I have a Markov Chain with state space $0, 1, 2, \ldots, 12$ and transition behavior given by:
- For $0 \le i \le 11$, the distribution of $X_{n+1}$ given $X_n = i$ is uniform on $i+1, i+2, \ldots , 12$.
- $P(12, 0) = 1$.
I am trying to build the transition matrix, $\mathbb{P}$, knowing that the probability of moving from state $12 \rightarrow 0$ is $1$, but how do I find the other transition probabilities to complete $\mathbb{P}$?
If $\ p_{ij}=P\left(X_{n+1}=j\left|X_n=i\right.\right)\ $, then $$ p_{ij}=\cases{0& if $\ 0\le j\le i< 12\ $ or $\ i=12, j\ne0\ $,\\ \frac{1}{12-i}& if $\ i<j\le12\ $,\\ 1 & if $\ i=12, j=0\ $.} $$ So the transition matrix is $$ \hspace{-1em}\pmatrix{0&\frac{1}{12}&\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12} &\frac{1}{12}\\ 0&0 &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11} &\frac{1}{11}\\ 0&0 &0 &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10} &\frac{1}{10}\\ 0&0 &0 &0 &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9} &\frac{1}{9}\\ 0&0&0 &0 &0 &\frac{1}{8} &\frac{1}{8} &\frac{1}{8} &\frac{1}{8} &\frac{1}{8} &\frac{1}{8} &\frac{1}{8} &\frac{1}{8}\\ 0&0&0&0 &0 &0 &\frac{1}{7} &\frac{1}{7} &\frac{1}{7} &\frac{1}{7} &\frac{1}{7} &\frac{1}{7} &\frac{1}{7}\\ 0&0&0&0&0 &0 &0 &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6}\\ 0&0&0&0&0&0 &0 &0 &\frac{1}{5} &\frac{1}{5} &\frac{1}{5} &\frac{1}{5} &\frac{1}{5}\\ 0&0&0&0&0&0&0 &0 &0 &\frac{1}{4} &\frac{1}{4} &\frac{1}{4} &\frac{1}{4}\\ 0&0&0&0&0&0&0&0 &0 &0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3}\\ 0&0&0&0&0&0&0&0&0 &0 &0 &\frac{1}{2} &\frac{1}{2}\\ 0&0&0&0&0&0&0&0&0&0 &0 &0 &1\\ 1&0&0&0&0&0&0&0&0&0&0 &0 &0}\ . $$