Construction of positive recurrent Markov chain Let $\{X_i\}_{i\geq 1}$ be i.i.d. with values in $\mathbb N_0$. Define a Markov chain via the following transition matrix:
$$p(0,n) = \mathbb P(X_1 = n-1) \qquad p(m,n) = \mathbb P\left(\sum_{k=1}^m X_k = n\right)$$
Under what conditions is this Markov chain positive recurrent?
I tried to find conditions under which the chain is irreducible and has an invariant distribution, but couldn't pin down the calculation. It would then follow that it is positive recurrent.
 A: Your Markov chain is the total population of a branching process with offspring distribution $X$, 
except that when the population goes extinct (hits state 0) it regenerates a random number of 
ancestors who again start to grow family trees. Standard results on the extinction of 
branching processes gets you pretty far.  
Let's assume that $\mathbb{P}(X=0)>0$, $\mathbb{P}(X=1)>0$, and $\mathbb{P}(X>1)>0$ so that the chain 
is irreducible on the state space $\mathbb N_0$. 
When $\mathbb{E}(X)<1$, then the expected time to extinction starting with one individual 
 is finite; $\mathbb{E}_1(T_0)<\infty$. Then 
\begin{eqnarray*}
\mathbb{E}_0(T_0)&=&1+\sum p(0,n)\,\mathbb{E}_n(T_0)\\[5pt]
                 &\leq&1+\sum p(0,n)\, n \,\mathbb{E}_1(T_0)\\[5pt]
                 &=&1+(\mathbb{E}(X)+1)\, \mathbb{E}_1(T_0)<\infty.
\end{eqnarray*}
Therefore the state 0 is positive recurrent and hence the whole chain.  
If $\mathbb{E}(X)>1$, the chain is transient. The population will grow to $\infty$ 
with probability one. 
If $\mathbb{E}(X)=1$, extinction is guaranteed so the chain is recurrent.
When $\mathbb{E}(X^2)<\infty$ the chain is  null since $\mathbb{E}_1(T_0)=\infty$. 
I'm not sure about the case  $\mathbb{E}(X)=1$ and $\mathbb{E}(X^2)=\infty$.  

Footnote 1: Where does the equation come from?
Let's start with some boundary theory for Markov chains. 
Let $(X_n)$ be a Markov chain
 with state space $\cal S$ and let $B\subset {\cal S}$. Define
$$V_B=\inf(n\geq 0: X_n\in B),\qquad T_B=\inf(n\geq 1: X_n\in B),$$
and 
$$h(x)=\mathbb{E}_x(V_B),\qquad f(x)=\mathbb{E}_x(T_B).$$
Notice that $h(x)=f(x)$ for $x\notin B$. Using the shift operator 
we  write $T_B=1+V_B\circ\theta_1$ so 
$$f(x)=\mathbb{E}_x(1+V_B\circ\theta_1)=1+\mathbb{E}_x(h(X_1)).$$
In your problem (with $B=\{0\}$), since there are no transitions 
from $0$ to itself, we get $$\mathbb{P}_0(f(X_1)=h(X_1))=1,$$ and hence
$$f(0)=\mathbb{E}_0(T_0)=1 +\mathbb{E}_0(f(X_1))=1 +\sum_n p(0,n)\,\mathbb{E}_n(T_0).$$ 
