Consider a discrete–time, time–homogeneous Markov chain on states 1,…,n\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$1, \ldots , n$$\end{document} whose transition matrix is irreducible. Denote the mean first passage times by mjk,j,k=1,…,n,\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$m_{jk}, j,k=1,\ldots , n,$$\end{document} and stationary distribution vector entries by vk,k=1,…,n\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$v_k, k=1, \ldots , n$$\end{document}. A result of Kemeny reveals that the quantity ∑k=1nmjkvk\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$\sum _{k=1}^n m_{jk}v_k$$\end{document}, which is the expected number of steps needed to arrive at a randomly chosen destination state starting from state j, is–surprisingly–independent of the initial state j. In this note, we consider ∑k=1nmjkvk\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$\sum _{k=1}^n m_{jk}v_k$$\end{document} from the perspective of algebraic combinatorics and provide an intuitive explanation for its independence on the initial state j. The all minors matrix tree theorem is the key tool employed.