\u02c6 a 168.83\ue001 0.1\ue003 6.21 168.21 101(1) 100(2) \ue000 1 101 \ue002 \ue000 \ue002 \ue002 z\u02c6 z\u02c6 a \ue000 168.83\ue000 0.18\ue003 6.21 169.95 101(2) 100(3) 2 101 \ue002 \ue002 \ue002 z\u02c6 z\u02c6 z\u02c6 ... \ue002 169.95. 101(2) 101(3) 101(11)
b) By defnition \ ztu ate \ue001\ue000 0 at0 0.28 2 at\\ue001 u zt\ue000 1.1 \ue000 1 1 1 p\ue000 d
\ue006 j t 1
\ue000 z\u02c6 z\u02c6 ( l) \ue002 (l \ 1 t\ue000
1 j \ue002
\ue000
j)u
e
0
0
0
e\ue001 0 0 \ue006 j at 1 l j
1
\
u
q
1 j \ue002
\ u e 0 0 0 \ u e 0 0 0 \ue001
\ue003 \ue002 u z\u02c6 z101 \ue001 1.1a101 \ue000 0.28a100 \ue002 174\ 1.1 (1) 101
e
0
0
1
e \ue000 00.28 \ue005 0 3.73 1\ue002 \ 168.21 u e 6.21
0
0
5
\ue002 \ue000 z\u02c6 z\u02c6 a101 \ue002 0.28 169.95 101(2) 101(1) \ue002\ue001 z\u02c6 z\u02c6 1) for l \ue004 3 101( l) 101( l
\ zu a e 01.1 0 3\ue000 \ ua e of0Exercise 0 2 \ 5.2 u at\ue001 5.5 For the model 0.28 1 t t t-2
e
0
0
1
et e(1),..., t origin . (a) Write down expressions for the forecast errors (6), form the same t et (b) Calculate and plot the autocorrelations oferies the sof forecast errors et (c) Calculate and plot the correlations between
(3).
e j and j \ue002 (2) ( 1...6. ) for t
a) \ue003
eat\ue001 0\ue000 00.18 2at\ue001 \ u 0 e\ue002 \0u e 10 0\ue000 2j \\ue002 u0.18 e 0 0for 1 j \ue004 1, 0\ue000 0.1, 2) zt \ atu 0.1 \ue005 1 1 j (ie: \ue000 2 j \ue002
\ue000 \ue000 \ 0u\ u\ue000 e\ue001 01 0 et ( l) \ue002 at\ue000 \ue000 0 02a \ u e 00 10... e l0 1at\ lu 1e 0 \ue000 10a1 l t\ elu 02 t\ue000
0
\
u
e
0
0
0
1 l \ue001
\ue002 \ue001 0.1at\ lu 1e \ue000 0.18a at\ue000 0 0\ue005 0 \ u l 2e 0 0 1 t\ lu ej 0 0 0 \ u e 0 0 1 j \ue002
b) \ue005
\ue007
\ue000 0 0 3t\ue001 \ \ue001 u0 \ue003 e 0 \ u00.1, e 01 0 j3\ue003 \0.18 ue0 0j 1 \ue006 0.18a (ie: 1, \ue001 for 2) zt \ at u 0.1eat\ue001 \ue001 1 1 j 2 j \ue003
1 l \ue001 \ue008 \ ui e 0i\ue001 0j 1 \ u e 0 0 1 \ue007 \ue003 i j \ue00b \ u e 0 0 4 \ u e 0 0 2 0 j l \ue00b \ue001 l 1 2 \ue003 [ ( ), ( )] e l e l \ue000 \ue009 t t\ue001 j \ue001 \ue007 0 i i \ue003 \ue00b 0\ue00a j \ue006 l \ue00b
4
5 08/04/14
j
ρ[et(3),
et-j(3)]
0
1.000
1
-0.113
2
0.173
o.w.
0.000
c) l 1
[et ( l ),et ( l
j)]
l 1 h 0
j
ρ[et(2),
et(j)]
1
-0.100
2
1.000
3
-0.115
4
0.155
5
0.153
6
0.151
i 0
i
j i
2 h
l
j 1
g 0
2 g
12
