data zlh;
  input cpi@@;
  date=intnx('month','1jan99'd,_n_-1);
  format date monyy.;
  cards;
99 98.8 98.7 98.2 97.8 97.8 97.9 98.6 98.7 99.2 99.4 99.1
99 99.8 100.7 99.8 99.7 100.1 100.5 100.5 100.3 100 100 101.3
101.5 101.2 100 100.8 101.6 101.7 101.4 101.5 101 99.9 100.2 99.7
99.7 99 100 99.2 98.7 98.9 99.2 99.1 99.3 99.3 99.2 99.3
99.6 100.4 100.2 100.9 101 100.7 100.3 100.5 100.9 101.1 101.8 103
103.2 103.2 102.1 103 103.8 104.4 105 105.3 105.3 105.2 104.3 102.8
102.4 101.9 103.9 102.7 101.8 101.8 101.6 101.8 101.3 100.9 101.2 101.3
101.6 101.9 100.9 100.8 101.2 101.4 101.5 101 101.3 101.5 101.4 101.9
102.8 102.2 102.7 103.3 103 103.4 104.4 105.6 106.5 106.2 106.5 106.9
106.5 107.1 108.7 108.3 108.5 107.7 107.1 106.3 104.9 104.6 104
;
run;
proc gplot data=Zlh;
 symbol1 i=join v=dot c=red;
 plot cpi*date=1;
run;
data cpi;
 set zlh;
 difcpi=dif(cpi);
proc gplot;
 plot cpi*date difcpi*date;
 symbol v=star c=black i=join;
run;
data cpi;
 set zlh;
 difcpi=dif12(dif(dif(cpi))) ;
proc gplot;
 plot cpi*date difcpi*date;
 symbol v=star c=black i=join;
run;就是这一步之后出现问题，我想对着两次差分后的序列值进行方差检验，看有没有出现过度差分的情况，但是，我

翻看原来上课的书上这个程序找不到，所以想请各位高手帮忙，设计一下分别对difcpi和dif12(dif(dif(cpi)))处理后的序列求方差

proc arima;

identify var=cpi(1);
 estimate p=1 noint ;
 forecast lead=12 id=date;
run;

谢谢各位了

[此贴子已经被作者于2009-3-9 10:46:34编辑过]