How can we calculate the stationary distribution for any directed graph, let's say this one. What are the steps?
This is the transition matrix , i did:
$$ A=\left(\begin{array}{cccc} 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right) $$
The transition matrix is $A=\left(\begin{array}{cccc} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right)$. By definition, a stationary distribution is $\pi=(a,b,c,d)^t\in\mathbb{R}^4$ such that $\pi=A^t\pi$ and $a+b+c+d=1$. Ie,
$$\begin{align*}\pi&=\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{3} & 1 & 1 & 0 \end{array}\right)\pi\\ \left(\begin{array}{c}a\\b\\c\\d\end{array}\right) &=\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{3} & 1 & 1 & 0 \end{array}\right)\left(\begin{array}{c}a\\b\\c\\d\end{array}\right)\\ \left(\begin{array}{c}a\\b\\c\\d\end{array}\right)&=\left(\begin{array}{c}d\\\frac{a}{3}\\\frac{a}{3}\\\frac{a}{3}+b+c\end{array}\right) \end{align*}$$
Now $1=a+b+c+d=a+\frac{a}{3}+\frac{a}{3}+a=\frac{8}{3}a\iff a=\frac{3}{8}\implies b=c=\frac{1}{8},d=\frac{3}{8}$.