作者Glamsight (安稳残忆)
看板Statistics
标题[问题] Stable vs. Stationary
时间Sun Jul 31 20:03:22 2016
各位先进好，小的想请问一个有关於stable与stationary的时间序列问题。
为了避免定义上的问题，我先说明一下使用的定义，主要是根据 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.)
接着是一个关於stable与stationarity的性质与该性质下面的一段话。
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.)
不知这是否表明了一个VAR(p)形式的随机过程有可能是stationary但不是stable？
从stable与stationary的定义来看，由於使用的字句差异太大，无法感受到其间的相似之处与差。
异。
想请问各位先进，是否有一个stationary VAR(p)但不是stable的例子呢？
或着是有更为明确的说法，让我可以比较两者间的差异？
谢谢！！
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)这种unstable process 08/01 13:27
2F：→ Glamsight: 取z=1/2，则有det(1-2*z)=det(0)=0。是stable吧？ 08/01 14:38
3F：→ kerwinhui: 看清楚stable的定义，正正是这根这process stable 08/01 15:55
4F：→ Glamsight: 我确认了一次，Stable的定义没有打错。det(1-2*0.5)=0 08/01 22:27
5F：→ Glamsight: 而根据定义|z|<=1不能有z使得det(1-2*z)=0，必须都 08/01 22:28
6F：→ Glamsight: det(1-2*z)!=0才是stable。那我找到一个z=1/2且|z|<=1 08/01 22:29
7F：→ Glamsight: 使得det(1-2*z)=0，应该是违反了stable的定义吧？ 08/01 22:30
8F：→ Glamsight: 还是说我哪里误会Lutkepohl的定义？ 08/01 22:31
9F：→ kerwinhui: 刚刚才看到被吃字了，应该是…令这process unstable 08/02 12:33
10F：→ kerwinhui: stable的目的是令u(t)不要影响u(t+T), T>>1 08/02 12:37
11F：→ kerwinhui: deterministic case 是想要 y(t) (t→+∞) 收敛 08/02 12:39
12F：→ Glamsight: 请问一下，是怎麽知道该y是stationary?(已知unstable) 08/02 15:07
13F：→ kerwinhui: 最简单是知道另一种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
我查阅了Hamilton的Time Series Analysis(P.53)写到在AR(1)时，|A(1)|>=1则不会是
stationary。同时Cryer & Chan的Time Series Analysis(P.71)也写到在AR(1)时，
|A(1)|<1 iff stationary。似乎与您叙述的没有单根就是stationary有所不同？
※ 编辑: Glamsight (111.248.103.233), 08/02/2016 21:47:27