Limit distribution of Markov Chain 
I have to determine the limiting distribution for the MC
so I determine the equations first:
$\Pi_0 +\Pi_1 + \Pi_2 + \Pi_3 +\Pi_4 =1$
$\Pi_0 = \Pi_0q + \Pi_1q + \Pi_2q +\Pi_3q +\Pi_4$
$\Pi_1 = \Pi_0p$
$\Pi_2 = \Pi_1p$
$\Pi_3 = \Pi_2p$
$\Pi_4 = \Pi_3p$
and then I tried to solve the system and I got this:
$0= \Pi_0 (-1+q+pq+p^2q+p^3q+p^4) $
So I conclude $0= \Pi_0$ therefore $\Pi_1 = \Pi_2 = \Pi_3 =\Pi_4 =0$
but this doesn't match the answer of the book.
Can somebody tell me where is my mistake?
 A: In this Markov chain, you need find the vector ${\bf \Pi}$ that satisfies the following:
$$ \bf \Pi\ {\bf{P}} = \Pi$$
To start, compute $\bf\Pi\ P$, and we get
$$
{\bf\Pi\ P} = \left[\Pi _5+\Pi _1 q+\Pi _2 q+\Pi _3 q+\Pi _4 q,\ p \Pi _1,\ p \Pi _2,\ p \Pi _3,\ p \Pi _4\right].
$$
Then we set up the linear equations, as shown below:
$$
\begin{eqnarray}
\Pi _5+\Pi _1 q+\Pi _2 q+\Pi _3 q+\Pi _4 q & = & \Pi _1 \\
p \Pi _1 & =& \Pi _2 \\
p \Pi _2 & =&\Pi _3 \\
p \Pi _3& = &\Pi _4 \\
\Pi _1+\Pi _2+\Pi _3+\Pi _4+\Pi _5& =& 1
\end{eqnarray}
$$
An then once we solve them, we get:
$$\Pi _1 = -\frac{1}{p \left(p^2+p+1\right) (q-1)+q-2}$$
$$\Pi _2=-\frac{p}{p \left(p^2+p+1\right) (q-1)+q-2}$$
$$\Pi _3 = -\frac{p^2}{p \left(p^2+p+1\right) (q-1)+q-2}$$
$$\Pi _4= -\frac{p^3}{p \left(p^2+p+1\right) (q-1)+q-2}$$
$$\Pi _5= -\frac{p^4}{p \left(p^2+p+1\right) (q-1)+q-2}$$
A: You were correct in your setup, and getting to $$0= \Pi_0 (-1+q+pq+p^2q+p^3q+p^4).$$
However, you need to use the fact that $p+q=1$ to rewrite this.  You get
\begin{align*}
-1+q+pq+p^2q+p^3q+p^4 &= -p+pq+p^2q+p^3q+p^4\\
&=p(q-1)+p^2q+p^3q+p^4\\
&=-p^{2}+p^2q+p^3q+p^4\\
&=p^{2}(q-1)+p^3q+p^4\\
&=-p^{3}+p^3q+p^4\\
&=p^{3}(q-1) + p^{4}\\
&=-p^{4}+p^{4}\\
&=0.
\end{align*}
Hence, we can choose any non-zero value for $\Pi_{0}$ and normalize later.  By normalize I mean enforce the condition that $\Pi_0 +\Pi_1 + \Pi_2 + \Pi_3 +\Pi_4 =1.$  Without this condition we are simply finding a left eigenvector for $P$ corresponding to an eigenvalue of $1$.  Choosing $\Pi_{0} = 1$ we get
\begin{align*}
\Pi_{0}&=1\\
\Pi_{1}&=p\\
\Pi_{2}&=p^{2}\\
\Pi_{3}&=p^{3}\\
\Pi_{4}&=p^{4}.
\end{align*}
Note that we now have a vector $\Pi$ such that $\Pi P = \Pi$, however $\Pi$ as it is now is not a probability vector because we relaxed the condition that $\Pi_0 +\Pi_1 + \Pi_2 + \Pi_3 +\Pi_4 =1.$  To normalize $\Pi$ so it satisfies this condition, we divide $\Pi$ by the sum of its components so that the resulting vector is a probability vector.  Since currently $\Pi_0 +\Pi_1 + \Pi_2 + \Pi_3 +\Pi_4 =1 + p + p^{2} + p^{3} + p^{4}$, we divide each component of $\Pi$ by this to obtain the limiting distribution (which I'm also calling $\Pi$):
\begin{align*}
\Pi_{0}&=\frac{1}{1+p+p^{2}+p^{3}+p^{4}}\\
\Pi_{1}&=\frac{p}{1+p+p^{2}+p^{3}+p^{4}}\\
\Pi_{2}&=\frac{p^{2}}{1+p+p^{2}+p^{3}+p^{4}}\\
\Pi_{3}&=\frac{p^{3}}{1+p+p^{2}+p^{3}+p^{4}}\\
\Pi_{4}&=\frac{p^{4}}{1+p+p^{2}+p^{3}+p^{4}}.
\end{align*}
This division didn't change the fact that $\Pi P = \Pi$, but now if we add the components we get $1$ which solves the system of equations. 
