Very basic question on periodicity of markov chain

• Nov 19th 2010, 10:16 PM
noob mathematician
Very basic question on periodicity of markov chain
Suppose I have the following markov chain:

$$\left( \begin{array}{ccc} 0 & 0.5 & 0.5 \\ 0.5 & 0 & 0.5 \\ 0.5 & 0.5 & 0 \end{array} \right)$$

it seems that in this case that a state is able to return to itself either by 2 steps or 3 steps. Then what is the period of any single state? (all three states have the same periodicity though)

By the way, does a transient state has a period too?
• Nov 20th 2010, 02:08 AM
Moo
Hello,

If it were 2 or 3 steps (or a multiple), the periodicity of a state will be the gcd of 2 and 3. This is merely the definition of a period :) But here, you can have 5 steps : 0 - 1 - 2 - 1 - 2 - 0. So actually, the period will be 1.

Transience and periodicity have no relationship, there is no period associated to a transient state.
• Nov 20th 2010, 05:41 AM
noob mathematician
hi,

Thanks for your help.. I was still quite unsure though. So it's possible that a state has d=1 even though it can't reach back by one step?

State i is said to have a period d if (p_ii^n)=0 whenever n is not divisible by d, and d is the GCD. it's so confusing
• Nov 22nd 2010, 06:51 AM
Moo
I guess your definition of a period only holds for d>1.
The definition I've been taught for the period of a state is that it is the gcd of the following set : $\{n\in\mathbb N ~:~ p_{ii}^n>0\}$.

We can read in the English wikipedia that a state is of period 1 if it returns to the state irregularly (no pattern). Here it can return after 2,3,5,7,8,9, etc... the gcd of these first 5 numbers is 1.