Time Series Analysis
midterm report
2010/04/21
大氣所碩一 闕珮羽 R98229016
Data
• Vostok : Petit J.R. et al.,
Climate and Atmospheric History of the Past 420,000
years from the Vostok Ice Core, Antarctica, Nature,
399, pp.429-436.
資料全期平均及標準差
• Mean=-4.7116
• STD=2.7541
全期變動趨勢
某兩段的值
Mean= -5.1172
Std= 2.5399
Mean= -5.6417
Std= 2.4169
兩段時間平均數差異統計檢定
• 利用t test 去做檢定，查詢 matlab 的 function
• H = TTEST2(X,Y) performs a T-test of the hypothesis that two
independent samples, in the vectors X and Y, come from
distributions with equal means, and returns the result of the test in
H. H=0 indicates that the null hypothesis ("means are equal")
cannot be rejected at the 5% significance level. H=1 indicates that
the null hypothesis can be rejected at the 5% level. The data are
assumed to come from normal distributions with unknown, but
equal, variances. X and Y can have different lengths.
• 利用我取的兩段時間做出來 H = TTEST2(TS1,TS2)=0
• 所以表示 the null hypothesis cannot be rejected at the 5% level
兩段時間變異數差異檢定
• 利用F test 去做檢定，查詢 matlab 的 function
• H = VARTEST2(X,Y) performs an F test of the hypothesis that two
independent samples, in the vectors X and Y, come from normal
distributions with the same variance, against the alternative that
they come from normal distributions with different variances. The
result is H=0 if the null hypothesis ("variances are equal") cannot be
rejected at the 5% significance level, or H=1 if the null hypothesis
can be rejected at the 5% level. X and Y can have different lengths.
• 利用我取的兩段時間做出來 F = VARTEST2(TS1,TS2)=0
• 所以表示 the null hypothesis cannot be rejected at the 5% level
時間數列的時間迴歸分析
y=0.00464*x-5.7819
從一階可以看出是
增加的趨勢，二三
階發現有先下降後
來上升的趨勢，五
六階是整體的變化，
更高階雜訊會很多。
突變點分析
滑動平均
時間數列相關分析
• ACF 慢慢衰減，PACF在N>2後截斷。
→ AR(2) model的時間數列
建立時間數列的ARMA模式
•
•
•
•
利用ARMA model (p,q)=(1,4)時有最小的AIC
得到ARMA model的係數為
A(q) = 1 - 0.9819 q^-1
C(q) = 1 + 0.4214 q^-1 + 0.1409 q^-2 +
0.01416 q^-3 + 0.01416 q^-4
ARMA模式
預測出來的振幅較大
ARMA模式