 Solution

Find a stationary distribution $\mathbf{\pi }=(u,v,w{)}^T$ from $P\mathbf{\pi }=\mathbf{\pi }$ and $u+v+w=1$. The first two independent linear equations areThey give $u=9w/8$ and $v=7w/4$. Normalization gives $w=8/31$, hence

To justify the limit using the assumptions, distinct eigenvalues give an eigenbasis, over the complex numbers if necessary. Choose $\mathbf{\pi }$ as the eigenvector for 1 and let ${\mathbf{v}}_2,{\mathbf{v}}_3$ correspond to the other eigenvalues ${\lambda }_2,{\lambda }_3$. Expand the initial probability distribution asSince $1^{T}P=1^T$, each ${\lambda }_j\ne 1$ satisfiesMultiplying the expansion by $1^T$ therefore gives $\alpha =1$. It follows thatbecause $|{\lambda }_2|,|{\lambda }_3|<1$. The components of the limiting probability distribution are approximately $0.2903,0.4516,0.2581$.