时间序列分析ch3习题解答17-20

时间序列分析第七次作业（第三章17-20题）解答第三章P.1003.5习题17.某城市过去63年中每年降雪量数据(单位：mm)如表3-20所示。表3-20126.482.478.151.190.976.2104.587.4110.52569.353.539.863.646.772.979.683.680.760.37974.449.654.771.849.1103.951.682.483.677.879.389.685.558120.7110.565.439.940.188.771.48355.989.984.8105.2113.7124.7114.5115.6102.4101.489.871.570.998.355.566.178.4120.597110(1)判断该序列的平稳性与纯随机性。(2)如果序列平稳且非白噪声，选择适当模型拟合该序列的发展。(3)利用拟合模型，预测该城市未来5年的降雪量。解：(1)在SAS中输入以下程序：dataexample1;inputx@@;time=_n_;cards;126.482.478.151.190.976.2104.587.4110.52569.353.539.863.646.772.979.683.680.760.37974.449.654.771.849.1103.951.682.483.677.879.389.685.558120.7110.565.439.940.188.771.48355.989.984.8105.2113.7124.7114.5115.6102.4101.489.871.570.998.355.566.178.4120.597110;procgplotdata=example1;plotx*time=1;symbol1c=redI=joinv=star;procarimadata=example1;identifyvar=xnlag=8minicp=(0:5)q=(0:5);run;得时序图：可得自相关图：从时序图可以看出，序列基本上在一个数值上随机波动，可认为该序列平稳。从自相关图可以看出，该序列的自相关系数一直都比较小，始终在2倍标准差范围以内，可以认为该序列自始至终都在零轴附近波动，所以认为该序列平稳。白噪声检验结果为：表中，P值6阶的小于0.05，所以认为该序列具有非纯随机性。(2)上述程序可以中有（p和q分别从1到5）相对最有定阶输出结果为：所以该模型为AR(1)模型时拟合最好。再在上述程序中添加以下语句，再次运行：estimatep=1;得到：由于P值都小于0.05，所以认为所有参数显著，所以可以得到该AR(1)模型为：180.994100.31587tttxx。(3)在程序中加入语句：forecastlead=5id=timeout=results;可以得到预测信息如下：所以预测未来5年的降雪量依次为：90.1563mm,83.8882mm,81.9083mm,81.2829mm,81.0853mm。18.某地区连续74年的谷物产量（单位：千吨）如表3-21所示。表3-210.970.451.611.261.371.431.321.230.840.891.181.331.210.980.910.611.230.971.10.740.80.810.80.60.590.630.870.360.810.910.770.960.930.950.650.980.70.861.320.880.680.781.250.791.190.690.920.860.860.850.90.540.321.41.140.690.910.680.570.940.350.390.450.990.840.620.850.730.660.760.630.320.170.46(1)判断该序列的平稳性与纯随机性。(2)选择适当模型拟合该序列的发展。(3)利用拟合模型，预测该地区未来5年的谷物产量。解：(1).在SAS中输入以下程序：dataexample2;inputx@@;time=_n_;cards;0.970.451.611.261.371.431.321.230.840.891.181.331.210.980.910.611.230.971.100.740.800.810.800.600.590.630.870.360.810.910.770.960.930.950.650.980.700.861.320.880.680.781.250.791.190.690.920.860.860.850.900.540.321.401.140.690.910.680.570.940.350.390.450.990.840.620.850.730.660.760.630.320.170.46;procgplotdata=example2;plotx*time=1;symbol1c=redI=joinv=star;procarimadata=example2;identifyvar=xnlag=18minicp=(0:5)q=(0:5);run;得时序图：可得自相关图：从时序图可以看出，序列基本上在一个数值上随机波动，可认为该序列平稳。从自相关图可以看出，该序列的自相关系数一直都比较小，始终在2倍标准差范围以内，可以认为该序列自始至终都在零轴附近波动，所以认为该序列平稳。白噪声检验结果为：表中，P值都小于0.05，所以认为该序列具有非纯随机性。(2)上述程序可以中有（p和q分别从1到5）相对最有定阶输出结果为：所以该模型为AR(1)模型时拟合最好。再在上述程序中添加以下语句后，再次运行：estimatep=1;得到：由于P值都小于0.05，所以认为所有参数显著，所以可以得到该AR(1)模型为：10.849630.37235tttxx。(3).在程序中加入语句：forecastlead=5id=timeout=results;可以得到预测信息如下：所以预测得到未来5年的谷物产量依次为：0.7046千吨,0.7956千吨,0.8295千吨,0.8421千吨,0.8468千吨。19.现有201个连续的生产记录，如表3-22所示。表3-2281.989.47981.484.885.98880.382.683.580.285.287.283.584.382.984.782.981.583.487.781.879.685.877.989.785.486.380.783.890.584.582.486.78381.889.379.382.78879.687.883.679.583.388.486.684.679.78684.28384.883.681.885.988.283.587.283.787.38390.580.783.186.59077.584.784.687.280.586.182.685.484.782.881.983.686.88484.282.8838284.784.488.982.4838582.281.686.285.482.181.48585.884.283.586.58580.485.786.786.782.386.482.58279.586.780.591.781.683.985.684.878.489.98586.28385.484.484.586.285.683.285.783.580.182.288.6828585.285.384.382.389.784.883.180.687.486.883.586.284.182.384.886.683.578.188.881.983.38087.283.386.679.584.182.290.886.579.78187.281.684.484.482.288.980.985.187.18476.582.785.183.390.48180.379.88983.780.987.381.185.686.68086.683.383.182.386.780.2(1)判断该序列的平稳性与纯随机性。(2)如果序列平稳且非白噪声，选择适当模型拟合该序列的发展。(3)利用拟合模型，预测该序列下一时刻95%的置信区间。解：(1).在SAS中输入以下程序：dataexample3;inputx@@;time=_n_;cards;81.989.479.081.484.885.988.080.382.683.580.285.287.283.584.382.984.782.981.583.487.781.879.685.877.989.785.486.380.783.890.584.582.486.783.081.889.379.382.788.079.687.883.679.583.388.486.684.679.786.084.283.084.883.681.885.988.283.587.283.787.383.090.580.783.186.590.077.584.784.687.280.586.182.685.484.782.881.983.686.884.084.282.883.082.084.784.488.982.483.085.082.281.686.285.482.181.485.085.884.283.586.585.080.485.786.786.782.386.482.582.079.586.780.591.781.683.985.684.878.489.985.086.283.085.484.484.586.285.683.285.783.580.182.288.682.085.085.285.384.382.389.784.883.180.687.486.883.586.284.182.384.886.683.578.188.881.983.380.087.283.386.679.584.182.290.886.579.781.087.281.684.484.482.288.980.985.187.184.076.582.785.183.390.481.080.379.889.083.780.987.381.185.686.680.086.683.383.182.386.780.2;procgplotdata=example3;plotx*time=1;symbol1c=redI=joinv=star;procarimadata=example3;identifyvar=xnlag=18minicp=(0:5)q=(0:5);run;得时序图：可得自相关图：从时序图可以看出，序列基本上在一个数值上随机波动，可认为该序列平稳。从自相关图可以看出，该序列的自相关系数迅速递减为0，自始至终都在零轴附近波动，所以认为该序列平稳。白噪声检验结果为：表中，P值6阶的小于0.05，所以认为该序列具有非纯随机性。(2).上述程序可以中有（p和q分别从1到5）相对最有定阶输出结果为：所以该模型为MA(1)模型时拟合最好。再在上述程序中添加以下语句，再次运行：estimateq=1;得到：由于P值都小于0.05，所以认为所有参数显著，所以可以得到该MA(1)模型为：184.128890.47959tttx。(3).在程序中加入语句：forecastlead=1id=timeout=results;可以得到预测信息如下：所以预测得到该序列下一时刻95%的置信区间为：（80.4131，90.9580）。20．1971年9月—1993年6月澳大利亚季度常驻人口变动（单位：千人）情况如表3-23所示（行数据）。表3-2363.267.955.849.550.255.449.945.348.161.755.253.149.559.930.630.433.842.135.828.432.944.145.536.639.549.848.82937.334.247.637.339.247.643.94951.260