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I just started on a course with stochastic processes, so I am really new to this and I have a problem with the following exercise


A Markov chain $X_0, X_1, X_2$ has the transition probability matrix

$$P=\begin{Vmatrix} 0.7 & 0.2 & 0.1 \\ 0 & 0.6 & 0.4 \\ 0.5 & 0 & 0.5\end{Vmatrix}$$

Determine the limiting distribution


In my book it says that the limiting distribution $\pi$ is the unique nonnegative solution of the following equations. $$\pi_j=\sum_{k=0}^N \pi_kP_{kj}, \qquad for \qquad j=0,1,...,N \qquad (*)$$ $$\sum_{k=0}^N \pi_k=1 \qquad (**)$$

By this I get that I have to write a system of linear equations and this is the system I get $${\frac {7}{10}\pi_0}+{\frac {2}{5}\pi_1}+{\frac {1}{10}\pi_2}=\pi_0 \qquad (1)$$ $${\frac {0}{10}\pi_0}+{\frac {3}{5}\pi_1}+{\frac {4}{10}\pi_2}=\pi_1 \qquad (2)$$ $${\frac {3}{10}\pi_0}+{\frac {0}{5}\pi_1}+{\frac {1}{2}\pi_2}=\pi_2 \qquad (3)$$ $$\pi_0+\pi_1+\pi_2=1 \qquad (4)$$ I understand that my columns have to sum to 1 because of (*),(**) so in (3) I have 3/10 for $\pi_0$. I solve for (1),(2) and (4) and the solution I get is $\pi_0={\frac{5}{11}}, \pi_1={\frac{3}{11}}, \pi_2={\frac{3}{11}}$ which is not the answer to the exercise in my book. The answer my book gives is that the solution is $\pi_0={\frac{10}{21}}, \pi_1={\frac{5}{21}}, \pi_2={\frac{6}{21}}$ and I can't quite work out what I'm doing wrong?

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  • $\begingroup$ The matrix equation is $\pi P=\pi$, not $P \pi=\pi$. $\endgroup$ – Xi'an Sep 15 '15 at 12:57
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What needs to sum to 1 is the rows, not the columns, as the process needs to move somewhere from any state (or remain in its state).

Also, the $(1,2)$-element is 0.2, thus $2/10$, not $2/5$.

The stationary distribution can most easily (imo) be computed as the normalized eigenvector corresponding to the unit eigenvalue:

The invariant distribution can be written as $$ \pi ^{\prime }\left( P-I\right) =0^{\prime } $$ or, taking transposes, $$ \left( P^{\prime }-I\right)\pi =0 $$ Hence, $\pi$ is the eigenvector of $P'$ corresponding to the eigenvalue $\lambda _{i}=1$ (the implicit 1 in front of the identity matrix).

In R,

P <- matrix(c(7,0,5,2,6,0,1,4,5),3,3)/10
q <- eigen(t(P))$vectors[,1] 
(q/sum(q)) # normalize sum of probabilities to 1
[1] 0.4761905+0i 0.2380952+0i 0.2857143+0i
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  • $\begingroup$ I was a little fast on my fingers there on element (1,2), but how do you get that it can be written as that? Is it from * and **? $\endgroup$ – ilhano Sep 16 '15 at 7:41
  • $\begingroup$ it follows from writing * in matrix notation, by collecting the equations for all the $\pi_j$ in row vectors $\pi$ and $\pi P$ (or writing $\pi'$ for a row vector in my answer) $\endgroup$ – Christoph Hanck Sep 16 '15 at 8:18

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