怎么实现arma算法来预测时间序列数据,to: 高高手

wjcwjc 2005-05-25 04:55:36
预测分析时间序列数据,想用arma算法来做,请问怎么实现
谢谢
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ldxab 2010-06-08
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接上面:
K=0;T=X7;
for t=8:88
r=0;
for i=1:7
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(7)=K/(88-7) % 7阶模型残差方差 0.4331

K=0;T=X8;
for t=9:88
r=0;
for i=1:8
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(8)=K/(88-8) % 8阶模型残差方差0.4310

K=0;T=X9;
for t=10:88
r=0;
for i=1:9
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(9)=K/(88-9) %9阶模型残差方差 0.4297

K=0;T=X10;
for t=11:88
r=0;
for i=1:10
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(10)=K/(88-10) % 10阶模型残差方差 0.4317

U=10*U
for i=1:10
AIC2(i)=88*log(U(i))+2*(i) % AIC值分别为:172.6632 165.4660 153.2087 145.1442 140.7898 141.6824 142.9944 144.5601 146.3067 148.7036
end
%-----------------取使AIC值为最小值的阶次,判断模型阶次为5。用最小二乘法估计参数--------------------%


%------------------检验{at}是否为白噪声。求{at}的自相关系数,看其是否趋近于零-----------------------%
C=0;K=0;
for t=7:88
at=Y(t)-W(1)*Y(t-1)-W(2)*Y(t-2)-W(3)*Y(t-3)-W(4)*Y(t-4)-W(5)*Y(t-5)+Y(6)-W(1)*Y(5)-W(2)*Y(4)-W(3)*Y(3)-W(4)*Y(2)-W(5)*Y(1);
at1=Y(t-1)-W(1)*Y(t-2)-W(2)*Y(t-3)-W(3)*Y(t-4)-W(4)*Y(t-5)-W(5)*Y(t-6);
C=at*at1+C;
K=(at)^2+K;
end
p=C/K %若p接近于零,则{at}可看作是白噪声
%--------------------------------{at}的自相关系数,趋近于零,模型适用--------------------------------%


%------------AR(5)模型方程为------------------------------------------------------------------------%
% X(t)=W(1)*X(t-1)-W(2)*X(t-2)-W(3)*X(t-3)-W(4)*X(t-4)-W(5)*X(t-5)+at (at=0.4420)


%------------------------------------------后六年的数据 进行预测和效果检验----------------------------------------------%

%-----------------------------单步预测 预测当前时刻后的六个数据----------------------------------%
XT=[L(84:94)];
for t=6:11
m(t)=0;
for i=1:5
m(t)=W(i)*XT(t-i)+m(t);
end
end

m=m(6:11);

%-------------预测值进行反处理---------------%
m(1)=Yt(90)+m(1); %一次反差分
z1(1)=P(90)+m(1); %二次反差分
m(2)=Yt(91)+m(2);
z1(2)=P(91)+m(2);
m(3)=Yt(92)+m(3);
z1(3)=P(92)+m(3);
m(4)=Yt(93)+m(4);
z1(4)=P(93)+m(4);
m(5)=Yt(94)+m(5);
z1(5)=P(94)+m(5);
m(6)=Yt(95)+m(6);
z1(6)=P(95)+m(6);
z1 % 单步预测的向后6个预测值:z1= 13.9423 13.4101 13.3588 12.9856 13.2594 12.9552

%---------------------------绘制数据模型逼近曲线-----------------------------------%
for t=6:88
r=0;
for i=1:5
r=W(i)*Y(t-i)+r;
end
at= Y(t)-r;
end

figure;
for t=6:88
y(t)=0;
for i=1:5
y(t)=W(i)*Y(t-i)+y(t);
end
y(t)=y(t)+at;
y(t)=Yt(t+1)-y(t);
y(t)=P(t+1)-y(t);
end
plot(y,'r-*'); % 样本数据模型逼近曲线
hold on;
plot(91:96,z1,'r-*');
hold on;
plot(P,'--'); % 原样本曲线
title('AR(5)模型样本逼近预测曲线');
pause
%-----------------------------绘制数据模型逼近曲线-----------------------------------%

%-------------------------预测误差分析------------------------%
%-----------计算单步预测绝对误差-------------%
D=[13.70 13.66 13.27 13.56 13.14 14.19 ];
for i=1:6
e1(i)=D(i)-z1(i);
PE1(i)= (e1(i)/D(i))*100;
end
e1 % 单步预测的绝对误差 e1 = -0.2423 0.2499 -0.0888 0.5744 -0.1194 1.2348
PE1

%------单步预测平均绝对误差-------------------%
mae1=sum(abs(e1)) /6 % mae1 = 0.2681

%------单步预测平均绝对百分比误差-------------------%
MAPE1=sum(abs(PE1))/6


%------绘制预测结果和实际值的比较图-----------%
figure;
plot(1:6,D,'-+') ;
hold on;
plot(z1,'r-*');
title('向前一步预测值和实际值对比图');
hold off;
pause
%--------------------------------单步预测 预测当前时刻后的六个数据---------------------------------%



%----------------------------------多步预测 目的是向前六步预测--------------------------------------%
Xt=[ Y(84) Y(85) Y(86) Y(87) Y(88)]; %取当前时刻之前的6个数据

Z(1)=W(1)*Xt(5)+W(2)*Xt(4)+W(3)*Xt(3)-W(4)*Xt(2)-W(5)*Xt(1)
%------求向前l步的预测值
%预测步数小于5时
for l=2:5
K(l)=0;
for i=1:l-1
K(l)=W(i)*Z(l-i)+K(l);
end
G(l)=0;
for j=l:5
G(l)=W(j)*Xt(5+l-j)+G(l);
end
Z(l)=K(l)+G(l);
end
%预测步数大于5时(向前6步预测)
for l=6:6
K(l)=0;
for i=1:5
K(l)=W(i)*Z(l-i)+K(l);
end
Z(l)=K(l);
end

%----预测值进行反标准化处理
r=Z*v+Ux % 0.0581 0.0844 0.0156 0.0319 0.0632 0.0652
r(1)=Yt(90)+r(1); %一次反差分
z(1)=P(90)+r(1) %二次反差分
for i=2:6
r(i)=r(i-1)+r(i);
z(i)=z(i-1)+r(i)
end

%---------------------------- 预测误差分析 ------------------------------%
%-------计算绝对误差和相对误差
D=[13.70 13.66 13.27 13.56 13.14 14.19 ]; % 预测值 z =14.0281 13.9606 13.9087 13.8887 13.9318 14.0403
for i=1:6
e6(i)=D(i)-z(i);
PE6(i)= (e6(i)/D(i))*100;
end
e6 % 多步预测的绝对误差 e = -0.3281 -0.3006 -0.6387 -0.3287 -0.7918 0.1497
PE6 % 多步预测的相对误差
1-abs(PE6) % 准确率

%------多步预测平均绝对误差
mae6=sum(abs(e6)) /6

%------多步预测平均绝对百分比误差
MAPE6=sum(abs(PE6))/6

%------绘制预测结果和实际值的比较图
figure;
plot(1:6,D,'-+')
hold on;
plot(z,'r-*');
title('向前六步预测值和实际值对比图');
hold off;


////////////////////////
x =

Columns 1 through 6

-0.5386 0.0522 -0.1024 0.1529 -0.1005 0.0879

Columns 7 through 12

-0.0663 0.0127 -0.0147 0.1122 -0.1924 0.1290

Columns 13 through 18

-0.0523 0.1399 -0.1931 0.0557 -0.0059 0.1092

Columns 19 through 20

-0.1111 0.0548


X2 =

-0.7190
-0.3350


X22 =

-0.3350


X3 =

-0.8494
-0.6149
-0.3892


X33 =

-0.3892


X4 =

-0.9306
-0.7432
-0.5664
-0.2087


X44 =

-0.2087


X5 =

-0.9784
-0.8729
-0.7366
-0.4218
-0.2290


X55 =

-0.2290


X6 =

-1.0038
-0.9196
-0.8182
-0.5185
-0.3374
-0.1109


X66 =

-0.1109


X7 =

-1.0132
-0.9482
-0.8622
-0.5879
-0.4154
-0.1959
-0.0848


X77 =

-0.0848


X8 =

-1.0213
-0.9671
-0.9023
-0.6446
-0.4986
-0.2875
-0.1826
-0.0965


X88 =

-0.0965


X9 =

-1.0310
-0.9854
-0.9310
-0.6944
-0.5630
-0.3775
-0.2791
-0.1985
-0.0998


X99 =

-0.0998


X10 =

-1.0228
-0.9691
-0.9082
-0.6636
-0.5170
-0.3209
-0.2031
-0.1181
-0.0157
0.0816


X1010 =

0.0816


X101 =

-1.0145
-0.9708
-0.9203
-0.6844
-0.5499
-0.3738
-0.2711
-0.2110
-0.1149
-0.0231
-0.1024


X1111 =

-0.1024


X12 =

-1.0204
-0.9721
-0.9269
-0.6965
-0.5655
-0.3953
-0.3027
-0.2504
-0.1678
-0.0789
-0.1607
-0.0575


X1212 =

-0.0575


X =

Columns 1 through 6

-0.5386 -0.3350 -0.3892 -0.2087 -0.2290 -0.1109

Columns 7 through 12

-0.0848 -0.0965 -0.0998 0.0816 -0.1024 -0.0575
ldxab 2010-06-08
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clear;
%--------------------------------油价序列零均值化后的数据如下----------------------------------------%:
P=[ 19.5900 14.9100 15.7400 15.4000 13.0600 19.0700 15.2800 15.8200 12.7700 12.0500...
11.6900 13.8500 13.8500 10.0700 9.1700 10.7900 13.4400 21.1700 18.6400 13.2100...
15.5400 21.9400 23.1100 18.6400 14.9400 16.9000 15.4600 11.1500 13.1300 12.4800...
12.9500 12.5900 10.5800 10.5800 12.3900 15.5300 13.0600 10.2200 16.3300 19.7200...
21.3100 18.8400 24.8400 15.6700 15.5700 12.7300 13.5600 15.5400 17.2200 12.1400...
11.0700 12.0200 11.5500 6.9200 10.3300 8.3800 12.1100 11.4600 12.7500 13.3200...
13.0000 11.9000 11.7900 12.5500 11.8400 11.2500 11.1500 10.9900 11.7000 14.0100...
17.5100 17.2700 16.9000 15.7900 15.4500 6.2400 16.7100 16.7700 16.6400 17.8000...
16.8700 16.1300 15.7600 15.6600 15.5400 15.3000 15.0500 14.6900 14.3900 14.1800...
13.70 13.66 13.27 13.56 13.14 14.19 ];
F=[ 19.5900 14.9100 15.7400 15.4000 13.0600 19.0700 15.2800 15.8200 12.7700 12.0500...
11.6900 13.8500 13.8500 10.0700 9.1700 10.7900 13.4400 21.1700 18.6400 13.2100...
15.5400 21.9400 23.1100 18.6400 14.9400 16.9000 15.4600 11.1500 13.1300 12.4800...
12.9500 12.5900 10.5800 10.5800 12.3900 15.5300 13.0600 10.2200 16.3300 19.7200...
21.3100 18.8400 24.8400 15.6700 15.5700 12.7300 13.5600 15.5400 17.2200 12.1400...
11.0700 12.0200 11.5500 6.9200 10.3300 8.3800 12.1100 11.4600 12.7500 13.3200...
13.0000 11.9000 11.7900 12.5500 11.8400 11.2500 11.1500 10.9900 11.7000 14.0100...
17.5100 17.2700 16.9000 15.7900 15.4500 6.2400 16.7100 16.7700 16.6400 17.8000...
16.8700 16.1300 15.7600 15.6600 15.5400 15.3000 15.0500 14.6900 14.3900 14.180];

%----------------------由于时间序列有不平稳趋势,进行两次差分运算,消除趋势性----------------------%
for i=2:96
Yt(i)=P(i)-P(i-1);
end
for i=3:96
L(i)=Yt(i)-Yt(i-1);
end
figure;
L=L(3:96);
Y=L(1:88);
plot(P);
title('原数据序列图');
hold on;
pause
plot(Y,'r');
title('两次差分后的序列图和原数对比图');
pause
%--------------------------------------对数据标准化处理----------------------------------------------%
Ux=sum(Y)/88 % 求序列均值
yt=Y-Ux;
b=0;
for i=1:88
b=yt(i)^2/88+b;
end
v=sqrt(b) % 求序列方差
Y=(Y-Ux)/v; % 标准化处理公式
f=F(1:88);
t=1:88;
figure;
plot(t,f,t,Y,'r')
title('原始数据和标准化处理后对比图');
xlabel('时间t'),ylabel('油价y');
legend('原始数据 F ','标准化后数据Y ');
pause
%--------------------------------------对数据标准化处理----------------------------------------------%


%------------------------检验预处理后的数据是否符合AR建模要求,计算自相关和偏相关系数---------------%
%---------------------------------------计算自相关系数-----------------------------------%
R0=0;
for i=1:88
R0=Y(i)^2/88+R0;
end
R0
for k=1:20
R(k)=0;
for i=k+1:88
R(k)=Y(i)*Y(i-k)/88+R(k);
end
R %自协方差函数R
end
x=R/R0 %自相关系数x
figure;
plot(x)
title('自相关系数分析图');
pause
%-----------------------------------计算自相关系数-------------------------------------%

%-----------------------解Y-W方程,其系数矩阵是Toeplit
矩阵。求得偏相关函数X-----------------------%
X1=x(1);
X11=x(1);
B=[x(1) x(2)]';
x2=[1 x(1)];
A=toeplitz(x2);
X2=A\B
X22=X2(2)

B=[x(1) x(2) x(3)]';
x3=[1 x(1) x(2)];
A=toeplitz(x3);
X3=A\B
X33=X3(3)

B=[x(1) x(2) x(3) x(4)]';
x4=[1 x(1) x(2) x(3)];
A=toeplitz(x4);
X4=A\B
X44=X4(4)

B=[x(1) x(2) x(3) x(4) x(5)]';
x5=[1 x(1) x(2) x(3) x(4)];
A=toeplitz(x5);
X5=A\B
X55=X5(5)

B=[x(1) x(2) x(3) x(4) x(5) x(6)]';
x6=[1 x(1) x(2) x(3) x(4) x(5)];
A=toeplitz(x6);
X6=A\B
X66=X6(6)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7)]';
x7=[1 x(1) x(2) x(3) x(4) x(5) x(6)];
A=toeplitz(x7);
X7=A\B
X77=X7(7)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8)]';
x8=[1 x(1) x(2) x(3) x(4) x(5) x(6) x(7)];
A=toeplitz(x8);
X8=A\B
X88=X8(8)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9)]';
x9=[1 x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8)];
A=toeplitz(x9);
X9=A\B
X99=X9(9)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9) x(10)]';
x10=[1 x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9)];
A=toeplitz(x10);
X10=A\B
X1010=X10(10)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9) x(10) x(11)]';
x11=[1 x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9) x(10)];
A=toeplitz(x11);
X101=A\B
X1111=X101(11)

B=[x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9) x(10) x(11) x(12)]';
x12=[1 x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(8) x(9) x(10) x(11)];
A=toeplitz(x12);
X12=A\B
X1212=X12(12)

X=[X11 X22 X33 X44 X55 X66 X77 X88 X99 X1010 X1111 X1212]
%-----------------------------------解Y-W方程,得偏相关函数X-------------------------------------%
figure;
plot(X);
title('偏相关函数图');
pause

%-----根据偏相关函数截尾性,初判模型阶次为5。用最小二乘法估计参数,计算10阶以内的模型残差方差和AIC值,应用AIC准则为模型定阶------%
S=[R0 R(1) R(2) R(3) R(4)];
G=toeplitz(S);
W=inv(G)*[R(1:5)]' % 参数W(i) 与X5相同

K=0;
for t=6:88
r=0;
for i=1:5
r=W(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(5)=K/(88-5) % 5阶模型残差方差 0.4420

K=0;T=X1;
for t=2:88
at=Y(t)-T(1)*Y(t-1);
K=(at)^2+K;
end
U(1)=K/(89-1) % 1阶模型残差方差0.6954

K=0;T=X2;
for t=3:88
r=0;
for i=1:2
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(2)=K/(88-2) % 2阶模型残差方差 0.6264

K=0;T=X3;
for t=4:88
r=0;
for i=1:3
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(3)=K/(88-3) % 3阶模型残差方差 0.5327

K=0;T=X4;
for t=5:88
r=0;
for i=1:4
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(4)=K/(88-4) % 4阶模型残差方差 0.4751

K=0;T=X6;
for t=7:88
r=0;
for i=1:6
r=T(i)*Y(t-i)+r;
end
at= Y(t)-r;
K=(at)^2+K;
end
U(6)=K/(88-6) % 6阶模型残差方差 0.4365

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