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st: transfer function model and regression: EQUIVALENT???


From   Frank Zhang <zhangmailer@yahoo.com>
To   statalist@hsphsun2.harvard.edu
Subject   st: transfer function model and regression: EQUIVALENT???
Date   Thu, 27 Apr 2006 11:50:51 -0700 (PDT)

Dear Statalisters, 

I have a question about transfer function model with
ARIMA and REGRESSION precedures. 
If the model specification is the same, in my view,
both of the following procedures should have the same
results.

proc reg data=one;
   model y= y1 x ; 
run;
proc arima data=one;
     identify var=y crosscorr=x;
     estimate p=1 input=(x );
run;

But it turned out that the above two procedures give
quite different results. CONFUSING!
Can anybody tell me why? THANK YOU!
 
The data and code in SAS are as follows: 

---------------------------
data one; 
input x y;
y1=lag(y);
cards;
1	-0.109	53.8
2	0	53.6
3	0.178	53.5
4	0.339	53.5
5	0.373	53.4
6	0.441	53.1
7	0.461	52.7
8	0.348	52.4
9	0.127	52.2
10	-0.18	52
11	-0.588	52
12	-1.055	52.4
13	-1.421	53
14	-1.52	54
15	-1.302	54.9
16	-0.814	56
17	-0.475	56.8
18	-0.193	56.8
19	0.088	56.4
20	0.435	55.7
21	0.771	55
22	0.866	54.3
23	0.875	53.2
24	0.891	52.3
25	0.987	51.6
26	1.263	51.2
27	1.775	50.8
28	1.976	50.5
29	1.934	50
30	1.866	49.2
31	1.832	48.4
32	1.767	47.9
33	1.608	47.6
34	1.265	47.5
35	0.79	47.5
36	0.36	47.6
37	0.115	48.1
38	0.088	49
39	0.331	50
40	0.645	51.1
41	0.96	51.8
42	1.409	51.9
43	2.67	51.7
44	2.834	51.2
45	2.812	50
46	2.483	48.3
47	1.929	47
48	1.485	45.8
49	1.214	45.6
50	1.239	46
;
run;
proc arima data=one;
     identify var=y crosscorr=x;
     estimate p=1 input=(x );
run;
proc reg;
   model y= y1 x ; 
run;


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