A chain with stats 1,2,....,n has a matrix whose first and last rows are (q,p,0,...,0) and (0,...,0,q,p). In all other rows Pk,k+1 = p, Pk,k-1 = q. Find the stationary distribution.
I am completely stuck. Help would be greatly appreciated.
A chain with stats 1,2,....,n has a matrix whose first and last rows are (q,p,0,...,0) and (0,...,0,q,p). In all other rows Pk,k+1 = p, Pk,k-1 = q. Find the stationary distribution.
I am completely stuck. Help would be greatly appreciated.
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Hint. You need $\def\\#1{{\bf#1}}P\\v=\\v$, that is, $(P-I)\\v=\\0$. Now $$P-I=\pmatrix{-p&p&0&0&0&\cdots\cr q&-1&p&0&0&\cdots\cr 0&q&-1&p&0&\cdots\cr \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\cr}\ .$$ (I am assuming that $p+q=1$ although you didn't actually say so.) You want to multiply this by a vector $\\v=(v_1,v_2,v_3,\ldots)^T$ to get the zero vector as a result. Now if you look at row $1$ you see straight away that $v_1=v_2$. Then row $2$ will tell you about $v_3$ and so on. Once you have got to the end of the matrix you will need to find the specific values for $v_1,v_2,v_3,\ldots$ such that $\\v$ is a probability vector, that is, the entries add up to $1$.
Good luck!