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I need to know how to find the stationary distribution for this matrix: $$ Q= \begin{bmatrix} -2 & 2 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 &1\\ 0 & 0 & 2 & -2 \\ \end{bmatrix}$$ in three different ways.

I only know one, which is to replace the last column with $1$'s and then take the inverse. The last row will be the stationary distribution.

Thank you

Update: Sorry, I forgot a row. Thanks for the comments!

Akit
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    For this question to make sense, $Q$ ought to be a square matrix with entries $q_{ij} \in [0,1]$ and row sums equal to 1. – Hans Engler May 12 '13 at 02:52
  • Hello, and welcome to math.SE! I have attempted to improve the formatting of your question by introducing LaTeX. Please check that I have not introduced any errors. – user642796 May 12 '13 at 02:57
  • If you "replace the last column with 1s," you'll still have a $3\times4$ matrix; "and then take the inverse" doesn't make sense. – Gerry Myerson May 12 '13 at 06:32
  • Since Q has some negative entries, Q can only be the generator of a Markov process in continuous time (and not the transition matrix of a Markov chain in discrete time). But then Q ought to have zero sum lines. Akit: homework? – Did May 13 '13 at 13:50

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Since you are asking for different proofs, here is a proof by exhibition: the unique stationary distribution $\pi$ is $$ \pi=\left(\tfrac16,\tfrac13,\tfrac13,\tfrac16\right). $$ And a proof by uniformization: assume that the state space is $S=\{1,2,3,4\}$, duplicate state $2$ into states $2$ and $6$, and duplicate state $3$ into states $3$ and $5$. Thus, the transition rates from state $1$ are now $+1$ to state $2$ and $+1$ to state $6$. Likewise, the transition rates from state $4$ are now $+1$ to state $3$ and $+1$ to state $5$. The transition rates from states $2$ and $3$ are unchanged and the transition rates from states $5$ and $6$ are those from states $3$ and $2$ respectively.

Phew! Almost done... The resulting Markov chain performs a standard random walk on the state space $\bar S=\{1,2,3,4,5,6\}$, best viewed as the discrete circle $\mathbb Z/6\mathbb Z$. Its stationary distribution $\bar\pi$ is obviously uniform on $\bar S$ and $\pi$ is the projection of $\bar\pi$ on $S$. States $1$ and $4$ correspond to one state in $\bar S$ while states $2$ and $3$ correspond to two, hence the ratios $1:2:2:1$ in $\pi$.

And finally a proof by definition (probably the most direct in this context): solve $\pi Q=0$.

Did
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