平稳时间序列的分析,建模以及预测VIP免费

平稳时间序列的分析，建模以及预测题：某化工厂生产过程中的70 个产量数据obs BF 1 47 64 23 71 38 6 64 55 41 59 48 11 71 35 57 40 58 16 44 80 55 37 74 21 51 57 50 60 45 26 57 50 45 25 59 31 50 71 56 74 50 36 58 45 54 36 54 41 48 55 45 57 50 46 62 44 64 43 52 51 38 59 55 41 53 56 49 34 35 54 45 61 68 38 50 60 39 66 59 40 57 54 23 对此时间序列的分析，本人是基于Matlab 编程完成的，其具体情况如下：1、对数据进行零均值化% % 化工生产过程的产量数据分析% N=70; n=[1:N]; x=[47,64,23,71,38,64,55,41,59,48,71,35,57,40,58,44,80,55,37,74,51,57,50,60,45,57,50,45,25,59, 50,71,56,74,50,58,45,54,36,54,48,55,45,57,50,62,44,64,43,52,38,59,55,41,53,49,34,35,54,45,68, 38,50,60,39,59,40,57,54,23]; y=zeros(1,70); for nn=[1:70], y(nn+1)=y(nn)+x(nn); end y=y(nn+1)/N; x1=zeros(1,N); x1=x-y; 得出均值为： 51.1286，x1 数组即为零均值后的数组。2、相关函数及偏相关函数的计算（接上面程序）R1=zeros(1,N); x2=zeros(1,N); for nn=[1:70], for ii=[1:71-nn], x2(nn)=x2(nn)+x1(ii)*x1(nn+ii-1); end end for jj=[1:70], R1(jj)=x2(jj)/N; end R2=R1; R3=zeros(1,16); for jj=[1:16] R3(jj)=R2(jj)/139.7978; end h=zeros(1,15); g=zeros(1,15); A=zeros(15); for k=[1:14], for j=[1:k], A(1,1)=R3(2); h(k)=h(k)+R3(k+2-j)*A(k,j); g(k)=g(k)+R3(j+1)*A(k,j); A(k+1,k+1)=(R3(k+2)-h(k))/(1-g(k)); A(k+1,j)=A(k,j)-A(k+1,k+1)*A(k,k+1-j); end end B=A; figure(1) set(1,'Position',[10,35,350,650]) ii=[1:16]; plot(ii,R3,'b'); grid figure(2) jj=[1:16]; m=zeros(1,16); m(1)=1; for j=[2:16], m(j)=A(j-1,j-1); end plot(jj,m,'b'); grid 运行程序后得出的自相关函数偏相关函数数据如下表：0?r =139.7978, 0? =1 1?r-54.5041 1?-0.3899 11?-0.3899 2?r42.5536 2?0.3044 22?0.1797 3?r-23.1442 3?-0.1656 33?0.0023 4?r9.8864 4?0.0707 44?-0.0411 5?r-13.5659 5?-0.0970 55?-0.0693 6?r-6.5786 6?-0.0471 66?-0.1202 7?r4.9451 7?0.0354 77?0.0198 8?r-6.0754 8?-0.0435 88?0.0027 9?r-0.6705 9?-0.0048 99?-...