给定初始状态下的马尔可夫链平稳分布的意义是什么? [英] What is the significance of the stationary distribution of a markov chain given it's initial state?

问题描述

让X_n成为MC，P不定期

假设我们有一个固定距离(pi_0，...，pi_n)并且P(X_0 = i)= 0.2，这说明什么了吗?

更清楚:

我之所以问是因为，卡林说，当平稳距离不是极限距离时，P(X_n = i)取决于初始分布.这到底是什么意思?

解决方案

您的标题问题需要冗长的答案；我只需要提供一些参考即可，以使您阅读更多有关马尔可夫链和遍历理论的信息.但是，您的具体问题是:

"...当固定距离不是极限距离时，P(X_n = i)取决于初始分布.这到底是什么意思?"

可以用一个简单的例子来回答.考虑具有两个状态的马尔可夫链，

假设转移矩阵P =

[0.4，0.6]

[0.6，0.4]

如果我告诉您在t时刻您当前处于状态A，然后问您在下一个状态(在t + 1时)将处于哪种状态，则可以将P乘以[1，0]并解释结果为[0.4，0.6]，表示您有40％的心态仍将保持状态A，有60％的心态将最终进入状态B.

现在，如果我告诉您在时间t处在状态A，并问您在时间t + 999处在什么状态，该怎么办?在过去的所有时间里，您如何在州周围跳来跳去有很多随机性，以至于您实际上无法从州A开始这一事实来密切关注链条".基本上，这些信息是混合的"直到您开始于A都没什么大不了.只需问问自己:如果我告诉您您开始于B州而不是A州，那么您在t + 999时对州的看法会如何变化?您无法想出观点上的差异；那就是不变性.

在数学上，给定时间t的状态，时间t + 999的转换矩阵为P ^(999).该矩阵的每一行都是相同的，因此将其乘以任何概率分布([x，y]，其中x + y = 1)将得出相同的答案.对于此问题，此极限"分布为[0.5，0.5]，这意味着在999个时间步长之后，无论999是多少，您都将确保50％的用户处于状态A，而50％的用户处于状态B.时间步长之前是从A开始的.固定分布是P的左特征向量，对应于特征值1.如果P ^ t的所有行都收敛为t-> inf，则称为P的极限分布.

那没有限制分布的马尔可夫链呢?考虑一个P，

[0，1]

[1，0]

此马尔可夫链具有触发器"周期性性质.如果在时间t处处于状态A，则100％确定在时间t + 1处将处于状态B，反之亦然.因此，如果我告诉您在时间t您处于状态A，您将知道在时间t + 999您将处于状态B，因为999为奇数.或者，如果我告诉您在时间t您处于状态B，那么在时间t + 999您将处于状态A.请注意对初始条件的依赖性.该马尔可夫链对起始信息敏感.它不会混在一起".从数学上讲，P ^ t不收敛为t-> inf.

尝试在代码中使用这些矩阵！

Let X_n be a MC, P not regular

Say we have a stationary dist (pi_0, ..., pi_n) and P(X_0 = i) = 0.2, does this say anything?

To be more clear:

I ask because Karlin says when a stationary dist is not a limiting dist, P(X_n = i) is dependent on the initial distribution. What does this exactly mean?

解决方案

Your title's question requires a lengthy answer; I'd have to just provide some references for you to read more on Markov chains and ergodic theory. However, your specific question:

"...when a stationary dist is not a limiting dist, P(X_n = i) is dependent on the initial distribution. What does this exactly mean?"

can be answered with a simple example. Consider a Markov chain with two states,

Suppose a transition matrix P =

[0.4, 0.6]

[0.6, 0.4]

If I told you that you are currently in state A at time t, and then asked what state you will be in next (at time t+1), you would left multiply P by [1, 0] and interpret the result [0.4, 0.6] as meaning you're 40% sure that you'll still be in state A and 60% sure you'll end up in state B.

Now what if I told you that you were in state A at time t and asked what state you'll be in at time t+999? In all that time, there is so much randomness in how you have bounced around the states that you really wouldn't be able to "follow the chain closely" from the fact that you started at state A. Basically, that information was "mixed around" by the Markov chain until it didn't really matter that you started at A. Just ask yourself: how would your opinion on your state at time t+999 change if I told you that you started in state B rather than state A? You aren't able to conjure up a difference in opinion; that is the invariance.

Mathematically, the transition matrix for time t+999 given the state at time t is P^(999). Every row of this matrix will be identical, so left multiplying by any probability distribution ([x, y] where x+y=1) will result in the same answer. For this problem, this "limiting" distribution is [0.5, 0.5], meaning that after 999 timesteps, you would be 50% sure you're in state A and 50% sure you're in state B regardless of the fact that 999 timesteps ago you started in A. The stationary distribution is the left eigenvector of P corresponding to an eigenvalue of 1. It is called the limiting distribution of P if all the rows of P^t converge to it as t -> inf.

What about a Markov chain that doesn't have a limiting distribution? Consider a P of,

[0, 1]

[1, 0]

This Markov chain has a "flip flopping" periodic nature. If at time t you are in state A, you are 100% sure you'll be in state B at time t+1, and vis-versa. So if I told you that you are in state A at time t, you would know that at time t+999 you will be in state B since 999 is odd. Alternatively, if I told you that you are in state B at time t, then at time t+999 you would expect to be in state A. Notice the dependence on initial condition. This Markov chain is sensitive to that starting information. It doesn't "mix away". Mathematically, P^t does not converge as t -> inf.

Try playing with these matrices in code!

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