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Consider the random walk induced by the Markov matrix \begin{equation} \begin{Vmatrix}P_{ij}\end{Vmatrix}= \begin{Vmatrix} r_0 & p_0 & 0 & 0 & \cdots \\ q_1 & r_1 & p_1 & 0 & \cdots \\ 0 & q_2 & r_2 & p_2 & \cdots \\ \end{Vmatrix} \end{equation} Let \begin{equation} \pi_0=1,\pi_n=\frac{p_0p_1\cdots p_{n-1}}{q_1q_2\cdots q_n}. \end{equation} Assume that the Markov chain is recurrent. Show that \begin{equation} \sum^\infty_{i=0}\pi_i=\infty\Longrightarrow\mathrm{null\;reccurent} \end{equation} \begin{equation} \sum^\infty_{i=0}\pi_i<\infty\Longrightarrow\mathrm{positive\;reccurent} \end{equation}

I have completely no idea. Please give me a hint.

1 Answers1

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If $\lim_{n\rightarrow\infty}P^n_{ii}=b_i>0$ for one $i$ in an aperiodic recurrent class, then $b_j>0$ for all $j$ in the class of $i$. In this case, we call the class positive recurrent or strongly ergodic. If each $b_i=0$ and the class is recurrent we speak of the class as null recurrent or weakly ergodic.

Theorem In a positive recurrent aperiodic class with states $j=0,1,2,\cdots$, \begin{equation} \lim_{n\rightarrow\infty}P^n_{jj}=b_j=\sum^\infty_{i=0}b_iP_{ij},\;\sum^\infty_{i=0}b_i=1 \end{equation} and the $b$'s are uniquely determined by the set of equations \begin{equation} b_i\geq0,\;\sum^\infty_{i=0}b_i=1,\;\mathrm{and}\;b_j=\sum^\infty_{i=0}b_iP_{ij} \end{equation}

We investigate the existence of a stationary probability distribution, i.e., we wish to determine the positive solutions of \begin{equation} (p_i+q_i)x_i=\sum^\infty_{j=0}x_jP_{ji}=p_{i-1}x_{i-1}+q_{i+1}x_{i+1},\;i=0,1,\cdots,\;(1) \end{equation} under the normalization \begin{equation} \sum^\infty_{i=0}x_i=1, \end{equation} where $p_{-1}=0$, and thus $x_0=q_1x_1+r_0x_0$. Using Equation (1) for $i=1$, we could determine $x_2$ in terms of $x_0$. Equation (1) for $i=2$ determines $x_3$ in terms of $x_0$, etc. It is immediately verified that \begin{equation} x_i=\frac{p_{i-1}p_{i-2}\cdots p_0}{q_iq_{i-1}\cdots q_1}x_0=x_0\pi_i,\;i\geq1, \end{equation} is a solution of (1), with $x_0$ still to be determined. Now since \begin{equation} 1=x_0+\sum^\infty_{i=1}x_0\pi_i=\sum^\infty_{i=0}x_0\pi_i, \end{equation} we have \begin{equation} x_0=\frac{1}{\sum^\infty_{i=0}x_0\pi_i} \end{equation} and so \begin{equation} x_0>0\;\mathrm{if\;and\;only\;if}\;\sum^\infty_{i=0}x_0\pi_i<\infty. \end{equation}

Combining this result, the definition, and the theorem, we get the desired sufficient condition.