# What's the period of this matrix?

Consider the matrix $$A = \begin{pmatrix} 0.1 & 0.3 & 0.4 & 0.2 \\ 0.2 & 0.4 & 0.0 & 0.4 \\ 0.0 & 0.3 & 0.5 & 0.2 \\ 0.5 & 0.3 & 0.2 & 0.0 \end{pmatrix}.$$ Note since $$A^2 = \begin{pmatrix} 0.17 & 0.33 & 0.28 & 0.22 \\ 0.30 & 0.34 & 0.16 & 0.20 \\ 0.16 & 0.33 & 0.29 & 0.22 \\ 0.11 & 0.33 & 0.30 & 0.26 \end{pmatrix},$$ the matrix $A$ is irreducible, as described in Definition 3 here. The confusing question now is, what's the period? Again according to the description linked above, the period of indices 1,2 and 3 is 1, since $(A)_{ii} > 0$ for $i = 1,2,3$. For index $4$, however, the period should be $2$ since $(A)_{44} = 0$ but $(A^2)_{44} > 0$. The link then says, "When A is irreducible, the period of every index is the same..." but I don't see how this is true in this case. Is the period 1 or 2? If it's 1, how does that not conflict with index 4's period or 2? If it's 2, that conflicts with the rest of the indices.

Since all entries of $A^2$ are strictly positive, so are all entries of $A^n$ for all $n > 2$. The period of index $4$ is the gcd of all $n$ such that $(A^n)_{44} > 0$, so that is $1$, not $2$.
• okay, that makes sense, but it goes against my intuition of 'period' for a Markov chain. If the matrix $A$ above is the transition matrix for a homogeneous Markov chain, the process cannot return to state $4$ in one step; rather, it can in at earliest 2 steps, right? So it seems the period of index $4$ should be $2$... – bcf Feb 7 '16 at 21:32
• Your intuition doesn't match the definition. The period is not the least number of steps the process can take to return, it's the gcd of the possible number of steps the process can take to return. In this case the process can't return in 1 step, but it can return in 2, 3, .... Since $\gcd(2,3) = 1$, the period is 1. – Robert Israel Feb 8 '16 at 2:35