作者Glamsight (安穩殘憶)
看板Statistics
標題[問題] Stable vs. Stationary
時間Sun Jul 31 20:03:22 2016
各位先進好，小的想請問一個有關於stable與stationary的時間序列問題。
為了避免定義上的問題，我先說明一下使用的定義，主要是根據 Lutkepohl[1]。
Definition Stationary Stochastic Process
A stochastic process is stationary if its first and second moments are time
invariant. In other words, a stochastic process y(t) is stationary if
(1) E[y(t)]=mu for all t;
(2) E[(y(t)-mu)*(y(t-h)-mu)^T]=E[(y(t)-mu)*(y(t+h)-mu)^T] for all t and
h=0,1,....
(This is Page 24 on Lutkepohl's text.)
Definition Staable VAR(p) Processes
A VAR(p) process, y(t)=A(1)*y(t-1)+...+A(p)*y(t-p)+u(t), is stable if the
polynomial defined by det(I(K)-A(1)*z-...-A(p)*z^p) has no roots in and on
the complex unit circle.
(This is Page 238 on Lutkepohl's text.)
Definition Stability Condition
Formally y(t) is stable if det(I(K)-A(1)*z-...-A(p)*z^p)!=0 for |z|<=1. This
condition is called the stability condition.
(This is Page 16 on Lutkepohl's text.)
接著是一個關於stable與stationarity的性質與該性質下面的一段話。
Proposition Stationarity Condition
A stable VAR(p) process y(t), t=0,+-1,+-2,..., is stationary.
(This is Proposition 2.1 on Lutkepohl's text.)
Because stability implies stationarity, the stability condition is often
referred to as stationarity condition in the time series literature. The
converse of Proposition 2.1 is not true. In other words, an unstable process
is not necessarily nonstationary. Because unstable stationary processes are
not of interest in the following, we will not discuss this possibility here.
(This is Page 25 on Lutkepohl's text.)
不知這是否表明了一個VAR(p)形式的隨機過程有可能是stationary但不是stable？
從stable與stationary的定義來看，由於使用的字句差異太大，無法感受到其間的相似之處與差。
異。
想請問各位先進，是否有一個stationary VAR(p)但不是stable的例子呢？
或著是有更為明確的說法，讓我可以比較兩者間的差異？
謝謝！！
References
[1] Lutkepohl, Helmut. New Introduction to Multiple Time Series Analysis.
Berlin: Springer, 2005. Print.
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1F：→ kerwinhui: 例如y(t)=2y(t-1)+u(t)這種unstable process 08/01 13:27
2F：→ Glamsight: 取z=1/2，則有det(1-2*z)=det(0)=0。是stable吧？ 08/01 14:38
3F：→ kerwinhui: 看清楚stable的定義，正正是這根這process stable 08/01 15:55
4F：→ Glamsight: 我確認了一次，Stable的定義沒有打錯。det(1-2*0.5)=0 08/01 22:27
5F：→ Glamsight: 而根據定義|z|<=1不能有z使得det(1-2*z)=0，必須都 08/01 22:28
6F：→ Glamsight: det(1-2*z)!=0才是stable。那我找到一個z=1/2且|z|<=1 08/01 22:29
7F：→ Glamsight: 使得det(1-2*z)=0，應該是違反了stable的定義吧？ 08/01 22:30
8F：→ Glamsight: 還是說我哪裡誤會Lutkepohl的定義？ 08/01 22:31
9F：→ kerwinhui: 剛剛才看到被吃字了，應該是…令這process unstable 08/02 12:33
10F：→ kerwinhui: stable的目的是令u(t)不要影響u(t+T), T>>1 08/02 12:37
11F：→ kerwinhui: deterministic case 是想要 y(t) (t→+∞) 收斂 08/02 12:39
12F：→ Glamsight: 請問一下，是怎麼知道該y是stationary?(已知unstable) 08/02 15:07
13F：→ kerwinhui: 最簡單是知道另一種formulation 08/02 16:31
14F：→ kerwinhui: stable = (all roots modulus > 1) 08/02 16:31
15F：→ kerwinhui: stationary = stable or (all roots modulus < 1) 08/02 16:31
我查閱了Hamilton的Time Series Analysis(P.53)寫到在AR(1)時，|A(1)|>=1則不會是
stationary。同時Cryer & Chan的Time Series Analysis(P.71)也寫到在AR(1)時，
|A(1)|<1 iff stationary。似乎與您敘述的沒有單根就是stationary有所不同？
※ 編輯: Glamsight (111.248.103.233), 08/02/2016 21:47:27