下面来看一个例子 AR(3) 模型:
y
t
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0.1
+
0.2
y
t
−
1
+
0.3
y
t
−
2
+
0.5
y
t
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3
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ε
t
y_t = 0.1 + 0.2y_{t - 1} + 0.3y_{t - 2} + 0.5y_{t - 3} + \varepsilon_t
yt=0.1+0.2yt−1+0.3yt−2+0.5yt−3+εt 齐次部分的特征方程为:
x
3
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0.2
x
2
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0.3
x
+
0.5
x^3 = 0.2x^2 + 0.3x + 0.5
x3=0.2x2+0.3x+0.5 解得:
x
1
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4
−
34
i
10
,
x
2
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−
4
+
34
i
10
,
x
3
=
1
x_1 = \frac{-4 - \sqrt{34}i}{10}, x_2 = \frac{-4 + \sqrt{34}i}{10}, x_3 = 1
x1=10−4−34
i,x2=10−4+34
i,x3=1 求模长:
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x
1
∣
=
∣
x
2
∣
=
16
100
+
34
100
=
0.7071
<
1
|x_1| = |x_2| = \sqrt{\frac{16}{100} + \frac{34}{100}} = 0.7071 < 1
∣x1∣=∣x2∣=10016+10034
=0.7071yt} 平稳的充分条件:
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α
1
∣
+
∣
α
2
∣
+
⋯
+
∣
α
p
∣
<
1
|\alpha_1| + |\alpha_2| + \cdots + |\alpha_p| < 1
∣α1∣+∣α2∣+⋯+∣αp∣ |