Wir berechnen den Kern von

${\displaystyle {}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}-{\begin{pmatrix}{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}\\{\frac {1}{2}}&{\frac {2}{3}}&0\\{\frac {1}{6}}&0&{\frac {2}{3}}\end{pmatrix}}={\begin{pmatrix}{\frac {2}{3}}&-{\frac {1}{3}}&-{\frac {1}{3}}\\-{\frac {1}{2}}&{\frac {1}{3}}&0\\-{\frac {1}{6}}&0&{\frac {1}{3}}\end{pmatrix}}\,.}$

Dieser wird von ${\displaystyle {}{\begin{pmatrix}2\\3\\1\end{pmatrix}}}$ erzeugt und die stationäre Verteilung ist

${\displaystyle {}{\begin{pmatrix}{\frac {2}{6}}\\{\frac {3}{6}}\\{\frac {1}{6}}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{3}}\\{\frac {1}{2}}\\{\frac {1}{6}}\end{pmatrix}}\,.}$